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

Optimal. Leaf size=199 \[ -\frac{2 a^2 \tan ^4(c+d x)}{7 d \sqrt{a+i a \tan (c+d x)}}+\frac{2 i a^2 \tan ^3(c+d x)}{7 d \sqrt{a+i a \tan (c+d x)}}+\frac{2 \sqrt{2} a^{3/2} \tanh ^{-1}\left (\frac{\sqrt{a+i a \tan (c+d x)}}{\sqrt{2} \sqrt{a}}\right )}{d}+\frac{16 a \tan ^2(c+d x) \sqrt{a+i a \tan (c+d x)}}{35 d}-\frac{76 (a+i a \tan (c+d x))^{3/2}}{105 d}-\frac{64 a \sqrt{a+i a \tan (c+d x)}}{35 d} \]

[Out]

(2*Sqrt[2]*a^(3/2)*ArcTanh[Sqrt[a + I*a*Tan[c + d*x]]/(Sqrt[2]*Sqrt[a])])/d + (((2*I)/7)*a^2*Tan[c + d*x]^3)/(
d*Sqrt[a + I*a*Tan[c + d*x]]) - (2*a^2*Tan[c + d*x]^4)/(7*d*Sqrt[a + I*a*Tan[c + d*x]]) - (64*a*Sqrt[a + I*a*T
an[c + d*x]])/(35*d) + (16*a*Tan[c + d*x]^2*Sqrt[a + I*a*Tan[c + d*x]])/(35*d) - (76*(a + I*a*Tan[c + d*x])^(3
/2))/(105*d)

________________________________________________________________________________________

Rubi [A]  time = 0.510279, antiderivative size = 199, normalized size of antiderivative = 1., number of steps used = 7, number of rules used = 7, integrand size = 26, \(\frac{\text{number of rules}}{\text{integrand size}}\) = 0.269, Rules used = {3556, 3595, 3597, 3592, 3527, 3480, 206} \[ -\frac{2 a^2 \tan ^4(c+d x)}{7 d \sqrt{a+i a \tan (c+d x)}}+\frac{2 i a^2 \tan ^3(c+d x)}{7 d \sqrt{a+i a \tan (c+d x)}}+\frac{2 \sqrt{2} a^{3/2} \tanh ^{-1}\left (\frac{\sqrt{a+i a \tan (c+d x)}}{\sqrt{2} \sqrt{a}}\right )}{d}+\frac{16 a \tan ^2(c+d x) \sqrt{a+i a \tan (c+d x)}}{35 d}-\frac{76 (a+i a \tan (c+d x))^{3/2}}{105 d}-\frac{64 a \sqrt{a+i a \tan (c+d x)}}{35 d} \]

Antiderivative was successfully verified.

[In]

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

[Out]

(2*Sqrt[2]*a^(3/2)*ArcTanh[Sqrt[a + I*a*Tan[c + d*x]]/(Sqrt[2]*Sqrt[a])])/d + (((2*I)/7)*a^2*Tan[c + d*x]^3)/(
d*Sqrt[a + I*a*Tan[c + d*x]]) - (2*a^2*Tan[c + d*x]^4)/(7*d*Sqrt[a + I*a*Tan[c + d*x]]) - (64*a*Sqrt[a + I*a*T
an[c + d*x]])/(35*d) + (16*a*Tan[c + d*x]^2*Sqrt[a + I*a*Tan[c + d*x]])/(35*d) - (76*(a + I*a*Tan[c + d*x])^(3
/2))/(105*d)

Rule 3556

Int[((a_) + (b_.)*tan[(e_.) + (f_.)*(x_)])^(m_)*((c_.) + (d_.)*tan[(e_.) + (f_.)*(x_)])^(n_), x_Symbol] :> Sim
p[(b^2*(a + b*Tan[e + f*x])^(m - 2)*(c + d*Tan[e + f*x])^(n + 1))/(d*f*(m + n - 1)), x] + Dist[a/(d*(m + n - 1
)), Int[(a + b*Tan[e + f*x])^(m - 2)*(c + d*Tan[e + f*x])^n*Simp[b*c*(m - 2) + a*d*(m + 2*n) + (a*c*(m - 2) +
b*d*(3*m + 2*n - 4))*Tan[e + f*x], x], x], x] /; FreeQ[{a, b, c, d, e, f, n}, x] && NeQ[b*c - a*d, 0] && EqQ[a
^2 + b^2, 0] && NeQ[c^2 + d^2, 0] && IntegerQ[2*m] && GtQ[m, 1] && NeQ[m + n - 1, 0] && (IntegerQ[m] || Intege
rsQ[2*m, 2*n])

Rule 3595

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

Rule 3597

Int[((a_) + (b_.)*tan[(e_.) + (f_.)*(x_)])^(m_)*((A_.) + (B_.)*tan[(e_.) + (f_.)*(x_)])*((c_.) + (d_.)*tan[(e_
.) + (f_.)*(x_)])^(n_), x_Symbol] :> Simp[(B*(a + b*Tan[e + f*x])^m*(c + d*Tan[e + f*x])^n)/(f*(m + n)), x] +
Dist[1/(a*(m + n)), Int[(a + b*Tan[e + f*x])^m*(c + d*Tan[e + f*x])^(n - 1)*Simp[a*A*c*(m + n) - B*(b*c*m + a*
d*n) + (a*A*d*(m + n) - B*(b*d*m - a*c*n))*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] && GtQ[n, 0]

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 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 ^3(c+d x) (a+i a \tan (c+d x))^{3/2} \, dx &=-\frac{2 a^2 \tan ^4(c+d x)}{7 d \sqrt{a+i a \tan (c+d x)}}+\frac{1}{7} (2 a) \int \frac{\tan ^3(c+d x) \left (\frac{15 a}{2}+\frac{13}{2} i a \tan (c+d x)\right )}{\sqrt{a+i a \tan (c+d x)}} \, dx\\ &=\frac{2 i a^2 \tan ^3(c+d x)}{7 d \sqrt{a+i a \tan (c+d x)}}-\frac{2 a^2 \tan ^4(c+d x)}{7 d \sqrt{a+i a \tan (c+d x)}}-\frac{2 \int \tan ^2(c+d x) \sqrt{a+i a \tan (c+d x)} \left (3 i a^2-4 a^2 \tan (c+d x)\right ) \, dx}{7 a}\\ &=\frac{2 i a^2 \tan ^3(c+d x)}{7 d \sqrt{a+i a \tan (c+d x)}}-\frac{2 a^2 \tan ^4(c+d x)}{7 d \sqrt{a+i a \tan (c+d x)}}+\frac{16 a \tan ^2(c+d x) \sqrt{a+i a \tan (c+d x)}}{35 d}-\frac{4 \int \tan (c+d x) \sqrt{a+i a \tan (c+d x)} \left (8 a^3+\frac{19}{2} i a^3 \tan (c+d x)\right ) \, dx}{35 a^2}\\ &=\frac{2 i a^2 \tan ^3(c+d x)}{7 d \sqrt{a+i a \tan (c+d x)}}-\frac{2 a^2 \tan ^4(c+d x)}{7 d \sqrt{a+i a \tan (c+d x)}}+\frac{16 a \tan ^2(c+d x) \sqrt{a+i a \tan (c+d x)}}{35 d}-\frac{76 (a+i a \tan (c+d x))^{3/2}}{105 d}-\frac{4 \int \sqrt{a+i a \tan (c+d x)} \left (-\frac{19 i a^3}{2}+8 a^3 \tan (c+d x)\right ) \, dx}{35 a^2}\\ &=\frac{2 i a^2 \tan ^3(c+d x)}{7 d \sqrt{a+i a \tan (c+d x)}}-\frac{2 a^2 \tan ^4(c+d x)}{7 d \sqrt{a+i a \tan (c+d x)}}-\frac{64 a \sqrt{a+i a \tan (c+d x)}}{35 d}+\frac{16 a \tan ^2(c+d x) \sqrt{a+i a \tan (c+d x)}}{35 d}-\frac{76 (a+i a \tan (c+d x))^{3/2}}{105 d}+(2 i a) \int \sqrt{a+i a \tan (c+d x)} \, dx\\ &=\frac{2 i a^2 \tan ^3(c+d x)}{7 d \sqrt{a+i a \tan (c+d x)}}-\frac{2 a^2 \tan ^4(c+d x)}{7 d \sqrt{a+i a \tan (c+d x)}}-\frac{64 a \sqrt{a+i a \tan (c+d x)}}{35 d}+\frac{16 a \tan ^2(c+d x) \sqrt{a+i a \tan (c+d x)}}{35 d}-\frac{76 (a+i a \tan (c+d x))^{3/2}}{105 d}+\frac{\left (4 a^2\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} \tanh ^{-1}\left (\frac{\sqrt{a+i a \tan (c+d x)}}{\sqrt{2} \sqrt{a}}\right )}{d}+\frac{2 i a^2 \tan ^3(c+d x)}{7 d \sqrt{a+i a \tan (c+d x)}}-\frac{2 a^2 \tan ^4(c+d x)}{7 d \sqrt{a+i a \tan (c+d x)}}-\frac{64 a \sqrt{a+i a \tan (c+d x)}}{35 d}+\frac{16 a \tan ^2(c+d x) \sqrt{a+i a \tan (c+d x)}}{35 d}-\frac{76 (a+i a \tan (c+d x))^{3/2}}{105 d}\\ \end{align*}

Mathematica [A]  time = 2.57582, size = 166, normalized size = 0.83 \[ \frac{(a+i a \tan (c+d x))^{3/2} \left (\frac{2 \sqrt{2} \sinh ^{-1}\left (e^{i (c+d x)}\right )}{\left (\frac{e^{i (c+d x)}}{1+e^{2 i (c+d x)}}\right )^{3/2} \left (1+e^{2 i (c+d x)}\right )^{3/2}}+\frac{1}{210} (-1+i \tan (c+d x)) \sec ^{\frac{5}{2}}(c+d x) (-7 i \sin (c+d x)+53 i \sin (3 (c+d x))+378 \cos (c+d x)+158 \cos (3 (c+d x)))\right )}{d \sec ^{\frac{3}{2}}(c+d x)} \]

Antiderivative was successfully verified.

[In]

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

[Out]

(((2*Sqrt[2]*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]^(5/2)*(378*Cos[c + d*x] + 158*Cos[3*(c + d*x)] - (7*I)*Sin[c + d*x] + (53*I)*Sin[
3*(c + d*x)])*(-1 + I*Tan[c + d*x]))/210)*(a + I*a*Tan[c + d*x])^(3/2))/(d*Sec[c + d*x]^(3/2))

________________________________________________________________________________________

Maple [A]  time = 0.019, size = 111, normalized size = 0.6 \begin{align*} -2\,{\frac{1}{{a}^{2}d} \left ( 1/7\, \left ( a+ia\tan \left ( dx+c \right ) \right ) ^{7/2}-1/5\,a \left ( a+ia\tan \left ( dx+c \right ) \right ) ^{5/2}+1/3\, \left ( a+ia\tan \left ( dx+c \right ) \right ) ^{3/2}{a}^{2}+{a}^{3}\sqrt{a+ia\tan \left ( dx+c \right ) }-{a}^{7/2}\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)^3*(a+I*a*tan(d*x+c))^(3/2),x)

[Out]

-2/d/a^2*(1/7*(a+I*a*tan(d*x+c))^(7/2)-1/5*a*(a+I*a*tan(d*x+c))^(5/2)+1/3*(a+I*a*tan(d*x+c))^(3/2)*a^2+a^3*(a+
I*a*tan(d*x+c))^(1/2)-a^(7/2)*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)^3*(a+I*a*tan(d*x+c))^(3/2),x, algorithm="maxima")

[Out]

Exception raised: ValueError

________________________________________________________________________________________

Fricas [B]  time = 2.25688, size = 1057, normalized size = 5.31 \begin{align*} -\frac{2 \, \sqrt{2}{\left (211 \, a e^{\left (6 i \, d x + 6 i \, c\right )} + 371 \, a e^{\left (4 i \, d x + 4 i \, c\right )} + 385 \, a e^{\left (2 i \, d x + 2 i \, c\right )} + 105 \, a\right )} \sqrt{\frac{a}{e^{\left (2 i \, d x + 2 i \, c\right )} + 1}} e^{\left (i \, d x + i \, c\right )} - 105 \, \sqrt{2}{\left (d e^{\left (6 i \, d x + 6 i \, c\right )} + 3 \, d e^{\left (4 i \, d x + 4 i \, c\right )} + 3 \, d e^{\left (2 i \, d x + 2 i \, c\right )} + d\right )} \sqrt{\frac{a^{3}}{d^{2}}} \log \left (\frac{{\left (\sqrt{2} \sqrt{\frac{a^{3}}{d^{2}}} d e^{\left (2 i \, d x + 2 i \, c\right )} + \sqrt{2}{\left (a e^{\left (2 i \, d x + 2 i \, c\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 )}\right )} e^{\left (-2 i \, d x - 2 i \, c\right )}}{a}\right ) + 105 \, \sqrt{2}{\left (d e^{\left (6 i \, d x + 6 i \, c\right )} + 3 \, d e^{\left (4 i \, d x + 4 i \, c\right )} + 3 \, d e^{\left (2 i \, d x + 2 i \, c\right )} + d\right )} \sqrt{\frac{a^{3}}{d^{2}}} \log \left (-\frac{{\left (\sqrt{2} \sqrt{\frac{a^{3}}{d^{2}}} d e^{\left (2 i \, d x + 2 i \, c\right )} - \sqrt{2}{\left (a e^{\left (2 i \, d x + 2 i \, c\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 )}\right )} e^{\left (-2 i \, d x - 2 i \, c\right )}}{a}\right )}{105 \,{\left (d e^{\left (6 i \, d x + 6 i \, c\right )} + 3 \, d e^{\left (4 i \, d x + 4 i \, c\right )} + 3 \, 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)^3*(a+I*a*tan(d*x+c))^(3/2),x, algorithm="fricas")

[Out]

-1/105*(2*sqrt(2)*(211*a*e^(6*I*d*x + 6*I*c) + 371*a*e^(4*I*d*x + 4*I*c) + 385*a*e^(2*I*d*x + 2*I*c) + 105*a)*
sqrt(a/(e^(2*I*d*x + 2*I*c) + 1))*e^(I*d*x + I*c) - 105*sqrt(2)*(d*e^(6*I*d*x + 6*I*c) + 3*d*e^(4*I*d*x + 4*I*
c) + 3*d*e^(2*I*d*x + 2*I*c) + d)*sqrt(a^3/d^2)*log((sqrt(2)*sqrt(a^3/d^2)*d*e^(2*I*d*x + 2*I*c) + sqrt(2)*(a*
e^(2*I*d*x + 2*I*c) + a)*sqrt(a/(e^(2*I*d*x + 2*I*c) + 1))*e^(I*d*x + I*c))*e^(-2*I*d*x - 2*I*c)/a) + 105*sqrt
(2)*(d*e^(6*I*d*x + 6*I*c) + 3*d*e^(4*I*d*x + 4*I*c) + 3*d*e^(2*I*d*x + 2*I*c) + d)*sqrt(a^3/d^2)*log(-(sqrt(2
)*sqrt(a^3/d^2)*d*e^(2*I*d*x + 2*I*c) - sqrt(2)*(a*e^(2*I*d*x + 2*I*c) + a)*sqrt(a/(e^(2*I*d*x + 2*I*c) + 1))*
e^(I*d*x + I*c))*e^(-2*I*d*x - 2*I*c)/a))/(d*e^(6*I*d*x + 6*I*c) + 3*d*e^(4*I*d*x + 4*I*c) + 3*d*e^(2*I*d*x +
2*I*c) + d)

________________________________________________________________________________________

Sympy [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)**3*(a+I*a*tan(d*x+c))**(3/2),x)

[Out]

Timed out

________________________________________________________________________________________

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

[Out]

Timed out