\(\int \frac {(a+i a \tan (c+d x))^{3/2} (A+B \tan (c+d x))}{\tan ^{\frac {5}{2}}(c+d x)} \, dx\) [165]

Optimal result

Integrand size = 38, antiderivative size = 137 \[ \int \frac {(a+i a \tan (c+d x))^{3/2} (A+B \tan (c+d x))}{\tan ^{\frac {5}{2}}(c+d x)} \, dx=\frac {(2+2 i) a^{3/2} (i A+B) \text {arctanh}\left (\frac {(1+i) \sqrt {a} \sqrt {\tan (c+d x)}}{\sqrt {a+i a \tan (c+d x)}}\right )}{d}-\frac {2 a A \sqrt {a+i a \tan (c+d x)}}{3 d \tan ^{\frac {3}{2}}(c+d x)}-\frac {2 a (4 i A+3 B) \sqrt {a+i a \tan (c+d x)}}{3 d \sqrt {\tan (c+d x)}} \]

[Out]

(2+2*I)*a^(3/2)*(I*A+B)*arctanh((1+I)*a^(1/2)*tan(d*x+c)^(1/2)/(a+I*a*tan(d*x+c))^(1/2))/d-2/3*a*(4*I*A+3*B)*(
a+I*a*tan(d*x+c))^(1/2)/d/tan(d*x+c)^(1/2)-2/3*a*A*(a+I*a*tan(d*x+c))^(1/2)/d/tan(d*x+c)^(3/2)

Rubi [A] (verified)

Time = 0.42 (sec) , antiderivative size = 137, normalized size of antiderivative = 1.00, number of steps used = 5, number of rules used = 5, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.132, Rules used = {3674, 3679, 12, 3625, 211} \[ \int \frac {(a+i a \tan (c+d x))^{3/2} (A+B \tan (c+d x))}{\tan ^{\frac {5}{2}}(c+d x)} \, dx=\frac {(2+2 i) a^{3/2} (B+i A) \text {arctanh}\left (\frac {(1+i) \sqrt {a} \sqrt {\tan (c+d x)}}{\sqrt {a+i a \tan (c+d x)}}\right )}{d}-\frac {2 a (3 B+4 i A) \sqrt {a+i a \tan (c+d x)}}{3 d \sqrt {\tan (c+d x)}}-\frac {2 a A \sqrt {a+i a \tan (c+d x)}}{3 d \tan ^{\frac {3}{2}}(c+d x)} \]

[In]

Int[((a + I*a*Tan[c + d*x])^(3/2)*(A + B*Tan[c + d*x]))/Tan[c + d*x]^(5/2),x]

[Out]

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

Rule 12

Int[(a_)*(u_), x_Symbol] :> Dist[a, Int[u, x], x] /; FreeQ[a, x] &&  !MatchQ[u, (b_)*(v_) /; FreeQ[b, x]]

Rule 211

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

Rule 3625

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 3674

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^2)*(B*c - A*d)*(a + b*Tan[e + f*x])^(m - 1)*((c + d*Tan[e + f*x]
)^(n + 1)/(d*f*(b*c + a*d)*(n + 1))), x] - Dist[a/(d*(b*c + a*d)*(n + 1)), Int[(a + b*Tan[e + f*x])^(m - 1)*(c
+ d*Tan[e + f*x])^(n + 1)*Simp[A*b*d*(m - n - 2) - B*(b*c*(m - 1) + a*d*(n + 1)) + (a*A*d*(m + n) - B*(a*c*(m
- 1) + b*d*(n + 1)))*Tan[e + f*x], x], x], x] /; FreeQ[{a, b, c, d, e, f, A, B}, x] && NeQ[b*c - a*d, 0] && E
qQ[a^2 + b^2, 0] && GtQ[m, 1] && LtQ[n, -1]

Rule 3679

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*d - B*c)*(a + b*Tan[e + f*x])^m*((c + d*Tan[e + f*x])^(n + 1)/(f*
(n + 1)*(c^2 + d^2))), x] - Dist[1/(a*(n + 1)*(c^2 + d^2)), Int[(a + b*Tan[e + f*x])^m*(c + d*Tan[e + f*x])^(n
+ 1)*Simp[A*(b*d*m - a*c*(n + 1)) - B*(b*c*m + a*d*(n + 1)) - a*(B*c - A*d)*(m + n + 1)*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] && LtQ[n, -1]

Rubi steps \begin{align*} \text {integral}& = -\frac {2 a A \sqrt {a+i a \tan (c+d x)}}{3 d \tan ^{\frac {3}{2}}(c+d x)}+\frac {2}{3} \int \frac {\sqrt {a+i a \tan (c+d x)} \left (\frac {1}{2} a (4 i A+3 B)-\frac {1}{2} a (2 A-3 i B) \tan (c+d x)\right )}{\tan ^{\frac {3}{2}}(c+d x)} \, dx \\ & = -\frac {2 a A \sqrt {a+i a \tan (c+d x)}}{3 d \tan ^{\frac {3}{2}}(c+d x)}-\frac {2 a (4 i A+3 B) \sqrt {a+i a \tan (c+d x)}}{3 d \sqrt {\tan (c+d x)}}+\frac {4 \int -\frac {3 a^2 (A-i B) \sqrt {a+i a \tan (c+d x)}}{2 \sqrt {\tan (c+d x)}} \, dx}{3 a} \\ & = -\frac {2 a A \sqrt {a+i a \tan (c+d x)}}{3 d \tan ^{\frac {3}{2}}(c+d x)}-\frac {2 a (4 i A+3 B) \sqrt {a+i a \tan (c+d x)}}{3 d \sqrt {\tan (c+d x)}}-(2 a (A-i B)) \int \frac {\sqrt {a+i a \tan (c+d x)}}{\sqrt {\tan (c+d x)}} \, dx \\ & = -\frac {2 a A \sqrt {a+i a \tan (c+d x)}}{3 d \tan ^{\frac {3}{2}}(c+d x)}-\frac {2 a (4 i A+3 B) \sqrt {a+i a \tan (c+d x)}}{3 d \sqrt {\tan (c+d x)}}+\frac {\left (4 a^3 (i A+B)\right ) \text {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 )}{d} \\ & = \frac {(2+2 i) a^{3/2} (i A+B) \text {arctanh}\left (\frac {(1+i) \sqrt {a} \sqrt {\tan (c+d x)}}{\sqrt {a+i a \tan (c+d x)}}\right )}{d}-\frac {2 a A \sqrt {a+i a \tan (c+d x)}}{3 d \tan ^{\frac {3}{2}}(c+d x)}-\frac {2 a (4 i A+3 B) \sqrt {a+i a \tan (c+d x)}}{3 d \sqrt {\tan (c+d x)}} \\ \end{align*}

Mathematica [B] (verified)

Both result and optimal contain complex but leaf count is larger than twice the leaf count of optimal. \(304\) vs. \(2(137)=274\).

Time = 7.01 (sec) , antiderivative size = 304, normalized size of antiderivative = 2.22 \[ \int \frac {(a+i a \tan (c+d x))^{3/2} (A+B \tan (c+d x))}{\tan ^{\frac {5}{2}}(c+d x)} \, dx=-\frac {2 A (a+i a \tan (c+d x))^{3/2}}{3 d \tan ^{\frac {3}{2}}(c+d x)}+(i A+B) \left (\frac {2 \sqrt {2} a \text {arctanh}\left (\frac {\sqrt {2} \sqrt {i a \tan (c+d x)}}{\sqrt {a+i a \tan (c+d x)}}\right ) \sqrt {i a \tan (c+d x)}}{d \sqrt {\tan (c+d x)}}-\frac {2 a^{3/2} \text {arcsinh}\left (\frac {\sqrt {i a \tan (c+d x)}}{\sqrt {a}}\right ) \sqrt {1+i \tan (c+d x)} \sqrt {i a \tan (c+d x)}}{d \sqrt {\tan (c+d x)} \sqrt {a+i a \tan (c+d x)}}+\frac {2 \sqrt [4]{-1} a \text {arcsinh}\left (\sqrt [4]{-1} \sqrt {\tan (c+d x)}\right ) \sqrt {a+i a \tan (c+d x)}}{d \sqrt {1+i \tan (c+d x)}}-\frac {2 a \sqrt {a+i a \tan (c+d x)}}{d \sqrt {\tan (c+d x)}}\right ) \]

[In]

Integrate[((a + I*a*Tan[c + d*x])^(3/2)*(A + B*Tan[c + d*x]))/Tan[c + d*x]^(5/2),x]

[Out]

(-2*A*(a + I*a*Tan[c + d*x])^(3/2))/(3*d*Tan[c + d*x]^(3/2)) + (I*A + B)*((2*Sqrt[2]*a*ArcTanh[(Sqrt[2]*Sqrt[I
*a*Tan[c + d*x]])/Sqrt[a + I*a*Tan[c + d*x]]]*Sqrt[I*a*Tan[c + d*x]])/(d*Sqrt[Tan[c + d*x]]) - (2*a^(3/2)*ArcS
inh[Sqrt[I*a*Tan[c + d*x]]/Sqrt[a]]*Sqrt[1 + I*Tan[c + d*x]]*Sqrt[I*a*Tan[c + d*x]])/(d*Sqrt[Tan[c + d*x]]*Sqr
t[a + I*a*Tan[c + d*x]]) + (2*(-1)^(1/4)*a*ArcSinh[(-1)^(1/4)*Sqrt[Tan[c + d*x]]]*Sqrt[a + I*a*Tan[c + d*x]])/
(d*Sqrt[1 + I*Tan[c + d*x]]) - (2*a*Sqrt[a + I*a*Tan[c + d*x]])/(d*Sqrt[Tan[c + d*x]]))

Maple [B] (verified)

Both result and optimal contain complex but leaf count of result is larger than twice the leaf count of optimal. 615 vs. \(2 (112 ) = 224\).

Time = 0.13 (sec) , antiderivative size = 616, normalized size of antiderivative = 4.50

method	result	size
derivativedivides	\(-\frac {\sqrt {a \left (1+i \tan \left (d x +c \right )\right )}\, a \left (-12 i B \ln \left (\frac {2 i a \tan \left (d x +c \right )+2 \sqrt {a \tan \left (d x +c \right ) \left (1+i \tan \left (d x +c \right )\right )}\, \sqrt {i a}+a}{2 \sqrt {i a}}\right ) \sqrt {-i a}\, a \left (\tan ^{2}\left (d x +c \right )\right )+3 i \sqrt {i a}\, \sqrt {2}\, \ln \left (\frac {2 \sqrt {2}\, \sqrt {-i a}\, \sqrt {a \tan \left (d x +c \right ) \left (1+i \tan \left (d x +c \right )\right )}-i a +3 a \tan \left (d x +c \right )}{\tan \left (d x +c \right )+i}\right ) a \left (\tan ^{2}\left (d x +c \right )\right )+16 i A \sqrt {i a}\, \sqrt {-i a}\, \sqrt {a \tan \left (d x +c \right ) \left (1+i \tan \left (d x +c \right )\right )}\, \tan \left (d x +c \right )+12 A \ln \left (\frac {2 i a \tan \left (d x +c \right )+2 \sqrt {a \tan \left (d x +c \right ) \left (1+i \tan \left (d x +c \right )\right )}\, \sqrt {i a}+a}{2 \sqrt {i a}}\right ) \sqrt {-i a}\, a \left (\tan ^{2}\left (d x +c \right )\right )+6 i \ln \left (\frac {2 i a \tan \left (d x +c \right )+2 \sqrt {a \tan \left (d x +c \right ) \left (1+i \tan \left (d x +c \right )\right )}\, \sqrt {i a}+a}{2 \sqrt {i a}}\right ) \sqrt {-i a}\, a \left (\tan ^{2}\left (d x +c \right )\right )-3 \sqrt {i a}\, \sqrt {2}\, \ln \left (\frac {2 \sqrt {2}\, \sqrt {-i a}\, \sqrt {a \tan \left (d x +c \right ) \left (1+i \tan \left (d x +c \right )\right )}-i a +3 a \tan \left (d x +c \right )}{\tan \left (d x +c \right )+i}\right ) a \left (\tan ^{2}\left (d x +c \right )\right )+12 B \sqrt {i a}\, \sqrt {-i a}\, \sqrt {a \tan \left (d x +c \right ) \left (1+i \tan \left (d x +c \right )\right )}\, \tan \left (d x +c \right )+6 \ln \left (\frac {2 i a \tan \left (d x +c \right )+2 \sqrt {a \tan \left (d x +c \right ) \left (1+i \tan \left (d x +c \right )\right )}\, \sqrt {i a}+a}{2 \sqrt {i a}}\right ) \sqrt {-i a}\, a \left (\tan ^{2}\left (d x +c \right )\right )+4 A \sqrt {i a}\, \sqrt {-i a}\, \sqrt {a \tan \left (d x +c \right ) \left (1+i \tan \left (d x +c \right )\right )}\right )}{6 d \tan \left (d x +c \right )^{\frac {3}{2}} \sqrt {i a}\, \sqrt {-i a}\, \sqrt {a \tan \left (d x +c \right ) \left (1+i \tan \left (d x +c \right )\right )}}\)	\(616\)
default	\(-\frac {\sqrt {a \left (1+i \tan \left (d x +c \right )\right )}\, a \left (-12 i B \ln \left (\frac {2 i a \tan \left (d x +c \right )+2 \sqrt {a \tan \left (d x +c \right ) \left (1+i \tan \left (d x +c \right )\right )}\, \sqrt {i a}+a}{2 \sqrt {i a}}\right ) \sqrt {-i a}\, a \left (\tan ^{2}\left (d x +c \right )\right )+3 i \sqrt {i a}\, \sqrt {2}\, \ln \left (\frac {2 \sqrt {2}\, \sqrt {-i a}\, \sqrt {a \tan \left (d x +c \right ) \left (1+i \tan \left (d x +c \right )\right )}-i a +3 a \tan \left (d x +c \right )}{\tan \left (d x +c \right )+i}\right ) a \left (\tan ^{2}\left (d x +c \right )\right )+16 i A \sqrt {i a}\, \sqrt {-i a}\, \sqrt {a \tan \left (d x +c \right ) \left (1+i \tan \left (d x +c \right )\right )}\, \tan \left (d x +c \right )+12 A \ln \left (\frac {2 i a \tan \left (d x +c \right )+2 \sqrt {a \tan \left (d x +c \right ) \left (1+i \tan \left (d x +c \right )\right )}\, \sqrt {i a}+a}{2 \sqrt {i a}}\right ) \sqrt {-i a}\, a \left (\tan ^{2}\left (d x +c \right )\right )+6 i \ln \left (\frac {2 i a \tan \left (d x +c \right )+2 \sqrt {a \tan \left (d x +c \right ) \left (1+i \tan \left (d x +c \right )\right )}\, \sqrt {i a}+a}{2 \sqrt {i a}}\right ) \sqrt {-i a}\, a \left (\tan ^{2}\left (d x +c \right )\right )-3 \sqrt {i a}\, \sqrt {2}\, \ln \left (\frac {2 \sqrt {2}\, \sqrt {-i a}\, \sqrt {a \tan \left (d x +c \right ) \left (1+i \tan \left (d x +c \right )\right )}-i a +3 a \tan \left (d x +c \right )}{\tan \left (d x +c \right )+i}\right ) a \left (\tan ^{2}\left (d x +c \right )\right )+12 B \sqrt {i a}\, \sqrt {-i a}\, \sqrt {a \tan \left (d x +c \right ) \left (1+i \tan \left (d x +c \right )\right )}\, \tan \left (d x +c \right )+6 \ln \left (\frac {2 i a \tan \left (d x +c \right )+2 \sqrt {a \tan \left (d x +c \right ) \left (1+i \tan \left (d x +c \right )\right )}\, \sqrt {i a}+a}{2 \sqrt {i a}}\right ) \sqrt {-i a}\, a \left (\tan ^{2}\left (d x +c \right )\right )+4 A \sqrt {i a}\, \sqrt {-i a}\, \sqrt {a \tan \left (d x +c \right ) \left (1+i \tan \left (d x +c \right )\right )}\right )}{6 d \tan \left (d x +c \right )^{\frac {3}{2}} \sqrt {i a}\, \sqrt {-i a}\, \sqrt {a \tan \left (d x +c \right ) \left (1+i \tan \left (d x +c \right )\right )}}\)	\(616\)
parts	\(-\frac {A \sqrt {a \left (1+i \tan \left (d x +c \right )\right )}\, a \left (3 i \sqrt {i a}\, \sqrt {2}\, \ln \left (\frac {2 \sqrt {2}\, \sqrt {-i a}\, \sqrt {a \tan \left (d x +c \right ) \left (1+i \tan \left (d x +c \right )\right )}-i a +3 a \tan \left (d x +c \right )}{\tan \left (d x +c \right )+i}\right ) a \left (\tan ^{2}\left (d x +c \right )\right )+12 i \ln \left (\frac {2 i a \tan \left (d x +c \right )+2 \sqrt {a \tan \left (d x +c \right ) \left (1+i \tan \left (d x +c \right )\right )}\, \sqrt {i a}+a}{2 \sqrt {i a}}\right ) \sqrt {-i a}\, a \left (\tan ^{2}\left (d x +c \right )\right )-3 \sqrt {i a}\, \sqrt {2}\, \ln \left (\frac {2 \sqrt {2}\, \sqrt {-i a}\, \sqrt {a \tan \left (d x +c \right ) \left (1+i \tan \left (d x +c \right )\right )}-i a +3 a \tan \left (d x +c \right )}{\tan \left (d x +c \right )+i}\right ) a \left (\tan ^{2}\left (d x +c \right )\right )+16 i \sqrt {a \tan \left (d x +c \right ) \left (1+i \tan \left (d x +c \right )\right )}\, \sqrt {i a}\, \sqrt {-i a}\, \tan \left (d x +c \right )+4 \sqrt {a \tan \left (d x +c \right ) \left (1+i \tan \left (d x +c \right )\right )}\, \sqrt {i a}\, \sqrt {-i a}\right )}{6 d \tan \left (d x +c \right )^{\frac {3}{2}} \sqrt {a \tan \left (d x +c \right ) \left (1+i \tan \left (d x +c \right )\right )}\, \sqrt {i a}\, \sqrt {-i a}}-\frac {B \sqrt {a \left (1+i \tan \left (d x +c \right )\right )}\, a \left (i \sqrt {i a}\, \sqrt {2}\, \ln \left (\frac {2 \sqrt {2}\, \sqrt {-i a}\, \sqrt {a \tan \left (d x +c \right ) \left (1+i \tan \left (d x +c \right )\right )}-i a +3 a \tan \left (d x +c \right )}{\tan \left (d x +c \right )+i}\right ) a \tan \left (d x +c \right )+\sqrt {i a}\, \sqrt {2}\, \ln \left (\frac {2 \sqrt {2}\, \sqrt {-i a}\, \sqrt {a \tan \left (d x +c \right ) \left (1+i \tan \left (d x +c \right )\right )}-i a +3 a \tan \left (d x +c \right )}{\tan \left (d x +c \right )+i}\right ) \tan \left (d x +c \right ) a +4 \ln \left (\frac {2 i a \tan \left (d x +c \right )+2 \sqrt {a \tan \left (d x +c \right ) \left (1+i \tan \left (d x +c \right )\right )}\, \sqrt {i a}+a}{2 \sqrt {i a}}\right ) \sqrt {-i a}\, a \tan \left (d x +c \right )+4 \sqrt {a \tan \left (d x +c \right ) \left (1+i \tan \left (d x +c \right )\right )}\, \sqrt {i a}\, \sqrt {-i a}\right )}{2 d \sqrt {\tan \left (d x +c \right )}\, \sqrt {i a}\, \sqrt {-i a}\, \sqrt {a \tan \left (d x +c \right ) \left (1+i \tan \left (d x +c \right )\right )}}\)	\(688\)

[In]

int((a+I*a*tan(d*x+c))^(3/2)*(A+B*tan(d*x+c))/tan(d*x+c)^(5/2),x,method=_RETURNVERBOSE)

[Out]

-1/6/d*(a*(1+I*tan(d*x+c)))^(1/2)*a/tan(d*x+c)^(3/2)*(-12*I*B*ln(1/2*(2*I*a*tan(d*x+c)+2*(a*tan(d*x+c)*(1+I*ta
n(d*x+c)))^(1/2)*(I*a)^(1/2)+a)/(I*a)^(1/2))*(-I*a)^(1/2)*a*tan(d*x+c)^2+3*I*(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*tan(d*x+c)^2+16*I*A*
(I*a)^(1/2)*(-I*a)^(1/2)*tan(d*x+c)*(a*tan(d*x+c)*(1+I*tan(d*x+c)))^(1/2)+12*A*ln(1/2*(2*I*a*tan(d*x+c)+2*(a*t
an(d*x+c)*(1+I*tan(d*x+c)))^(1/2)*(I*a)^(1/2)+a)/(I*a)^(1/2))*(-I*a)^(1/2)*a*tan(d*x+c)^2+6*I*ln(1/2*(2*I*a*ta
n(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)*a*tan(d*x+c)^2-3*(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*tan(d*x+c)^2+12*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)+6*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)*a*tan(d*x+
c)^2+4*A*(I*a)^(1/2)*(-I*a)^(1/2)*(a*tan(d*x+c)*(1+I*tan(d*x+c)))^(1/2))/(I*a)^(1/2)/(-I*a)^(1/2)/(a*tan(d*x+c
)*(1+I*tan(d*x+c)))^(1/2)

Fricas [B] (verification not implemented)

Both result and optimal contain complex but leaf count of result is larger than twice the leaf count of optimal. 524 vs. \(2 (103) = 206\).

Time = 0.26 (sec) , antiderivative size = 524, normalized size of antiderivative = 3.82 \[ \int \frac {(a+i a \tan (c+d x))^{3/2} (A+B \tan (c+d x))}{\tan ^{\frac {5}{2}}(c+d x)} \, dx=\frac {3 \, \sqrt {2} \sqrt {-\frac {{\left (i \, A^{2} + 2 \, A B - i \, B^{2}\right )} a^{3}}{d^{2}}} {\left (d e^{\left (4 i \, d x + 4 i \, c\right )} - 2 \, d e^{\left (2 i \, d x + 2 i \, c\right )} + d\right )} \log \left (\frac {{\left (i \, \sqrt {2} \sqrt {-\frac {{\left (i \, A^{2} + 2 \, A B - i \, B^{2}\right )} a^{3}}{d^{2}}} d e^{\left (i \, d x + i \, c\right )} + \sqrt {2} {\left ({\left (-i \, A - B\right )} a e^{\left (2 i \, d x + 2 i \, c\right )} + {\left (-i \, A - B\right )} a\right )} \sqrt {\frac {a}{e^{\left (2 i \, d x + 2 i \, c\right )} + 1}} \sqrt {\frac {-i \, e^{\left (2 i \, d x + 2 i \, c\right )} + i}{e^{\left (2 i \, d x + 2 i \, c\right )} + 1}}\right )} e^{\left (-i \, d x - i \, c\right )}}{{\left (-i \, A - B\right )} a}\right ) - 3 \, \sqrt {2} \sqrt {-\frac {{\left (i \, A^{2} + 2 \, A B - i \, B^{2}\right )} a^{3}}{d^{2}}} {\left (d e^{\left (4 i \, d x + 4 i \, c\right )} - 2 \, d e^{\left (2 i \, d x + 2 i \, c\right )} + d\right )} \log \left (\frac {{\left (-i \, \sqrt {2} \sqrt {-\frac {{\left (i \, A^{2} + 2 \, A B - i \, B^{2}\right )} a^{3}}{d^{2}}} d e^{\left (i \, d x + i \, c\right )} + \sqrt {2} {\left ({\left (-i \, A - B\right )} a e^{\left (2 i \, d x + 2 i \, c\right )} + {\left (-i \, A - B\right )} a\right )} \sqrt {\frac {a}{e^{\left (2 i \, d x + 2 i \, c\right )} + 1}} \sqrt {\frac {-i \, e^{\left (2 i \, d x + 2 i \, c\right )} + i}{e^{\left (2 i \, d x + 2 i \, c\right )} + 1}}\right )} e^{\left (-i \, d x - i \, c\right )}}{{\left (-i \, A - B\right )} a}\right ) + 2 \, \sqrt {2} {\left ({\left (5 \, A - 3 i \, B\right )} a e^{\left (5 i \, d x + 5 i \, c\right )} + 2 \, A a e^{\left (3 i \, d x + 3 i \, c\right )} - 3 \, {\left (A - i \, B\right )} a e^{\left (i \, d x + i \, c\right )}\right )} \sqrt {\frac {a}{e^{\left (2 i \, d x + 2 i \, c\right )} + 1}} \sqrt {\frac {-i \, e^{\left (2 i \, d x + 2 i \, c\right )} + i}{e^{\left (2 i \, d x + 2 i \, c\right )} + 1}}}{3 \, {\left (d e^{\left (4 i \, d x + 4 i \, c\right )} - 2 \, d e^{\left (2 i \, d x + 2 i \, c\right )} + d\right )}} \]

[In]

integrate((a+I*a*tan(d*x+c))^(3/2)*(A+B*tan(d*x+c))/tan(d*x+c)^(5/2),x, algorithm="fricas")

[Out]

1/3*(3*sqrt(2)*sqrt(-(I*A^2 + 2*A*B - I*B^2)*a^3/d^2)*(d*e^(4*I*d*x + 4*I*c) - 2*d*e^(2*I*d*x + 2*I*c) + d)*lo
g((I*sqrt(2)*sqrt(-(I*A^2 + 2*A*B - I*B^2)*a^3/d^2)*d*e^(I*d*x + I*c) + sqrt(2)*((-I*A - B)*a*e^(2*I*d*x + 2*I
*c) + (-I*A - B)*a)*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)/((-I*A - B)*a)) - 3*sqrt(2)*sqrt(-(I*A^2 + 2*A*B - I*B^2)*a^3/d^2)*(d*e^(4*I*d*x + 4*I*
c) - 2*d*e^(2*I*d*x + 2*I*c) + d)*log((-I*sqrt(2)*sqrt(-(I*A^2 + 2*A*B - I*B^2)*a^3/d^2)*d*e^(I*d*x + I*c) + s
qrt(2)*((-I*A - B)*a*e^(2*I*d*x + 2*I*c) + (-I*A - B)*a)*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)/((-I*A - B)*a)) + 2*sqrt(2)*((5*A - 3*I*B)*a*e^(5*
I*d*x + 5*I*c) + 2*A*a*e^(3*I*d*x + 3*I*c) - 3*(A - I*B)*a*e^(I*d*x + I*c))*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)))/(d*e^(4*I*d*x + 4*I*c) - 2*d*e^(2*I*d*x + 2*I*c)
+ d)

Sympy [F]

\[ \int \frac {(a+i a \tan (c+d x))^{3/2} (A+B \tan (c+d x))}{\tan ^{\frac {5}{2}}(c+d x)} \, dx=\int \frac {\left (i a \left (\tan {\left (c + d x \right )} - i\right )\right )^{\frac {3}{2}} \left (A + B \tan {\left (c + d x \right )}\right )}{\tan ^{\frac {5}{2}}{\left (c + d x \right )}}\, dx \]

[In]

integrate((a+I*a*tan(d*x+c))**(3/2)*(A+B*tan(d*x+c))/tan(d*x+c)**(5/2),x)

[Out]

Integral((I*a*(tan(c + d*x) - I))**(3/2)*(A + B*tan(c + d*x))/tan(c + d*x)**(5/2), x)

Maxima [F]

\[ \int \frac {(a+i a \tan (c+d x))^{3/2} (A+B \tan (c+d x))}{\tan ^{\frac {5}{2}}(c+d x)} \, dx=\int { \frac {{\left (B \tan \left (d x + c\right ) + A\right )} {\left (i \, a \tan \left (d x + c\right ) + a\right )}^{\frac {3}{2}}}{\tan \left (d x + c\right )^{\frac {5}{2}}} \,d x } \]

[In]

integrate((a+I*a*tan(d*x+c))^(3/2)*(A+B*tan(d*x+c))/tan(d*x+c)^(5/2),x, algorithm="maxima")

[Out]

integrate((B*tan(d*x + c) + A)*(I*a*tan(d*x + c) + a)^(3/2)/tan(d*x + c)^(5/2), x)

Giac [F(-2)]

Exception generated. \[ \int \frac {(a+i a \tan (c+d x))^{3/2} (A+B \tan (c+d x))}{\tan ^{\frac {5}{2}}(c+d x)} \, dx=\text {Exception raised: TypeError} \]

[In]

integrate((a+I*a*tan(d*x+c))^(3/2)*(A+B*tan(d*x+c))/tan(d*x+c)^(5/2),x, algorithm="giac")

[Out]

Exception raised: TypeError >> an error occurred running a Giac command:INPUT:sage2:=int(sage0,sageVARx):;OUTP
UT:Non regular value [0] was discarded and replaced randomly by 0=[78]Warning, replacing 78 by 64, a substitut
ion variabl

Mupad [F(-1)]

Timed out. \[ \int \frac {(a+i a \tan (c+d x))^{3/2} (A+B \tan (c+d x))}{\tan ^{\frac {5}{2}}(c+d x)} \, dx=\int \frac {\left (A+B\,\mathrm {tan}\left (c+d\,x\right )\right )\,{\left (a+a\,\mathrm {tan}\left (c+d\,x\right )\,1{}\mathrm {i}\right )}^{3/2}}{{\mathrm {tan}\left (c+d\,x\right )}^{5/2}} \,d x \]

[In]

int(((A + B*tan(c + d*x))*(a + a*tan(c + d*x)*1i)^(3/2))/tan(c + d*x)^(5/2),x)

[Out]

int(((A + B*tan(c + d*x))*(a + a*tan(c + d*x)*1i)^(3/2))/tan(c + d*x)^(5/2), x)