\(\int (a+i a \tan (c+d x))^{3/2} (A+B \tan (c+d x)) \, dx\) [77]

Optimal result

Integrand size = 28, antiderivative size = 107 \[ \int (a+i a \tan (c+d x))^{3/2} (A+B \tan (c+d x)) \, dx=-\frac {2 \sqrt {2} a^{3/2} (i A+B) \text {arctanh}\left (\frac {\sqrt {a+i a \tan (c+d x)}}{\sqrt {2} \sqrt {a}}\right )}{d}+\frac {2 a (i A+B) \sqrt {a+i a \tan (c+d x)}}{d}+\frac {2 B (a+i a \tan (c+d x))^{3/2}}{3 d} \]

[Out]

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

Rubi [A] (verified)

Time = 0.17 (sec) , antiderivative size = 107, normalized size of antiderivative = 1.00, number of steps used = 4, number of rules used = 4, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.143, Rules used = {3608, 3559, 3561, 212} \[ \int (a+i a \tan (c+d x))^{3/2} (A+B \tan (c+d x)) \, dx=-\frac {2 \sqrt {2} a^{3/2} (B+i A) \text {arctanh}\left (\frac {\sqrt {a+i a \tan (c+d x)}}{\sqrt {2} \sqrt {a}}\right )}{d}+\frac {2 a (B+i A) \sqrt {a+i a \tan (c+d x)}}{d}+\frac {2 B (a+i a \tan (c+d x))^{3/2}}{3 d} \]

[In]

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

[Out]

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

Rule 212

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

Rule 3559

Int[((a_) + (b_.)*tan[(c_.) + (d_.)*(x_)])^(n_), x_Symbol] :> Simp[b*((a + b*Tan[c + d*x])^(n - 1)/(d*(n - 1))
), x] + Dist[2*a, Int[(a + b*Tan[c + d*x])^(n - 1), x], x] /; FreeQ[{a, b, c, d}, x] && EqQ[a^2 + b^2, 0] && G
tQ[n, 1]

Rule 3561

Int[Sqrt[(a_) + (b_.)*tan[(c_.) + (d_.)*(x_)]], x_Symbol] :> Dist[-2*(b/d), Subst[Int[1/(2*a - x^2), x], x, Sq
rt[a + b*Tan[c + d*x]]], x] /; FreeQ[{a, b, c, d}, x] && EqQ[a^2 + b^2, 0]

Rule 3608

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

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

Mathematica [A] (verified)

Time = 0.56 (sec) , antiderivative size = 98, normalized size of antiderivative = 0.92 \[ \int (a+i a \tan (c+d x))^{3/2} (A+B \tan (c+d x)) \, dx=\frac {-6 i \sqrt {2} a^{3/2} (A-i B) \text {arctanh}\left (\frac {\sqrt {a+i a \tan (c+d x)}}{\sqrt {2} \sqrt {a}}\right )+2 a \sqrt {a+i a \tan (c+d x)} (3 i A+4 B+i B \tan (c+d x))}{3 d} \]

[In]

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

[Out]

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

Maple [A] (verified)

Time = 0.10 (sec) , antiderivative size = 99, normalized size of antiderivative = 0.93

method	result	size
derivativedivides	\(\frac {2 i \left (-\frac {i B \left (a +i a \tan \left (d x +c \right )\right )^{\frac {3}{2}}}{3}-i a B \sqrt {a +i a \tan \left (d x +c \right )}+a A \sqrt {a +i a \tan \left (d x +c \right )}-a^{\frac {3}{2}} \left (-i B +A \right ) \sqrt {2}\, \operatorname {arctanh}\left (\frac {\sqrt {a +i a \tan \left (d x +c \right )}\, \sqrt {2}}{2 \sqrt {a}}\right )\right )}{d}\)	\(99\)
default	\(\frac {2 i \left (-\frac {i B \left (a +i a \tan \left (d x +c \right )\right )^{\frac {3}{2}}}{3}-i a B \sqrt {a +i a \tan \left (d x +c \right )}+a A \sqrt {a +i a \tan \left (d x +c \right )}-a^{\frac {3}{2}} \left (-i B +A \right ) \sqrt {2}\, \operatorname {arctanh}\left (\frac {\sqrt {a +i a \tan \left (d x +c \right )}\, \sqrt {2}}{2 \sqrt {a}}\right )\right )}{d}\)	\(99\)
parts	\(\frac {2 i A a \left (\sqrt {a +i a \tan \left (d x +c \right )}-\sqrt {a}\, \sqrt {2}\, \operatorname {arctanh}\left (\frac {\sqrt {a +i a \tan \left (d x +c \right )}\, \sqrt {2}}{2 \sqrt {a}}\right )\right )}{d}+\frac {B \left (\frac {2 \left (a +i a \tan \left (d x +c \right )\right )^{\frac {3}{2}}}{3}+2 a \sqrt {a +i a \tan \left (d x +c \right )}-2 a^{\frac {3}{2}} \sqrt {2}\, \operatorname {arctanh}\left (\frac {\sqrt {a +i a \tan \left (d x +c \right )}\, \sqrt {2}}{2 \sqrt {a}}\right )\right )}{d}\)	\(126\)

[In]

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

[Out]

2*I/d*(-1/3*I*B*(a+I*a*tan(d*x+c))^(3/2)-I*a*B*(a+I*a*tan(d*x+c))^(1/2)+a*A*(a+I*a*tan(d*x+c))^(1/2)-a^(3/2)*(
A-I*B)*2^(1/2)*arctanh(1/2*(a+I*a*tan(d*x+c))^(1/2)*2^(1/2)/a^(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. 359 vs. \(2 (82) = 164\).

Time = 0.25 (sec) , antiderivative size = 359, normalized size of antiderivative = 3.36 \[ \int (a+i a \tan (c+d x))^{3/2} (A+B \tan (c+d x)) \, dx=\frac {3 \, \sqrt {2} \sqrt {-\frac {{\left (A^{2} - 2 i \, A B - B^{2}\right )} a^{3}}{d^{2}}} {\left (d e^{\left (2 i \, d x + 2 i \, c\right )} + d\right )} \log \left (\frac {4 \, {\left ({\left (-i \, A - B\right )} a^{2} e^{\left (i \, d x + i \, c\right )} + \sqrt {-\frac {{\left (A^{2} - 2 i \, A B - B^{2}\right )} a^{3}}{d^{2}}} {\left (d e^{\left (2 i \, d x + 2 i \, c\right )} + d\right )} \sqrt {\frac {a}{e^{\left (2 i \, d x + 2 i \, c\right )} + 1}}\right )} e^{\left (-i \, d x - i \, c\right )}}{{\left (-i \, A - B\right )} a}\right ) - 3 \, \sqrt {2} \sqrt {-\frac {{\left (A^{2} - 2 i \, A B - B^{2}\right )} a^{3}}{d^{2}}} {\left (d e^{\left (2 i \, d x + 2 i \, c\right )} + d\right )} \log \left (\frac {4 \, {\left ({\left (-i \, A - B\right )} a^{2} e^{\left (i \, d x + i \, c\right )} - \sqrt {-\frac {{\left (A^{2} - 2 i \, A B - B^{2}\right )} a^{3}}{d^{2}}} {\left (d e^{\left (2 i \, d x + 2 i \, c\right )} + d\right )} \sqrt {\frac {a}{e^{\left (2 i \, d x + 2 i \, c\right )} + 1}}\right )} e^{\left (-i \, d x - i \, c\right )}}{{\left (-i \, A - B\right )} a}\right ) - 2 \, \sqrt {2} {\left ({\left (-3 i \, A - 5 \, B\right )} a e^{\left (3 i \, d x + 3 i \, c\right )} + 3 \, {\left (-i \, A - B\right )} a e^{\left (i \, d x + i \, c\right )}\right )} \sqrt {\frac {a}{e^{\left (2 i \, d x + 2 i \, c\right )} + 1}}}{3 \, {\left (d e^{\left (2 i \, d x + 2 i \, c\right )} + d\right )}} \]

[In]

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

[Out]

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

Sympy [F]

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

[In]

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

[Out]

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

Maxima [A] (verification not implemented)

none

Time = 0.31 (sec) , antiderivative size = 111, normalized size of antiderivative = 1.04 \[ \int (a+i a \tan (c+d x))^{3/2} (A+B \tan (c+d x)) \, dx=\frac {i \, {\left (3 \, \sqrt {2} {\left (A - i \, B\right )} a^{\frac {5}{2}} \log \left (-\frac {\sqrt {2} \sqrt {a} - \sqrt {i \, a \tan \left (d x + c\right ) + a}}{\sqrt {2} \sqrt {a} + \sqrt {i \, a \tan \left (d x + c\right ) + a}}\right ) - 2 i \, {\left (i \, a \tan \left (d x + c\right ) + a\right )}^{\frac {3}{2}} B a + 6 \, \sqrt {i \, a \tan \left (d x + c\right ) + a} {\left (A - i \, B\right )} a^{2}\right )}}{3 \, a d} \]

[In]

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

[Out]

1/3*I*(3*sqrt(2)*(A - I*B)*a^(5/2)*log(-(sqrt(2)*sqrt(a) - sqrt(I*a*tan(d*x + c) + a))/(sqrt(2)*sqrt(a) + sqrt
(I*a*tan(d*x + c) + a))) - 2*I*(I*a*tan(d*x + c) + a)^(3/2)*B*a + 6*sqrt(I*a*tan(d*x + c) + a)*(A - I*B)*a^2)/
(a*d)

Giac [F]

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

[In]

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

[Out]

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

Mupad [B] (verification not implemented)

Time = 7.87 (sec) , antiderivative size = 139, normalized size of antiderivative = 1.30 \[ \int (a+i a \tan (c+d x))^{3/2} (A+B \tan (c+d x)) \, dx=\frac {2\,B\,{\left (a+a\,\mathrm {tan}\left (c+d\,x\right )\,1{}\mathrm {i}\right )}^{3/2}}{3\,d}+\frac {A\,a\,\sqrt {a+a\,\mathrm {tan}\left (c+d\,x\right )\,1{}\mathrm {i}}\,2{}\mathrm {i}}{d}+\frac {2\,B\,a\,\sqrt {a+a\,\mathrm {tan}\left (c+d\,x\right )\,1{}\mathrm {i}}}{d}+\frac {\sqrt {2}\,A\,{\left (-a\right )}^{3/2}\,\mathrm {atan}\left (\frac {\sqrt {2}\,\sqrt {a+a\,\mathrm {tan}\left (c+d\,x\right )\,1{}\mathrm {i}}}{2\,\sqrt {-a}}\right )\,2{}\mathrm {i}}{d}-\frac {2\,\sqrt {2}\,B\,a^{3/2}\,\mathrm {atanh}\left (\frac {\sqrt {2}\,\sqrt {a+a\,\mathrm {tan}\left (c+d\,x\right )\,1{}\mathrm {i}}}{2\,\sqrt {a}}\right )}{d} \]

[In]

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

[Out]

(2*B*(a + a*tan(c + d*x)*1i)^(3/2))/(3*d) + (A*a*(a + a*tan(c + d*x)*1i)^(1/2)*2i)/d + (2*B*a*(a + a*tan(c + d
*x)*1i)^(1/2))/d + (2^(1/2)*A*(-a)^(3/2)*atan((2^(1/2)*(a + a*tan(c + d*x)*1i)^(1/2))/(2*(-a)^(1/2)))*2i)/d -
(2*2^(1/2)*B*a^(3/2)*atanh((2^(1/2)*(a + a*tan(c + d*x)*1i)^(1/2))/(2*a^(1/2))))/d