3.110 \(\int \frac{\tan ^3(c+d x)}{\sqrt{a+i a \tan (c+d x)}} \, dx\)

Optimal. Leaf size=126 \[ -\frac{5 (a+i a \tan (c+d x))^{3/2}}{3 a^2 d}-\frac{\tan ^2(c+d x)}{d \sqrt{a+i a \tan (c+d x)}}+\frac{4 \sqrt{a+i a \tan (c+d x)}}{a d}+\frac{\tanh ^{-1}\left (\frac{\sqrt{a+i a \tan (c+d x)}}{\sqrt{2} \sqrt{a}}\right )}{\sqrt{2} \sqrt{a} d} \]

[Out]

ArcTanh[Sqrt[a + I*a*Tan[c + d*x]]/(Sqrt[2]*Sqrt[a])]/(Sqrt[2]*Sqrt[a]*d) - Tan[c + d*x]^2/(d*Sqrt[a + I*a*Tan
[c + d*x]]) + (4*Sqrt[a + I*a*Tan[c + d*x]])/(a*d) - (5*(a + I*a*Tan[c + d*x])^(3/2))/(3*a^2*d)

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Rubi [A]  time = 0.201004, antiderivative size = 126, 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, 3527, 3480, 206} \[ -\frac{5 (a+i a \tan (c+d x))^{3/2}}{3 a^2 d}-\frac{\tan ^2(c+d x)}{d \sqrt{a+i a \tan (c+d x)}}+\frac{4 \sqrt{a+i a \tan (c+d x)}}{a d}+\frac{\tanh ^{-1}\left (\frac{\sqrt{a+i a \tan (c+d x)}}{\sqrt{2} \sqrt{a}}\right )}{\sqrt{2} \sqrt{a} d} \]

Antiderivative was successfully verified.

[In]

Int[Tan[c + d*x]^3/Sqrt[a + I*a*Tan[c + d*x]],x]

[Out]

ArcTanh[Sqrt[a + I*a*Tan[c + d*x]]/(Sqrt[2]*Sqrt[a])]/(Sqrt[2]*Sqrt[a]*d) - Tan[c + d*x]^2/(d*Sqrt[a + I*a*Tan
[c + d*x]]) + (4*Sqrt[a + I*a*Tan[c + d*x]])/(a*d) - (5*(a + I*a*Tan[c + d*x])^(3/2))/(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 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 \frac{\tan ^3(c+d x)}{\sqrt{a+i a \tan (c+d x)}} \, dx &=-\frac{\tan ^2(c+d x)}{d \sqrt{a+i a \tan (c+d x)}}-\frac{\int \tan (c+d x) \sqrt{a+i a \tan (c+d x)} \left (-2 a+\frac{5}{2} i a \tan (c+d x)\right ) \, dx}{a^2}\\ &=-\frac{\tan ^2(c+d x)}{d \sqrt{a+i a \tan (c+d x)}}-\frac{5 (a+i a \tan (c+d x))^{3/2}}{3 a^2 d}-\frac{\int \left (-\frac{5 i a}{2}-2 a \tan (c+d x)\right ) \sqrt{a+i a \tan (c+d x)} \, dx}{a^2}\\ &=-\frac{\tan ^2(c+d x)}{d \sqrt{a+i a \tan (c+d x)}}+\frac{4 \sqrt{a+i a \tan (c+d x)}}{a d}-\frac{5 (a+i a \tan (c+d x))^{3/2}}{3 a^2 d}+\frac{i \int \sqrt{a+i a \tan (c+d x)} \, dx}{2 a}\\ &=-\frac{\tan ^2(c+d x)}{d \sqrt{a+i a \tan (c+d x)}}+\frac{4 \sqrt{a+i a \tan (c+d x)}}{a d}-\frac{5 (a+i a \tan (c+d x))^{3/2}}{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 )}{d}\\ &=\frac{\tanh ^{-1}\left (\frac{\sqrt{a+i a \tan (c+d x)}}{\sqrt{2} \sqrt{a}}\right )}{\sqrt{2} \sqrt{a} d}-\frac{\tan ^2(c+d x)}{d \sqrt{a+i a \tan (c+d x)}}+\frac{4 \sqrt{a+i a \tan (c+d x)}}{a d}-\frac{5 (a+i a \tan (c+d x))^{3/2}}{3 a^2 d}\\ \end{align*}

Mathematica [A]  time = 1.02713, size = 129, normalized size = 1.02 \[ \frac{18 e^{2 i (c+d x)}+7 e^{4 i (c+d x)}+3 e^{i (c+d x)} \left (1+e^{2 i (c+d x)}\right )^{3/2} \sinh ^{-1}\left (e^{i (c+d x)}\right )+3}{3 \sqrt{2} d \left (1+e^{2 i (c+d x)}\right )^2 \sqrt{\frac{a e^{2 i (c+d x)}}{1+e^{2 i (c+d x)}}}} \]

Antiderivative was successfully verified.

[In]

Integrate[Tan[c + d*x]^3/Sqrt[a + I*a*Tan[c + d*x]],x]

[Out]

(3 + 18*E^((2*I)*(c + d*x)) + 7*E^((4*I)*(c + d*x)) + 3*E^(I*(c + d*x))*(1 + E^((2*I)*(c + d*x)))^(3/2)*ArcSin
h[E^(I*(c + d*x))])/(3*Sqrt[2]*d*Sqrt[(a*E^((2*I)*(c + d*x)))/(1 + E^((2*I)*(c + d*x)))]*(1 + E^((2*I)*(c + d*
x)))^2)

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Maple [A]  time = 0.044, size = 93, normalized size = 0.7 \begin{align*} -2\,{\frac{1}{{a}^{2}d} \left ( 1/3\, \left ( a+ia\tan \left ( dx+c \right ) \right ) ^{3/2}-a\sqrt{a+ia\tan \left ( dx+c \right ) }-1/2\,{\frac{{a}^{2}}{\sqrt{a+ia\tan \left ( dx+c \right ) }}}-1/4\,{a}^{3/2}\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))^(1/2),x)

[Out]

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

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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))^(1/2),x, algorithm="maxima")

[Out]

Exception raised: ValueError

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Fricas [B]  time = 2.30391, size = 878, normalized size = 6.97 \begin{align*} \frac{2 \, \sqrt{2} \sqrt{\frac{a}{e^{\left (2 i \, d x + 2 i \, c\right )} + 1}}{\left (7 \, e^{\left (4 i \, d x + 4 i \, c\right )} + 18 \, e^{\left (2 i \, d x + 2 i \, c\right )} + 3\right )} e^{\left (i \, d x + i \, c\right )} + 3 \, \sqrt{2}{\left (a d e^{\left (4 i \, d x + 4 i \, c\right )} + a d e^{\left (2 i \, d x + 2 i \, c\right )}\right )} \sqrt{\frac{1}{a d^{2}}} \log \left ({\left (\sqrt{2} a d \sqrt{\frac{1}{a 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{2}{\left (a d e^{\left (4 i \, d x + 4 i \, c\right )} + a d e^{\left (2 i \, d x + 2 i \, c\right )}\right )} \sqrt{\frac{1}{a d^{2}}} \log \left (-{\left (\sqrt{2} a d \sqrt{\frac{1}{a 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 )}{12 \,{\left (a d e^{\left (4 i \, d x + 4 i \, c\right )} + a d e^{\left (2 i \, d x + 2 i \, c\right )}\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))^(1/2),x, algorithm="fricas")

[Out]

1/12*(2*sqrt(2)*sqrt(a/(e^(2*I*d*x + 2*I*c) + 1))*(7*e^(4*I*d*x + 4*I*c) + 18*e^(2*I*d*x + 2*I*c) + 3)*e^(I*d*
x + I*c) + 3*sqrt(2)*(a*d*e^(4*I*d*x + 4*I*c) + a*d*e^(2*I*d*x + 2*I*c))*sqrt(1/(a*d^2))*log((sqrt(2)*a*d*sqrt
(1/(a*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(2)*(a*d*e^(4*I*d*x + 4*I*c) + a*d*e^(2*I*d*x + 2*I*c))*sqrt(1/(a*d^2))*log
(-(sqrt(2)*a*d*sqrt(1/(a*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)))/(a*d*e^(4*I*d*x + 4*I*c) + a*d*e^(2*I*d*x + 2*I*c))

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

[Out]

Integral(tan(c + d*x)**3/sqrt(a*(I*tan(c + d*x) + 1)), 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}}{\sqrt{i \, a \tan \left (d x + c\right ) + a}}\,{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))^(1/2),x, algorithm="giac")

[Out]

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