\(\int \frac {A+B \tan (c+d x)}{(a+i a \tan (c+d x))^{3/2}} \, dx\) [100]

Optimal result

Integrand size = 28, antiderivative size = 121 \[ \int \frac {A+B \tan (c+d x)}{(a+i a \tan (c+d x))^{3/2}} \, dx=-\frac {(i A+B) \text {arctanh}\left (\frac {\sqrt {a+i a \tan (c+d x)}}{\sqrt {2} \sqrt {a}}\right )}{2 \sqrt {2} a^{3/2} d}+\frac {i A-B}{3 d (a+i a \tan (c+d x))^{3/2}}+\frac {i A+B}{2 a d \sqrt {a+i a \tan (c+d x)}} \]

[Out]

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

Rubi [A] (verified)

Time = 0.18 (sec) , antiderivative size = 121, 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 = {3607, 3560, 3561, 212} \[ \int \frac {A+B \tan (c+d x)}{(a+i a \tan (c+d x))^{3/2}} \, dx=-\frac {(B+i A) \text {arctanh}\left (\frac {\sqrt {a+i a \tan (c+d x)}}{\sqrt {2} \sqrt {a}}\right )}{2 \sqrt {2} a^{3/2} d}+\frac {-B+i A}{3 d (a+i a \tan (c+d x))^{3/2}}+\frac {B+i A}{2 a d \sqrt {a+i a \tan (c+d x)}} \]

[In]

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

[Out]

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

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 3560

Int[((a_) + (b_.)*tan[(c_.) + (d_.)*(x_)])^(n_), x_Symbol] :> Simp[a*((a + b*Tan[c + d*x])^n/(2*b*d*n)), x] +
Dist[1/(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] && LtQ[n
, 0]

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 3607

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

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

Mathematica [C] (verified)

Result contains higher order function than in optimal. Order 5 vs. order 3 in optimal.

Time = 0.62 (sec) , antiderivative size = 76, normalized size of antiderivative = 0.63 \[ \int \frac {A+B \tan (c+d x)}{(a+i a \tan (c+d x))^{3/2}} \, dx=\frac {2 i A-2 B-3 (A-i B) \operatorname {Hypergeometric2F1}\left (-\frac {1}{2},1,\frac {1}{2},\frac {1}{2} (1+i \tan (c+d x))\right ) (-i+\tan (c+d x))}{6 d (a+i a \tan (c+d x))^{3/2}} \]

[In]

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

[Out]

((2*I)*A - 2*B - 3*(A - I*B)*Hypergeometric2F1[-1/2, 1, 1/2, (1 + I*Tan[c + d*x])/2]*(-I + Tan[c + d*x]))/(6*d
*(a + I*a*Tan[c + d*x])^(3/2))

Maple [A] (verified)

Time = 0.13 (sec) , antiderivative size = 96, normalized size of antiderivative = 0.79

method	result	size
derivativedivides	\(\frac {2 i \left (-\frac {-\frac {A}{2}-\frac {i B}{2}}{3 \left (a +i a \tan \left (d x +c \right )\right )^{\frac {3}{2}}}-\frac {i B -A}{4 a \sqrt {a +i a \tan \left (d x +c \right )}}-\frac {\left (-i B +A \right ) \sqrt {2}\, \operatorname {arctanh}\left (\frac {\sqrt {a +i a \tan \left (d x +c \right )}\, \sqrt {2}}{2 \sqrt {a}}\right )}{8 a^{\frac {3}{2}}}\right )}{d}\)	\(96\)
default	\(\frac {2 i \left (-\frac {-\frac {A}{2}-\frac {i B}{2}}{3 \left (a +i a \tan \left (d x +c \right )\right )^{\frac {3}{2}}}-\frac {i B -A}{4 a \sqrt {a +i a \tan \left (d x +c \right )}}-\frac {\left (-i B +A \right ) \sqrt {2}\, \operatorname {arctanh}\left (\frac {\sqrt {a +i a \tan \left (d x +c \right )}\, \sqrt {2}}{2 \sqrt {a}}\right )}{8 a^{\frac {3}{2}}}\right )}{d}\)	\(96\)
parts	\(\frac {2 i A a \left (-\frac {\sqrt {2}\, \operatorname {arctanh}\left (\frac {\sqrt {a +i a \tan \left (d x +c \right )}\, \sqrt {2}}{2 \sqrt {a}}\right )}{8 a^{\frac {5}{2}}}+\frac {1}{4 a^{2} \sqrt {a +i a \tan \left (d x +c \right )}}+\frac {1}{6 a \left (a +i a \tan \left (d x +c \right )\right )^{\frac {3}{2}}}\right )}{d}+\frac {B \left (-\frac {\sqrt {2}\, \operatorname {arctanh}\left (\frac {\sqrt {a +i a \tan \left (d x +c \right )}\, \sqrt {2}}{2 \sqrt {a}}\right )}{4 a^{\frac {3}{2}}}-\frac {1}{3 \left (a +i a \tan \left (d x +c \right )\right )^{\frac {3}{2}}}+\frac {1}{2 a \sqrt {a +i a \tan \left (d x +c \right )}}\right )}{d}\)	\(152\)

[In]

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

[Out]

2*I/d*(-1/3*(-1/2*A-1/2*I*B)/(a+I*a*tan(d*x+c))^(3/2)-1/4/a*(-A+I*B)/(a+I*a*tan(d*x+c))^(1/2)-1/8*(A-I*B)/a^(3
/2)*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. 372 vs. \(2 (90) = 180\).

Time = 0.26 (sec) , antiderivative size = 372, normalized size of antiderivative = 3.07 \[ \int \frac {A+B \tan (c+d x)}{(a+i a \tan (c+d x))^{3/2}} \, dx=\frac {{\left (3 \, \sqrt {\frac {1}{2}} a^{2} d \sqrt {-\frac {A^{2} - 2 i \, A B - B^{2}}{a^{3} d^{2}}} e^{\left (3 i \, d x + 3 i \, c\right )} \log \left (-\frac {4 \, {\left (\sqrt {2} \sqrt {\frac {1}{2}} {\left (a^{2} d e^{\left (2 i \, d x + 2 i \, c\right )} + a^{2} d\right )} \sqrt {\frac {a}{e^{\left (2 i \, d x + 2 i \, c\right )} + 1}} \sqrt {-\frac {A^{2} - 2 i \, A B - B^{2}}{a^{3} d^{2}}} + {\left (-i \, A - B\right )} a e^{\left (i \, d x + i \, c\right )}\right )} e^{\left (-i \, d x - i \, c\right )}}{i \, A + B}\right ) - 3 \, \sqrt {\frac {1}{2}} a^{2} d \sqrt {-\frac {A^{2} - 2 i \, A B - B^{2}}{a^{3} d^{2}}} e^{\left (3 i \, d x + 3 i \, c\right )} \log \left (\frac {4 \, {\left (\sqrt {2} \sqrt {\frac {1}{2}} {\left (a^{2} d e^{\left (2 i \, d x + 2 i \, c\right )} + a^{2} d\right )} \sqrt {\frac {a}{e^{\left (2 i \, d x + 2 i \, c\right )} + 1}} \sqrt {-\frac {A^{2} - 2 i \, A B - B^{2}}{a^{3} d^{2}}} - {\left (-i \, A - B\right )} a e^{\left (i \, d x + i \, c\right )}\right )} e^{\left (-i \, d x - i \, c\right )}}{i \, A + B}\right ) - \sqrt {2} {\left (2 \, {\left (-2 i \, A - B\right )} e^{\left (4 i \, d x + 4 i \, c\right )} - {\left (5 i \, A + B\right )} e^{\left (2 i \, d x + 2 i \, c\right )} - i \, A + B\right )} \sqrt {\frac {a}{e^{\left (2 i \, d x + 2 i \, c\right )} + 1}}\right )} e^{\left (-3 i \, d x - 3 i \, c\right )}}{12 \, a^{2} d} \]

[In]

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

[Out]

1/12*(3*sqrt(1/2)*a^2*d*sqrt(-(A^2 - 2*I*A*B - B^2)/(a^3*d^2))*e^(3*I*d*x + 3*I*c)*log(-4*(sqrt(2)*sqrt(1/2)*(
a^2*d*e^(2*I*d*x + 2*I*c) + a^2*d)*sqrt(a/(e^(2*I*d*x + 2*I*c) + 1))*sqrt(-(A^2 - 2*I*A*B - B^2)/(a^3*d^2)) +
(-I*A - B)*a*e^(I*d*x + I*c))*e^(-I*d*x - I*c)/(I*A + B)) - 3*sqrt(1/2)*a^2*d*sqrt(-(A^2 - 2*I*A*B - B^2)/(a^3
*d^2))*e^(3*I*d*x + 3*I*c)*log(4*(sqrt(2)*sqrt(1/2)*(a^2*d*e^(2*I*d*x + 2*I*c) + a^2*d)*sqrt(a/(e^(2*I*d*x + 2
*I*c) + 1))*sqrt(-(A^2 - 2*I*A*B - B^2)/(a^3*d^2)) - (-I*A - B)*a*e^(I*d*x + I*c))*e^(-I*d*x - I*c)/(I*A + B))
- sqrt(2)*(2*(-2*I*A - B)*e^(4*I*d*x + 4*I*c) - (5*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)))*e^(-3*I*d*x - 3*I*c)/(a^2*d)

Sympy [F]

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

[In]

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

[Out]

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

Maxima [A] (verification not implemented)

none

Time = 0.31 (sec) , antiderivative size = 113, normalized size of antiderivative = 0.93 \[ \int \frac {A+B \tan (c+d x)}{(a+i a \tan (c+d x))^{3/2}} \, dx=\frac {i \, {\left (\frac {3 \, \sqrt {2} {\left (A - i \, B\right )} \log \left (-\frac {\sqrt {2} \sqrt {a} - \sqrt {i \, a \tan \left (d x + c\right ) + a}}{\sqrt {2} \sqrt {a} + \sqrt {i \, a \tan \left (d x + c\right ) + a}}\right )}{\sqrt {a}} + \frac {4 \, {\left (3 \, {\left (i \, a \tan \left (d x + c\right ) + a\right )} {\left (A - i \, B\right )} + 2 \, {\left (A + i \, B\right )} a\right )}}{{\left (i \, a \tan \left (d x + c\right ) + a\right )}^{\frac {3}{2}}}\right )}}{24 \, a d} \]

[In]

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

[Out]

1/24*I*(3*sqrt(2)*(A - I*B)*log(-(sqrt(2)*sqrt(a) - sqrt(I*a*tan(d*x + c) + a))/(sqrt(2)*sqrt(a) + sqrt(I*a*ta
n(d*x + c) + a)))/sqrt(a) + 4*(3*(I*a*tan(d*x + c) + a)*(A - I*B) + 2*(A + I*B)*a)/(I*a*tan(d*x + c) + a)^(3/2
))/(a*d)

Giac [F]

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

[In]

integrate((A+B*tan(d*x+c))/(a+I*a*tan(d*x+c))^(3/2),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 = 7.56 (sec) , antiderivative size = 162, normalized size of antiderivative = 1.34 \[ \int \frac {A+B \tan (c+d x)}{(a+i a \tan (c+d x))^{3/2}} \, dx=\frac {\frac {A\,1{}\mathrm {i}}{3\,d}+\frac {A\,\left (a+a\,\mathrm {tan}\left (c+d\,x\right )\,1{}\mathrm {i}\right )\,1{}\mathrm {i}}{2\,a\,d}}{{\left (a+a\,\mathrm {tan}\left (c+d\,x\right )\,1{}\mathrm {i}\right )}^{3/2}}-\frac {\frac {B}{3}-\frac {B\,\left (a+a\,\mathrm {tan}\left (c+d\,x\right )\,1{}\mathrm {i}\right )}{2\,a}}{d\,{\left (a+a\,\mathrm {tan}\left (c+d\,x\right )\,1{}\mathrm {i}\right )}^{3/2}}-\frac {\sqrt {2}\,A\,\mathrm {atan}\left (\frac {\sqrt {2}\,\sqrt {a+a\,\mathrm {tan}\left (c+d\,x\right )\,1{}\mathrm {i}}}{2\,\sqrt {-a}}\right )\,1{}\mathrm {i}}{4\,{\left (-a\right )}^{3/2}\,d}-\frac {\sqrt {2}\,B\,\mathrm {atanh}\left (\frac {\sqrt {2}\,\sqrt {a+a\,\mathrm {tan}\left (c+d\,x\right )\,1{}\mathrm {i}}}{2\,\sqrt {a}}\right )}{4\,a^{3/2}\,d} \]

[In]

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

[Out]

((A*1i)/(3*d) + (A*(a + a*tan(c + d*x)*1i)*1i)/(2*a*d))/(a + a*tan(c + d*x)*1i)^(3/2) - (B/3 - (B*(a + a*tan(c
+ d*x)*1i))/(2*a))/(d*(a + a*tan(c + d*x)*1i)^(3/2)) - (2^(1/2)*A*atan((2^(1/2)*(a + a*tan(c + d*x)*1i)^(1/2)
)/(2*(-a)^(1/2)))*1i)/(4*(-a)^(3/2)*d) - (2^(1/2)*B*atanh((2^(1/2)*(a + a*tan(c + d*x)*1i)^(1/2))/(2*a^(1/2)))
)/(4*a^(3/2)*d)