3.1151 \(\int \sqrt{a+i a \tan (e+f x)} (c+d \tan (e+f x))^{5/2} \, dx\)

Optimal. Leaf size=257 \[ -\frac{\sqrt [4]{-1} \sqrt{a} \sqrt{d} \left (15 c^2-10 i c d-7 d^2\right ) \tanh ^{-1}\left (\frac{(-1)^{3/4} \sqrt{d} \sqrt{a+i a \tan (e+f x)}}{\sqrt{a} \sqrt{c+d \tan (e+f x)}}\right )}{4 f}+\frac{d \sqrt{a+i a \tan (e+f x)} (c+d \tan (e+f x))^{3/2}}{2 f}+\frac{d (7 c-i d) \sqrt{a+i a \tan (e+f x)} \sqrt{c+d \tan (e+f x)}}{4 f}-\frac{i \sqrt{2} \sqrt{a} (c-i d)^{5/2} \tanh ^{-1}\left (\frac{\sqrt{2} \sqrt{a} \sqrt{c+d \tan (e+f x)}}{\sqrt{c-i d} \sqrt{a+i a \tan (e+f x)}}\right )}{f} \]

[Out]

-((-1)^(1/4)*Sqrt[a]*Sqrt[d]*(15*c^2 - (10*I)*c*d - 7*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*f) - (I*Sqrt[2]*Sqrt[a]*(c - I*d)^(5/2)*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 + ((7*c - I*d)*d*Sqrt[a + I*a*Tan[e +
f*x]]*Sqrt[c + d*Tan[e + f*x]])/(4*f) + (d*Sqrt[a + I*a*Tan[e + f*x]]*(c + d*Tan[e + f*x])^(3/2))/(2*f)

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Rubi [A]  time = 1.04507, antiderivative size = 257, 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 = {3560, 3597, 3601, 3544, 208, 3599, 63, 217, 206} \[ -\frac{\sqrt [4]{-1} \sqrt{a} \sqrt{d} \left (15 c^2-10 i c d-7 d^2\right ) \tanh ^{-1}\left (\frac{(-1)^{3/4} \sqrt{d} \sqrt{a+i a \tan (e+f x)}}{\sqrt{a} \sqrt{c+d \tan (e+f x)}}\right )}{4 f}+\frac{d \sqrt{a+i a \tan (e+f x)} (c+d \tan (e+f x))^{3/2}}{2 f}+\frac{d (7 c-i d) \sqrt{a+i a \tan (e+f x)} \sqrt{c+d \tan (e+f x)}}{4 f}-\frac{i \sqrt{2} \sqrt{a} (c-i d)^{5/2} \tanh ^{-1}\left (\frac{\sqrt{2} \sqrt{a} \sqrt{c+d \tan (e+f x)}}{\sqrt{c-i d} \sqrt{a+i a \tan (e+f x)}}\right )}{f} \]

Antiderivative was successfully verified.

[In]

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

[Out]

-((-1)^(1/4)*Sqrt[a]*Sqrt[d]*(15*c^2 - (10*I)*c*d - 7*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*f) - (I*Sqrt[2]*Sqrt[a]*(c - I*d)^(5/2)*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 + ((7*c - I*d)*d*Sqrt[a + I*a*Tan[e +
f*x]]*Sqrt[c + d*Tan[e + f*x]])/(4*f) + (d*Sqrt[a + I*a*Tan[e + f*x]]*(c + d*Tan[e + f*x])^(3/2))/(2*f)

Rule 3560

Int[((a_) + (b_.)*tan[(e_.) + (f_.)*(x_)])^(m_)*((c_.) + (d_.)*tan[(e_.) + (f_.)*(x_)])^(n_), x_Symbol] :> Sim
p[(d*(a + b*Tan[e + f*x])^m*(c + d*Tan[e + f*x])^(n - 1))/(f*(m + n - 1)), x] - Dist[1/(a*(m + n - 1)), Int[(a
+ b*Tan[e + f*x])^m*(c + d*Tan[e + f*x])^(n - 2)*Simp[d*(b*c*m + a*d*(-1 + n)) - a*c^2*(m + n - 1) + d*(b*d*m
- a*c*(m + 2*n - 2))*Tan[e + f*x], x], x], x] /; FreeQ[{a, b, c, d, e, f, m}, x] && NeQ[b*c - a*d, 0] && EqQ[
a^2 + b^2, 0] && NeQ[c^2 + d^2, 0] && GtQ[n, 1] && NeQ[m + n - 1, 0] && (IntegerQ[n] || IntegersQ[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 \sqrt{a+i a \tan (e+f x)} (c+d \tan (e+f x))^{5/2} \, dx &=\frac{d \sqrt{a+i a \tan (e+f x)} (c+d \tan (e+f x))^{3/2}}{2 f}-\frac{\int \sqrt{a+i a \tan (e+f x)} \sqrt{c+d \tan (e+f x)} \left (-\frac{1}{2} a \left (4 c^2-i c d-3 d^2\right )-\frac{1}{2} a (7 c-i d) d \tan (e+f x)\right ) \, dx}{2 a}\\ &=\frac{(7 c-i d) d \sqrt{a+i a \tan (e+f x)} \sqrt{c+d \tan (e+f x)}}{4 f}+\frac{d \sqrt{a+i a \tan (e+f x)} (c+d \tan (e+f x))^{3/2}}{2 f}-\frac{\int \frac{\sqrt{a+i a \tan (e+f x)} \left (-\frac{1}{4} a^2 \left (8 c^3-9 i c^2 d-14 c d^2+i d^3\right )-\frac{1}{4} a^2 d \left (15 c^2-10 i c d-7 d^2\right ) \tan (e+f x)\right )}{\sqrt{c+d \tan (e+f x)}} \, dx}{2 a^2}\\ &=\frac{(7 c-i d) d \sqrt{a+i a \tan (e+f x)} \sqrt{c+d \tan (e+f x)}}{4 f}+\frac{d \sqrt{a+i a \tan (e+f x)} (c+d \tan (e+f x))^{3/2}}{2 f}+(c-i d)^3 \int \frac{\sqrt{a+i a \tan (e+f x)}}{\sqrt{c+d \tan (e+f x)}} \, dx+\frac{\left (d \left (15 i c^2+10 c d-7 i d^2\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 a}\\ &=\frac{(7 c-i d) d \sqrt{a+i a \tan (e+f x)} \sqrt{c+d \tan (e+f x)}}{4 f}+\frac{d \sqrt{a+i a \tan (e+f x)} (c+d \tan (e+f x))^{3/2}}{2 f}+\frac{\left (2 a^2 (i c+d)^3\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 d \left (15 i c^2+10 c d-7 i d^2\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 f}\\ &=-\frac{i \sqrt{2} \sqrt{a} (c-i d)^{5/2} \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{(7 c-i d) d \sqrt{a+i a \tan (e+f x)} \sqrt{c+d \tan (e+f x)}}{4 f}+\frac{d \sqrt{a+i a \tan (e+f x)} (c+d \tan (e+f x))^{3/2}}{2 f}+\frac{\left (d \left (15 c^2-10 i c d-7 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 f}\\ &=-\frac{i \sqrt{2} \sqrt{a} (c-i d)^{5/2} \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{(7 c-i d) d \sqrt{a+i a \tan (e+f x)} \sqrt{c+d \tan (e+f x)}}{4 f}+\frac{d \sqrt{a+i a \tan (e+f x)} (c+d \tan (e+f x))^{3/2}}{2 f}+\frac{\left (d \left (15 c^2-10 i c d-7 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 f}\\ &=-\frac{\sqrt [4]{-1} \sqrt{a} \sqrt{d} \left (15 c^2-10 i c d-7 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 f}-\frac{i \sqrt{2} \sqrt{a} (c-i d)^{5/2} \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{(7 c-i d) d \sqrt{a+i a \tan (e+f x)} \sqrt{c+d \tan (e+f x)}}{4 f}+\frac{d \sqrt{a+i a \tan (e+f x)} (c+d \tan (e+f x))^{3/2}}{2 f}\\ \end{align*}

Mathematica [B]  time = 5.14576, size = 539, normalized size = 2.1 \[ \frac{\left (\frac{1}{8}+\frac{i}{8}\right ) \sqrt{a+i a \tan (e+f x)} \left ((1-i) d \sqrt{c+d \tan (e+f x)} (9 c+2 d \tan (e+f x)-i d)-\frac{\cos (e+f x) \left (\sqrt{d} \left (15 c^2-10 i c d-7 d^2\right ) \left (\log \left (\frac{(2+2 i) e^{\frac{i e}{2}} \left ((1+i) \sqrt{d} \sqrt{1+e^{2 i (e+f x)}} \sqrt{c-\frac{i d \left (-1+e^{2 i (e+f x)}\right )}{1+e^{2 i (e+f x)}}}-i c e^{i (e+f x)}+c-d e^{i (e+f x)}+i d\right )}{d^{3/2} \left (-15 c^2+10 i c d+7 d^2\right ) \left (e^{i (e+f x)}+i\right )}\right )-\log \left (-\frac{(2+2 i) e^{\frac{i e}{2}} \left ((1+i) \sqrt{d} \sqrt{1+e^{2 i (e+f x)}} \sqrt{c-\frac{i d \left (-1+e^{2 i (e+f x)}\right )}{1+e^{2 i (e+f x)}}}+i c e^{i (e+f x)}+c+d e^{i (e+f x)}+i d\right )}{d^{3/2} \left (-15 c^2+10 i c d+7 d^2\right ) \left (e^{i (e+f x)}-i\right )}\right )\right )+(8+8 i) (c-i d)^{5/2} \log \left (2 \left (i \sqrt{c-i d} \sin (e+f x)+\sqrt{c-i d} \cos (e+f x)+\sqrt{i \sin (2 (e+f x))+\cos (2 (e+f x))+1} \sqrt{c+d \tan (e+f x)}\right )\right )\right )}{\sqrt{i \sin (2 (e+f x))+\cos (2 (e+f x))+1}}\right )}{f} \]

Warning: Unable to verify antiderivative.

[In]

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

[Out]

((1/8 + I/8)*Sqrt[a + I*a*Tan[e + f*x]]*(-((Cos[e + f*x]*(Sqrt[d]*(15*c^2 - (10*I)*c*d - 7*d^2)*(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)))]))/(d^(3/2)*(-15*c^2 + (10*I)*c*d + 7*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)))]))/(d^(3/2)*(-15*c^2 + (10*I)*c*d + 7*d^2)*(-I + E^(I*(e + f*x))))]) + (8 + 8*I)*(c - I*d)^(5/2)*Log[2*(S
qrt[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)]]*Sqr
t[c + d*Tan[e + f*x]])]))/Sqrt[1 + Cos[2*(e + f*x)] + I*Sin[2*(e + f*x)]]) + (1 - I)*d*Sqrt[c + d*Tan[e + f*x]
]*(9*c - I*d + 2*d*Tan[e + f*x])))/f

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Maple [B]  time = 0.111, size = 2130, normalized size = 8.3 \begin{align*} \text{result too large to display} \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

1/16/f*(a*(1+I*tan(f*x+e)))^(1/2)*(c+d*tan(f*x+e))^(1/2)*(-6*I*(a*(c+d*tan(f*x+e))*(1+I*tan(f*x+e)))^(1/2)*(I*
a*d)^(1/2)*2^(1/2)*(-a*(I*d-c))^(1/2)*tan(f*x+e)*d^3-8*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^4+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)*tan(f*x+e)*a*c^3*d-18*I*(a*(c+d*tan(f*x+e))*(1+I*tan(f*x+e)))^(1
/2)*(I*a*d)^(1/2)*2^(1/2)*(-a*(I*d-c))^(1/2)*tan(f*x+e)*c^2*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)
*tan(f*x+e)*a*c*d^3-16*I*(a*(c+d*tan(f*x+e))*(1+I*tan(f*x+e)))^(1/2)*(I*a*d)^(1/2)*2^(1/2)*(-a*(I*d-c))^(1/2)*
c*d^2-4*I*(a*(c+d*tan(f*x+e))*(1+I*tan(f*x+e)))^(1/2)*(I*a*d)^(1/2)*2^(1/2)*(-a*(I*d-c))^(1/2)*tan(f*x+e)^2*c*
d^2-15*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^3*d-7*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)*tan(f*x+e)*a*d^4+15*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)*tan(f*x+e)*a*c^3*d+3*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)*tan(f*x+e)*a*c*d^3+4*(a*(c+d*tan(f*x+e
))*(1+I*tan(f*x+e)))^(1/2)*(I*a*d)^(1/2)*2^(1/2)*(-a*(I*d-c))^(1/2)*tan(f*x+e)^2*d^3+5*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)*tan(f*x+e)*a*c^2*d^2+8*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^4-3*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^3+5*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*d^2-7*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^4+12*(a*(c+d*tan
(f*x+e))*(1+I*tan(f*x+e)))^(1/2)*(I*a*d)^(1/2)*2^(1/2)*(-a*(I*d-c))^(1/2)*tan(f*x+e)*c*d^2-8*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))/(ta
n(f*x+e)+I))*(I*a*d)^(1/2)*tan(f*x+e)*a*c^4+8*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)*tan(f*x+e)*a*d^4-18*
(a*(c+d*tan(f*x+e))*(1+I*tan(f*x+e)))^(1/2)*(I*a*d)^(1/2)*2^(1/2)*(-a*(I*d-c))^(1/2)*c^2*d-2*(a*(c+d*tan(f*x+e
))*(1+I*tan(f*x+e)))^(1/2)*(I*a*d)^(1/2)*2^(1/2)*(-a*(I*d-c))^(1/2)*d^3+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^3*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*c*d^3)*2^(1/2)/(a*(c+d*tan(f*x+e))*(1+I*tan(
f*x+e)))^(1/2)/(I*a*d)^(1/2)/(I*c-d)/(-tan(f*x+e)+I)/(-a*(I*d-c))^(1/2)

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Maxima [F]  time = 0., size = 0, normalized size = 0. \begin{align*} \int \sqrt{i \, a \tan \left (f x + e\right ) + a}{\left (d \tan \left (f x + e\right ) + c\right )}^{\frac{5}{2}}\,{d x} \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

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

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Fricas [B]  time = 1.93097, size = 2977, normalized size = 11.58 \begin{align*} \text{result too large to display} \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

1/8*(2*sqrt(2)*(9*c*d + I*d^2 + (9*c*d - 3*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))*sqrt(a/(e^(2*I*f*x + 2*I*e) + 1))*e^(I*f*x + I*e) - (f*e^(2*I*f*x + 2*I*e)
+ f)*sqrt((225*I*a*c^4*d + 300*a*c^3*d^2 - 310*I*a*c^2*d^3 - 140*a*c*d^4 + 49*I*a*d^5)/f^2)*log((sqrt(2)*(15*c
^2 - 10*I*c*d - 7*d^2 + (15*c^2 - 10*I*c*d - 7*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) + 2*I*f*sqrt((225*I*a*c
^4*d + 300*a*c^3*d^2 - 310*I*a*c^2*d^3 - 140*a*c*d^4 + 49*I*a*d^5)/f^2)*e^(2*I*f*x + 2*I*e))*e^(-2*I*f*x - 2*I
*e)/(15*c^2 - 10*I*c*d - 7*d^2)) + (f*e^(2*I*f*x + 2*I*e) + f)*sqrt((225*I*a*c^4*d + 300*a*c^3*d^2 - 310*I*a*c
^2*d^3 - 140*a*c*d^4 + 49*I*a*d^5)/f^2)*log((sqrt(2)*(15*c^2 - 10*I*c*d - 7*d^2 + (15*c^2 - 10*I*c*d - 7*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) - 2*I*f*sqrt((225*I*a*c^4*d + 300*a*c^3*d^2 - 310*I*a*c^2*d^3 - 140*a*c*d^4
+ 49*I*a*d^5)/f^2)*e^(2*I*f*x + 2*I*e))*e^(-2*I*f*x - 2*I*e)/(15*c^2 - 10*I*c*d - 7*d^2)) + 4*(f*e^(2*I*f*x +
2*I*e) + f)*sqrt(-(2*a*c^5 - 10*I*a*c^4*d - 20*a*c^3*d^2 + 20*I*a*c^2*d^3 + 10*a*c*d^4 - 2*I*a*d^5)/f^2)*log(
(sqrt(2)*(c^2 - 2*I*c*d - d^2 + (c^2 - 2*I*c*d - 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) + I*f*sqrt(-(2*a*c^5
- 10*I*a*c^4*d - 20*a*c^3*d^2 + 20*I*a*c^2*d^3 + 10*a*c*d^4 - 2*I*a*d^5)/f^2)*e^(2*I*f*x + 2*I*e))*e^(-2*I*f*x
- 2*I*e)/(c^2 - 2*I*c*d - d^2)) - 4*(f*e^(2*I*f*x + 2*I*e) + f)*sqrt(-(2*a*c^5 - 10*I*a*c^4*d - 20*a*c^3*d^2
+ 20*I*a*c^2*d^3 + 10*a*c*d^4 - 2*I*a*d^5)/f^2)*log((sqrt(2)*(c^2 - 2*I*c*d - d^2 + (c^2 - 2*I*c*d - 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) - I*f*sqrt(-(2*a*c^5 - 10*I*a*c^4*d - 20*a*c^3*d^2 + 20*I*a*c^2*d^3 + 10*a*c*d^
4 - 2*I*a*d^5)/f^2)*e^(2*I*f*x + 2*I*e))*e^(-2*I*f*x - 2*I*e)/(c^2 - 2*I*c*d - d^2)))/(f*e^(2*I*f*x + 2*I*e) +
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((a+I*a*tan(f*x+e))**(1/2)*(c+d*tan(f*x+e))**(5/2),x)

[Out]

Timed out

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Giac [A]  time = 5.47632, size = 383, normalized size = 1.49 \begin{align*} \frac{{\left (a^{2} c^{2} - 2 i \,{\left (i \, a \tan \left (f x + e\right ) + a\right )} a c d + 2 i \, a^{2} c d -{\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}\right )} \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} d^{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 \,{\left (i \, a \tan \left (f x + e\right ) + a\right )} a^{2} d^{2} + 2 i \, a^{3} d^{2}} \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

(a^2*c^2 - 2*I*(I*a*tan(f*x + e) + a)*a*c*d + 2*I*a^2*c*d - (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)*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)*d^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*(I*a*ta
n(f*x + e) + a)*a^2*d^2 + 2*I*a^3*d^2)