3.1152 \(\int \frac{(c+d \tan (e+f x))^{5/2}}{\sqrt{a+i a \tan (e+f x)}} \, dx\)

Optimal. Leaf size=250 \[ \frac{\sqrt [4]{-1} d^{3/2} (-d+5 i c) \tanh ^{-1}\left (\frac{(-1)^{3/4} \sqrt{d} \sqrt{a+i a \tan (e+f x)}}{\sqrt{a} \sqrt{c+d \tan (e+f x)}}\right )}{\sqrt{a} f}+\frac{(-d+i c) (c+d \tan (e+f x))^{3/2}}{f \sqrt{a+i a \tan (e+f x)}}-\frac{d (c+2 i d) \sqrt{a+i a \tan (e+f x)} \sqrt{c+d \tan (e+f x)}}{a f}-\frac{i (c-i d)^{5/2} \tanh ^{-1}\left (\frac{\sqrt{2} \sqrt{a} \sqrt{c+d \tan (e+f x)}}{\sqrt{c-i d} \sqrt{a+i a \tan (e+f x)}}\right )}{\sqrt{2} \sqrt{a} f} \]

[Out]

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

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Rubi [A]  time = 1.02507, antiderivative size = 250, 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 = {3558, 3597, 3601, 3544, 208, 3599, 63, 217, 206} \[ \frac{\sqrt [4]{-1} d^{3/2} (-d+5 i c) \tanh ^{-1}\left (\frac{(-1)^{3/4} \sqrt{d} \sqrt{a+i a \tan (e+f x)}}{\sqrt{a} \sqrt{c+d \tan (e+f x)}}\right )}{\sqrt{a} f}+\frac{(-d+i c) (c+d \tan (e+f x))^{3/2}}{f \sqrt{a+i a \tan (e+f x)}}-\frac{d (c+2 i d) \sqrt{a+i a \tan (e+f x)} \sqrt{c+d \tan (e+f x)}}{a f}-\frac{i (c-i d)^{5/2} \tanh ^{-1}\left (\frac{\sqrt{2} \sqrt{a} \sqrt{c+d \tan (e+f x)}}{\sqrt{c-i d} \sqrt{a+i a \tan (e+f x)}}\right )}{\sqrt{2} \sqrt{a} f} \]

Antiderivative was successfully verified.

[In]

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

[Out]

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

Rule 3558

Int[((a_) + (b_.)*tan[(e_.) + (f_.)*(x_)])^(m_)*((c_.) + (d_.)*tan[(e_.) + (f_.)*(x_)])^(n_), x_Symbol] :> -Si
mp[((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 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 \frac{(c+d \tan (e+f x))^{5/2}}{\sqrt{a+i a \tan (e+f x)}} \, dx &=\frac{(i c-d) (c+d \tan (e+f x))^{3/2}}{f \sqrt{a+i a \tan (e+f x)}}-\frac{\int \sqrt{a+i a \tan (e+f x)} \sqrt{c+d \tan (e+f x)} \left (-\frac{1}{2} a \left (c^2-4 i c d+3 d^2\right )+a (c+2 i d) d \tan (e+f x)\right ) \, dx}{a^2}\\ &=-\frac{(c+2 i d) d \sqrt{a+i a \tan (e+f x)} \sqrt{c+d \tan (e+f x)}}{a f}+\frac{(i c-d) (c+d \tan (e+f x))^{3/2}}{f \sqrt{a+i a \tan (e+f x)}}-\frac{\int \frac{\sqrt{a+i a \tan (e+f x)} \left (-\frac{1}{2} a^2 \left (c^3-3 i c^2 d+2 c d^2+2 i d^3\right )+\frac{1}{2} a^2 (5 i c-d) d^2 \tan (e+f x)\right )}{\sqrt{c+d \tan (e+f x)}} \, dx}{a^3}\\ &=-\frac{(c+2 i d) d \sqrt{a+i a \tan (e+f x)} \sqrt{c+d \tan (e+f x)}}{a f}+\frac{(i c-d) (c+d \tan (e+f x))^{3/2}}{f \sqrt{a+i a \tan (e+f x)}}+\frac{(c-i d)^3 \int \frac{\sqrt{a+i a \tan (e+f x)}}{\sqrt{c+d \tan (e+f x)}} \, dx}{2 a}+\frac{\left ((5 c+i d) d^2\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}{2 a^2}\\ &=-\frac{(c+2 i d) d \sqrt{a+i a \tan (e+f x)} \sqrt{c+d \tan (e+f x)}}{a f}+\frac{(i c-d) (c+d \tan (e+f x))^{3/2}}{f \sqrt{a+i a \tan (e+f x)}}+\frac{\left ((5 c+i d) d^2\right ) \operatorname{Subst}\left (\int \frac{1}{\sqrt{a+i a x} \sqrt{c+d x}} \, dx,x,\tan (e+f x)\right )}{2 f}+\frac{\left (a (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{i (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 )}{\sqrt{2} \sqrt{a} f}-\frac{(c+2 i d) d \sqrt{a+i a \tan (e+f x)} \sqrt{c+d \tan (e+f x)}}{a f}+\frac{(i c-d) (c+d \tan (e+f x))^{3/2}}{f \sqrt{a+i a \tan (e+f x)}}-\frac{\left ((5 i c-d) d^2\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 )}{a f}\\ &=-\frac{i (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 )}{\sqrt{2} \sqrt{a} f}-\frac{(c+2 i d) d \sqrt{a+i a \tan (e+f x)} \sqrt{c+d \tan (e+f x)}}{a f}+\frac{(i c-d) (c+d \tan (e+f x))^{3/2}}{f \sqrt{a+i a \tan (e+f x)}}-\frac{\left ((5 i c-d) d^2\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 )}{a f}\\ &=\frac{\sqrt [4]{-1} (5 i c-d) d^{3/2} \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 )}{\sqrt{a} f}-\frac{i (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 )}{\sqrt{2} \sqrt{a} f}-\frac{(c+2 i d) d \sqrt{a+i a \tan (e+f x)} \sqrt{c+d \tan (e+f x)}}{a f}+\frac{(i c-d) (c+d \tan (e+f x))^{3/2}}{f \sqrt{a+i a \tan (e+f x)}}\\ \end{align*}

Mathematica [B]  time = 7.01667, size = 549, normalized size = 2.2 \[ \frac{\left (\frac{1}{2}+\frac{i}{2}\right ) (\cos (f x)+i \sin (f x)) \left ((1+i) (\cos (f x)-i \sin (f x)) \sqrt{c+d \tan (e+f x)} \left (c^2+2 i c d-i d^2 \tan (e+f x)-2 d^2\right )-\frac{(\cos (e)+i \sin (e)) \left (d^{3/2} (d-5 i c) \left (\log \left (\frac{(1+i) e^{\frac{i e}{2}} \left (-(1+i) \sqrt{d} \sqrt{1+e^{2 i (e+f x)}} \sqrt{c-\frac{i d \left (-1+e^{2 i (e+f x)}\right )}{1+e^{2 i (e+f x)}}}+i c \left (e^{i (e+f x)}+i\right )+d e^{i (e+f x)}-i d\right )}{d^{5/2} (d-5 i c) \left (e^{i (e+f x)}+i\right )}\right )-\log \left (\frac{(1+i) e^{\frac{i e}{2}} \left ((1+i) \sqrt{d} \sqrt{1+e^{2 i (e+f x)}} \sqrt{c-\frac{i d \left (-1+e^{2 i (e+f x)}\right )}{1+e^{2 i (e+f x)}}}+i c e^{i (e+f x)}+c+d e^{i (e+f x)}+i d\right )}{d^{5/2} (d-5 i c) \left (e^{i (e+f x)}-i\right )}\right )\right )+(1+i) (c-i d)^{5/2} \log \left (2 \left (i \sqrt{c-i d} \sin (e+f x)+\sqrt{c-i d} \cos (e+f x)+\sqrt{i \sin (2 (e+f x))+\cos (2 (e+f x))+1} \sqrt{c+d \tan (e+f x)}\right )\right )\right )}{\sqrt{i \sin (2 (e+f x))+\cos (2 (e+f x))+1}}\right )}{f \sqrt{a+i a \tan (e+f x)}} \]

Warning: Unable to verify antiderivative.

[In]

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

[Out]

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

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Maple [B]  time = 0.095, size = 2647, normalized size = 10.6 \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))^(5/2)/(a+I*a*tan(f*x+e))^(1/2),x)

[Out]

-1/4/f*(c+d*tan(f*x+e))^(1/2)*(a*(1+I*tan(f*x+e)))^(1/2)*(2*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)*2^(1
/2)*(-a*(I*d-c))^(1/2)*tan(f*x+e)*c*d^2-24*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))*tan(f*x+e)*a*c*d^3+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)*2^(1/2)*(-a*(I*d-c))^(1/2)*tan(f*x+e)^2*d^3-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)*2^(1/2)*(-a*(I*d-c)
)^(1/2)*tan(f*x+e)^2*c*d^2+2*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)*2^(1/2)*(-a*(I*d-c))^(1/2)*tan(f*x+e)
*c^2*d+2*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)*2^(1/2)*(-a*(I*d-c))^(1/2)*tan(f*x+e)*c^3-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)*2^(1/2)*(-a*(I*d-c))^(1/2)*c^2*d+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
)*2^(1/2)*(-a*(I*d-c))^(1/2)*tan(f*x+e)^2*c^2*d+10*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))*a*c^2*d^2-4*I*tan(f*x+e)*c^3*(I*a*d)^(1/2)*(a*(c+d*tan
(f*x+e))*(1+I*tan(f*x+e)))^(1/2)-12*I*(a*(c+d*tan(f*x+e))*(1+I*tan(f*x+e)))^(1/2)*(I*a*d)^(1/2)*c^2*d+12*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))*ta
n(f*x+e)^2*a*c*d^3+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)*2^(1/2)*(-a*(I*d-c))^(1/2)*c^3-20*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))*tan(f*x+e)
*a*c^2*d^2+12*(a*(c+d*tan(f*x+e))*(1+I*tan(f*x+e)))^(1/2)*(I*a*d)^(1/2)*tan(f*x+e)*c^2*d-4*(a*(c+d*tan(f*x+e))
*(1+I*tan(f*x+e)))^(1/2)*(I*a*d)^(1/2)*tan(f*x+e)^2*c*d^2+2*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))*tan(f*x+e)^2*a*d^4-4*I*(a*(c+d*tan(f*x+e))*(1
+I*tan(f*x+e)))^(1/2)*(I*a*d)^(1/2)*tan(f*x+e)^2*d^3-4*c^3*(I*a*d)^(1/2)*(a*(c+d*tan(f*x+e))*(1+I*tan(f*x+e)))
^(1/2)-2*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))*a*d^4+4*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))*tan(f*x+e)*a*d^4-12*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))*a*c*d^3-12*(a*(c+d*tan(f*x+e))*(1+I*tan(f*x+e)))^(1/2)*(I*a*d)
^(1/2)*tan(f*x+e)*d^3+16*(a*(c+d*tan(f*x+e))*(1+I*tan(f*x+e)))^(1/2)*(I*a*d)^(1/2)*c*d^2+8*I*(a*(c+d*tan(f*x+e
))*(1+I*tan(f*x+e)))^(1/2)*(I*a*d)^(1/2)*d^3+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)*2^(1/2)*(-a*(I*d-c))^
(1/2)*c*d^2-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)*2^(1/2)*(-a*(I*d-c))^(1/2)*tan(f*x+e)^2*c^3-10*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))*ta
n(f*x+e)^2*a*c^2*d^2+20*I*(a*(c+d*tan(f*x+e))*(1+I*tan(f*x+e)))^(1/2)*(I*a*d)^(1/2)*tan(f*x+e)*c*d^2+2*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)*2^(1/2)*(-a*(I*d-c))^(1/2)*tan(f*x+e)*d^3-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)*2^(1/2)*(-a*(I*d-c))^(1/2)*d^3)/a/(a*(c+d*tan(f*x+e))*(1+I*tan(f*x+e)))^(1/2)/(I*c-d)/(-tan(f*x
+e)+I)^2/(I*a*d)^(1/2)

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

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))^(1/2),x, algorithm="maxima")

[Out]

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

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Fricas [B]  time = 2.97373, size = 2992, normalized size = 11.97 \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))^(5/2)/(a+I*a*tan(f*x+e))^(1/2),x, algorithm="fricas")

[Out]

-1/4*(a*f*sqrt(-(2*c^5 - 10*I*c^4*d - 20*c^3*d^2 + 20*I*c^2*d^3 + 10*c*d^4 - 2*I*d^5)/(a*f^2))*e^(2*I*f*x + 2*
I*e)*log(-(I*a*f*sqrt(-(2*c^5 - 10*I*c^4*d - 20*c^3*d^2 + 20*I*c^2*d^3 + 10*c*d^4 - 2*I*d^5)/(a*f^2))*e^(2*I*f
*x + 2*I*e) - 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))*e^(-I*f
*x - I*e)/(c^2 - 2*I*c*d - d^2)) - a*f*sqrt(-(2*c^5 - 10*I*c^4*d - 20*c^3*d^2 + 20*I*c^2*d^3 + 10*c*d^4 - 2*I*
d^5)/(a*f^2))*e^(2*I*f*x + 2*I*e)*log(-(-I*a*f*sqrt(-(2*c^5 - 10*I*c^4*d - 20*c^3*d^2 + 20*I*c^2*d^3 + 10*c*d^
4 - 2*I*d^5)/(a*f^2))*e^(2*I*f*x + 2*I*e) - 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))*e^(-I*f*x - I*e)/(c^2 - 2*I*c*d - d^2)) + a*f*sqrt((-25*I*c^2*d^3 + 10*c*d^4 + I*d^5)/(
a*f^2))*e^(2*I*f*x + 2*I*e)*log(-4*(2*sqrt(2)*(5*I*c*d^3 - d^4 + (5*I*c*d^3 - d^4)*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) + ((a*c*d - 3*I*a*d^2)*f*e^(2*I*f*x + 2*I*e) + (a*c*d + I*a*d^2)*f)*sqrt((-25*I*c^2*d^3 + 10*c*d^4 + I
*d^5)/(a*f^2)))/(5*c^4 - 4*I*c^3*d + 6*c^2*d^2 - 4*I*c*d^3 + d^4 + (5*c^4 - 4*I*c^3*d + 6*c^2*d^2 - 4*I*c*d^3
+ d^4)*e^(2*I*f*x + 2*I*e))) - a*f*sqrt((-25*I*c^2*d^3 + 10*c*d^4 + I*d^5)/(a*f^2))*e^(2*I*f*x + 2*I*e)*log(-4
*(2*sqrt(2)*(5*I*c*d^3 - d^4 + (5*I*c*d^3 - d^4)*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) - ((a*c*d - 3*I*a*d^2)*f*e
^(2*I*f*x + 2*I*e) + (a*c*d + I*a*d^2)*f)*sqrt((-25*I*c^2*d^3 + 10*c*d^4 + I*d^5)/(a*f^2)))/(5*c^4 - 4*I*c^3*d
+ 6*c^2*d^2 - 4*I*c*d^3 + d^4 + (5*c^4 - 4*I*c^3*d + 6*c^2*d^2 - 4*I*c*d^3 + d^4)*e^(2*I*f*x + 2*I*e))) - sqr
t(2)*(2*I*c^2 - 4*c*d - 2*I*d^2 + (2*I*c^2 - 4*c*d - 6*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))*e^(-2*I*f*x
- 2*I*e)/(a*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))**(5/2)/(a+I*a*tan(f*x+e))**(1/2),x)

[Out]

Timed out

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

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))^(1/2),x, algorithm="giac")

[Out]

Exception raised: TypeError