3.12.16 \(\int \frac {(c+d \tan (e+f x))^{5/2}}{a+i a \tan (e+f x)} \, dx\) [1116]

Optimal. Leaf size=185 \[ -\frac {i (c-i d)^{5/2} \tanh ^{-1}\left (\frac {\sqrt {c+d \tan (e+f x)}}{\sqrt {c-i d}}\right )}{2 a f}+\frac {(c+i d)^{3/2} (i c+4 d) \tanh ^{-1}\left (\frac {\sqrt {c+d \tan (e+f x)}}{\sqrt {c+i d}}\right )}{2 a f}-\frac {(c+5 i d) d \sqrt {c+d \tan (e+f x)}}{2 a f}+\frac {(i c-d) (c+d \tan (e+f x))^{3/2}}{2 f (a+i a \tan (e+f x))} \]

[Out]

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

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Rubi [A]
time = 0.29, antiderivative size = 185, 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 = {3631, 3609, 3620, 3618, 65, 214} \begin {gather*} \frac {(-d+i c) (c+d \tan (e+f x))^{3/2}}{2 f (a+i a \tan (e+f x))}-\frac {d (c+5 i d) \sqrt {c+d \tan (e+f x)}}{2 a f}-\frac {i (c-i d)^{5/2} \tanh ^{-1}\left (\frac {\sqrt {c+d \tan (e+f x)}}{\sqrt {c-i d}}\right )}{2 a f}+\frac {(c+i d)^{3/2} (4 d+i c) \tanh ^{-1}\left (\frac {\sqrt {c+d \tan (e+f x)}}{\sqrt {c+i d}}\right )}{2 a f} \end {gather*}

Antiderivative was successfully verified.

[In]

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

[Out]

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

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 3609

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

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 3631

Int[((c_.) + (d_.)*tan[(e_.) + (f_.)*(x_)])^(n_)/((a_) + (b_.)*tan[(e_.) + (f_.)*(x_)]), x_Symbol] :> Simp[(b*
c - a*d)*((c + d*Tan[e + f*x])^(n - 1)/(2*a*f*(a + b*Tan[e + f*x]))), x] + Dist[1/(2*a^2), Int[(c + d*Tan[e +
f*x])^(n - 2)*Simp[a*c^2 + a*d^2*(n - 1) - b*c*d*n - d*(a*c*(n - 2) + b*d*n)*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] && GtQ[n, 1]

Rubi steps

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

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Mathematica [A]
time = 2.15, size = 260, normalized size = 1.41 \begin {gather*} \frac {\sec (e+f x) (\cos (f x)+i \sin (f x)) \left (\frac {2 \left (-i \sqrt {-c+i d} (c+i d)^2 (c-4 i d) \text {ArcTan}\left (\frac {\sqrt {c+d \tan (e+f x)}}{\sqrt {-c-i d}}\right )-\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 (e)+i \sin (e))}{\sqrt {-c-i d} \sqrt {-c+i d}}+2 (i \cos (f x)+\sin (f x)) \left (\left (c^2+2 i c d-5 d^2\right ) \cos (e+f x)-4 i d^2 \sin (e+f x)\right ) \sqrt {c+d \tan (e+f x)}\right )}{4 f (a+i a \tan (e+f x))} \end {gather*}

Antiderivative was successfully verified.

[In]

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

[Out]

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

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Maple [A]
time = 0.32, size = 237, normalized size = 1.28

method	result	size
derivativedivides	\(\frac {2 d^{2} \left (-i \sqrt {c +d \tan \left (f x +e \right )}+\frac {\left (i c^{3}-3 i c \,d^{2}+3 c^{2} d -d^{3}\right ) \arctan \left (\frac {\sqrt {c +d \tan \left (f x +e \right )}}{\sqrt {i d -c}}\right )}{4 d^{2} \sqrt {i d -c}}+\frac {-\frac {d \left (3 i c^{2} d -i d^{3}+c^{3}-3 c \,d^{2}\right ) \sqrt {c +d \tan \left (f x +e \right )}}{\left (i d +c \right ) \left (-d \tan \left (f x +e \right )+i d \right )}-\frac {\left (i c^{4}+9 i c^{2} d^{2}-4 i d^{4}+c^{3} d -11 c \,d^{3}\right ) \arctan \left (\frac {\sqrt {c +d \tan \left (f x +e \right )}}{\sqrt {-i d -c}}\right )}{\left (i d +c \right ) \sqrt {-i d -c}}}{4 d^{2}}\right )}{f a}\)	\(237\)
default	\(\frac {2 d^{2} \left (-i \sqrt {c +d \tan \left (f x +e \right )}+\frac {\left (i c^{3}-3 i c \,d^{2}+3 c^{2} d -d^{3}\right ) \arctan \left (\frac {\sqrt {c +d \tan \left (f x +e \right )}}{\sqrt {i d -c}}\right )}{4 d^{2} \sqrt {i d -c}}+\frac {-\frac {d \left (3 i c^{2} d -i d^{3}+c^{3}-3 c \,d^{2}\right ) \sqrt {c +d \tan \left (f x +e \right )}}{\left (i d +c \right ) \left (-d \tan \left (f x +e \right )+i d \right )}-\frac {\left (i c^{4}+9 i c^{2} d^{2}-4 i d^{4}+c^{3} d -11 c \,d^{3}\right ) \arctan \left (\frac {\sqrt {c +d \tan \left (f x +e \right )}}{\sqrt {-i d -c}}\right )}{\left (i d +c \right ) \sqrt {-i d -c}}}{4 d^{2}}\right )}{f a}\)	\(237\)

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)),x,method=_RETURNVERBOSE)

[Out]

2/f/a*d^2*(-I*(c+d*tan(f*x+e))^(1/2)+1/4*(-3*I*c*d^2-d^3+I*c^3+3*c^2*d)/d^2/(I*d-c)^(1/2)*arctan((c+d*tan(f*x+
e))^(1/2)/(I*d-c)^(1/2))+1/4/d^2*(-d*(3*I*c^2*d-I*d^3+c^3-3*c*d^2)/(c+I*d)*(c+d*tan(f*x+e))^(1/2)/(-d*tan(f*x+
e)+I*d)-(c^3*d-11*c*d^3+I*c^4+9*I*c^2*d^2-4*I*d^4)/(c+I*d)/(-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)),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. 1053 vs. \(2 (146) = 292\).
time = 0.99, size = 1053, normalized size = 5.69 \begin {gather*} \frac {{\left (a 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^{2} f^{2}}} e^{\left (2 i \, f x + 2 i \, e\right )} \log \left (\frac {2 \, {\left (c^{3} - 2 i \, c^{2} d - c d^{2} + {\left (i \, a f e^{\left (2 i \, f x + 2 i \, e\right )} + i \, a 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^{2} 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 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^{2} f^{2}}} e^{\left (2 i \, f x + 2 i \, e\right )} \log \left (\frac {2 \, {\left (c^{3} - 2 i \, c^{2} d - c d^{2} + {\left (-i \, a f e^{\left (2 i \, f x + 2 i \, e\right )} - i \, a 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^{2} 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 f \sqrt {-\frac {c^{5} - 5 i \, c^{4} d + 5 \, c^{3} d^{2} - 25 i \, c^{2} d^{3} + 40 \, c d^{4} + 16 i \, d^{5}}{a^{2} f^{2}}} e^{\left (2 i \, f x + 2 i \, e\right )} \log \left (\frac {{\left (i \, c^{3} + 2 \, c^{2} d + 7 i \, c d^{2} - 4 \, d^{3} + {\left (a f e^{\left (2 i \, f x + 2 i \, e\right )} + a 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 + 5 \, c^{3} d^{2} - 25 i \, c^{2} d^{3} + 40 \, c d^{4} + 16 i \, d^{5}}{a^{2} f^{2}}} + {\left (i \, c^{3} + 3 \, c^{2} d + 4 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 )}}{2 \, a f}\right ) - a f \sqrt {-\frac {c^{5} - 5 i \, c^{4} d + 5 \, c^{3} d^{2} - 25 i \, c^{2} d^{3} + 40 \, c d^{4} + 16 i \, d^{5}}{a^{2} f^{2}}} e^{\left (2 i \, f x + 2 i \, e\right )} \log \left (\frac {{\left (i \, c^{3} + 2 \, c^{2} d + 7 i \, c d^{2} - 4 \, d^{3} - {\left (a f e^{\left (2 i \, f x + 2 i \, e\right )} + a 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 + 5 \, c^{3} d^{2} - 25 i \, c^{2} d^{3} + 40 \, c d^{4} + 16 i \, d^{5}}{a^{2} f^{2}}} + {\left (i \, c^{3} + 3 \, c^{2} d + 4 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 )}}{2 \, a f}\right ) + 2 \, {\left (i \, c^{2} - 2 \, c d - i \, d^{2} + {\left (i \, c^{2} - 2 \, c d - 9 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 (-2 i \, f x - 2 i \, e\right )}}{8 \, a 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)),x, algorithm="fricas")

[Out]

1/8*(a*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^2*f^2))*e^(2*I*f*x + 2*I*e)*
log(2*(c^3 - 2*I*c^2*d - c*d^2 + (I*a*f*e^(2*I*f*x + 2*I*e) + I*a*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^2*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
*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^2*f^2))*e^(2*I*f*x + 2*I*e)*log(2*
(c^3 - 2*I*c^2*d - c*d^2 + (-I*a*f*e^(2*I*f*x + 2*I*e) - I*a*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^2*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*f*sq
rt(-(c^5 - 5*I*c^4*d + 5*c^3*d^2 - 25*I*c^2*d^3 + 40*c*d^4 + 16*I*d^5)/(a^2*f^2))*e^(2*I*f*x + 2*I*e)*log(1/2*
(I*c^3 + 2*c^2*d + 7*I*c*d^2 - 4*d^3 + (a*f*e^(2*I*f*x + 2*I*e) + a*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 + 5*c^3*d^2 - 25*I*c^2*d^3 + 40*c*d^4 + 16*I*d^5)/(a
^2*f^2)) + (I*c^3 + 3*c^2*d + 4*I*c*d^2)*e^(2*I*f*x + 2*I*e))*e^(-2*I*f*x - 2*I*e)/(a*f)) - a*f*sqrt(-(c^5 - 5
*I*c^4*d + 5*c^3*d^2 - 25*I*c^2*d^3 + 40*c*d^4 + 16*I*d^5)/(a^2*f^2))*e^(2*I*f*x + 2*I*e)*log(1/2*(I*c^3 + 2*c
^2*d + 7*I*c*d^2 - 4*d^3 - (a*f*e^(2*I*f*x + 2*I*e) + a*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 + 5*c^3*d^2 - 25*I*c^2*d^3 + 40*c*d^4 + 16*I*d^5)/(a^2*f^2)) + (
I*c^3 + 3*c^2*d + 4*I*c*d^2)*e^(2*I*f*x + 2*I*e))*e^(-2*I*f*x - 2*I*e)/(a*f)) + 2*(I*c^2 - 2*c*d - I*d^2 + (I*
c^2 - 2*c*d - 9*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^(-2*I*f*x - 2*I*e)/(a*f)

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

[Out]

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

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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. 464 vs. \(2 (146) = 292\).
time = 0.62, size = 464, normalized size = 2.51 \begin {gather*} -\frac {2 i \, \sqrt {d \tan \left (f x + e\right ) + c} d^{2}}{a f} + \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 )}{a \sqrt {2 \, c + 2 \, \sqrt {c^{2} + d^{2}}} f {\left (-\frac {i \, d}{c + \sqrt {c^{2} + d^{2}}} + 1\right )}} + \frac {{\left (-i \, c^{3} - 2 \, c^{2} d - 7 i \, c d^{2} + 4 \, 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 )}{a \sqrt {-2 \, c + 2 \, \sqrt {c^{2} + d^{2}}} f {\left (\frac {i \, d}{c - \sqrt {c^{2} + d^{2}}} + 1\right )}} + \frac {\sqrt {d \tan \left (f x + e\right ) + c} c^{2} d + 2 i \, \sqrt {d \tan \left (f x + e\right ) + c} c d^{2} - \sqrt {d \tan \left (f x + e\right ) + c} d^{3}}{2 \, {\left (d \tan \left (f x + e\right ) - i \, d\right )} a 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)),x, algorithm="giac")

[Out]

-2*I*sqrt(d*tan(f*x + e) + c)*d^2/(a*f) + (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*sqrt(2*c + 2*sqrt(c^2 + d^2))*f*(-I*d/(c + s
qrt(c^2 + d^2)) + 1)) + (-I*c^3 - 2*c^2*d - 7*I*c*d^2 + 4*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 - sqr
t(c^2 + d^2)*sqrt(-2*c + 2*sqrt(c^2 + d^2))))/(a*sqrt(-2*c + 2*sqrt(c^2 + d^2))*f*(I*d/(c - sqrt(c^2 + d^2)) +
1)) + 1/2*(sqrt(d*tan(f*x + e) + c)*c^2*d + 2*I*sqrt(d*tan(f*x + e) + c)*c*d^2 - sqrt(d*tan(f*x + e) + c)*d^3
)/((d*tan(f*x + e) - I*d)*a*f)

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Mupad [B]
time = 9.36, size = 2500, normalized size = 13.51 \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),x)

[Out]

log((a*f*(12*d^11 - c*d^10*25i + 75*c^2*d^9 - c^3*d^8*35i + 113*c^4*d^7 + c^5*d^6*5i + 49*c^6*d^5 + c^7*d^4*15
i - c^8*d^3))/2 - (((a*f*(a^2*d^6*f^2*80i + 16*a^2*c*d^5*f^2 + a^2*c^2*d^4*f^2*80i + 16*a^2*c^3*d^3*f^2))/2 +
64*a^4*c*d^2*f^4*(c + d*tan(e + f*x))^(1/2)*(-(720*c*d^10 + d^11*240i + 640*c^3*d^8 - c^4*d^7*400i - 48*c^5*d^
6 - c^6*d^5*160i + 32*c^7*d^4 + a^2*f^2*((((400*c^4*d^7 - 240*d^11 + 160*c^6*d^5)*1i)/(a^2*f^2) - (720*c*d^10
+ 640*c^3*d^8 - 48*c^5*d^6 + 32*c^7*d^4)/(a^2*f^2))^2 - 4*(256*d^6 + 256*c^2*d^4)*(((40*c*d^15 + 150*c^3*d^13
+ 200*c^5*d^11 + 100*c^7*d^9 - 10*c^11*d^5)*1i)/(a^4*f^4) - (35*c^4*d^12 - 31*c^2*d^14 - 16*d^16 + 130*c^6*d^1
0 + 110*c^8*d^8 + 29*c^10*d^6 - c^12*d^4)/(a^4*f^4)))^(1/2))/(512*a^2*f^2*(d^6 + c^2*d^4)))^(1/2))*(-(720*c*d^
10 + d^11*240i + 640*c^3*d^8 - c^4*d^7*400i - 48*c^5*d^6 - c^6*d^5*160i + 32*c^7*d^4 + a^2*f^2*((((400*c^4*d^7
- 240*d^11 + 160*c^6*d^5)*1i)/(a^2*f^2) - (720*c*d^10 + 640*c^3*d^8 - 48*c^5*d^6 + 32*c^7*d^4)/(a^2*f^2))^2 -
4*(256*d^6 + 256*c^2*d^4)*(((40*c*d^15 + 150*c^3*d^13 + 200*c^5*d^11 + 100*c^7*d^9 - 10*c^11*d^5)*1i)/(a^4*f^
4) - (35*c^4*d^12 - 31*c^2*d^14 - 16*d^16 + 130*c^6*d^10 + 110*c^8*d^8 + 29*c^10*d^6 - c^12*d^4)/(a^4*f^4)))^(
1/2))/(512*a^2*f^2*(d^6 + c^2*d^4)))^(1/2) + 2*a^2*f^2*(c + d*tan(e + f*x))^(1/2)*(c*d^7*50i - 17*d^8 + 80*c^2
*d^6 - 5*c^4*d^4 - c^5*d^3*10i + 2*c^6*d^2))*(-(720*c*d^10 + d^11*240i + 640*c^3*d^8 - c^4*d^7*400i - 48*c^5*d
^6 - c^6*d^5*160i + 32*c^7*d^4 + a^2*f^2*((((400*c^4*d^7 - 240*d^11 + 160*c^6*d^5)*1i)/(a^2*f^2) - (720*c*d^10
+ 640*c^3*d^8 - 48*c^5*d^6 + 32*c^7*d^4)/(a^2*f^2))^2 - 4*(256*d^6 + 256*c^2*d^4)*(((40*c*d^15 + 150*c^3*d^13
+ 200*c^5*d^11 + 100*c^7*d^9 - 10*c^11*d^5)*1i)/(a^4*f^4) - (35*c^4*d^12 - 31*c^2*d^14 - 16*d^16 + 130*c^6*d^
10 + 110*c^8*d^8 + 29*c^10*d^6 - c^12*d^4)/(a^4*f^4)))^(1/2))/(512*a^2*f^2*(d^6 + c^2*d^4)))^(1/2))*(-(720*c*d
^10 + d^11*240i + 640*c^3*d^8 - c^4*d^7*400i - 48*c^5*d^6 - c^6*d^5*160i + 32*c^7*d^4 + a^2*f^2*((((400*c^4*d^
7 - 240*d^11 + 160*c^6*d^5)*1i)/(a^2*f^2) - (720*c*d^10 + 640*c^3*d^8 - 48*c^5*d^6 + 32*c^7*d^4)/(a^2*f^2))^2
- 4*(256*d^6 + 256*c^2*d^4)*(((40*c*d^15 + 150*c^3*d^13 + 200*c^5*d^11 + 100*c^7*d^9 - 10*c^11*d^5)*1i)/(a^4*f
^4) - (35*c^4*d^12 - 31*c^2*d^14 - 16*d^16 + 130*c^6*d^10 + 110*c^8*d^8 + 29*c^10*d^6 - c^12*d^4)/(a^4*f^4)))^
(1/2))/(512*a^2*f^2*(d^6 + c^2*d^4)))^(1/2) + log((a*f*(12*d^11 - c*d^10*25i + 75*c^2*d^9 - c^3*d^8*35i + 113*
c^4*d^7 + c^5*d^6*5i + 49*c^6*d^5 + c^7*d^4*15i - c^8*d^3))/2 - (((a*f*(a^2*d^6*f^2*80i + 16*a^2*c*d^5*f^2 + a
^2*c^2*d^4*f^2*80i + 16*a^2*c^3*d^3*f^2))/2 + 64*a^4*c*d^2*f^4*(c + d*tan(e + f*x))^(1/2)*(-(720*c*d^10 + d^11
*240i + 640*c^3*d^8 - c^4*d^7*400i - 48*c^5*d^6 - c^6*d^5*160i + 32*c^7*d^4 - a^2*f^2*((((400*c^4*d^7 - 240*d^
11 + 160*c^6*d^5)*1i)/(a^2*f^2) - (720*c*d^10 + 640*c^3*d^8 - 48*c^5*d^6 + 32*c^7*d^4)/(a^2*f^2))^2 - 4*(256*d
^6 + 256*c^2*d^4)*(((40*c*d^15 + 150*c^3*d^13 + 200*c^5*d^11 + 100*c^7*d^9 - 10*c^11*d^5)*1i)/(a^4*f^4) - (35*
c^4*d^12 - 31*c^2*d^14 - 16*d^16 + 130*c^6*d^10 + 110*c^8*d^8 + 29*c^10*d^6 - c^12*d^4)/(a^4*f^4)))^(1/2))/(51
2*a^2*f^2*(d^6 + c^2*d^4)))^(1/2))*(-(720*c*d^10 + d^11*240i + 640*c^3*d^8 - c^4*d^7*400i - 48*c^5*d^6 - c^6*d
^5*160i + 32*c^7*d^4 - a^2*f^2*((((400*c^4*d^7 - 240*d^11 + 160*c^6*d^5)*1i)/(a^2*f^2) - (720*c*d^10 + 640*c^3
*d^8 - 48*c^5*d^6 + 32*c^7*d^4)/(a^2*f^2))^2 - 4*(256*d^6 + 256*c^2*d^4)*(((40*c*d^15 + 150*c^3*d^13 + 200*c^5
*d^11 + 100*c^7*d^9 - 10*c^11*d^5)*1i)/(a^4*f^4) - (35*c^4*d^12 - 31*c^2*d^14 - 16*d^16 + 130*c^6*d^10 + 110*c
^8*d^8 + 29*c^10*d^6 - c^12*d^4)/(a^4*f^4)))^(1/2))/(512*a^2*f^2*(d^6 + c^2*d^4)))^(1/2) + 2*a^2*f^2*(c + d*ta
n(e + f*x))^(1/2)*(c*d^7*50i - 17*d^8 + 80*c^2*d^6 - 5*c^4*d^4 - c^5*d^3*10i + 2*c^6*d^2))*(-(720*c*d^10 + d^1
1*240i + 640*c^3*d^8 - c^4*d^7*400i - 48*c^5*d^6 - c^6*d^5*160i + 32*c^7*d^4 - a^2*f^2*((((400*c^4*d^7 - 240*d
^11 + 160*c^6*d^5)*1i)/(a^2*f^2) - (720*c*d^10 + 640*c^3*d^8 - 48*c^5*d^6 + 32*c^7*d^4)/(a^2*f^2))^2 - 4*(256*
d^6 + 256*c^2*d^4)*(((40*c*d^15 + 150*c^3*d^13 + 200*c^5*d^11 + 100*c^7*d^9 - 10*c^11*d^5)*1i)/(a^4*f^4) - (35
*c^4*d^12 - 31*c^2*d^14 - 16*d^16 + 130*c^6*d^10 + 110*c^8*d^8 + 29*c^10*d^6 - c^12*d^4)/(a^4*f^4)))^(1/2))/(5
12*a^2*f^2*(d^6 + c^2*d^4)))^(1/2))*(-(720*c*d^10 + d^11*240i + 640*c^3*d^8 - c^4*d^7*400i - 48*c^5*d^6 - c^6*
d^5*160i + 32*c^7*d^4 - a^2*f^2*((((400*c^4*d^7 - 240*d^11 + 160*c^6*d^5)*1i)/(a^2*f^2) - (720*c*d^10 + 640*c^
3*d^8 - 48*c^5*d^6 + 32*c^7*d^4)/(a^2*f^2))^2 - 4*(256*d^6 + 256*c^2*d^4)*(((40*c*d^15 + 150*c^3*d^13 + 200*c^
5*d^11 + 100*c^7*d^9 - 10*c^11*d^5)*1i)/(a^4*f^4) - (35*c^4*d^12 - 31*c^2*d^14 - 16*d^16 + 130*c^6*d^10 + 110*
c^8*d^8 + 29*c^10*d^6 - c^12*d^4)/(a^4*f^4)))^(1/2))/(512*a^2*f^2*(d^6 + c^2*d^4)))^(1/2) - log((a*f*(12*d^11
- c*d^10*25i + 75*c^2*d^9 - c^3*d^8*35i + 113*c^4*d^7 + c^5*d^6*5i + 49*c^6*d^5 + c^7*d^4*15i - c^8*d^3))/2 -
(((a*f*(a^2*d^6*f^2*80i + 16*a^2*c*d^5*f^2 + a^2*c^2*d^4*f^2*80i + 16*a^2*c^3*d^3*f^2))/2 - 64*a^4*c*d^2*f^4*(
-(720*c*d^10 + d^11*240i + 640*c^3*d^8 - c^4*d^...

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