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

Optimal. Leaf size=137 \[ -\frac{2 \sqrt{2} a^{3/2} (A-i B) \tanh ^{-1}\left (\frac{\sqrt{a+i a \tan (c+d x)}}{\sqrt{2} \sqrt{a}}\right )}{d}+\frac{2 a (A-i B) \sqrt{a+i a \tan (c+d x)}}{d}+\frac{2 A (a+i a \tan (c+d x))^{3/2}}{3 d}-\frac{2 i B (a+i a \tan (c+d x))^{5/2}}{5 a d} \]

[Out]

(-2*Sqrt[2]*a^(3/2)*(A - I*B)*ArcTanh[Sqrt[a + I*a*Tan[c + d*x]]/(Sqrt[2]*Sqrt[a])])/d + (2*a*(A - I*B)*Sqrt[a
+ I*a*Tan[c + d*x]])/d + (2*A*(a + I*a*Tan[c + d*x])^(3/2))/(3*d) - (((2*I)/5)*B*(a + I*a*Tan[c + d*x])^(5/2)
)/(a*d)

________________________________________________________________________________________

Rubi [A]  time = 0.176307, antiderivative size = 137, normalized size of antiderivative = 1., number of steps used = 5, number of rules used = 5, integrand size = 34, \(\frac{\text{number of rules}}{\text{integrand size}}\) = 0.147, Rules used = {3592, 3527, 3478, 3480, 206} \[ -\frac{2 \sqrt{2} a^{3/2} (A-i B) \tanh ^{-1}\left (\frac{\sqrt{a+i a \tan (c+d x)}}{\sqrt{2} \sqrt{a}}\right )}{d}+\frac{2 a (A-i B) \sqrt{a+i a \tan (c+d x)}}{d}+\frac{2 A (a+i a \tan (c+d x))^{3/2}}{3 d}-\frac{2 i B (a+i a \tan (c+d x))^{5/2}}{5 a d} \]

Antiderivative was successfully verified.

[In]

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

[Out]

(-2*Sqrt[2]*a^(3/2)*(A - I*B)*ArcTanh[Sqrt[a + I*a*Tan[c + d*x]]/(Sqrt[2]*Sqrt[a])])/d + (2*a*(A - I*B)*Sqrt[a
+ I*a*Tan[c + d*x]])/d + (2*A*(a + I*a*Tan[c + d*x])^(3/2))/(3*d) - (((2*I)/5)*B*(a + I*a*Tan[c + d*x])^(5/2)
)/(a*d)

Rule 3592

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

Rule 3527

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]

Rule 3478

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

Mathematica [A]  time = 3.64747, size = 204, normalized size = 1.49 \[ \frac{(a+i a \tan (c+d x))^{3/2} (A+B \tan (c+d x)) \left (\frac{1}{15} (\tan (c+d x)+i) \sec ^{\frac{3}{2}}(c+d x) ((5 A-6 i B) \sin (2 (c+d x))+(-21 B-20 i A) \cos (2 (c+d x))-20 i A-15 B)-\frac{2 \sqrt{2} (A-i B) \sinh ^{-1}\left (e^{i (c+d x)}\right )}{\left (\frac{e^{i (c+d x)}}{1+e^{2 i (c+d x)}}\right )^{3/2} \left (1+e^{2 i (c+d x)}\right )^{3/2}}\right )}{d \sec ^{\frac{5}{2}}(c+d x) (A \cos (c+d x)+B \sin (c+d x))} \]

Antiderivative was successfully verified.

[In]

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

[Out]

((a + I*a*Tan[c + d*x])^(3/2)*(A + B*Tan[c + d*x])*((-2*Sqrt[2]*(A - I*B)*ArcSinh[E^(I*(c + d*x))])/((E^(I*(c
+ d*x))/(1 + E^((2*I)*(c + d*x))))^(3/2)*(1 + E^((2*I)*(c + d*x)))^(3/2)) + (Sec[c + d*x]^(3/2)*((-20*I)*A - 1
5*B + ((-20*I)*A - 21*B)*Cos[2*(c + d*x)] + (5*A - (6*I)*B)*Sin[2*(c + d*x)])*(I + Tan[c + d*x]))/15))/(d*Sec[
c + d*x]^(5/2)*(A*Cos[c + d*x] + B*Sin[c + d*x]))

________________________________________________________________________________________

Maple [A]  time = 0.018, size = 123, normalized size = 0.9 \begin{align*} 2\,{\frac{1}{ad} \left ( -i/5B \left ( a+ia\tan \left ( dx+c \right ) \right ) ^{5/2}+1/3\,A \left ( a+ia\tan \left ( dx+c \right ) \right ) ^{3/2}a-i{a}^{2}B\sqrt{a+ia\tan \left ( dx+c \right ) }+{a}^{2}A\sqrt{a+ia\tan \left ( dx+c \right ) }-{a}^{5/2} \left ( A-iB \right ) \sqrt{2}{\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+I*a*tan(d*x+c))^(3/2)*(A+B*tan(d*x+c)),x)

[Out]

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

________________________________________________________________________________________

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

[Out]

Exception raised: ValueError

________________________________________________________________________________________

Fricas [B]  time = 1.75105, size = 1180, normalized size = 8.61 \begin{align*} \frac{4 \, \sqrt{2}{\left ({\left (25 \, A - 27 i \, B\right )} a e^{\left (4 i \, d x + 4 i \, c\right )} + 10 \,{\left (4 \, A - 3 i \, B\right )} a e^{\left (2 i \, d x + 2 i \, c\right )} + 15 \,{\left (A - i \, B\right )} a\right )} \sqrt{\frac{a}{e^{\left (2 i \, d x + 2 i \, c\right )} + 1}} e^{\left (i \, d x + i \, c\right )} - 15 \, \sqrt{\frac{{\left (8 \, A^{2} - 16 i \, A B - 8 \, 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 (\sqrt{2}{\left ({\left (2 i \, A + 2 \, B\right )} a e^{\left (2 i \, d x + 2 i \, c\right )} +{\left (2 i \, A + 2 \, B\right )} a\right )} \sqrt{\frac{a}{e^{\left (2 i \, d x + 2 i \, c\right )} + 1}} e^{\left (i \, d x + i \, c\right )} + i \, \sqrt{\frac{{\left (8 \, A^{2} - 16 i \, A B - 8 \, B^{2}\right )} a^{3}}{d^{2}}} d e^{\left (2 i \, d x + 2 i \, c\right )}\right )} e^{\left (-2 i \, d x - 2 i \, c\right )}}{{\left (2 i \, A + 2 \, B\right )} a}\right ) + 15 \, \sqrt{\frac{{\left (8 \, A^{2} - 16 i \, A B - 8 \, 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 (\sqrt{2}{\left ({\left (2 i \, A + 2 \, B\right )} a e^{\left (2 i \, d x + 2 i \, c\right )} +{\left (2 i \, A + 2 \, B\right )} a\right )} \sqrt{\frac{a}{e^{\left (2 i \, d x + 2 i \, c\right )} + 1}} e^{\left (i \, d x + i \, c\right )} - i \, \sqrt{\frac{{\left (8 \, A^{2} - 16 i \, A B - 8 \, B^{2}\right )} a^{3}}{d^{2}}} d e^{\left (2 i \, d x + 2 i \, c\right )}\right )} e^{\left (-2 i \, d x - 2 i \, c\right )}}{{\left (2 i \, A + 2 \, B\right )} a}\right )}{30 \,{\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 )}} \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

1/30*(4*sqrt(2)*((25*A - 27*I*B)*a*e^(4*I*d*x + 4*I*c) + 10*(4*A - 3*I*B)*a*e^(2*I*d*x + 2*I*c) + 15*(A - I*B)
*a)*sqrt(a/(e^(2*I*d*x + 2*I*c) + 1))*e^(I*d*x + I*c) - 15*sqrt((8*A^2 - 16*I*A*B - 8*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((sqrt(2)*((2*I*A + 2*B)*a*e^(2*I*d*x + 2*I*c) + (2*I*A + 2*B)*
a)*sqrt(a/(e^(2*I*d*x + 2*I*c) + 1))*e^(I*d*x + I*c) + I*sqrt((8*A^2 - 16*I*A*B - 8*B^2)*a^3/d^2)*d*e^(2*I*d*x
+ 2*I*c))*e^(-2*I*d*x - 2*I*c)/((2*I*A + 2*B)*a)) + 15*sqrt((8*A^2 - 16*I*A*B - 8*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((sqrt(2)*((2*I*A + 2*B)*a*e^(2*I*d*x + 2*I*c) + (2*I*A + 2*B)*a)*
sqrt(a/(e^(2*I*d*x + 2*I*c) + 1))*e^(I*d*x + I*c) - I*sqrt((8*A^2 - 16*I*A*B - 8*B^2)*a^3/d^2)*d*e^(2*I*d*x +
2*I*c))*e^(-2*I*d*x - 2*I*c)/((2*I*A + 2*B)*a)))/(d*e^(4*I*d*x + 4*I*c) + 2*d*e^(2*I*d*x + 2*I*c) + d)

________________________________________________________________________________________

Sympy [F]  time = 0., size = 0, normalized size = 0. \begin{align*} \int \left (a \left (i \tan{\left (c + d x \right )} + 1\right )\right )^{\frac{3}{2}} \left (A + B \tan{\left (c + d x \right )}\right ) \tan{\left (c + d x \right )}\, dx \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

Integral((a*(I*tan(c + d*x) + 1))**(3/2)*(A + B*tan(c + d*x))*tan(c + d*x), x)

________________________________________________________________________________________

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

[Out]

Timed out