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

Optimal. Leaf size=119 \[ -\frac{(A-i B) \tanh ^{-1}\left (\frac{\sqrt{a+i a \tan (c+d x)}}{\sqrt{2} \sqrt{a}}\right )}{2 \sqrt{2} a^{3/2} d}-\frac{A+i B}{3 d (a+i a \tan (c+d x))^{3/2}}+\frac{A+3 i B}{2 a d \sqrt{a+i a \tan (c+d x)}} \]

[Out]

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

________________________________________________________________________________________

Rubi [A]  time = 0.185539, antiderivative size = 119, normalized size of antiderivative = 1., number of steps used = 4, number of rules used = 4, integrand size = 34, \(\frac{\text{number of rules}}{\text{integrand size}}\) = 0.118, Rules used = {3590, 3526, 3480, 206} \[ -\frac{(A-i B) \tanh ^{-1}\left (\frac{\sqrt{a+i a \tan (c+d x)}}{\sqrt{2} \sqrt{a}}\right )}{2 \sqrt{2} a^{3/2} d}-\frac{A+i B}{3 d (a+i a \tan (c+d x))^{3/2}}+\frac{A+3 i B}{2 a d \sqrt{a+i a \tan (c+d x)}} \]

Antiderivative was successfully verified.

[In]

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

[Out]

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

Rule 3590

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

Rule 3526

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]

Rule 3480

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

Mathematica [A]  time = 2.34392, size = 145, normalized size = 1.22 \[ \frac{\sqrt{1+e^{2 i (c+d x)}} \left (B \left (-1+8 e^{2 i (c+d x)}\right )-i A \left (-1+2 e^{2 i (c+d x)}\right )\right )+3 (B+i A) e^{3 i (c+d x)} \sinh ^{-1}\left (e^{i (c+d x)}\right )}{3 a d \left (1+e^{2 i (c+d x)}\right )^{3/2} (\tan (c+d x)-i) \sqrt{a+i a \tan (c+d x)}} \]

Antiderivative was successfully verified.

[In]

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

[Out]

(Sqrt[1 + E^((2*I)*(c + d*x))]*((-I)*A*(-1 + 2*E^((2*I)*(c + d*x))) + B*(-1 + 8*E^((2*I)*(c + d*x)))) + 3*(I*A
+ B)*E^((3*I)*(c + d*x))*ArcSinh[E^(I*(c + d*x))])/(3*a*d*(1 + E^((2*I)*(c + d*x)))^(3/2)*(-I + Tan[c + d*x])
*Sqrt[a + I*a*Tan[c + d*x]])

________________________________________________________________________________________

Maple [A]  time = 0.024, size = 96, normalized size = 0.8 \begin{align*} 2\,{\frac{1}{ad} \left ( -{\frac{-A/4-3/4\,iB}{\sqrt{a+ia\tan \left ( dx+c \right ) }}}-1/6\,{\frac{a \left ( A+iB \right ) }{ \left ( a+ia\tan \left ( dx+c \right ) \right ) ^{3/2}}}-1/2\,{\frac{ \left ( A/4-i/4B \right ) \sqrt{2}}{\sqrt{a}}{\it Artanh} \left ( 1/2\,{\frac{\sqrt{a+ia\tan \left ( dx+c \right ) }\sqrt{2}}{\sqrt{a}}} \right ) } \right ) } \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

2/d/a*(-(-1/4*A-3/4*I*B)/(a+I*a*tan(d*x+c))^(1/2)-1/6*a*(A+I*B)/(a+I*a*tan(d*x+c))^(3/2)-1/2*(1/4*A-1/4*I*B)*2
^(1/2)/a^(1/2)*arctanh(1/2*(a+I*a*tan(d*x+c))^(1/2)*2^(1/2)/a^(1/2)))

________________________________________________________________________________________

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

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

Exception raised: ValueError

________________________________________________________________________________________

Fricas [B]  time = 2.02055, size = 1027, normalized size = 8.63 \begin{align*} -\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 (4 i \, d x + 4 i \, c\right )} \log \left (\frac{{\left (2 i \, \sqrt{\frac{1}{2}} a^{2} d \sqrt{\frac{A^{2} - 2 i \, A B - B^{2}}{a^{3} d^{2}}} e^{\left (2 i \, d x + 2 i \, c\right )} + \sqrt{2}{\left ({\left (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}} 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 (4 i \, d x + 4 i \, c\right )} \log \left (\frac{{\left (-2 i \, \sqrt{\frac{1}{2}} a^{2} d \sqrt{\frac{A^{2} - 2 i \, A B - B^{2}}{a^{3} d^{2}}} e^{\left (2 i \, d x + 2 i \, c\right )} + \sqrt{2}{\left ({\left (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}} 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 (A + 4 i \, B\right )} e^{\left (4 i \, d x + 4 i \, c\right )} +{\left (A + 7 i \, B\right )} e^{\left (2 i \, d x + 2 i \, c\right )} - A - i \, B\right )} \sqrt{\frac{a}{e^{\left (2 i \, d x + 2 i \, c\right )} + 1}} e^{\left (i \, d x + i \, c\right )}\right )} e^{\left (-4 i \, d x - 4 i \, c\right )}}{12 \, a^{2} d} \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate(tan(d*x+c)*(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^(4*I*d*x + 4*I*c)*log((2*I*sqrt(1/2)*a^2*d*sq
rt((A^2 - 2*I*A*B - B^2)/(a^3*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)*sq
rt(a/(e^(2*I*d*x + 2*I*c) + 1))*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^(4*I*d*x + 4*I*c)*log((-2*I*sqrt(1/2)*a^2*d*sqrt((A^2 - 2*I*A*B - B^2)/(a^3*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))*e^(I*
d*x + I*c))*e^(-I*d*x - I*c)/(I*A + B)) - sqrt(2)*(2*(A + 4*I*B)*e^(4*I*d*x + 4*I*c) + (A + 7*I*B)*e^(2*I*d*x
+ 2*I*c) - A - I*B)*sqrt(a/(e^(2*I*d*x + 2*I*c) + 1))*e^(I*d*x + I*c))*e^(-4*I*d*x - 4*I*c)/(a^2*d)

________________________________________________________________________________________

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

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

Exception raised: AttributeError

________________________________________________________________________________________

Giac [F]  time = 0., size = 0, normalized size = 0. \begin{align*} \int \frac{{\left (B \tan \left (d x + c\right ) + A\right )} \tan \left (d x + c\right )}{{\left (i \, a \tan \left (d x + c\right ) + a\right )}^{\frac{3}{2}}}\,{d x} \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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