3.190 \(\int \frac{\tan ^{\frac{5}{2}}(c+d x) (A+B \tan (c+d x))}{(a+i a \tan (c+d x))^{5/2}} \, dx\)

Optimal. Leaf size=249 \[ -\frac{(-7 B+i A) \sqrt{\tan (c+d x)}}{4 a^2 d \sqrt{a+i a \tan (c+d x)}}+\frac{\left (\frac{1}{8}+\frac{i}{8}\right ) (A-i B) \tanh ^{-1}\left (\frac{(1+i) \sqrt{a} \sqrt{\tan (c+d x)}}{\sqrt{a+i a \tan (c+d x)}}\right )}{a^{5/2} d}+\frac{2 \sqrt [4]{-1} B \tan ^{-1}\left (\frac{(-1)^{3/4} \sqrt{a} \sqrt{\tan (c+d x)}}{\sqrt{a+i a \tan (c+d x)}}\right )}{a^{5/2} d}+\frac{(-B+i A) \tan ^{\frac{5}{2}}(c+d x)}{5 d (a+i a \tan (c+d x))^{5/2}}+\frac{(A+3 i B) \tan ^{\frac{3}{2}}(c+d x)}{6 a d (a+i a \tan (c+d x))^{3/2}} \]

[Out]

(2*(-1)^(1/4)*B*ArcTan[((-1)^(3/4)*Sqrt[a]*Sqrt[Tan[c + d*x]])/Sqrt[a + I*a*Tan[c + d*x]]])/(a^(5/2)*d) + ((1/
8 + I/8)*(A - I*B)*ArcTanh[((1 + I)*Sqrt[a]*Sqrt[Tan[c + d*x]])/Sqrt[a + I*a*Tan[c + d*x]]])/(a^(5/2)*d) + ((I
*A - B)*Tan[c + d*x]^(5/2))/(5*d*(a + I*a*Tan[c + d*x])^(5/2)) + ((A + (3*I)*B)*Tan[c + d*x]^(3/2))/(6*a*d*(a
+ I*a*Tan[c + d*x])^(3/2)) - ((I*A - 7*B)*Sqrt[Tan[c + d*x]])/(4*a^2*d*Sqrt[a + I*a*Tan[c + d*x]])

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Rubi [A]  time = 0.855009, antiderivative size = 249, normalized size of antiderivative = 1., number of steps used = 10, number of rules used = 8, integrand size = 38, \(\frac{\text{number of rules}}{\text{integrand size}}\) = 0.21, Rules used = {3595, 3601, 3544, 205, 3599, 63, 217, 203} \[ -\frac{(-7 B+i A) \sqrt{\tan (c+d x)}}{4 a^2 d \sqrt{a+i a \tan (c+d x)}}+\frac{\left (\frac{1}{8}+\frac{i}{8}\right ) (A-i B) \tanh ^{-1}\left (\frac{(1+i) \sqrt{a} \sqrt{\tan (c+d x)}}{\sqrt{a+i a \tan (c+d x)}}\right )}{a^{5/2} d}+\frac{2 \sqrt [4]{-1} B \tan ^{-1}\left (\frac{(-1)^{3/4} \sqrt{a} \sqrt{\tan (c+d x)}}{\sqrt{a+i a \tan (c+d x)}}\right )}{a^{5/2} d}+\frac{(-B+i A) \tan ^{\frac{5}{2}}(c+d x)}{5 d (a+i a \tan (c+d x))^{5/2}}+\frac{(A+3 i B) \tan ^{\frac{3}{2}}(c+d x)}{6 a d (a+i a \tan (c+d x))^{3/2}} \]

Antiderivative was successfully verified.

[In]

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

[Out]

(2*(-1)^(1/4)*B*ArcTan[((-1)^(3/4)*Sqrt[a]*Sqrt[Tan[c + d*x]])/Sqrt[a + I*a*Tan[c + d*x]]])/(a^(5/2)*d) + ((1/
8 + I/8)*(A - I*B)*ArcTanh[((1 + I)*Sqrt[a]*Sqrt[Tan[c + d*x]])/Sqrt[a + I*a*Tan[c + d*x]]])/(a^(5/2)*d) + ((I
*A - B)*Tan[c + d*x]^(5/2))/(5*d*(a + I*a*Tan[c + d*x])^(5/2)) + ((A + (3*I)*B)*Tan[c + d*x]^(3/2))/(6*a*d*(a
+ I*a*Tan[c + d*x])^(3/2)) - ((I*A - 7*B)*Sqrt[Tan[c + d*x]])/(4*a^2*d*Sqrt[a + I*a*Tan[c + d*x]])

Rule 3595

Int[((a_) + (b_.)*tan[(e_.) + (f_.)*(x_)])^(m_)*((A_.) + (B_.)*tan[(e_.) + (f_.)*(x_)])*((c_.) + (d_.)*tan[(e_
.) + (f_.)*(x_)])^(n_), x_Symbol] :> -Simp[((A*b - a*B)*(a + b*Tan[e + f*x])^m*(c + d*Tan[e + f*x])^n)/(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 - 1)*Simp[A*(a*c*m + b*d*n
) - B*(b*c*m + a*d*n) - d*(b*B*(m - n) - a*A*(m + n))*Tan[e + f*x], x], x], x] /; FreeQ[{a, b, c, d, e, f, A,
B}, x] && NeQ[b*c - a*d, 0] && EqQ[a^2 + b^2, 0] && LtQ[m, 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 205

Int[((a_) + (b_.)*(x_)^2)^(-1), x_Symbol] :> Simp[(Rt[a/b, 2]*ArcTan[x/Rt[a/b, 2]])/a, x] /; FreeQ[{a, b}, x]
&& PosQ[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 203

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

Rubi steps

\begin{align*} \int \frac{\tan ^{\frac{5}{2}}(c+d x) (A+B \tan (c+d x))}{(a+i a \tan (c+d x))^{5/2}} \, dx &=\frac{(i A-B) \tan ^{\frac{5}{2}}(c+d x)}{5 d (a+i a \tan (c+d x))^{5/2}}-\frac{\int \frac{\tan ^{\frac{3}{2}}(c+d x) \left (\frac{5}{2} a (i A-B)+5 i a B \tan (c+d x)\right )}{(a+i a \tan (c+d x))^{3/2}} \, dx}{5 a^2}\\ &=\frac{(i A-B) \tan ^{\frac{5}{2}}(c+d x)}{5 d (a+i a \tan (c+d x))^{5/2}}+\frac{(A+3 i B) \tan ^{\frac{3}{2}}(c+d x)}{6 a d (a+i a \tan (c+d x))^{3/2}}+\frac{\int \frac{\sqrt{\tan (c+d x)} \left (-\frac{15}{4} a^2 (A+3 i B)-15 a^2 B \tan (c+d x)\right )}{\sqrt{a+i a \tan (c+d x)}} \, dx}{15 a^4}\\ &=\frac{(i A-B) \tan ^{\frac{5}{2}}(c+d x)}{5 d (a+i a \tan (c+d x))^{5/2}}+\frac{(A+3 i B) \tan ^{\frac{3}{2}}(c+d x)}{6 a d (a+i a \tan (c+d x))^{3/2}}-\frac{(i A-7 B) \sqrt{\tan (c+d x)}}{4 a^2 d \sqrt{a+i a \tan (c+d x)}}-\frac{\int \frac{\sqrt{a+i a \tan (c+d x)} \left (-\frac{15}{8} a^3 (i A-7 B)-15 i a^3 B \tan (c+d x)\right )}{\sqrt{\tan (c+d x)}} \, dx}{15 a^6}\\ &=\frac{(i A-B) \tan ^{\frac{5}{2}}(c+d x)}{5 d (a+i a \tan (c+d x))^{5/2}}+\frac{(A+3 i B) \tan ^{\frac{3}{2}}(c+d x)}{6 a d (a+i a \tan (c+d x))^{3/2}}-\frac{(i A-7 B) \sqrt{\tan (c+d x)}}{4 a^2 d \sqrt{a+i a \tan (c+d x)}}-\frac{B \int \frac{(a-i a \tan (c+d x)) \sqrt{a+i a \tan (c+d x)}}{\sqrt{\tan (c+d x)}} \, dx}{a^4}+\frac{(i A+B) \int \frac{\sqrt{a+i a \tan (c+d x)}}{\sqrt{\tan (c+d x)}} \, dx}{8 a^3}\\ &=\frac{(i A-B) \tan ^{\frac{5}{2}}(c+d x)}{5 d (a+i a \tan (c+d x))^{5/2}}+\frac{(A+3 i B) \tan ^{\frac{3}{2}}(c+d x)}{6 a d (a+i a \tan (c+d x))^{3/2}}-\frac{(i A-7 B) \sqrt{\tan (c+d x)}}{4 a^2 d \sqrt{a+i a \tan (c+d x)}}+\frac{(A-i B) \operatorname{Subst}\left (\int \frac{1}{-i a-2 a^2 x^2} \, dx,x,\frac{\sqrt{\tan (c+d x)}}{\sqrt{a+i a \tan (c+d x)}}\right )}{4 a d}-\frac{B \operatorname{Subst}\left (\int \frac{1}{\sqrt{x} \sqrt{a+i a x}} \, dx,x,\tan (c+d x)\right )}{a^2 d}\\ &=\frac{\left (\frac{1}{8}+\frac{i}{8}\right ) (A-i B) \tanh ^{-1}\left (\frac{(1+i) \sqrt{a} \sqrt{\tan (c+d x)}}{\sqrt{a+i a \tan (c+d x)}}\right )}{a^{5/2} d}+\frac{(i A-B) \tan ^{\frac{5}{2}}(c+d x)}{5 d (a+i a \tan (c+d x))^{5/2}}+\frac{(A+3 i B) \tan ^{\frac{3}{2}}(c+d x)}{6 a d (a+i a \tan (c+d x))^{3/2}}-\frac{(i A-7 B) \sqrt{\tan (c+d x)}}{4 a^2 d \sqrt{a+i a \tan (c+d x)}}-\frac{(2 B) \operatorname{Subst}\left (\int \frac{1}{\sqrt{a+i a x^2}} \, dx,x,\sqrt{\tan (c+d x)}\right )}{a^2 d}\\ &=\frac{\left (\frac{1}{8}+\frac{i}{8}\right ) (A-i B) \tanh ^{-1}\left (\frac{(1+i) \sqrt{a} \sqrt{\tan (c+d x)}}{\sqrt{a+i a \tan (c+d x)}}\right )}{a^{5/2} d}+\frac{(i A-B) \tan ^{\frac{5}{2}}(c+d x)}{5 d (a+i a \tan (c+d x))^{5/2}}+\frac{(A+3 i B) \tan ^{\frac{3}{2}}(c+d x)}{6 a d (a+i a \tan (c+d x))^{3/2}}-\frac{(i A-7 B) \sqrt{\tan (c+d x)}}{4 a^2 d \sqrt{a+i a \tan (c+d x)}}-\frac{(2 B) \operatorname{Subst}\left (\int \frac{1}{1-i a x^2} \, dx,x,\frac{\sqrt{\tan (c+d x)}}{\sqrt{a+i a \tan (c+d x)}}\right )}{a^2 d}\\ &=\frac{2 \sqrt [4]{-1} B \tan ^{-1}\left (\frac{(-1)^{3/4} \sqrt{a} \sqrt{\tan (c+d x)}}{\sqrt{a+i a \tan (c+d x)}}\right )}{a^{5/2} d}+\frac{\left (\frac{1}{8}+\frac{i}{8}\right ) (A-i B) \tanh ^{-1}\left (\frac{(1+i) \sqrt{a} \sqrt{\tan (c+d x)}}{\sqrt{a+i a \tan (c+d x)}}\right )}{a^{5/2} d}+\frac{(i A-B) \tan ^{\frac{5}{2}}(c+d x)}{5 d (a+i a \tan (c+d x))^{5/2}}+\frac{(A+3 i B) \tan ^{\frac{3}{2}}(c+d x)}{6 a d (a+i a \tan (c+d x))^{3/2}}-\frac{(i A-7 B) \sqrt{\tan (c+d x)}}{4 a^2 d \sqrt{a+i a \tan (c+d x)}}\\ \end{align*}

Mathematica [A]  time = 7.55064, size = 275, normalized size = 1.1 \[ \frac{e^{-2 i (c+d x)} \sqrt{\tan (c+d x)} \sec ^2(c+d x) \left (\sqrt{-1+e^{2 i (c+d x)}} \left (i A \left (-11 e^{2 i (c+d x)}+23 e^{4 i (c+d x)}+3\right )-3 B \left (-7 e^{2 i (c+d x)}+41 e^{4 i (c+d x)}+1\right )\right )-15 i (A-i B) e^{5 i (c+d x)} \tanh ^{-1}\left (\frac{e^{i (c+d x)}}{\sqrt{-1+e^{2 i (c+d x)}}}\right )+120 \sqrt{2} B e^{5 i (c+d x)} \tanh ^{-1}\left (\frac{\sqrt{2} e^{i (c+d x)}}{\sqrt{-1+e^{2 i (c+d x)}}}\right )\right )}{60 a^2 d \sqrt{-1+e^{2 i (c+d x)}} (\tan (c+d x)-i)^2 \sqrt{a+i a \tan (c+d x)}} \]

Antiderivative was successfully verified.

[In]

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

[Out]

((Sqrt[-1 + E^((2*I)*(c + d*x))]*(I*A*(3 - 11*E^((2*I)*(c + d*x)) + 23*E^((4*I)*(c + d*x))) - 3*B*(1 - 7*E^((2
*I)*(c + d*x)) + 41*E^((4*I)*(c + d*x)))) - (15*I)*(A - I*B)*E^((5*I)*(c + d*x))*ArcTanh[E^(I*(c + d*x))/Sqrt[
-1 + E^((2*I)*(c + d*x))]] + 120*Sqrt[2]*B*E^((5*I)*(c + d*x))*ArcTanh[(Sqrt[2]*E^(I*(c + d*x)))/Sqrt[-1 + E^(
(2*I)*(c + d*x))]])*Sec[c + d*x]^2*Sqrt[Tan[c + d*x]])/(60*a^2*d*E^((2*I)*(c + d*x))*Sqrt[-1 + E^((2*I)*(c + d
*x))]*(-I + Tan[c + d*x])^2*Sqrt[a + I*a*Tan[c + d*x]])

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Maple [B]  time = 0.073, size = 1542, normalized size = 6.2 \begin{align*} \text{result too large to display} \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

int(tan(d*x+c)^(5/2)*(A+B*tan(d*x+c))/(a+I*a*tan(d*x+c))^(5/2),x)

[Out]

-1/240/d*tan(d*x+c)^(1/2)*(a*(1+I*tan(d*x+c)))^(1/2)/a^3*(148*A*(I*a)^(1/2)*(-I*a)^(1/2)*(a*tan(d*x+c)*(1+I*ta
n(d*x+c)))^(1/2)*tan(d*x+c)^3-220*A*(a*tan(d*x+c)*(1+I*tan(d*x+c)))^(1/2)*(-I*a)^(1/2)*(I*a)^(1/2)*tan(d*x+c)-
15*A*(I*a)^(1/2)*2^(1/2)*ln(-(-2*2^(1/2)*(-I*a)^(1/2)*(a*tan(d*x+c)*(1+I*tan(d*x+c)))^(1/2)+I*a-3*a*tan(d*x+c)
)/(tan(d*x+c)+I))*tan(d*x+c)^4*a+1548*B*(a*tan(d*x+c)*(1+I*tan(d*x+c)))^(1/2)*(-I*a)^(1/2)*(I*a)^(1/2)*tan(d*x
+c)^2+240*B*(-I*a)^(1/2)*ln(1/2*(2*I*a*tan(d*x+c)+2*(a*tan(d*x+c)*(1+I*tan(d*x+c)))^(1/2)*(I*a)^(1/2)+a)/(I*a)
^(1/2))*a-420*B*(a*tan(d*x+c)*(1+I*tan(d*x+c)))^(1/2)*(-I*a)^(1/2)*(I*a)^(1/2)+240*B*ln(1/2*(2*I*a*tan(d*x+c)+
2*(a*tan(d*x+c)*(1+I*tan(d*x+c)))^(1/2)*(I*a)^(1/2)+a)/(I*a)^(1/2))*(-I*a)^(1/2)*tan(d*x+c)^4*a-1440*B*ln(1/2*
(2*I*a*tan(d*x+c)+2*(a*tan(d*x+c)*(1+I*tan(d*x+c)))^(1/2)*(I*a)^(1/2)+a)/(I*a)^(1/2))*(-I*a)^(1/2)*tan(d*x+c)^
2*a-15*A*2^(1/2)*(I*a)^(1/2)*ln(-(-2*2^(1/2)*(-I*a)^(1/2)*(a*tan(d*x+c)*(1+I*tan(d*x+c)))^(1/2)+I*a-3*a*tan(d*
x+c))/(tan(d*x+c)+I))*a+960*I*B*ln(1/2*(2*I*a*tan(d*x+c)+2*(a*tan(d*x+c)*(1+I*tan(d*x+c)))^(1/2)*(I*a)^(1/2)+a
)/(I*a)^(1/2))*(-I*a)^(1/2)*tan(d*x+c)*a+60*B*(I*a)^(1/2)*2^(1/2)*ln(-(-2*2^(1/2)*(-I*a)^(1/2)*(a*tan(d*x+c)*(
1+I*tan(d*x+c)))^(1/2)+I*a-3*a*tan(d*x+c))/(tan(d*x+c)+I))*tan(d*x+c)^3*a+90*A*(I*a)^(1/2)*2^(1/2)*ln(-(-2*2^(
1/2)*(-I*a)^(1/2)*(a*tan(d*x+c)*(1+I*tan(d*x+c)))^(1/2)+I*a-3*a*tan(d*x+c))/(tan(d*x+c)+I))*tan(d*x+c)^2*a-60*
B*2^(1/2)*(I*a)^(1/2)*ln(-(-2*2^(1/2)*(-I*a)^(1/2)*(a*tan(d*x+c)*(1+I*tan(d*x+c)))^(1/2)+I*a-3*a*tan(d*x+c))/(
tan(d*x+c)+I))*tan(d*x+c)*a+60*I*A*(I*a)^(1/2)*2^(1/2)*ln(-(-2*2^(1/2)*(-I*a)^(1/2)*(a*tan(d*x+c)*(1+I*tan(d*x
+c)))^(1/2)+I*a-3*a*tan(d*x+c))/(tan(d*x+c)+I))*tan(d*x+c)^3*a-90*I*B*(I*a)^(1/2)*2^(1/2)*ln(-(-2*2^(1/2)*(-I*
a)^(1/2)*(a*tan(d*x+c)*(1+I*tan(d*x+c)))^(1/2)+I*a-3*a*tan(d*x+c))/(tan(d*x+c)+I))*tan(d*x+c)^2*a+15*I*B*(I*a)
^(1/2)*2^(1/2)*ln(-(-2*2^(1/2)*(-I*a)^(1/2)*(a*tan(d*x+c)*(1+I*tan(d*x+c)))^(1/2)+I*a-3*a*tan(d*x+c))/(tan(d*x
+c)+I))*a+588*I*B*(I*a)^(1/2)*(-I*a)^(1/2)*(a*tan(d*x+c)*(1+I*tan(d*x+c)))^(1/2)*tan(d*x+c)^3+15*I*B*(I*a)^(1/
2)*2^(1/2)*ln(-(-2*2^(1/2)*(-I*a)^(1/2)*(a*tan(d*x+c)*(1+I*tan(d*x+c)))^(1/2)+I*a-3*a*tan(d*x+c))/(tan(d*x+c)+
I))*tan(d*x+c)^4*a-960*I*B*ln(1/2*(2*I*a*tan(d*x+c)+2*(a*tan(d*x+c)*(1+I*tan(d*x+c)))^(1/2)*(I*a)^(1/2)+a)/(I*
a)^(1/2))*(-I*a)^(1/2)*tan(d*x+c)^3*a-60*I*A*(I*a)^(1/2)*2^(1/2)*ln(-(-2*2^(1/2)*(-I*a)^(1/2)*(a*tan(d*x+c)*(1
+I*tan(d*x+c)))^(1/2)+I*a-3*a*tan(d*x+c))/(tan(d*x+c)+I))*tan(d*x+c)*a-308*I*A*(I*a)^(1/2)*(-I*a)^(1/2)*(a*tan
(d*x+c)*(1+I*tan(d*x+c)))^(1/2)*tan(d*x+c)^2+60*I*A*(I*a)^(1/2)*(-I*a)^(1/2)*(a*tan(d*x+c)*(1+I*tan(d*x+c)))^(
1/2)-1380*I*B*(I*a)^(1/2)*(-I*a)^(1/2)*(a*tan(d*x+c)*(1+I*tan(d*x+c)))^(1/2)*tan(d*x+c))/(a*tan(d*x+c)*(1+I*ta
n(d*x+c)))^(1/2)/(-tan(d*x+c)+I)^4/(I*a)^(1/2)/(-I*a)^(1/2)

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

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

Exception raised: RuntimeError

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Fricas [B]  time = 2.9785, size = 2205, normalized size = 8.86 \begin{align*} \text{result too large to display} \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

1/120*(15*sqrt(1/2)*a^3*d*sqrt((I*A^2 + 2*A*B - I*B^2)/(a^5*d^2))*e^(6*I*d*x + 6*I*c)*log((2*sqrt(1/2)*a^3*d*s
qrt((I*A^2 + 2*A*B - I*B^2)/(a^5*d^2))*e^(2*I*d*x + 2*I*c) + sqrt(2)*((I*A + B)*e^(2*I*d*x + 2*I*c) + I*A + B)
*sqrt(a/(e^(2*I*d*x + 2*I*c) + 1))*sqrt((-I*e^(2*I*d*x + 2*I*c) + I)/(e^(2*I*d*x + 2*I*c) + 1))*e^(I*d*x + I*c
))*e^(-I*d*x - I*c)/(4*I*A + 4*B)) - 15*sqrt(1/2)*a^3*d*sqrt((I*A^2 + 2*A*B - I*B^2)/(a^5*d^2))*e^(6*I*d*x + 6
*I*c)*log(-(2*sqrt(1/2)*a^3*d*sqrt((I*A^2 + 2*A*B - I*B^2)/(a^5*d^2))*e^(2*I*d*x + 2*I*c) - sqrt(2)*((I*A + B)
*e^(2*I*d*x + 2*I*c) + I*A + B)*sqrt(a/(e^(2*I*d*x + 2*I*c) + 1))*sqrt((-I*e^(2*I*d*x + 2*I*c) + I)/(e^(2*I*d*
x + 2*I*c) + 1))*e^(I*d*x + I*c))*e^(-I*d*x - I*c)/(4*I*A + 4*B)) - 30*a^3*d*sqrt(-4*I*B^2/(a^5*d^2))*e^(6*I*d
*x + 6*I*c)*log(52/605*(4*sqrt(2)*(B*e^(2*I*d*x + 2*I*c) + B)*sqrt(a/(e^(2*I*d*x + 2*I*c) + 1))*sqrt((-I*e^(2*
I*d*x + 2*I*c) + I)/(e^(2*I*d*x + 2*I*c) + 1))*e^(I*d*x + I*c) + (3*a^3*d*e^(2*I*d*x + 2*I*c) - a^3*d)*sqrt(-4
*I*B^2/(a^5*d^2)))/(B*e^(2*I*d*x + 2*I*c) + B)) + 30*a^3*d*sqrt(-4*I*B^2/(a^5*d^2))*e^(6*I*d*x + 6*I*c)*log(52
/605*(4*sqrt(2)*(B*e^(2*I*d*x + 2*I*c) + B)*sqrt(a/(e^(2*I*d*x + 2*I*c) + 1))*sqrt((-I*e^(2*I*d*x + 2*I*c) + I
)/(e^(2*I*d*x + 2*I*c) + 1))*e^(I*d*x + I*c) - (3*a^3*d*e^(2*I*d*x + 2*I*c) - a^3*d)*sqrt(-4*I*B^2/(a^5*d^2)))
/(B*e^(2*I*d*x + 2*I*c) + B)) + sqrt(2)*((-23*I*A + 123*B)*e^(6*I*d*x + 6*I*c) + (-12*I*A + 102*B)*e^(4*I*d*x
+ 4*I*c) + (8*I*A - 18*B)*e^(2*I*d*x + 2*I*c) - 3*I*A + 3*B)*sqrt(a/(e^(2*I*d*x + 2*I*c) + 1))*sqrt((-I*e^(2*I
*d*x + 2*I*c) + I)/(e^(2*I*d*x + 2*I*c) + 1))*e^(I*d*x + I*c))*e^(-6*I*d*x - 6*I*c)/(a^3*d)

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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(tan(d*x+c)**(5/2)*(A+B*tan(d*x+c))/(a+I*a*tan(d*x+c))**(5/2),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(tan(d*x+c)^(5/2)*(A+B*tan(d*x+c))/(a+I*a*tan(d*x+c))^(5/2),x, algorithm="giac")

[Out]

Exception raised: TypeError