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

Optimal. Leaf size=133 \[ -\frac{7 \sqrt{a+i a \tan (c+d x)}}{3 a^2 d}+\frac{\tanh ^{-1}\left (\frac{\sqrt{a+i a \tan (c+d x)}}{\sqrt{2} \sqrt{a}}\right )}{2 \sqrt{2} a^{3/2} d}-\frac{\tan ^2(c+d x)}{3 d (a+i a \tan (c+d x))^{3/2}}-\frac{11}{6 a d \sqrt{a+i a \tan (c+d x)}} \]

[Out]

ArcTanh[Sqrt[a + I*a*Tan[c + d*x]]/(Sqrt[2]*Sqrt[a])]/(2*Sqrt[2]*a^(3/2)*d) - Tan[c + d*x]^2/(3*d*(a + I*a*Tan
[c + d*x])^(3/2)) - 11/(6*a*d*Sqrt[a + I*a*Tan[c + d*x]]) - (7*Sqrt[a + I*a*Tan[c + d*x]])/(3*a^2*d)

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Rubi [A]  time = 0.214151, antiderivative size = 133, normalized size of antiderivative = 1., number of steps used = 5, number of rules used = 5, integrand size = 26, \(\frac{\text{number of rules}}{\text{integrand size}}\) = 0.192, Rules used = {3558, 3592, 3526, 3480, 206} \[ -\frac{7 \sqrt{a+i a \tan (c+d x)}}{3 a^2 d}+\frac{\tanh ^{-1}\left (\frac{\sqrt{a+i a \tan (c+d x)}}{\sqrt{2} \sqrt{a}}\right )}{2 \sqrt{2} a^{3/2} d}-\frac{\tan ^2(c+d x)}{3 d (a+i a \tan (c+d x))^{3/2}}-\frac{11}{6 a d \sqrt{a+i a \tan (c+d x)}} \]

Antiderivative was successfully verified.

[In]

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

[Out]

ArcTanh[Sqrt[a + I*a*Tan[c + d*x]]/(Sqrt[2]*Sqrt[a])]/(2*Sqrt[2]*a^(3/2)*d) - Tan[c + d*x]^2/(3*d*(a + I*a*Tan
[c + d*x])^(3/2)) - 11/(6*a*d*Sqrt[a + I*a*Tan[c + d*x]]) - (7*Sqrt[a + I*a*Tan[c + d*x]])/(3*a^2*d)

Rule 3558

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

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 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 ^3(c+d x)}{(a+i a \tan (c+d x))^{3/2}} \, dx &=-\frac{\tan ^2(c+d x)}{3 d (a+i a \tan (c+d x))^{3/2}}-\frac{\int \frac{\tan (c+d x) \left (-2 a+\frac{7}{2} i a \tan (c+d x)\right )}{\sqrt{a+i a \tan (c+d x)}} \, dx}{3 a^2}\\ &=-\frac{\tan ^2(c+d x)}{3 d (a+i a \tan (c+d x))^{3/2}}-\frac{7 \sqrt{a+i a \tan (c+d x)}}{3 a^2 d}-\frac{\int \frac{-\frac{7 i a}{2}-2 a \tan (c+d x)}{\sqrt{a+i a \tan (c+d x)}} \, dx}{3 a^2}\\ &=-\frac{\tan ^2(c+d x)}{3 d (a+i a \tan (c+d x))^{3/2}}-\frac{11}{6 a d \sqrt{a+i a \tan (c+d x)}}-\frac{7 \sqrt{a+i a \tan (c+d x)}}{3 a^2 d}+\frac{i \int \sqrt{a+i a \tan (c+d x)} \, dx}{4 a^2}\\ &=-\frac{\tan ^2(c+d x)}{3 d (a+i a \tan (c+d x))^{3/2}}-\frac{11}{6 a d \sqrt{a+i a \tan (c+d x)}}-\frac{7 \sqrt{a+i a \tan (c+d x)}}{3 a^2 d}+\frac{\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{\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{\tan ^2(c+d x)}{3 d (a+i a \tan (c+d x))^{3/2}}-\frac{11}{6 a d \sqrt{a+i a \tan (c+d x)}}-\frac{7 \sqrt{a+i a \tan (c+d x)}}{3 a^2 d}\\ \end{align*}

Mathematica [A]  time = 1.11942, size = 123, normalized size = 0.92 \[ \frac{e^{-2 i (c+d x)} \left (-13 e^{2 i (c+d x)}-38 e^{4 i (c+d x)}+3 e^{3 i (c+d x)} \sqrt{1+e^{2 i (c+d x)}} \sinh ^{-1}\left (e^{i (c+d x)}\right )+1\right )}{6 a d \left (1+e^{2 i (c+d x)}\right ) \sqrt{a+i a \tan (c+d x)}} \]

Antiderivative was successfully verified.

[In]

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

[Out]

(1 - 13*E^((2*I)*(c + d*x)) - 38*E^((4*I)*(c + d*x)) + 3*E^((3*I)*(c + d*x))*Sqrt[1 + E^((2*I)*(c + d*x))]*Arc
Sinh[E^(I*(c + d*x))])/(6*a*d*E^((2*I)*(c + d*x))*(1 + E^((2*I)*(c + d*x)))*Sqrt[a + I*a*Tan[c + d*x]])

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Maple [A]  time = 0.022, size = 91, normalized size = 0.7 \begin{align*} -2\,{\frac{1}{{a}^{2}d} \left ( \sqrt{a+ia\tan \left ( dx+c \right ) }-1/8\,\sqrt{a}\sqrt{2}{\it Artanh} \left ( 1/2\,{\frac{\sqrt{a+ia\tan \left ( dx+c \right ) }\sqrt{2}}{\sqrt{a}}} \right ) -1/6\,{\frac{{a}^{2}}{ \left ( a+ia\tan \left ( dx+c \right ) \right ) ^{3/2}}}+5/4\,{\frac{a}{\sqrt{a+ia\tan \left ( dx+c \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*((a+I*a*tan(d*x+c))^(1/2)-1/8*a^(1/2)*2^(1/2)*arctanh(1/2*(a+I*a*tan(d*x+c))^(1/2)*2^(1/2)/a^(1/2))-1
/6/(a+I*a*tan(d*x+c))^(3/2)*a^2+5/4/(a+I*a*tan(d*x+c))^(1/2)*a)

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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

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Fricas [B]  time = 2.24119, size = 807, normalized size = 6.07 \begin{align*} \frac{{\left (3 \, \sqrt{\frac{1}{2}} a^{2} d \sqrt{\frac{1}{a^{3} d^{2}}} e^{\left (4 i \, d x + 4 i \, c\right )} \log \left ({\left (2 \, \sqrt{\frac{1}{2}} a^{2} d \sqrt{\frac{1}{a^{3} d^{2}}} e^{\left (2 i \, d x + 2 i \, c\right )} + \sqrt{2} \sqrt{\frac{a}{e^{\left (2 i \, d x + 2 i \, c\right )} + 1}}{\left (e^{\left (2 i \, d x + 2 i \, c\right )} + 1\right )} e^{\left (i \, d x + i \, c\right )}\right )} e^{\left (-i \, d x - i \, c\right )}\right ) - 3 \, \sqrt{\frac{1}{2}} a^{2} d \sqrt{\frac{1}{a^{3} d^{2}}} e^{\left (4 i \, d x + 4 i \, c\right )} \log \left (-{\left (2 \, \sqrt{\frac{1}{2}} a^{2} d \sqrt{\frac{1}{a^{3} d^{2}}} e^{\left (2 i \, d x + 2 i \, c\right )} - \sqrt{2} \sqrt{\frac{a}{e^{\left (2 i \, d x + 2 i \, c\right )} + 1}}{\left (e^{\left (2 i \, d x + 2 i \, c\right )} + 1\right )} e^{\left (i \, d x + i \, c\right )}\right )} e^{\left (-i \, d x - i \, c\right )}\right ) - \sqrt{2} \sqrt{\frac{a}{e^{\left (2 i \, d x + 2 i \, c\right )} + 1}}{\left (38 \, e^{\left (4 i \, d x + 4 i \, c\right )} + 13 \, e^{\left (2 i \, d x + 2 i \, c\right )} - 1\right )} 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)^3/(a+I*a*tan(d*x+c))^(3/2),x, algorithm="fricas")

[Out]

1/12*(3*sqrt(1/2)*a^2*d*sqrt(1/(a^3*d^2))*e^(4*I*d*x + 4*I*c)*log((2*sqrt(1/2)*a^2*d*sqrt(1/(a^3*d^2))*e^(2*I*
d*x + 2*I*c) + sqrt(2)*sqrt(a/(e^(2*I*d*x + 2*I*c) + 1))*(e^(2*I*d*x + 2*I*c) + 1)*e^(I*d*x + I*c))*e^(-I*d*x
- I*c)) - 3*sqrt(1/2)*a^2*d*sqrt(1/(a^3*d^2))*e^(4*I*d*x + 4*I*c)*log(-(2*sqrt(1/2)*a^2*d*sqrt(1/(a^3*d^2))*e^
(2*I*d*x + 2*I*c) - sqrt(2)*sqrt(a/(e^(2*I*d*x + 2*I*c) + 1))*(e^(2*I*d*x + 2*I*c) + 1)*e^(I*d*x + I*c))*e^(-I
*d*x - I*c)) - sqrt(2)*sqrt(a/(e^(2*I*d*x + 2*I*c) + 1))*(38*e^(4*I*d*x + 4*I*c) + 13*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)

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

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

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Giac [F]  time = 0., size = 0, normalized size = 0. \begin{align*} \int \frac{\tan \left (d x + c\right )^{3}}{{\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)^3/(a+I*a*tan(d*x+c))^(3/2),x, algorithm="giac")

[Out]

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