3.12.14 \(\int (a+i a \tan (e+f x))^2 (c+d \tan (e+f x))^{5/2} \, dx\) [1114]

Optimal. Leaf size=166 \[ -\frac {4 i a^2 (c-i d)^{5/2} \tanh ^{-1}\left (\frac {\sqrt {c+d \tan (e+f x)}}{\sqrt {c-i d}}\right )}{f}+\frac {4 i a^2 (c-i d)^2 \sqrt {c+d \tan (e+f x)}}{f}+\frac {4 a^2 (i c+d) (c+d \tan (e+f x))^{3/2}}{3 f}+\frac {4 i a^2 (c+d \tan (e+f x))^{5/2}}{5 f}-\frac {2 a^2 (c+d \tan (e+f x))^{7/2}}{7 d f} \]

[Out]

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

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Rubi [A]
time = 0.27, antiderivative size = 166, normalized size of antiderivative = 1.00, number of steps used = 7, number of rules used = 5, integrand size = 30, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.167, Rules used = {3624, 3609, 3618, 65, 214} \begin {gather*} -\frac {2 a^2 (c+d \tan (e+f x))^{7/2}}{7 d f}+\frac {4 i a^2 (c+d \tan (e+f x))^{5/2}}{5 f}+\frac {4 a^2 (d+i c) (c+d \tan (e+f x))^{3/2}}{3 f}+\frac {4 i a^2 (c-i d)^2 \sqrt {c+d \tan (e+f x)}}{f}-\frac {4 i a^2 (c-i d)^{5/2} \tanh ^{-1}\left (\frac {\sqrt {c+d \tan (e+f x)}}{\sqrt {c-i d}}\right )}{f} \end {gather*}

Antiderivative was successfully verified.

[In]

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

[Out]

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

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 3624

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

Rubi steps

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

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Mathematica [A]
time = 7.39, size = 271, normalized size = 1.63 \begin {gather*} \frac {a^2 (\cos (e+f x)+i \sin (e+f x))^2 \left (-4 i (c-i d)^{5/2} e^{-2 i e} \tanh ^{-1}\left (\frac {\sqrt {c-\frac {i d \left (-1+e^{2 i (e+f x)}\right )}{1+e^{2 i (e+f x)}}}}{\sqrt {c-i d}}\right )-\frac {\sec ^2(e+f x) (\cos (2 e)-i \sin (2 e)) \sqrt {c+d \tan (e+f x)} \left (15 c^3-322 i c^2 d-445 c d^2+168 i d^3+d \left (45 c^2-154 i c d-55 d^2\right ) \tan (e+f x)+\cos (2 (e+f x)) \left (15 c^3-322 i c^2 d-535 c d^2+252 i d^3+d \left (45 c^2-154 i c d-85 d^2\right ) \tan (e+f x)\right )\right )}{105 d}\right )}{f (\cos (f x)+i \sin (f x))^2} \end {gather*}

Antiderivative was successfully verified.

[In]

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

[Out]

(a^2*(Cos[e + f*x] + I*Sin[e + f*x])^2*(((-4*I)*(c - I*d)^(5/2)*ArcTanh[Sqrt[c - (I*d*(-1 + E^((2*I)*(e + f*x)
)))/(1 + E^((2*I)*(e + f*x)))]/Sqrt[c - I*d]])/E^((2*I)*e) - (Sec[e + f*x]^2*(Cos[2*e] - I*Sin[2*e])*Sqrt[c +
d*Tan[e + f*x]]*(15*c^3 - (322*I)*c^2*d - 445*c*d^2 + (168*I)*d^3 + d*(45*c^2 - (154*I)*c*d - 55*d^2)*Tan[e +
f*x] + Cos[2*(e + f*x)]*(15*c^3 - (322*I)*c^2*d - 535*c*d^2 + (252*I)*d^3 + d*(45*c^2 - (154*I)*c*d - 85*d^2)*
Tan[e + f*x])))/(105*d)))/(f*(Cos[f*x] + I*Sin[f*x])^2)

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Maple [B] Both result and optimal contain complex but leaf count of result is larger than twice the leaf count of optimal. 1002 vs. \(2 (139 ) = 278\).
time = 0.28, size = 1003, normalized size = 6.04

method	result	size
derivativedivides	\(\frac {2 a^{2} \left (\frac {2 i d \left (c +d \tan \left (f x +e \right )\right )^{\frac {5}{2}}}{5}-\frac {\left (c +d \tan \left (f x +e \right )\right )^{\frac {7}{2}}}{7}+\frac {2 i c d \left (c +d \tan \left (f x +e \right )\right )^{\frac {3}{2}}}{3}+2 i c^{2} d \sqrt {c +d \tan \left (f x +e \right )}-2 i d^{3} \sqrt {c +d \tan \left (f x +e \right )}+\frac {2 d^{2} \left (c +d \tan \left (f x +e \right )\right )^{\frac {3}{2}}}{3}+4 c \,d^{2} \sqrt {c +d \tan \left (f x +e \right )}-2 d \left (\frac {\frac {\left (i c^{3} \sqrt {c^{2}+d^{2}}-3 i c \,d^{2} \sqrt {c^{2}+d^{2}}+i c^{4}-i d^{4}+3 c^{2} d \sqrt {c^{2}+d^{2}}-d^{3} \sqrt {c^{2}+d^{2}}+2 c^{3} d +2 c \,d^{3}\right ) \ln \left (d \tan \left (f x +e \right )+c +\sqrt {c +d \tan \left (f x +e \right )}\, \sqrt {2 \sqrt {c^{2}+d^{2}}+2 c}+\sqrt {c^{2}+d^{2}}\right )}{2}+\frac {2 \left (i \sqrt {2 \sqrt {c^{2}+d^{2}}+2 c}\, c^{4}-i \sqrt {2 \sqrt {c^{2}+d^{2}}+2 c}\, d^{4}+2 \sqrt {2 \sqrt {c^{2}+d^{2}}+2 c}\, c^{3} d +2 \sqrt {2 \sqrt {c^{2}+d^{2}}+2 c}\, c \,d^{3}-\frac {\left (i c^{3} \sqrt {c^{2}+d^{2}}-3 i c \,d^{2} \sqrt {c^{2}+d^{2}}+i c^{4}-i d^{4}+3 c^{2} d \sqrt {c^{2}+d^{2}}-d^{3} \sqrt {c^{2}+d^{2}}+2 c^{3} d +2 c \,d^{3}\right ) \sqrt {2 \sqrt {c^{2}+d^{2}}+2 c}}{2}\right ) \arctan \left (\frac {2 \sqrt {c +d \tan \left (f x +e \right )}+\sqrt {2 \sqrt {c^{2}+d^{2}}+2 c}}{\sqrt {2 \sqrt {c^{2}+d^{2}}-2 c}}\right )}{\sqrt {2 \sqrt {c^{2}+d^{2}}-2 c}}}{2 \sqrt {2 \sqrt {c^{2}+d^{2}}+2 c}\, \sqrt {c^{2}+d^{2}}}+\frac {\frac {\left (-i c^{3} \sqrt {c^{2}+d^{2}}+3 i c \,d^{2} \sqrt {c^{2}+d^{2}}-i c^{4}+i d^{4}-3 c^{2} d \sqrt {c^{2}+d^{2}}+d^{3} \sqrt {c^{2}+d^{2}}-2 c^{3} d -2 c \,d^{3}\right ) \ln \left (d \tan \left (f x +e \right )+c -\sqrt {c +d \tan \left (f x +e \right )}\, \sqrt {2 \sqrt {c^{2}+d^{2}}+2 c}+\sqrt {c^{2}+d^{2}}\right )}{2}+\frac {2 \left (i \sqrt {2 \sqrt {c^{2}+d^{2}}+2 c}\, c^{4}-i \sqrt {2 \sqrt {c^{2}+d^{2}}+2 c}\, d^{4}+2 \sqrt {2 \sqrt {c^{2}+d^{2}}+2 c}\, c^{3} d +2 \sqrt {2 \sqrt {c^{2}+d^{2}}+2 c}\, c \,d^{3}+\frac {\left (-i c^{3} \sqrt {c^{2}+d^{2}}+3 i c \,d^{2} \sqrt {c^{2}+d^{2}}-i c^{4}+i d^{4}-3 c^{2} d \sqrt {c^{2}+d^{2}}+d^{3} \sqrt {c^{2}+d^{2}}-2 c^{3} d -2 c \,d^{3}\right ) \sqrt {2 \sqrt {c^{2}+d^{2}}+2 c}}{2}\right ) \arctan \left (\frac {2 \sqrt {c +d \tan \left (f x +e \right )}-\sqrt {2 \sqrt {c^{2}+d^{2}}+2 c}}{\sqrt {2 \sqrt {c^{2}+d^{2}}-2 c}}\right )}{\sqrt {2 \sqrt {c^{2}+d^{2}}-2 c}}}{2 \sqrt {2 \sqrt {c^{2}+d^{2}}+2 c}\, \sqrt {c^{2}+d^{2}}}\right )\right )}{f d}\)	\(1003\)
default	\(\frac {2 a^{2} \left (\frac {2 i d \left (c +d \tan \left (f x +e \right )\right )^{\frac {5}{2}}}{5}-\frac {\left (c +d \tan \left (f x +e \right )\right )^{\frac {7}{2}}}{7}+\frac {2 i c d \left (c +d \tan \left (f x +e \right )\right )^{\frac {3}{2}}}{3}+2 i c^{2} d \sqrt {c +d \tan \left (f x +e \right )}-2 i d^{3} \sqrt {c +d \tan \left (f x +e \right )}+\frac {2 d^{2} \left (c +d \tan \left (f x +e \right )\right )^{\frac {3}{2}}}{3}+4 c \,d^{2} \sqrt {c +d \tan \left (f x +e \right )}-2 d \left (\frac {\frac {\left (i c^{3} \sqrt {c^{2}+d^{2}}-3 i c \,d^{2} \sqrt {c^{2}+d^{2}}+i c^{4}-i d^{4}+3 c^{2} d \sqrt {c^{2}+d^{2}}-d^{3} \sqrt {c^{2}+d^{2}}+2 c^{3} d +2 c \,d^{3}\right ) \ln \left (d \tan \left (f x +e \right )+c +\sqrt {c +d \tan \left (f x +e \right )}\, \sqrt {2 \sqrt {c^{2}+d^{2}}+2 c}+\sqrt {c^{2}+d^{2}}\right )}{2}+\frac {2 \left (i \sqrt {2 \sqrt {c^{2}+d^{2}}+2 c}\, c^{4}-i \sqrt {2 \sqrt {c^{2}+d^{2}}+2 c}\, d^{4}+2 \sqrt {2 \sqrt {c^{2}+d^{2}}+2 c}\, c^{3} d +2 \sqrt {2 \sqrt {c^{2}+d^{2}}+2 c}\, c \,d^{3}-\frac {\left (i c^{3} \sqrt {c^{2}+d^{2}}-3 i c \,d^{2} \sqrt {c^{2}+d^{2}}+i c^{4}-i d^{4}+3 c^{2} d \sqrt {c^{2}+d^{2}}-d^{3} \sqrt {c^{2}+d^{2}}+2 c^{3} d +2 c \,d^{3}\right ) \sqrt {2 \sqrt {c^{2}+d^{2}}+2 c}}{2}\right ) \arctan \left (\frac {2 \sqrt {c +d \tan \left (f x +e \right )}+\sqrt {2 \sqrt {c^{2}+d^{2}}+2 c}}{\sqrt {2 \sqrt {c^{2}+d^{2}}-2 c}}\right )}{\sqrt {2 \sqrt {c^{2}+d^{2}}-2 c}}}{2 \sqrt {2 \sqrt {c^{2}+d^{2}}+2 c}\, \sqrt {c^{2}+d^{2}}}+\frac {\frac {\left (-i c^{3} \sqrt {c^{2}+d^{2}}+3 i c \,d^{2} \sqrt {c^{2}+d^{2}}-i c^{4}+i d^{4}-3 c^{2} d \sqrt {c^{2}+d^{2}}+d^{3} \sqrt {c^{2}+d^{2}}-2 c^{3} d -2 c \,d^{3}\right ) \ln \left (d \tan \left (f x +e \right )+c -\sqrt {c +d \tan \left (f x +e \right )}\, \sqrt {2 \sqrt {c^{2}+d^{2}}+2 c}+\sqrt {c^{2}+d^{2}}\right )}{2}+\frac {2 \left (i \sqrt {2 \sqrt {c^{2}+d^{2}}+2 c}\, c^{4}-i \sqrt {2 \sqrt {c^{2}+d^{2}}+2 c}\, d^{4}+2 \sqrt {2 \sqrt {c^{2}+d^{2}}+2 c}\, c^{3} d +2 \sqrt {2 \sqrt {c^{2}+d^{2}}+2 c}\, c \,d^{3}+\frac {\left (-i c^{3} \sqrt {c^{2}+d^{2}}+3 i c \,d^{2} \sqrt {c^{2}+d^{2}}-i c^{4}+i d^{4}-3 c^{2} d \sqrt {c^{2}+d^{2}}+d^{3} \sqrt {c^{2}+d^{2}}-2 c^{3} d -2 c \,d^{3}\right ) \sqrt {2 \sqrt {c^{2}+d^{2}}+2 c}}{2}\right ) \arctan \left (\frac {2 \sqrt {c +d \tan \left (f x +e \right )}-\sqrt {2 \sqrt {c^{2}+d^{2}}+2 c}}{\sqrt {2 \sqrt {c^{2}+d^{2}}-2 c}}\right )}{\sqrt {2 \sqrt {c^{2}+d^{2}}-2 c}}}{2 \sqrt {2 \sqrt {c^{2}+d^{2}}+2 c}\, \sqrt {c^{2}+d^{2}}}\right )\right )}{f d}\)	\(1003\)

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

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

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Maxima [F]
time = 0.00, size = 0, normalized size = 0.00 \begin {gather*} \text {Failed to integrate} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

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

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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. 941 vs. \(2 (137) = 274\).
time = 1.02, size = 941, normalized size = 5.67 \begin {gather*} -\frac {105 \, {\left (d f e^{\left (6 i \, f x + 6 i \, e\right )} + 3 \, d f e^{\left (4 i \, f x + 4 i \, e\right )} + 3 \, d f e^{\left (2 i \, f x + 2 i \, e\right )} + d f\right )} \sqrt {-\frac {a^{4} c^{5} - 5 i \, a^{4} c^{4} d - 10 \, a^{4} c^{3} d^{2} + 10 i \, a^{4} c^{2} d^{3} + 5 \, a^{4} c d^{4} - i \, a^{4} d^{5}}{f^{2}}} \log \left (\frac {2 \, {\left (a^{2} c^{3} - 2 i \, a^{2} c^{2} d - a^{2} c d^{2} - {\left (i \, f e^{\left (2 i \, f x + 2 i \, e\right )} + i \, 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 {a^{4} c^{5} - 5 i \, a^{4} c^{4} d - 10 \, a^{4} c^{3} d^{2} + 10 i \, a^{4} c^{2} d^{3} + 5 \, a^{4} c d^{4} - i \, a^{4} d^{5}}{f^{2}}} + {\left (a^{2} c^{3} - 3 i \, a^{2} c^{2} d - 3 \, a^{2} c d^{2} + i \, a^{2} d^{3}\right )} e^{\left (2 i \, f x + 2 i \, e\right )}\right )} e^{\left (-2 i \, f x - 2 i \, e\right )}}{a^{2} c^{2} - 2 i \, a^{2} c d - a^{2} d^{2}}\right ) - 105 \, {\left (d f e^{\left (6 i \, f x + 6 i \, e\right )} + 3 \, d f e^{\left (4 i \, f x + 4 i \, e\right )} + 3 \, d f e^{\left (2 i \, f x + 2 i \, e\right )} + d f\right )} \sqrt {-\frac {a^{4} c^{5} - 5 i \, a^{4} c^{4} d - 10 \, a^{4} c^{3} d^{2} + 10 i \, a^{4} c^{2} d^{3} + 5 \, a^{4} c d^{4} - i \, a^{4} d^{5}}{f^{2}}} \log \left (\frac {2 \, {\left (a^{2} c^{3} - 2 i \, a^{2} c^{2} d - a^{2} c d^{2} - {\left (-i \, f e^{\left (2 i \, f x + 2 i \, e\right )} - i \, 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 {a^{4} c^{5} - 5 i \, a^{4} c^{4} d - 10 \, a^{4} c^{3} d^{2} + 10 i \, a^{4} c^{2} d^{3} + 5 \, a^{4} c d^{4} - i \, a^{4} d^{5}}{f^{2}}} + {\left (a^{2} c^{3} - 3 i \, a^{2} c^{2} d - 3 \, a^{2} c d^{2} + i \, a^{2} d^{3}\right )} e^{\left (2 i \, f x + 2 i \, e\right )}\right )} e^{\left (-2 i \, f x - 2 i \, e\right )}}{a^{2} c^{2} - 2 i \, a^{2} c d - a^{2} d^{2}}\right ) + 2 \, {\left (15 \, a^{2} c^{3} - 277 i \, a^{2} c^{2} d - 381 \, a^{2} c d^{2} + 167 i \, a^{2} d^{3} + {\left (15 \, a^{2} c^{3} - 367 i \, a^{2} c^{2} d - 689 \, a^{2} c d^{2} + 337 i \, a^{2} d^{3}\right )} e^{\left (6 i \, f x + 6 i \, e\right )} + {\left (45 \, a^{2} c^{3} - 1011 i \, a^{2} c^{2} d - 1579 \, a^{2} c d^{2} + 613 i \, a^{2} d^{3}\right )} e^{\left (4 i \, f x + 4 i \, e\right )} + {\left (45 \, a^{2} c^{3} - 921 i \, a^{2} c^{2} d - 1271 \, a^{2} c d^{2} + 563 i \, a^{2} d^{3}\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}}}{105 \, {\left (d f e^{\left (6 i \, f x + 6 i \, e\right )} + 3 \, d f e^{\left (4 i \, f x + 4 i \, e\right )} + 3 \, d f e^{\left (2 i \, f x + 2 i \, e\right )} + d f\right )}} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

-1/105*(105*(d*f*e^(6*I*f*x + 6*I*e) + 3*d*f*e^(4*I*f*x + 4*I*e) + 3*d*f*e^(2*I*f*x + 2*I*e) + d*f)*sqrt(-(a^4
*c^5 - 5*I*a^4*c^4*d - 10*a^4*c^3*d^2 + 10*I*a^4*c^2*d^3 + 5*a^4*c*d^4 - I*a^4*d^5)/f^2)*log(2*(a^2*c^3 - 2*I*
a^2*c^2*d - a^2*c*d^2 - (I*f*e^(2*I*f*x + 2*I*e) + I*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(-(a^4*c^5 - 5*I*a^4*c^4*d - 10*a^4*c^3*d^2 + 10*I*a^4*c^2*d^3 + 5*a^4*c*d^4 - I*a^4*d
^5)/f^2) + (a^2*c^3 - 3*I*a^2*c^2*d - 3*a^2*c*d^2 + I*a^2*d^3)*e^(2*I*f*x + 2*I*e))*e^(-2*I*f*x - 2*I*e)/(a^2*
c^2 - 2*I*a^2*c*d - a^2*d^2)) - 105*(d*f*e^(6*I*f*x + 6*I*e) + 3*d*f*e^(4*I*f*x + 4*I*e) + 3*d*f*e^(2*I*f*x +
2*I*e) + d*f)*sqrt(-(a^4*c^5 - 5*I*a^4*c^4*d - 10*a^4*c^3*d^2 + 10*I*a^4*c^2*d^3 + 5*a^4*c*d^4 - I*a^4*d^5)/f^
2)*log(2*(a^2*c^3 - 2*I*a^2*c^2*d - a^2*c*d^2 - (-I*f*e^(2*I*f*x + 2*I*e) - I*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(-(a^4*c^5 - 5*I*a^4*c^4*d - 10*a^4*c^3*d^2 + 10*I*a^4*c^2*d^
3 + 5*a^4*c*d^4 - I*a^4*d^5)/f^2) + (a^2*c^3 - 3*I*a^2*c^2*d - 3*a^2*c*d^2 + I*a^2*d^3)*e^(2*I*f*x + 2*I*e))*e
^(-2*I*f*x - 2*I*e)/(a^2*c^2 - 2*I*a^2*c*d - a^2*d^2)) + 2*(15*a^2*c^3 - 277*I*a^2*c^2*d - 381*a^2*c*d^2 + 167
*I*a^2*d^3 + (15*a^2*c^3 - 367*I*a^2*c^2*d - 689*a^2*c*d^2 + 337*I*a^2*d^3)*e^(6*I*f*x + 6*I*e) + (45*a^2*c^3
- 1011*I*a^2*c^2*d - 1579*a^2*c*d^2 + 613*I*a^2*d^3)*e^(4*I*f*x + 4*I*e) + (45*a^2*c^3 - 921*I*a^2*c^2*d - 127
1*a^2*c*d^2 + 563*I*a^2*d^3)*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)))/(d*f*e^(6*I*f*x + 6*I*e) + 3*d*f*e^(4*I*f*x + 4*I*e) + 3*d*f*e^(2*I*f*x + 2*I*e) + d*f)

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Sympy [F]
time = 0.00, size = 0, normalized size = 0.00 \begin {gather*} - a^{2} \left (\int \left (- c^{2} \sqrt {c + d \tan {\left (e + f x \right )}}\right )\, dx + \int c^{2} \sqrt {c + d \tan {\left (e + f x \right )}} \tan ^{2}{\left (e + f x \right )}\, dx + \int \left (- d^{2} \sqrt {c + d \tan {\left (e + f x \right )}} \tan ^{2}{\left (e + f x \right )}\right )\, dx + \int d^{2} \sqrt {c + d \tan {\left (e + f x \right )}} \tan ^{4}{\left (e + f x \right )}\, dx + \int \left (- 2 i c^{2} \sqrt {c + d \tan {\left (e + f x \right )}} \tan {\left (e + f x \right )}\right )\, dx + \int \left (- 2 i d^{2} \sqrt {c + d \tan {\left (e + f x \right )}} \tan ^{3}{\left (e + f x \right )}\right )\, dx + \int \left (- 2 c d \sqrt {c + d \tan {\left (e + f x \right )}} \tan {\left (e + f x \right )}\right )\, dx + \int 2 c d \sqrt {c + d \tan {\left (e + f x \right )}} \tan ^{3}{\left (e + f x \right )}\, dx + \int \left (- 4 i c d \sqrt {c + d \tan {\left (e + f x \right )}} \tan ^{2}{\left (e + f x \right )}\right )\, dx\right ) \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

-a**2*(Integral(-c**2*sqrt(c + d*tan(e + f*x)), x) + Integral(c**2*sqrt(c + d*tan(e + f*x))*tan(e + f*x)**2, x
) + Integral(-d**2*sqrt(c + d*tan(e + f*x))*tan(e + f*x)**2, x) + Integral(d**2*sqrt(c + d*tan(e + f*x))*tan(e
+ f*x)**4, x) + Integral(-2*I*c**2*sqrt(c + d*tan(e + f*x))*tan(e + f*x), x) + Integral(-2*I*d**2*sqrt(c + d*
tan(e + f*x))*tan(e + f*x)**3, x) + Integral(-2*c*d*sqrt(c + d*tan(e + f*x))*tan(e + f*x), x) + Integral(2*c*d
*sqrt(c + d*tan(e + f*x))*tan(e + f*x)**3, x) + Integral(-4*I*c*d*sqrt(c + d*tan(e + f*x))*tan(e + f*x)**2, x)
)

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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. 374 vs. \(2 (137) = 274\).
time = 1.12, size = 374, normalized size = 2.25 \begin {gather*} -\frac {8 \, {\left (-i \, a^{2} c^{3} - 3 \, a^{2} c^{2} d + 3 i \, a^{2} c d^{2} + a^{2} 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 )}{\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 (15 \, {\left (d \tan \left (f x + e\right ) + c\right )}^{\frac {7}{2}} a^{2} d^{6} f^{6} - 42 i \, {\left (d \tan \left (f x + e\right ) + c\right )}^{\frac {5}{2}} a^{2} d^{7} f^{6} - 70 i \, {\left (d \tan \left (f x + e\right ) + c\right )}^{\frac {3}{2}} a^{2} c d^{7} f^{6} - 210 i \, \sqrt {d \tan \left (f x + e\right ) + c} a^{2} c^{2} d^{7} f^{6} - 70 \, {\left (d \tan \left (f x + e\right ) + c\right )}^{\frac {3}{2}} a^{2} d^{8} f^{6} - 420 \, \sqrt {d \tan \left (f x + e\right ) + c} a^{2} c d^{8} f^{6} + 210 i \, \sqrt {d \tan \left (f x + e\right ) + c} a^{2} d^{9} f^{6}\right )}}{105 \, d^{7} f^{7}} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

-8*(-I*a^2*c^3 - 3*a^2*c^2*d + 3*I*a^2*c*d^2 + a^2*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))))/(sqrt(-2*c + 2*sqrt(c^2 + d^2))*f*(-I*d/(c - sqrt(c^2 + d^2)) + 1)) - 2
/105*(15*(d*tan(f*x + e) + c)^(7/2)*a^2*d^6*f^6 - 42*I*(d*tan(f*x + e) + c)^(5/2)*a^2*d^7*f^6 - 70*I*(d*tan(f*
x + e) + c)^(3/2)*a^2*c*d^7*f^6 - 210*I*sqrt(d*tan(f*x + e) + c)*a^2*c^2*d^7*f^6 - 70*(d*tan(f*x + e) + c)^(3/
2)*a^2*d^8*f^6 - 420*sqrt(d*tan(f*x + e) + c)*a^2*c*d^8*f^6 + 210*I*sqrt(d*tan(f*x + e) + c)*a^2*d^9*f^6)/(d^7
*f^7)

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Mupad [B]
time = 19.22, size = 257, normalized size = 1.55 \begin {gather*} -\left (\frac {2\,a^2\,\left (c-d\,1{}\mathrm {i}\right )}{5\,d\,f}-\frac {2\,a^2\,\left (c+d\,1{}\mathrm {i}\right )}{5\,d\,f}\right )\,{\left (c+d\,\mathrm {tan}\left (e+f\,x\right )\right )}^{5/2}-{\left (c-d\,1{}\mathrm {i}\right )}^2\,\left (\frac {2\,a^2\,\left (c-d\,1{}\mathrm {i}\right )}{d\,f}-\frac {2\,a^2\,\left (c+d\,1{}\mathrm {i}\right )}{d\,f}\right )\,\sqrt {c+d\,\mathrm {tan}\left (e+f\,x\right )}-\frac {\left (c-d\,1{}\mathrm {i}\right )\,\left (\frac {2\,a^2\,\left (c-d\,1{}\mathrm {i}\right )}{d\,f}-\frac {2\,a^2\,\left (c+d\,1{}\mathrm {i}\right )}{d\,f}\right )\,{\left (c+d\,\mathrm {tan}\left (e+f\,x\right )\right )}^{3/2}}{3}-\frac {2\,a^2\,{\left (c+d\,\mathrm {tan}\left (e+f\,x\right )\right )}^{7/2}}{7\,d\,f}-\frac {\sqrt {4{}\mathrm {i}}\,a^2\,\mathrm {atan}\left (\frac {\sqrt {4{}\mathrm {i}}\,{\left (d+c\,1{}\mathrm {i}\right )}^{5/2}\,\sqrt {c+d\,\mathrm {tan}\left (e+f\,x\right )}\,1{}\mathrm {i}}{2\,\left (-c^3\,1{}\mathrm {i}-3\,c^2\,d+c\,d^2\,3{}\mathrm {i}+d^3\right )}\right )\,{\left (d+c\,1{}\mathrm {i}\right )}^{5/2}\,2{}\mathrm {i}}{f} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

- ((2*a^2*(c - d*1i))/(5*d*f) - (2*a^2*(c + d*1i))/(5*d*f))*(c + d*tan(e + f*x))^(5/2) - (c - d*1i)^2*((2*a^2*
(c - d*1i))/(d*f) - (2*a^2*(c + d*1i))/(d*f))*(c + d*tan(e + f*x))^(1/2) - ((c - d*1i)*((2*a^2*(c - d*1i))/(d*
f) - (2*a^2*(c + d*1i))/(d*f))*(c + d*tan(e + f*x))^(3/2))/3 - (2*a^2*(c + d*tan(e + f*x))^(7/2))/(7*d*f) - (4
i^(1/2)*a^2*atan((4i^(1/2)*(c*1i + d)^(5/2)*(c + d*tan(e + f*x))^(1/2)*1i)/(2*(c*d^2*3i - 3*c^2*d - c^3*1i + d
^3)))*(c*1i + d)^(5/2)*2i)/f

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