3.12.17 \(\int \frac {(c+d \tan (e+f x))^{5/2}}{(a+i a \tan (e+f x))^2} \, dx\) [1117]

Optimal. Leaf size=217 \[ -\frac {i (c-i d)^{5/2} \tanh ^{-1}\left (\frac {\sqrt {c+d \tan (e+f x)}}{\sqrt {c-i d}}\right )}{4 a^2 f}+\frac {\sqrt {c+i d} \left (2 i c^2+6 c d-7 i d^2\right ) \tanh ^{-1}\left (\frac {\sqrt {c+d \tan (e+f x)}}{\sqrt {c+i d}}\right )}{8 a^2 f}+\frac {(c+i d) (2 i c+5 d) \sqrt {c+d \tan (e+f x)}}{8 a^2 f (1+i \tan (e+f x))}+\frac {(i c-d) (c+d \tan (e+f x))^{3/2}}{4 f (a+i a \tan (e+f x))^2} \]

[Out]

-1/4*I*(c-I*d)^(5/2)*arctanh((c+d*tan(f*x+e))^(1/2)/(c-I*d)^(1/2))/a^2/f+1/8*(2*I*c^2+6*c*d-7*I*d^2)*arctanh((
c+d*tan(f*x+e))^(1/2)/(c+I*d)^(1/2))*(c+I*d)^(1/2)/a^2/f+1/8*(c+I*d)*(2*I*c+5*d)*(c+d*tan(f*x+e))^(1/2)/a^2/f/
(1+I*tan(f*x+e))+1/4*(I*c-d)*(c+d*tan(f*x+e))^(3/2)/f/(a+I*a*tan(f*x+e))^2

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Rubi [A]
time = 0.42, antiderivative size = 217, normalized size of antiderivative = 1.00, number of steps used = 9, number of rules used = 6, integrand size = 30, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.200, Rules used = {3639, 3676, 3620, 3618, 65, 214} \begin {gather*} \frac {\sqrt {c+i d} \left (2 i c^2+6 c d-7 i d^2\right ) \tanh ^{-1}\left (\frac {\sqrt {c+d \tan (e+f x)}}{\sqrt {c+i d}}\right )}{8 a^2 f}+\frac {(c+i d) (5 d+2 i c) \sqrt {c+d \tan (e+f x)}}{8 a^2 f (1+i \tan (e+f x))}-\frac {i (c-i d)^{5/2} \tanh ^{-1}\left (\frac {\sqrt {c+d \tan (e+f x)}}{\sqrt {c-i d}}\right )}{4 a^2 f}+\frac {(-d+i c) (c+d \tan (e+f x))^{3/2}}{4 f (a+i a \tan (e+f x))^2} \end {gather*}

Antiderivative was successfully verified.

[In]

Int[(c + d*Tan[e + f*x])^(5/2)/(a + I*a*Tan[e + f*x])^2,x]

[Out]

((-1/4*I)*(c - I*d)^(5/2)*ArcTanh[Sqrt[c + d*Tan[e + f*x]]/Sqrt[c - I*d]])/(a^2*f) + (Sqrt[c + I*d]*((2*I)*c^2
+ 6*c*d - (7*I)*d^2)*ArcTanh[Sqrt[c + d*Tan[e + f*x]]/Sqrt[c + I*d]])/(8*a^2*f) + ((c + I*d)*((2*I)*c + 5*d)*
Sqrt[c + d*Tan[e + f*x]])/(8*a^2*f*(1 + I*Tan[e + f*x])) + ((I*c - d)*(c + d*Tan[e + f*x])^(3/2))/(4*f*(a + I*
a*Tan[e + f*x])^2)

Rule 65

Int[((a_.) + (b_.)*(x_))^(m_)*((c_.) + (d_.)*(x_))^(n_), x_Symbol] :> With[{p = Denominator[m]}, Dist[p/b, Sub
st[Int[x^(p*(m + 1) - 1)*(c - a*(d/b) + d*(x^p/b))^n, x], x, (a + b*x)^(1/p)], x]] /; FreeQ[{a, b, c, d}, x] &
& NeQ[b*c - a*d, 0] && LtQ[-1, m, 0] && LeQ[-1, n, 0] && LeQ[Denominator[n], Denominator[m]] && IntLinearQ[a,
b, c, d, m, n, x]

Rule 214

Int[((a_) + (b_.)*(x_)^2)^(-1), x_Symbol] :> Simp[(Rt[-a/b, 2]/a)*ArcTanh[x/Rt[-a/b, 2]], x] /; FreeQ[{a, b},
x] && NegQ[a/b]

Rule 3618

Int[((a_.) + (b_.)*tan[(e_.) + (f_.)*(x_)])^(m_)*((c_) + (d_.)*tan[(e_.) + (f_.)*(x_)]), x_Symbol] :> Dist[c*(
d/f), Subst[Int[(a + (b/d)*x)^m/(d^2 + c*x), x], x, d*Tan[e + f*x]], x] /; FreeQ[{a, b, c, d, e, f, m}, x] &&
NeQ[b*c - a*d, 0] && NeQ[a^2 + b^2, 0] && EqQ[c^2 + d^2, 0]

Rule 3620

Int[((a_.) + (b_.)*tan[(e_.) + (f_.)*(x_)])^(m_)*((c_.) + (d_.)*tan[(e_.) + (f_.)*(x_)]), x_Symbol] :> Dist[(c
+ I*d)/2, Int[(a + b*Tan[e + f*x])^m*(1 - I*Tan[e + f*x]), x], x] + Dist[(c - I*d)/2, Int[(a + b*Tan[e + f*x]
)^m*(1 + I*Tan[e + f*x]), x], x] /; FreeQ[{a, b, c, d, e, f, m}, x] && NeQ[b*c - a*d, 0] && NeQ[a^2 + b^2, 0]
&& NeQ[c^2 + d^2, 0] &&  !IntegerQ[m]

Rule 3639

Int[((a_) + (b_.)*tan[(e_.) + (f_.)*(x_)])^(m_)*((c_.) + (d_.)*tan[(e_.) + (f_.)*(x_)])^(n_), x_Symbol] :> Sim
p[(-(b*c - a*d))*(a + b*Tan[e + f*x])^m*((c + d*Tan[e + f*x])^(n - 1)/(2*a*f*m)), x] + Dist[1/(2*a^2*m), Int[(
a + b*Tan[e + f*x])^(m + 1)*(c + d*Tan[e + f*x])^(n - 2)*Simp[c*(a*c*m + b*d*(n - 1)) - d*(b*c*m + a*d*(n - 1)
) - d*(b*d*(m - n + 1) - a*c*(m + n - 1))*Tan[e + f*x], x], x], x] /; FreeQ[{a, b, c, d, e, f}, x] && NeQ[b*c
- a*d, 0] && EqQ[a^2 + b^2, 0] && NeQ[c^2 + d^2, 0] && LtQ[m, 0] && GtQ[n, 1] && (IntegerQ[m] || IntegersQ[2*m
, 2*n])

Rule 3676

Int[((a_) + (b_.)*tan[(e_.) + (f_.)*(x_)])^(m_)*((A_.) + (B_.)*tan[(e_.) + (f_.)*(x_)])*((c_.) + (d_.)*tan[(e_
.) + (f_.)*(x_)])^(n_), x_Symbol] :> Simp[(-(A*b - a*B))*(a + b*Tan[e + f*x])^m*((c + d*Tan[e + f*x])^n/(2*a*f
*m)), x] + Dist[1/(2*a^2*m), Int[(a + b*Tan[e + f*x])^(m + 1)*(c + d*Tan[e + f*x])^(n - 1)*Simp[A*(a*c*m + b*d
*n) - B*(b*c*m + a*d*n) - d*(b*B*(m - n) - a*A*(m + n))*Tan[e + f*x], x], x], x] /; FreeQ[{a, b, c, d, e, f, A
, B}, x] && NeQ[b*c - a*d, 0] && EqQ[a^2 + b^2, 0] && LtQ[m, 0] && GtQ[n, 0]

Rubi steps

\begin {align*} \int \frac {(c+d \tan (e+f x))^{5/2}}{(a+i a \tan (e+f x))^2} \, dx &=\frac {(i c-d) (c+d \tan (e+f x))^{3/2}}{4 f (a+i a \tan (e+f x))^2}-\frac {\int \frac {\sqrt {c+d \tan (e+f x)} \left (-\frac {1}{2} a \left (4 c^2-7 i c d+3 d^2\right )-\frac {1}{2} a (c-7 i d) d \tan (e+f x)\right )}{a+i a \tan (e+f x)} \, dx}{4 a^2}\\ &=\frac {(c+i d) (2 i c+5 d) \sqrt {c+d \tan (e+f x)}}{8 a^2 f (1+i \tan (e+f x))}+\frac {(i c-d) (c+d \tan (e+f x))^{3/2}}{4 f (a+i a \tan (e+f x))^2}+\frac {\int \frac {\frac {1}{2} a^2 \left (4 c^3-10 i c^2 d-7 c d^2-5 i d^3\right )+\frac {1}{2} a^2 d \left (2 c^2-5 i c d-9 d^2\right ) \tan (e+f x)}{\sqrt {c+d \tan (e+f x)}} \, dx}{8 a^4}\\ &=\frac {(c+i d) (2 i c+5 d) \sqrt {c+d \tan (e+f x)}}{8 a^2 f (1+i \tan (e+f x))}+\frac {(i c-d) (c+d \tan (e+f x))^{3/2}}{4 f (a+i a \tan (e+f x))^2}+\frac {(c-i d)^3 \int \frac {1+i \tan (e+f x)}{\sqrt {c+d \tan (e+f x)}} \, dx}{8 a^2}+\frac {\left ((c+i d) \left (2 c^2-6 i c d-7 d^2\right )\right ) \int \frac {1-i \tan (e+f x)}{\sqrt {c+d \tan (e+f x)}} \, dx}{16 a^2}\\ &=\frac {(c+i d) (2 i c+5 d) \sqrt {c+d \tan (e+f x)}}{8 a^2 f (1+i \tan (e+f x))}+\frac {(i c-d) (c+d \tan (e+f x))^{3/2}}{4 f (a+i a \tan (e+f x))^2}-\frac {(i c+d)^3 \text {Subst}\left (\int \frac {1}{(-1+x) \sqrt {c-i d x}} \, dx,x,i \tan (e+f x)\right )}{8 a^2 f}-\frac {\left ((c+i d) \left (2 i c^2+6 c d-7 i d^2\right )\right ) \text {Subst}\left (\int \frac {1}{(-1+x) \sqrt {c+i d x}} \, dx,x,-i \tan (e+f x)\right )}{16 a^2 f}\\ &=\frac {(c+i d) (2 i c+5 d) \sqrt {c+d \tan (e+f x)}}{8 a^2 f (1+i \tan (e+f x))}+\frac {(i c-d) (c+d \tan (e+f x))^{3/2}}{4 f (a+i a \tan (e+f x))^2}-\frac {(c-i d)^3 \text {Subst}\left (\int \frac {1}{-1-\frac {i c}{d}+\frac {i x^2}{d}} \, dx,x,\sqrt {c+d \tan (e+f x)}\right )}{4 a^2 d f}-\frac {\left ((c+i d) \left (2 c^2-6 i c d-7 d^2\right )\right ) \text {Subst}\left (\int \frac {1}{-1+\frac {i c}{d}-\frac {i x^2}{d}} \, dx,x,\sqrt {c+d \tan (e+f x)}\right )}{8 a^2 d f}\\ &=-\frac {i (c-i d)^{5/2} \tanh ^{-1}\left (\frac {\sqrt {c+d \tan (e+f x)}}{\sqrt {c-i d}}\right )}{4 a^2 f}+\frac {\sqrt {c+i d} \left (2 i c^2+6 c d-7 i d^2\right ) \tanh ^{-1}\left (\frac {\sqrt {c+d \tan (e+f x)}}{\sqrt {c+i d}}\right )}{8 a^2 f}+\frac {(c+i d) (2 i c+5 d) \sqrt {c+d \tan (e+f x)}}{8 a^2 f (1+i \tan (e+f x))}+\frac {(i c-d) (c+d \tan (e+f x))^{3/2}}{4 f (a+i a \tan (e+f x))^2}\\ \end {align*}

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Mathematica [A]
time = 2.12, size = 291, normalized size = 1.34 \begin {gather*} \frac {\sec ^2(e+f x) (\cos (f x)+i \sin (f x))^2 \left (\frac {2 \left (-i \sqrt {-c+i d} \left (2 c^3-4 i c^2 d-c d^2-7 i d^3\right ) \text {ArcTan}\left (\frac {\sqrt {c+d \tan (e+f x)}}{\sqrt {-c-i d}}\right )-2 \sqrt {-c-i d} (i c+d)^3 \text {ArcTan}\left (\frac {\sqrt {c+d \tan (e+f x)}}{\sqrt {-c+i d}}\right )\right ) (\cos (2 e)+i \sin (2 e))}{\sqrt {-c-i d} \sqrt {-c+i d}}+2 (c+i d) \cos (e+f x) (\cos (2 f x)-i \sin (2 f x)) ((4 i c+5 d) \cos (e+f x)+(-2 c+7 i d) \sin (e+f x)) \sqrt {c+d \tan (e+f x)}\right )}{16 f (a+i a \tan (e+f x))^2} \end {gather*}

Antiderivative was successfully verified.

[In]

Integrate[(c + d*Tan[e + f*x])^(5/2)/(a + I*a*Tan[e + f*x])^2,x]

[Out]

(Sec[e + f*x]^2*(Cos[f*x] + I*Sin[f*x])^2*((2*((-I)*Sqrt[-c + I*d]*(2*c^3 - (4*I)*c^2*d - c*d^2 - (7*I)*d^3)*A
rcTan[Sqrt[c + d*Tan[e + f*x]]/Sqrt[-c - I*d]] - 2*Sqrt[-c - I*d]*(I*c + d)^3*ArcTan[Sqrt[c + d*Tan[e + f*x]]/
Sqrt[-c + I*d]])*(Cos[2*e] + I*Sin[2*e]))/(Sqrt[-c - I*d]*Sqrt[-c + I*d]) + 2*(c + I*d)*Cos[e + f*x]*(Cos[2*f*
x] - I*Sin[2*f*x])*(((4*I)*c + 5*d)*Cos[e + f*x] + (-2*c + (7*I)*d)*Sin[e + f*x])*Sqrt[c + d*Tan[e + f*x]]))/(
16*f*(a + I*a*Tan[e + f*x])^2)

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Maple [A]
time = 0.36, size = 302, normalized size = 1.39

method	result	size
derivativedivides	\(\frac {2 d^{3} \left (-\frac {i \left (i d -c \right )^{\frac {5}{2}} \arctan \left (\frac {\sqrt {c +d \tan \left (f x +e \right )}}{\sqrt {i d -c}}\right )}{8 d^{3}}+\frac {i \left (\frac {-\frac {d \left (2 i c^{4}+15 i c^{2} d^{2}-7 i d^{4}+c^{3} d -19 c \,d^{3}\right ) \left (c +d \tan \left (f x +e \right )\right )^{\frac {3}{2}}}{2 \left (2 i c d +c^{2}-d^{2}\right )}+\frac {d \left (2 i c^{5}+8 i c^{3} d^{2}-18 i c \,d^{4}-3 c^{4} d -22 c^{2} d^{3}+5 d^{5}\right ) \sqrt {c +d \tan \left (f x +e \right )}}{4 i c d +2 c^{2}-2 d^{2}}}{\left (-d \tan \left (f x +e \right )+i d \right )^{2}}+\frac {\left (5 i c^{2} d^{3}-7 i d^{5}-2 c^{5}-5 c^{3} d^{2}-15 c \,d^{4}\right ) \arctan \left (\frac {\sqrt {c +d \tan \left (f x +e \right )}}{\sqrt {-i d -c}}\right )}{2 \left (2 i c d +c^{2}-d^{2}\right ) \sqrt {-i d -c}}\right )}{8 d^{3}}\right )}{f \,a^{2}}\)	\(302\)
default	\(\frac {2 d^{3} \left (-\frac {i \left (i d -c \right )^{\frac {5}{2}} \arctan \left (\frac {\sqrt {c +d \tan \left (f x +e \right )}}{\sqrt {i d -c}}\right )}{8 d^{3}}+\frac {i \left (\frac {-\frac {d \left (2 i c^{4}+15 i c^{2} d^{2}-7 i d^{4}+c^{3} d -19 c \,d^{3}\right ) \left (c +d \tan \left (f x +e \right )\right )^{\frac {3}{2}}}{2 \left (2 i c d +c^{2}-d^{2}\right )}+\frac {d \left (2 i c^{5}+8 i c^{3} d^{2}-18 i c \,d^{4}-3 c^{4} d -22 c^{2} d^{3}+5 d^{5}\right ) \sqrt {c +d \tan \left (f x +e \right )}}{4 i c d +2 c^{2}-2 d^{2}}}{\left (-d \tan \left (f x +e \right )+i d \right )^{2}}+\frac {\left (5 i c^{2} d^{3}-7 i d^{5}-2 c^{5}-5 c^{3} d^{2}-15 c \,d^{4}\right ) \arctan \left (\frac {\sqrt {c +d \tan \left (f x +e \right )}}{\sqrt {-i d -c}}\right )}{2 \left (2 i c d +c^{2}-d^{2}\right ) \sqrt {-i d -c}}\right )}{8 d^{3}}\right )}{f \,a^{2}}\)	\(302\)

Verification of antiderivative is not currently implemented for this CAS.

[In]

int((c+d*tan(f*x+e))^(5/2)/(a+I*a*tan(f*x+e))^2,x,method=_RETURNVERBOSE)

[Out]

2/f/a^2*d^3*(-1/8*I*(I*d-c)^(5/2)/d^3*arctan((c+d*tan(f*x+e))^(1/2)/(I*d-c)^(1/2))+1/8*I/d^3*((-1/2*d*(2*I*c^4
+15*I*c^2*d^2-7*I*d^4+c^3*d-19*c*d^3)/(2*I*c*d+c^2-d^2)*(c+d*tan(f*x+e))^(3/2)+1/2*d*(2*I*c^5+8*I*c^3*d^2-18*I
*c*d^4-3*c^4*d-22*c^2*d^3+5*d^5)/(2*I*c*d+c^2-d^2)*(c+d*tan(f*x+e))^(1/2))/(-d*tan(f*x+e)+I*d)^2+1/2*(-5*c^3*d
^2-15*c*d^4+5*I*c^2*d^3-7*I*d^5-2*c^5)/(2*I*c*d+c^2-d^2)/(-I*d-c)^(1/2)*arctan((c+d*tan(f*x+e))^(1/2)/(-I*d-c)
^(1/2))))

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Maxima [F(-2)]
time = 0.00, size = 0, normalized size = 0.00 \begin {gather*} \text {Exception raised: RuntimeError} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate((c+d*tan(f*x+e))^(5/2)/(a+I*a*tan(f*x+e))^2,x, algorithm="maxima")

[Out]

Exception raised: RuntimeError >> ECL says: Error executing code in Maxima: expt: undefined: 0 to a negative e
xponent.

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Fricas [B] Both result and optimal contain complex but leaf count of result is larger than twice the leaf count of optimal. 1112 vs. \(2 (173) = 346\).
time = 0.98, size = 1112, normalized size = 5.12 \begin {gather*} \frac {{\left (2 \, a^{2} f \sqrt {-\frac {c^{5} - 5 i \, c^{4} d - 10 \, c^{3} d^{2} + 10 i \, c^{2} d^{3} + 5 \, c d^{4} - i \, d^{5}}{a^{4} f^{2}}} e^{\left (4 i \, f x + 4 i \, e\right )} \log \left (\frac {2 \, {\left (c^{3} - 2 i \, c^{2} d - c d^{2} + {\left (i \, a^{2} f e^{\left (2 i \, f x + 2 i \, e\right )} + i \, a^{2} f\right )} \sqrt {\frac {{\left (c - i \, d\right )} e^{\left (2 i \, f x + 2 i \, e\right )} + c + i \, d}{e^{\left (2 i \, f x + 2 i \, e\right )} + 1}} \sqrt {-\frac {c^{5} - 5 i \, c^{4} d - 10 \, c^{3} d^{2} + 10 i \, c^{2} d^{3} + 5 \, c d^{4} - i \, d^{5}}{a^{4} f^{2}}} + {\left (c^{3} - 3 i \, c^{2} d - 3 \, c d^{2} + i \, d^{3}\right )} e^{\left (2 i \, f x + 2 i \, e\right )}\right )} e^{\left (-2 i \, f x - 2 i \, e\right )}}{c^{2} - 2 i \, c d - d^{2}}\right ) - 2 \, a^{2} f \sqrt {-\frac {c^{5} - 5 i \, c^{4} d - 10 \, c^{3} d^{2} + 10 i \, c^{2} d^{3} + 5 \, c d^{4} - i \, d^{5}}{a^{4} f^{2}}} e^{\left (4 i \, f x + 4 i \, e\right )} \log \left (\frac {2 \, {\left (c^{3} - 2 i \, c^{2} d - c d^{2} + {\left (-i \, a^{2} f e^{\left (2 i \, f x + 2 i \, e\right )} - i \, a^{2} f\right )} \sqrt {\frac {{\left (c - i \, d\right )} e^{\left (2 i \, f x + 2 i \, e\right )} + c + i \, d}{e^{\left (2 i \, f x + 2 i \, e\right )} + 1}} \sqrt {-\frac {c^{5} - 5 i \, c^{4} d - 10 \, c^{3} d^{2} + 10 i \, c^{2} d^{3} + 5 \, c d^{4} - i \, d^{5}}{a^{4} f^{2}}} + {\left (c^{3} - 3 i \, c^{2} d - 3 \, c d^{2} + i \, d^{3}\right )} e^{\left (2 i \, f x + 2 i \, e\right )}\right )} e^{\left (-2 i \, f x - 2 i \, e\right )}}{c^{2} - 2 i \, c d - d^{2}}\right ) + a^{2} f \sqrt {-\frac {4 \, c^{5} - 20 i \, c^{4} d - 40 \, c^{3} d^{2} + 20 i \, c^{2} d^{3} - 35 \, c d^{4} + 49 i \, d^{5}}{a^{4} f^{2}}} e^{\left (4 i \, f x + 4 i \, e\right )} \log \left (\frac {{\left (2 i \, c^{3} + 4 \, c^{2} d - i \, c d^{2} + 7 \, d^{3} + {\left (a^{2} f e^{\left (2 i \, f x + 2 i \, e\right )} + a^{2} f\right )} \sqrt {\frac {{\left (c - i \, d\right )} e^{\left (2 i \, f x + 2 i \, e\right )} + c + i \, d}{e^{\left (2 i \, f x + 2 i \, e\right )} + 1}} \sqrt {-\frac {4 \, c^{5} - 20 i \, c^{4} d - 40 \, c^{3} d^{2} + 20 i \, c^{2} d^{3} - 35 \, c d^{4} + 49 i \, d^{5}}{a^{4} f^{2}}} + {\left (2 i \, c^{3} + 6 \, c^{2} d - 7 i \, c d^{2}\right )} e^{\left (2 i \, f x + 2 i \, e\right )}\right )} e^{\left (-2 i \, f x - 2 i \, e\right )}}{8 \, a^{2} f}\right ) - a^{2} f \sqrt {-\frac {4 \, c^{5} - 20 i \, c^{4} d - 40 \, c^{3} d^{2} + 20 i \, c^{2} d^{3} - 35 \, c d^{4} + 49 i \, d^{5}}{a^{4} f^{2}}} e^{\left (4 i \, f x + 4 i \, e\right )} \log \left (\frac {{\left (2 i \, c^{3} + 4 \, c^{2} d - i \, c d^{2} + 7 \, d^{3} - {\left (a^{2} f e^{\left (2 i \, f x + 2 i \, e\right )} + a^{2} f\right )} \sqrt {\frac {{\left (c - i \, d\right )} e^{\left (2 i \, f x + 2 i \, e\right )} + c + i \, d}{e^{\left (2 i \, f x + 2 i \, e\right )} + 1}} \sqrt {-\frac {4 \, c^{5} - 20 i \, c^{4} d - 40 \, c^{3} d^{2} + 20 i \, c^{2} d^{3} - 35 \, c d^{4} + 49 i \, d^{5}}{a^{4} f^{2}}} + {\left (2 i \, c^{3} + 6 \, c^{2} d - 7 i \, c d^{2}\right )} e^{\left (2 i \, f x + 2 i \, e\right )}\right )} e^{\left (-2 i \, f x - 2 i \, e\right )}}{8 \, a^{2} f}\right ) + 2 \, {\left (i \, c^{2} - 2 \, c d - i \, d^{2} - 3 \, {\left (-i \, c^{2} - c d - 2 i \, d^{2}\right )} e^{\left (4 i \, f x + 4 i \, e\right )} + {\left (4 i \, c^{2} + c d + 5 i \, d^{2}\right )} e^{\left (2 i \, f x + 2 i \, e\right )}\right )} \sqrt {\frac {{\left (c - i \, d\right )} e^{\left (2 i \, f x + 2 i \, e\right )} + c + i \, d}{e^{\left (2 i \, f x + 2 i \, e\right )} + 1}}\right )} e^{\left (-4 i \, f x - 4 i \, e\right )}}{32 \, a^{2} f} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate((c+d*tan(f*x+e))^(5/2)/(a+I*a*tan(f*x+e))^2,x, algorithm="fricas")

[Out]

1/32*(2*a^2*f*sqrt(-(c^5 - 5*I*c^4*d - 10*c^3*d^2 + 10*I*c^2*d^3 + 5*c*d^4 - I*d^5)/(a^4*f^2))*e^(4*I*f*x + 4*
I*e)*log(2*(c^3 - 2*I*c^2*d - c*d^2 + (I*a^2*f*e^(2*I*f*x + 2*I*e) + I*a^2*f)*sqrt(((c - I*d)*e^(2*I*f*x + 2*I
*e) + c + I*d)/(e^(2*I*f*x + 2*I*e) + 1))*sqrt(-(c^5 - 5*I*c^4*d - 10*c^3*d^2 + 10*I*c^2*d^3 + 5*c*d^4 - I*d^5
)/(a^4*f^2)) + (c^3 - 3*I*c^2*d - 3*c*d^2 + I*d^3)*e^(2*I*f*x + 2*I*e))*e^(-2*I*f*x - 2*I*e)/(c^2 - 2*I*c*d -
d^2)) - 2*a^2*f*sqrt(-(c^5 - 5*I*c^4*d - 10*c^3*d^2 + 10*I*c^2*d^3 + 5*c*d^4 - I*d^5)/(a^4*f^2))*e^(4*I*f*x +
4*I*e)*log(2*(c^3 - 2*I*c^2*d - c*d^2 + (-I*a^2*f*e^(2*I*f*x + 2*I*e) - I*a^2*f)*sqrt(((c - I*d)*e^(2*I*f*x +
2*I*e) + c + I*d)/(e^(2*I*f*x + 2*I*e) + 1))*sqrt(-(c^5 - 5*I*c^4*d - 10*c^3*d^2 + 10*I*c^2*d^3 + 5*c*d^4 - I*
d^5)/(a^4*f^2)) + (c^3 - 3*I*c^2*d - 3*c*d^2 + I*d^3)*e^(2*I*f*x + 2*I*e))*e^(-2*I*f*x - 2*I*e)/(c^2 - 2*I*c*d
- d^2)) + a^2*f*sqrt(-(4*c^5 - 20*I*c^4*d - 40*c^3*d^2 + 20*I*c^2*d^3 - 35*c*d^4 + 49*I*d^5)/(a^4*f^2))*e^(4*
I*f*x + 4*I*e)*log(1/8*(2*I*c^3 + 4*c^2*d - I*c*d^2 + 7*d^3 + (a^2*f*e^(2*I*f*x + 2*I*e) + a^2*f)*sqrt(((c - I
*d)*e^(2*I*f*x + 2*I*e) + c + I*d)/(e^(2*I*f*x + 2*I*e) + 1))*sqrt(-(4*c^5 - 20*I*c^4*d - 40*c^3*d^2 + 20*I*c^
2*d^3 - 35*c*d^4 + 49*I*d^5)/(a^4*f^2)) + (2*I*c^3 + 6*c^2*d - 7*I*c*d^2)*e^(2*I*f*x + 2*I*e))*e^(-2*I*f*x - 2
*I*e)/(a^2*f)) - a^2*f*sqrt(-(4*c^5 - 20*I*c^4*d - 40*c^3*d^2 + 20*I*c^2*d^3 - 35*c*d^4 + 49*I*d^5)/(a^4*f^2))
*e^(4*I*f*x + 4*I*e)*log(1/8*(2*I*c^3 + 4*c^2*d - I*c*d^2 + 7*d^3 - (a^2*f*e^(2*I*f*x + 2*I*e) + a^2*f)*sqrt((
(c - I*d)*e^(2*I*f*x + 2*I*e) + c + I*d)/(e^(2*I*f*x + 2*I*e) + 1))*sqrt(-(4*c^5 - 20*I*c^4*d - 40*c^3*d^2 + 2
0*I*c^2*d^3 - 35*c*d^4 + 49*I*d^5)/(a^4*f^2)) + (2*I*c^3 + 6*c^2*d - 7*I*c*d^2)*e^(2*I*f*x + 2*I*e))*e^(-2*I*f
*x - 2*I*e)/(a^2*f)) + 2*(I*c^2 - 2*c*d - I*d^2 - 3*(-I*c^2 - c*d - 2*I*d^2)*e^(4*I*f*x + 4*I*e) + (4*I*c^2 +
c*d + 5*I*d^2)*e^(2*I*f*x + 2*I*e))*sqrt(((c - I*d)*e^(2*I*f*x + 2*I*e) + c + I*d)/(e^(2*I*f*x + 2*I*e) + 1)))
*e^(-4*I*f*x - 4*I*e)/(a^2*f)

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Sympy [F]
time = 0.00, size = 0, normalized size = 0.00 \begin {gather*} - \frac {\int \frac {c^{2} \sqrt {c + d \tan {\left (e + f x \right )}}}{\tan ^{2}{\left (e + f x \right )} - 2 i \tan {\left (e + f x \right )} - 1}\, dx + \int \frac {d^{2} \sqrt {c + d \tan {\left (e + f x \right )}} \tan ^{2}{\left (e + f x \right )}}{\tan ^{2}{\left (e + f x \right )} - 2 i \tan {\left (e + f x \right )} - 1}\, dx + \int \frac {2 c d \sqrt {c + d \tan {\left (e + f x \right )}} \tan {\left (e + f x \right )}}{\tan ^{2}{\left (e + f x \right )} - 2 i \tan {\left (e + f x \right )} - 1}\, dx}{a^{2}} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate((c+d*tan(f*x+e))**(5/2)/(a+I*a*tan(f*x+e))**2,x)

[Out]

-(Integral(c**2*sqrt(c + d*tan(e + f*x))/(tan(e + f*x)**2 - 2*I*tan(e + f*x) - 1), x) + Integral(d**2*sqrt(c +
d*tan(e + f*x))*tan(e + f*x)**2/(tan(e + f*x)**2 - 2*I*tan(e + f*x) - 1), x) + Integral(2*c*d*sqrt(c + d*tan(
e + f*x))*tan(e + f*x)/(tan(e + f*x)**2 - 2*I*tan(e + f*x) - 1), x))/a**2

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Giac [B] Both result and optimal contain complex but leaf count of result is larger than twice the leaf count of optimal. 520 vs. \(2 (173) = 346\).
time = 0.78, size = 520, normalized size = 2.40 \begin {gather*} \frac {{\left (c^{3} - 3 i \, c^{2} d - 3 \, c d^{2} + i \, d^{3}\right )} \arctan \left (-\frac {2 \, {\left (i \, \sqrt {d \tan \left (f x + e\right ) + c} c + i \, \sqrt {c^{2} + d^{2}} \sqrt {d \tan \left (f x + e\right ) + c}\right )}}{\sqrt {2 \, c + 2 \, \sqrt {c^{2} + d^{2}}} c - i \, \sqrt {2 \, c + 2 \, \sqrt {c^{2} + d^{2}}} d + \sqrt {c^{2} + d^{2}} \sqrt {2 \, c + 2 \, \sqrt {c^{2} + d^{2}}}}\right )}{2 \, a^{2} \sqrt {2 \, c + 2 \, \sqrt {c^{2} + d^{2}}} f {\left (-\frac {i \, d}{c + \sqrt {c^{2} + d^{2}}} + 1\right )}} - \frac {{\left (2 i \, c^{3} + 4 \, c^{2} d - i \, c d^{2} + 7 \, d^{3}\right )} \arctan \left (\frac {2 \, {\left (\sqrt {d \tan \left (f x + e\right ) + c} c - \sqrt {c^{2} + d^{2}} \sqrt {d \tan \left (f x + e\right ) + c}\right )}}{c \sqrt {-2 \, c + 2 \, \sqrt {c^{2} + d^{2}}} + i \, \sqrt {-2 \, c + 2 \, \sqrt {c^{2} + d^{2}}} d - \sqrt {c^{2} + d^{2}} \sqrt {-2 \, c + 2 \, \sqrt {c^{2} + d^{2}}}}\right )}{4 \, a^{2} \sqrt {-2 \, c + 2 \, \sqrt {c^{2} + d^{2}}} f {\left (\frac {i \, d}{c - \sqrt {c^{2} + d^{2}}} + 1\right )}} + \frac {2 \, {\left (d \tan \left (f x + e\right ) + c\right )}^{\frac {3}{2}} c^{2} d - 2 \, \sqrt {d \tan \left (f x + e\right ) + c} c^{3} d - 5 i \, {\left (d \tan \left (f x + e\right ) + c\right )}^{\frac {3}{2}} c d^{2} + i \, \sqrt {d \tan \left (f x + e\right ) + c} c^{2} d^{2} + 7 \, {\left (d \tan \left (f x + e\right ) + c\right )}^{\frac {3}{2}} d^{3} - 8 \, \sqrt {d \tan \left (f x + e\right ) + c} c d^{3} - 5 i \, \sqrt {d \tan \left (f x + e\right ) + c} d^{4}}{8 \, {\left (d \tan \left (f x + e\right ) - i \, d\right )}^{2} a^{2} f} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate((c+d*tan(f*x+e))^(5/2)/(a+I*a*tan(f*x+e))^2,x, algorithm="giac")

[Out]

1/2*(c^3 - 3*I*c^2*d - 3*c*d^2 + I*d^3)*arctan(-2*(I*sqrt(d*tan(f*x + e) + c)*c + I*sqrt(c^2 + d^2)*sqrt(d*tan
(f*x + e) + c))/(sqrt(2*c + 2*sqrt(c^2 + d^2))*c - I*sqrt(2*c + 2*sqrt(c^2 + d^2))*d + sqrt(c^2 + d^2)*sqrt(2*
c + 2*sqrt(c^2 + d^2))))/(a^2*sqrt(2*c + 2*sqrt(c^2 + d^2))*f*(-I*d/(c + sqrt(c^2 + d^2)) + 1)) - 1/4*(2*I*c^3
+ 4*c^2*d - I*c*d^2 + 7*d^3)*arctan(2*(sqrt(d*tan(f*x + e) + c)*c - sqrt(c^2 + d^2)*sqrt(d*tan(f*x + e) + c))
/(c*sqrt(-2*c + 2*sqrt(c^2 + d^2)) + I*sqrt(-2*c + 2*sqrt(c^2 + d^2))*d - sqrt(c^2 + d^2)*sqrt(-2*c + 2*sqrt(c
^2 + d^2))))/(a^2*sqrt(-2*c + 2*sqrt(c^2 + d^2))*f*(I*d/(c - sqrt(c^2 + d^2)) + 1)) + 1/8*(2*(d*tan(f*x + e) +
c)^(3/2)*c^2*d - 2*sqrt(d*tan(f*x + e) + c)*c^3*d - 5*I*(d*tan(f*x + e) + c)^(3/2)*c*d^2 + I*sqrt(d*tan(f*x +
e) + c)*c^2*d^2 + 7*(d*tan(f*x + e) + c)^(3/2)*d^3 - 8*sqrt(d*tan(f*x + e) + c)*c*d^3 - 5*I*sqrt(d*tan(f*x +
e) + c)*d^4)/((d*tan(f*x + e) - I*d)^2*a^2*f)

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Mupad [B]
time = 9.19, size = 2500, normalized size = 11.52 \begin {gather*} \text {Too large to display} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

int((c + d*tan(e + f*x))^(5/2)/(a + a*tan(e + f*x)*1i)^2,x)

[Out]

- atan((((a^2*f*(a^4*d^6*f^2*640i - 256*a^4*c*d^5*f^2 + a^4*c^2*d^4*f^2*640i - 256*a^4*c^3*d^3*f^2) - 4096*a^8
*c*d^2*f^4*(c + d*tan(e + f*x))^(1/2)*(-(d^11*45i - 15*c*d^10 + c^2*d^9*105i - 95*c^3*d^8 + c^4*d^7*20i - 72*c
^5*d^6 - c^6*d^5*40i + 8*c^7*d^4 + a^4*f^2*(4*(256*d^6 + 256*c^2*d^4)*((((105*c^3*d^13)/16 - (35*c*d^15)/8 + (
295*c^5*d^11)/16 + (5*c^7*d^9)/16 - (105*c^9*d^7)/16 + (5*c^11*d^5)/8)*1i)/(a^8*f^4) - ((49*d^16)/64 - (149*c^
2*d^14)/16 - (235*c^4*d^12)/32 + (215*c^6*d^10)/16 + (505*c^8*d^8)/64 - (11*c^10*d^6)/4 + (c^12*d^4)/16)/(a^8*
f^4)) + (((45*d^11 + 105*c^2*d^9 + 20*c^4*d^7 - 40*c^6*d^5)*1i)/(a^4*f^2) - (15*c*d^10 + 95*c^3*d^8 + 72*c^5*d
^6 - 8*c^7*d^4)/(a^4*f^2))^2)^(1/2))/(512*a^4*f^2*(d^6 + c^2*d^4)))^(1/2))*(-(d^11*45i - 15*c*d^10 + c^2*d^9*1
05i - 95*c^3*d^8 + c^4*d^7*20i - 72*c^5*d^6 - c^6*d^5*40i + 8*c^7*d^4 + a^4*f^2*(4*(256*d^6 + 256*c^2*d^4)*(((
(105*c^3*d^13)/16 - (35*c*d^15)/8 + (295*c^5*d^11)/16 + (5*c^7*d^9)/16 - (105*c^9*d^7)/16 + (5*c^11*d^5)/8)*1i
)/(a^8*f^4) - ((49*d^16)/64 - (149*c^2*d^14)/16 - (235*c^4*d^12)/32 + (215*c^6*d^10)/16 + (505*c^8*d^8)/64 - (
11*c^10*d^6)/4 + (c^12*d^4)/16)/(a^8*f^4)) + (((45*d^11 + 105*c^2*d^9 + 20*c^4*d^7 - 40*c^6*d^5)*1i)/(a^4*f^2)
- (15*c*d^10 + 95*c^3*d^8 + 72*c^5*d^6 - 8*c^7*d^4)/(a^4*f^2))^2)^(1/2))/(512*a^4*f^2*(d^6 + c^2*d^4)))^(1/2)
+ 8*a^4*f^2*(c + d*tan(e + f*x))^(1/2)*(c*d^7*10i + 53*d^8 - 5*c^2*d^6 - c^3*d^5*60i + 80*c^4*d^4 + c^5*d^3*4
0i - 8*c^6*d^2))*(-(d^11*45i - 15*c*d^10 + c^2*d^9*105i - 95*c^3*d^8 + c^4*d^7*20i - 72*c^5*d^6 - c^6*d^5*40i
+ 8*c^7*d^4 + a^4*f^2*(4*(256*d^6 + 256*c^2*d^4)*((((105*c^3*d^13)/16 - (35*c*d^15)/8 + (295*c^5*d^11)/16 + (5
*c^7*d^9)/16 - (105*c^9*d^7)/16 + (5*c^11*d^5)/8)*1i)/(a^8*f^4) - ((49*d^16)/64 - (149*c^2*d^14)/16 - (235*c^4
*d^12)/32 + (215*c^6*d^10)/16 + (505*c^8*d^8)/64 - (11*c^10*d^6)/4 + (c^12*d^4)/16)/(a^8*f^4)) + (((45*d^11 +
105*c^2*d^9 + 20*c^4*d^7 - 40*c^6*d^5)*1i)/(a^4*f^2) - (15*c*d^10 + 95*c^3*d^8 + 72*c^5*d^6 - 8*c^7*d^4)/(a^4*
f^2))^2)^(1/2))/(512*a^4*f^2*(d^6 + c^2*d^4)))^(1/2)*1i - ((a^2*f*(a^4*d^6*f^2*640i - 256*a^4*c*d^5*f^2 + a^4*
c^2*d^4*f^2*640i - 256*a^4*c^3*d^3*f^2) + 4096*a^8*c*d^2*f^4*(c + d*tan(e + f*x))^(1/2)*(-(d^11*45i - 15*c*d^1
0 + c^2*d^9*105i - 95*c^3*d^8 + c^4*d^7*20i - 72*c^5*d^6 - c^6*d^5*40i + 8*c^7*d^4 + a^4*f^2*(4*(256*d^6 + 256
*c^2*d^4)*((((105*c^3*d^13)/16 - (35*c*d^15)/8 + (295*c^5*d^11)/16 + (5*c^7*d^9)/16 - (105*c^9*d^7)/16 + (5*c^
11*d^5)/8)*1i)/(a^8*f^4) - ((49*d^16)/64 - (149*c^2*d^14)/16 - (235*c^4*d^12)/32 + (215*c^6*d^10)/16 + (505*c^
8*d^8)/64 - (11*c^10*d^6)/4 + (c^12*d^4)/16)/(a^8*f^4)) + (((45*d^11 + 105*c^2*d^9 + 20*c^4*d^7 - 40*c^6*d^5)*
1i)/(a^4*f^2) - (15*c*d^10 + 95*c^3*d^8 + 72*c^5*d^6 - 8*c^7*d^4)/(a^4*f^2))^2)^(1/2))/(512*a^4*f^2*(d^6 + c^2
*d^4)))^(1/2))*(-(d^11*45i - 15*c*d^10 + c^2*d^9*105i - 95*c^3*d^8 + c^4*d^7*20i - 72*c^5*d^6 - c^6*d^5*40i +
8*c^7*d^4 + a^4*f^2*(4*(256*d^6 + 256*c^2*d^4)*((((105*c^3*d^13)/16 - (35*c*d^15)/8 + (295*c^5*d^11)/16 + (5*c
^7*d^9)/16 - (105*c^9*d^7)/16 + (5*c^11*d^5)/8)*1i)/(a^8*f^4) - ((49*d^16)/64 - (149*c^2*d^14)/16 - (235*c^4*d
^12)/32 + (215*c^6*d^10)/16 + (505*c^8*d^8)/64 - (11*c^10*d^6)/4 + (c^12*d^4)/16)/(a^8*f^4)) + (((45*d^11 + 10
5*c^2*d^9 + 20*c^4*d^7 - 40*c^6*d^5)*1i)/(a^4*f^2) - (15*c*d^10 + 95*c^3*d^8 + 72*c^5*d^6 - 8*c^7*d^4)/(a^4*f^
2))^2)^(1/2))/(512*a^4*f^2*(d^6 + c^2*d^4)))^(1/2) - 8*a^4*f^2*(c + d*tan(e + f*x))^(1/2)*(c*d^7*10i + 53*d^8
- 5*c^2*d^6 - c^3*d^5*60i + 80*c^4*d^4 + c^5*d^3*40i - 8*c^6*d^2))*(-(d^11*45i - 15*c*d^10 + c^2*d^9*105i - 95
*c^3*d^8 + c^4*d^7*20i - 72*c^5*d^6 - c^6*d^5*40i + 8*c^7*d^4 + a^4*f^2*(4*(256*d^6 + 256*c^2*d^4)*((((105*c^3
*d^13)/16 - (35*c*d^15)/8 + (295*c^5*d^11)/16 + (5*c^7*d^9)/16 - (105*c^9*d^7)/16 + (5*c^11*d^5)/8)*1i)/(a^8*f
^4) - ((49*d^16)/64 - (149*c^2*d^14)/16 - (235*c^4*d^12)/32 + (215*c^6*d^10)/16 + (505*c^8*d^8)/64 - (11*c^10*
d^6)/4 + (c^12*d^4)/16)/(a^8*f^4)) + (((45*d^11 + 105*c^2*d^9 + 20*c^4*d^7 - 40*c^6*d^5)*1i)/(a^4*f^2) - (15*c
*d^10 + 95*c^3*d^8 + 72*c^5*d^6 - 8*c^7*d^4)/(a^4*f^2))^2)^(1/2))/(512*a^4*f^2*(d^6 + c^2*d^4)))^(1/2)*1i)/(((
a^2*f*(a^4*d^6*f^2*640i - 256*a^4*c*d^5*f^2 + a^4*c^2*d^4*f^2*640i - 256*a^4*c^3*d^3*f^2) - 4096*a^8*c*d^2*f^4
*(c + d*tan(e + f*x))^(1/2)*(-(d^11*45i - 15*c*d^10 + c^2*d^9*105i - 95*c^3*d^8 + c^4*d^7*20i - 72*c^5*d^6 - c
^6*d^5*40i + 8*c^7*d^4 + a^4*f^2*(4*(256*d^6 + 256*c^2*d^4)*((((105*c^3*d^13)/16 - (35*c*d^15)/8 + (295*c^5*d^
11)/16 + (5*c^7*d^9)/16 - (105*c^9*d^7)/16 + (5*c^11*d^5)/8)*1i)/(a^8*f^4) - ((49*d^16)/64 - (149*c^2*d^14)/16
- (235*c^4*d^12)/32 + (215*c^6*d^10)/16 + (505*c^8*d^8)/64 - (11*c^10*d^6)/4 + (c^12*d^4)/16)/(a^8*f^4)) + ((
(45*d^11 + 105*c^2*d^9 + 20*c^4*d^7 - 40*c^6*d^5)*1i)/(a^4*f^2) - (15*c*d^10 + 95*c^3*d^8 + 72*c^5*d^6 - 8*c^7
*d^4)/(a^4*f^2))^2)^(1/2))/(512*a^4*f^2*(d^6 + c^2*d^4)))^(1/2))*(-(d^11*45i - 15*c*d^10 + c^2*d^9*105i - 95*c
^3*d^8 + c^4*d^7*20i - 72*c^5*d^6 - c^6*d^5*40i + 8*c^7*d^4 + a^4*f^2*(4*(256*d^6 + 256*c^2*d^4)*((((105*c^3*d
^13)/16 - (35*c*d^15)/8 + (295*c^5*d^11)/16 + (...

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