3.1137 \(\int (a+i a \tan (e+f x))^{5/2} \sqrt{c+d \tan (e+f x)} \, dx\)

Optimal. Leaf size=263 \[ -\frac{\sqrt [4]{-1} a^{5/2} \left (c^2+10 i c d+23 d^2\right ) \tanh ^{-1}\left (\frac{(-1)^{3/4} \sqrt{d} \sqrt{a+i a \tan (e+f x)}}{\sqrt{a} \sqrt{c+d \tan (e+f x)}}\right )}{4 d^{3/2} f}-\frac{a^2 \sqrt{a+i a \tan (e+f x)} (c+d \tan (e+f x))^{3/2}}{2 d f}+\frac{a^2 (c+9 i d) \sqrt{a+i a \tan (e+f x)} \sqrt{c+d \tan (e+f x)}}{4 d f}-\frac{4 i \sqrt{2} a^{5/2} \sqrt{c-i d} \tanh ^{-1}\left (\frac{\sqrt{2} \sqrt{a} \sqrt{c+d \tan (e+f x)}}{\sqrt{c-i d} \sqrt{a+i a \tan (e+f x)}}\right )}{f} \]

[Out]

-((-1)^(1/4)*a^(5/2)*(c^2 + (10*I)*c*d + 23*d^2)*ArcTanh[((-1)^(3/4)*Sqrt[d]*Sqrt[a + I*a*Tan[e + f*x]])/(Sqrt
[a]*Sqrt[c + d*Tan[e + f*x]])])/(4*d^(3/2)*f) - ((4*I)*Sqrt[2]*a^(5/2)*Sqrt[c - I*d]*ArcTanh[(Sqrt[2]*Sqrt[a]*
Sqrt[c + d*Tan[e + f*x]])/(Sqrt[c - I*d]*Sqrt[a + I*a*Tan[e + f*x]])])/f + (a^2*(c + (9*I)*d)*Sqrt[a + I*a*Tan
[e + f*x]]*Sqrt[c + d*Tan[e + f*x]])/(4*d*f) - (a^2*Sqrt[a + I*a*Tan[e + f*x]]*(c + d*Tan[e + f*x])^(3/2))/(2*
d*f)

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Rubi [A]  time = 1.0044, antiderivative size = 263, normalized size of antiderivative = 1., number of steps used = 9, number of rules used = 9, integrand size = 32, \(\frac{\text{number of rules}}{\text{integrand size}}\) = 0.281, Rules used = {3556, 3597, 3601, 3544, 208, 3599, 63, 217, 206} \[ -\frac{\sqrt [4]{-1} a^{5/2} \left (c^2+10 i c d+23 d^2\right ) \tanh ^{-1}\left (\frac{(-1)^{3/4} \sqrt{d} \sqrt{a+i a \tan (e+f x)}}{\sqrt{a} \sqrt{c+d \tan (e+f x)}}\right )}{4 d^{3/2} f}-\frac{a^2 \sqrt{a+i a \tan (e+f x)} (c+d \tan (e+f x))^{3/2}}{2 d f}+\frac{a^2 (c+9 i d) \sqrt{a+i a \tan (e+f x)} \sqrt{c+d \tan (e+f x)}}{4 d f}-\frac{4 i \sqrt{2} a^{5/2} \sqrt{c-i d} \tanh ^{-1}\left (\frac{\sqrt{2} \sqrt{a} \sqrt{c+d \tan (e+f x)}}{\sqrt{c-i d} \sqrt{a+i a \tan (e+f x)}}\right )}{f} \]

Antiderivative was successfully verified.

[In]

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

[Out]

-((-1)^(1/4)*a^(5/2)*(c^2 + (10*I)*c*d + 23*d^2)*ArcTanh[((-1)^(3/4)*Sqrt[d]*Sqrt[a + I*a*Tan[e + f*x]])/(Sqrt
[a]*Sqrt[c + d*Tan[e + f*x]])])/(4*d^(3/2)*f) - ((4*I)*Sqrt[2]*a^(5/2)*Sqrt[c - I*d]*ArcTanh[(Sqrt[2]*Sqrt[a]*
Sqrt[c + d*Tan[e + f*x]])/(Sqrt[c - I*d]*Sqrt[a + I*a*Tan[e + f*x]])])/f + (a^2*(c + (9*I)*d)*Sqrt[a + I*a*Tan
[e + f*x]]*Sqrt[c + d*Tan[e + f*x]])/(4*d*f) - (a^2*Sqrt[a + I*a*Tan[e + f*x]]*(c + d*Tan[e + f*x])^(3/2))/(2*
d*f)

Rule 3556

Int[((a_) + (b_.)*tan[(e_.) + (f_.)*(x_)])^(m_)*((c_.) + (d_.)*tan[(e_.) + (f_.)*(x_)])^(n_), x_Symbol] :> Sim
p[(b^2*(a + b*Tan[e + f*x])^(m - 2)*(c + d*Tan[e + f*x])^(n + 1))/(d*f*(m + n - 1)), x] + Dist[a/(d*(m + n - 1
)), Int[(a + b*Tan[e + f*x])^(m - 2)*(c + d*Tan[e + f*x])^n*Simp[b*c*(m - 2) + a*d*(m + 2*n) + (a*c*(m - 2) +
b*d*(3*m + 2*n - 4))*Tan[e + f*x], x], x], x] /; FreeQ[{a, b, c, d, e, f, n}, x] && NeQ[b*c - a*d, 0] && EqQ[a
^2 + b^2, 0] && NeQ[c^2 + d^2, 0] && IntegerQ[2*m] && GtQ[m, 1] && NeQ[m + n - 1, 0] && (IntegerQ[m] || Intege
rsQ[2*m, 2*n])

Rule 3597

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

Rule 3601

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

Rule 3544

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

Rule 208

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

Rule 3599

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

Rule 63

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 217

Int[1/Sqrt[(a_) + (b_.)*(x_)^2], x_Symbol] :> Subst[Int[1/(1 - b*x^2), x], x, x/Sqrt[a + b*x^2]] /; FreeQ[{a,
b}, x] &&  !GtQ[a, 0]

Rule 206

Int[((a_) + (b_.)*(x_)^2)^(-1), x_Symbol] :> Simp[(1*ArcTanh[(Rt[-b, 2]*x)/Rt[a, 2]])/(Rt[a, 2]*Rt[-b, 2]), x]
/; FreeQ[{a, b}, x] && NegQ[a/b] && (GtQ[a, 0] || LtQ[b, 0])

Rubi steps

\begin{align*} \int (a+i a \tan (e+f x))^{5/2} \sqrt{c+d \tan (e+f x)} \, dx &=-\frac{a^2 \sqrt{a+i a \tan (e+f x)} (c+d \tan (e+f x))^{3/2}}{2 d f}+\frac{a \int \sqrt{a+i a \tan (e+f x)} \left (\frac{1}{2} a (i c+7 d)+\frac{1}{2} a (c+9 i d) \tan (e+f x)\right ) \sqrt{c+d \tan (e+f x)} \, dx}{2 d}\\ &=\frac{a^2 (c+9 i d) \sqrt{a+i a \tan (e+f x)} \sqrt{c+d \tan (e+f x)}}{4 d f}-\frac{a^2 \sqrt{a+i a \tan (e+f x)} (c+d \tan (e+f x))^{3/2}}{2 d f}+\frac{\int \frac{\sqrt{a+i a \tan (e+f x)} \left (\frac{1}{4} a^2 \left (22 c d+i \left (c^2-9 d^2\right )\right )+\frac{1}{4} a^2 \left (c^2+10 i c d+23 d^2\right ) \tan (e+f x)\right )}{\sqrt{c+d \tan (e+f x)}} \, dx}{2 d}\\ &=\frac{a^2 (c+9 i d) \sqrt{a+i a \tan (e+f x)} \sqrt{c+d \tan (e+f x)}}{4 d f}-\frac{a^2 \sqrt{a+i a \tan (e+f x)} (c+d \tan (e+f x))^{3/2}}{2 d f}+\left (4 a^2 (c-i d)\right ) \int \frac{\sqrt{a+i a \tan (e+f x)}}{\sqrt{c+d \tan (e+f x)}} \, dx-\frac{\left (a \left (10 c d-i \left (c^2+23 d^2\right )\right )\right ) \int \frac{(a-i a \tan (e+f x)) \sqrt{a+i a \tan (e+f x)}}{\sqrt{c+d \tan (e+f x)}} \, dx}{8 d}\\ &=\frac{a^2 (c+9 i d) \sqrt{a+i a \tan (e+f x)} \sqrt{c+d \tan (e+f x)}}{4 d f}-\frac{a^2 \sqrt{a+i a \tan (e+f x)} (c+d \tan (e+f x))^{3/2}}{2 d f}-\frac{\left (8 a^4 (i c+d)\right ) \operatorname{Subst}\left (\int \frac{1}{a c-i a d-2 a^2 x^2} \, dx,x,\frac{\sqrt{c+d \tan (e+f x)}}{\sqrt{a+i a \tan (e+f x)}}\right )}{f}-\frac{\left (a^3 \left (10 c d-i \left (c^2+23 d^2\right )\right )\right ) \operatorname{Subst}\left (\int \frac{1}{\sqrt{a+i a x} \sqrt{c+d x}} \, dx,x,\tan (e+f x)\right )}{8 d f}\\ &=-\frac{4 i \sqrt{2} a^{5/2} \sqrt{c-i d} \tanh ^{-1}\left (\frac{\sqrt{2} \sqrt{a} \sqrt{c+d \tan (e+f x)}}{\sqrt{c-i d} \sqrt{a+i a \tan (e+f x)}}\right )}{f}+\frac{a^2 (c+9 i d) \sqrt{a+i a \tan (e+f x)} \sqrt{c+d \tan (e+f x)}}{4 d f}-\frac{a^2 \sqrt{a+i a \tan (e+f x)} (c+d \tan (e+f x))^{3/2}}{2 d f}+\frac{\left (a^2 \left (c^2+10 i c d+23 d^2\right )\right ) \operatorname{Subst}\left (\int \frac{1}{\sqrt{c+i d-\frac{i d x^2}{a}}} \, dx,x,\sqrt{a+i a \tan (e+f x)}\right )}{4 d f}\\ &=-\frac{4 i \sqrt{2} a^{5/2} \sqrt{c-i d} \tanh ^{-1}\left (\frac{\sqrt{2} \sqrt{a} \sqrt{c+d \tan (e+f x)}}{\sqrt{c-i d} \sqrt{a+i a \tan (e+f x)}}\right )}{f}+\frac{a^2 (c+9 i d) \sqrt{a+i a \tan (e+f x)} \sqrt{c+d \tan (e+f x)}}{4 d f}-\frac{a^2 \sqrt{a+i a \tan (e+f x)} (c+d \tan (e+f x))^{3/2}}{2 d f}+\frac{\left (a^2 \left (c^2+10 i c d+23 d^2\right )\right ) \operatorname{Subst}\left (\int \frac{1}{1+\frac{i d x^2}{a}} \, dx,x,\frac{\sqrt{a+i a \tan (e+f x)}}{\sqrt{c+d \tan (e+f x)}}\right )}{4 d f}\\ &=-\frac{\sqrt [4]{-1} a^{5/2} \left (c^2+10 i c d+23 d^2\right ) \tanh ^{-1}\left (\frac{(-1)^{3/4} \sqrt{d} \sqrt{a+i a \tan (e+f x)}}{\sqrt{a} \sqrt{c+d \tan (e+f x)}}\right )}{4 d^{3/2} f}-\frac{4 i \sqrt{2} a^{5/2} \sqrt{c-i d} \tanh ^{-1}\left (\frac{\sqrt{2} \sqrt{a} \sqrt{c+d \tan (e+f x)}}{\sqrt{c-i d} \sqrt{a+i a \tan (e+f x)}}\right )}{f}+\frac{a^2 (c+9 i d) \sqrt{a+i a \tan (e+f x)} \sqrt{c+d \tan (e+f x)}}{4 d f}-\frac{a^2 \sqrt{a+i a \tan (e+f x)} (c+d \tan (e+f x))^{3/2}}{2 d f}\\ \end{align*}

Mathematica [B]  time = 7.26643, size = 589, normalized size = 2.24 \[ \frac{\left (\frac{1}{8}+\frac{i}{8}\right ) \cos ^2(e+f x) (a+i a \tan (e+f x))^{5/2} \left (\frac{(1+i) (\sin (2 e)+i \cos (2 e)) \sqrt{c+d \tan (e+f x)} (c+2 d \tan (e+f x)-9 i d)}{d}-\frac{(\cos (2 e)-i \sin (2 e)) \cos (e+f x) \left (\left (c^2+10 i c d+23 d^2\right ) \left (\log \left (\frac{(2+2 i) e^{\frac{i e}{2}} \left (-(1+i) \sqrt{d} \sqrt{1+e^{2 i (e+f x)}} \sqrt{c-\frac{i d \left (-1+e^{2 i (e+f x)}\right )}{1+e^{2 i (e+f x)}}}+i c \left (e^{i (e+f x)}+i\right )+d e^{i (e+f x)}-i d\right )}{\sqrt{d} \left (c^2+10 i c d+23 d^2\right ) \left (e^{i (e+f x)}+i\right )}\right )-\log \left (\frac{(2+2 i) e^{\frac{i e}{2}} \left ((1+i) \sqrt{d} \sqrt{1+e^{2 i (e+f x)}} \sqrt{c-\frac{i d \left (-1+e^{2 i (e+f x)}\right )}{1+e^{2 i (e+f x)}}}+i c e^{i (e+f x)}+c+d e^{i (e+f x)}+i d\right )}{\sqrt{d} \left (c^2+10 i c d+23 d^2\right ) \left (e^{i (e+f x)}-i\right )}\right )\right )+(32+32 i) d^{3/2} \sqrt{c-i d} \log \left (2 \left (i \sqrt{c-i d} \sin (e+f x)+\sqrt{c-i d} \cos (e+f x)+\sqrt{i \sin (2 (e+f x))+\cos (2 (e+f x))+1} \sqrt{c+d \tan (e+f x)}\right )\right )\right )}{d^{3/2} \sqrt{i \sin (2 (e+f x))+\cos (2 (e+f x))+1}}\right )}{f (\cos (f x)+i \sin (f x))^2} \]

Warning: Unable to verify antiderivative.

[In]

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

[Out]

((1/8 + I/8)*Cos[e + f*x]^2*(a + I*a*Tan[e + f*x])^(5/2)*(-((Cos[e + f*x]*((c^2 + (10*I)*c*d + 23*d^2)*(Log[((
2 + 2*I)*E^((I/2)*e)*((-I)*d + d*E^(I*(e + f*x)) + I*c*(I + E^(I*(e + f*x))) - (1 + I)*Sqrt[d]*Sqrt[1 + E^((2*
I)*(e + f*x))]*Sqrt[c - (I*d*(-1 + E^((2*I)*(e + f*x))))/(1 + E^((2*I)*(e + f*x)))]))/(Sqrt[d]*(c^2 + (10*I)*c
*d + 23*d^2)*(I + E^(I*(e + f*x))))] - Log[((2 + 2*I)*E^((I/2)*e)*(c + I*d + I*c*E^(I*(e + f*x)) + d*E^(I*(e +
f*x)) + (1 + I)*Sqrt[d]*Sqrt[1 + E^((2*I)*(e + f*x))]*Sqrt[c - (I*d*(-1 + E^((2*I)*(e + f*x))))/(1 + E^((2*I)
*(e + f*x)))]))/(Sqrt[d]*(c^2 + (10*I)*c*d + 23*d^2)*(-I + E^(I*(e + f*x))))]) + (32 + 32*I)*Sqrt[c - I*d]*d^(
3/2)*Log[2*(Sqrt[c - I*d]*Cos[e + f*x] + I*Sqrt[c - I*d]*Sin[e + f*x] + Sqrt[1 + Cos[2*(e + f*x)] + I*Sin[2*(e
+ f*x)]]*Sqrt[c + d*Tan[e + f*x]])])*(Cos[2*e] - I*Sin[2*e]))/(d^(3/2)*Sqrt[1 + Cos[2*(e + f*x)] + I*Sin[2*(e
+ f*x)]])) + ((1 + I)*(I*Cos[2*e] + Sin[2*e])*Sqrt[c + d*Tan[e + f*x]]*(c - (9*I)*d + 2*d*Tan[e + f*x]))/d))/
(f*(Cos[f*x] + I*Sin[f*x])^2)

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Maple [B]  time = 0.201, size = 1079, normalized size = 4.1 \begin{align*} \text{result too large to display} \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

1/16/f*(c+d*tan(f*x+e))^(1/2)*(a*(1+I*tan(f*x+e)))^(1/2)*a^2*(-10*ln(1/2*(2*I*a*tan(f*x+e)*d+I*a*c+2*(a*(c+d*t
an(f*x+e))*(1+I*tan(f*x+e)))^(1/2)*(I*a*d)^(1/2)+a*d)/(I*a*d)^(1/2))*2^(1/2)*(-a*(I*d-c))^(1/2)*a*c*d+23*I*ln(
1/2*(2*I*a*tan(f*x+e)*d+I*a*c+2*(a*(c+d*tan(f*x+e))*(1+I*tan(f*x+e)))^(1/2)*(I*a*d)^(1/2)+a*d)/(I*a*d)^(1/2))*
2^(1/2)*(-a*(I*d-c))^(1/2)*a*d^2+18*I*(a*(c+d*tan(f*x+e))*(1+I*tan(f*x+e)))^(1/2)*2^(1/2)*(-a*(I*d-c))^(1/2)*(
I*a*d)^(1/2)*d+I*ln(1/2*(2*I*a*tan(f*x+e)*d+I*a*c+2*(a*(c+d*tan(f*x+e))*(1+I*tan(f*x+e)))^(1/2)*(I*a*d)^(1/2)+
a*d)/(I*a*d)^(1/2))*2^(1/2)*(-a*(I*d-c))^(1/2)*a*c^2-4*(a*(c+d*tan(f*x+e))*(1+I*tan(f*x+e)))^(1/2)*2^(1/2)*(-a
*(I*d-c))^(1/2)*(I*a*d)^(1/2)*tan(f*x+e)*d+32*ln(1/2*(2*I*a*tan(f*x+e)*d+I*a*c+2*(a*(c+d*tan(f*x+e))*(1+I*tan(
f*x+e)))^(1/2)*(I*a*d)^(1/2)+a*d)/(I*a*d)^(1/2))*2^(1/2)*(-a*(I*d-c))^(1/2)*a*d^2-2*(a*(c+d*tan(f*x+e))*(1+I*t
an(f*x+e)))^(1/2)*2^(1/2)*(-a*(I*d-c))^(1/2)*(I*a*d)^(1/2)*c+16*I*ln((3*a*c+I*a*tan(f*x+e)*c-I*a*d+3*a*tan(f*x
+e)*d+2*2^(1/2)*(-a*(I*d-c))^(1/2)*(a*(c+d*tan(f*x+e))*(1+I*tan(f*x+e)))^(1/2))/(tan(f*x+e)+I))*(I*a*d)^(1/2)*
a*c*d+32*I*ln(1/2*(2*I*a*tan(f*x+e)*d+I*a*c+2*(a*(c+d*tan(f*x+e))*(1+I*tan(f*x+e)))^(1/2)*(I*a*d)^(1/2)+a*d)/(
I*a*d)^(1/2))*2^(1/2)*(-a*(I*d-c))^(1/2)*a*c*d+16*I*ln((3*a*c+I*a*tan(f*x+e)*c-I*a*d+3*a*tan(f*x+e)*d+2*2^(1/2
)*(-a*(I*d-c))^(1/2)*(a*(c+d*tan(f*x+e))*(1+I*tan(f*x+e)))^(1/2))/(tan(f*x+e)+I))*(I*a*d)^(1/2)*a*d^2-16*ln((3
*a*c+I*a*tan(f*x+e)*c-I*a*d+3*a*tan(f*x+e)*d+2*2^(1/2)*(-a*(I*d-c))^(1/2)*(a*(c+d*tan(f*x+e))*(1+I*tan(f*x+e))
)^(1/2))/(tan(f*x+e)+I))*(I*a*d)^(1/2)*a*c*d+16*ln((3*a*c+I*a*tan(f*x+e)*c-I*a*d+3*a*tan(f*x+e)*d+2*2^(1/2)*(-
a*(I*d-c))^(1/2)*(a*(c+d*tan(f*x+e))*(1+I*tan(f*x+e)))^(1/2))/(tan(f*x+e)+I))*(I*a*d)^(1/2)*a*d^2)/(a*(c+d*tan
(f*x+e))*(1+I*tan(f*x+e)))^(1/2)/d/(I*a*d)^(1/2)*2^(1/2)/(-a*(I*d-c))^(1/2)

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

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate((c+d*tan(f*x+e))^(1/2)*(a+I*a*tan(f*x+e))^(5/2),x, algorithm="maxima")

[Out]

Exception raised: ValueError

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Fricas [B]  time = 2.72703, size = 2708, normalized size = 10.3 \begin{align*} \text{result too large to display} \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate((c+d*tan(f*x+e))^(1/2)*(a+I*a*tan(f*x+e))^(5/2),x, algorithm="fricas")

[Out]

-1/8*(2*sqrt(2)*(a^2*c - 7*I*a^2*d + (a^2*c - 11*I*a^2*d)*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))*sqrt(a/(e^(2*I*f*x + 2*I*e) + 1))*e^(I*f*x + I*e) + (d*f*e^(2*I*f*x
+ 2*I*e) + d*f)*sqrt((I*a^5*c^4 - 20*a^5*c^3*d - 54*I*a^5*c^2*d^2 - 460*a^5*c*d^3 + 529*I*a^5*d^4)/(d^3*f^2))
*log((2*I*d^2*f*sqrt((I*a^5*c^4 - 20*a^5*c^3*d - 54*I*a^5*c^2*d^2 - 460*a^5*c*d^3 + 529*I*a^5*d^4)/(d^3*f^2))*
e^(2*I*f*x + 2*I*e) + sqrt(2)*(a^2*c^2 + 10*I*a^2*c*d + 23*a^2*d^2 + (a^2*c^2 + 10*I*a^2*c*d + 23*a^2*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))*sqrt(a/(e^(2*I*f*x
+ 2*I*e) + 1))*e^(I*f*x + I*e))*e^(-2*I*f*x - 2*I*e)/(a^2*c^2 + 10*I*a^2*c*d + 23*a^2*d^2)) - (d*f*e^(2*I*f*x
+ 2*I*e) + d*f)*sqrt((I*a^5*c^4 - 20*a^5*c^3*d - 54*I*a^5*c^2*d^2 - 460*a^5*c*d^3 + 529*I*a^5*d^4)/(d^3*f^2))
*log((-2*I*d^2*f*sqrt((I*a^5*c^4 - 20*a^5*c^3*d - 54*I*a^5*c^2*d^2 - 460*a^5*c*d^3 + 529*I*a^5*d^4)/(d^3*f^2))
*e^(2*I*f*x + 2*I*e) + sqrt(2)*(a^2*c^2 + 10*I*a^2*c*d + 23*a^2*d^2 + (a^2*c^2 + 10*I*a^2*c*d + 23*a^2*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))*sqrt(a/(e^(2*I*f*
x + 2*I*e) + 1))*e^(I*f*x + I*e))*e^(-2*I*f*x - 2*I*e)/(a^2*c^2 + 10*I*a^2*c*d + 23*a^2*d^2)) - 4*(d*f*e^(2*I*
f*x + 2*I*e) + d*f)*sqrt(-(32*a^5*c - 32*I*a^5*d)/f^2)*log(1/4*(4*sqrt(2)*(a^2*e^(2*I*f*x + 2*I*e) + a^2)*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/(e^(2*I*f*x + 2*I*e) + 1))*e^(I*f
*x + I*e) + I*f*sqrt(-(32*a^5*c - 32*I*a^5*d)/f^2)*e^(2*I*f*x + 2*I*e))*e^(-2*I*f*x - 2*I*e)/a^2) + 4*(d*f*e^(
2*I*f*x + 2*I*e) + d*f)*sqrt(-(32*a^5*c - 32*I*a^5*d)/f^2)*log(1/4*(4*sqrt(2)*(a^2*e^(2*I*f*x + 2*I*e) + a^2)*
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/(e^(2*I*f*x + 2*I*e) + 1))*e^
(I*f*x + I*e) - I*f*sqrt(-(32*a^5*c - 32*I*a^5*d)/f^2)*e^(2*I*f*x + 2*I*e))*e^(-2*I*f*x - 2*I*e)/a^2))/(d*f*e^
(2*I*f*x + 2*I*e) + d*f)

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Sympy [F(-1)]  time = 0., size = 0, normalized size = 0. \begin{align*} \text{Timed out} \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate((c+d*tan(f*x+e))**(1/2)*(a+I*a*tan(f*x+e))**(5/2),x)

[Out]

Timed out

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Giac [A]  time = 5.7527, size = 275, normalized size = 1.05 \begin{align*} \frac{i \, \sqrt{2 \, a^{2} c + 2 \, \sqrt{a^{2} c^{2} +{\left (i \, a \tan \left (f x + e\right ) + a\right )}^{2} d^{2} - 2 \,{\left (i \, a \tan \left (f x + e\right ) + a\right )} a d^{2} + a^{2} d^{2}} a}{\left (i \, a \tan \left (f x + e\right ) + a\right )}^{2}{\left (\frac{-i \,{\left (i \, a \tan \left (f x + e\right ) + a\right )} a d + i \, a^{2} d}{a^{2} c + \sqrt{a^{4} c^{2} +{\left (i \, a \tan \left (f x + e\right ) + a\right )}^{2} a^{2} d^{2} - 2 \,{\left (i \, a \tan \left (f x + e\right ) + a\right )} a^{3} d^{2} + a^{4} d^{2}}} + 1\right )} \log \left (\sqrt{i \, a \tan \left (f x + e\right ) + a}\right )}{i \, a \tan \left (f x + e\right ) - a} \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate((c+d*tan(f*x+e))^(1/2)*(a+I*a*tan(f*x+e))^(5/2),x, algorithm="giac")

[Out]

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