3.282 \(\int \tan (c+d x) (a+i a \tan (c+d x))^{4/3} \, dx\)

Optimal. Leaf size=192 \[ -\frac{\sqrt [3]{2} \sqrt{3} a^{4/3} \tan ^{-1}\left (\frac{\sqrt [3]{a}+2^{2/3} \sqrt [3]{a+i a \tan (c+d x)}}{\sqrt{3} \sqrt [3]{a}}\right )}{d}+\frac{3 a^{4/3} \log \left (\sqrt [3]{2} \sqrt [3]{a}-\sqrt [3]{a+i a \tan (c+d x)}\right )}{2^{2/3} d}+\frac{a^{4/3} \log (\cos (c+d x))}{2^{2/3} d}+\frac{i a^{4/3} x}{2^{2/3}}+\frac{3 a \sqrt [3]{a+i a \tan (c+d x)}}{d}+\frac{3 (a+i a \tan (c+d x))^{4/3}}{4 d} \]

[Out]

(I*a^(4/3)*x)/2^(2/3) - (2^(1/3)*Sqrt[3]*a^(4/3)*ArcTan[(a^(1/3) + 2^(2/3)*(a + I*a*Tan[c + d*x])^(1/3))/(Sqrt
[3]*a^(1/3))])/d + (a^(4/3)*Log[Cos[c + d*x]])/(2^(2/3)*d) + (3*a^(4/3)*Log[2^(1/3)*a^(1/3) - (a + I*a*Tan[c +
d*x])^(1/3)])/(2^(2/3)*d) + (3*a*(a + I*a*Tan[c + d*x])^(1/3))/d + (3*(a + I*a*Tan[c + d*x])^(4/3))/(4*d)

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Rubi [A]  time = 0.147637, antiderivative size = 192, normalized size of antiderivative = 1., number of steps used = 7, number of rules used = 7, integrand size = 24, \(\frac{\text{number of rules}}{\text{integrand size}}\) = 0.292, Rules used = {3527, 3478, 3481, 57, 617, 204, 31} \[ -\frac{\sqrt [3]{2} \sqrt{3} a^{4/3} \tan ^{-1}\left (\frac{\sqrt [3]{a}+2^{2/3} \sqrt [3]{a+i a \tan (c+d x)}}{\sqrt{3} \sqrt [3]{a}}\right )}{d}+\frac{3 a^{4/3} \log \left (\sqrt [3]{2} \sqrt [3]{a}-\sqrt [3]{a+i a \tan (c+d x)}\right )}{2^{2/3} d}+\frac{a^{4/3} \log (\cos (c+d x))}{2^{2/3} d}+\frac{i a^{4/3} x}{2^{2/3}}+\frac{3 a \sqrt [3]{a+i a \tan (c+d x)}}{d}+\frac{3 (a+i a \tan (c+d x))^{4/3}}{4 d} \]

Antiderivative was successfully verified.

[In]

Int[Tan[c + d*x]*(a + I*a*Tan[c + d*x])^(4/3),x]

[Out]

(I*a^(4/3)*x)/2^(2/3) - (2^(1/3)*Sqrt[3]*a^(4/3)*ArcTan[(a^(1/3) + 2^(2/3)*(a + I*a*Tan[c + d*x])^(1/3))/(Sqrt
[3]*a^(1/3))])/d + (a^(4/3)*Log[Cos[c + d*x]])/(2^(2/3)*d) + (3*a^(4/3)*Log[2^(1/3)*a^(1/3) - (a + I*a*Tan[c +
d*x])^(1/3)])/(2^(2/3)*d) + (3*a*(a + I*a*Tan[c + d*x])^(1/3))/d + (3*(a + I*a*Tan[c + d*x])^(4/3))/(4*d)

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 3481

Int[((a_) + (b_.)*tan[(c_.) + (d_.)*(x_)])^(n_), x_Symbol] :> -Dist[b/d, Subst[Int[(a + x)^(n - 1)/(a - x), x]
, x, b*Tan[c + d*x]], x] /; FreeQ[{a, b, c, d, n}, x] && EqQ[a^2 + b^2, 0]

Rule 57

Int[1/(((a_.) + (b_.)*(x_))*((c_.) + (d_.)*(x_))^(2/3)), x_Symbol] :> With[{q = Rt[(b*c - a*d)/b, 3]}, -Simp[L
og[RemoveContent[a + b*x, x]]/(2*b*q^2), x] + (-Dist[3/(2*b*q), Subst[Int[1/(q^2 + q*x + x^2), x], x, (c + d*x
)^(1/3)], x] - Dist[3/(2*b*q^2), Subst[Int[1/(q - x), x], x, (c + d*x)^(1/3)], x])] /; FreeQ[{a, b, c, d}, x]
&& PosQ[(b*c - a*d)/b]

Rule 617

Int[((a_) + (b_.)*(x_) + (c_.)*(x_)^2)^(-1), x_Symbol] :> With[{q = 1 - 4*Simplify[(a*c)/b^2]}, Dist[-2/b, Sub
st[Int[1/(q - x^2), x], x, 1 + (2*c*x)/b], x] /; RationalQ[q] && (EqQ[q^2, 1] ||  !RationalQ[b^2 - 4*a*c])] /;
FreeQ[{a, b, c}, x] && NeQ[b^2 - 4*a*c, 0]

Rule 204

Int[((a_) + (b_.)*(x_)^2)^(-1), x_Symbol] :> -Simp[ArcTan[(Rt[-b, 2]*x)/Rt[-a, 2]]/(Rt[-a, 2]*Rt[-b, 2]), x] /
; FreeQ[{a, b}, x] && PosQ[a/b] && (LtQ[a, 0] || LtQ[b, 0])

Rule 31

Int[((a_) + (b_.)*(x_))^(-1), x_Symbol] :> Simp[Log[RemoveContent[a + b*x, x]]/b, x] /; FreeQ[{a, b}, x]

Rubi steps

\begin{align*} \int \tan (c+d x) (a+i a \tan (c+d x))^{4/3} \, dx &=\frac{3 (a+i a \tan (c+d x))^{4/3}}{4 d}-i \int (a+i a \tan (c+d x))^{4/3} \, dx\\ &=\frac{3 a \sqrt [3]{a+i a \tan (c+d x)}}{d}+\frac{3 (a+i a \tan (c+d x))^{4/3}}{4 d}-(2 i a) \int \sqrt [3]{a+i a \tan (c+d x)} \, dx\\ &=\frac{3 a \sqrt [3]{a+i a \tan (c+d x)}}{d}+\frac{3 (a+i a \tan (c+d x))^{4/3}}{4 d}-\frac{\left (2 a^2\right ) \operatorname{Subst}\left (\int \frac{1}{(a-x) (a+x)^{2/3}} \, dx,x,i a \tan (c+d x)\right )}{d}\\ &=\frac{i a^{4/3} x}{2^{2/3}}+\frac{a^{4/3} \log (\cos (c+d x))}{2^{2/3} d}+\frac{3 a \sqrt [3]{a+i a \tan (c+d x)}}{d}+\frac{3 (a+i a \tan (c+d x))^{4/3}}{4 d}-\frac{\left (3 a^{4/3}\right ) \operatorname{Subst}\left (\int \frac{1}{\sqrt [3]{2} \sqrt [3]{a}-x} \, dx,x,\sqrt [3]{a+i a \tan (c+d x)}\right )}{2^{2/3} d}-\frac{\left (3 a^{5/3}\right ) \operatorname{Subst}\left (\int \frac{1}{2^{2/3} a^{2/3}+\sqrt [3]{2} \sqrt [3]{a} x+x^2} \, dx,x,\sqrt [3]{a+i a \tan (c+d x)}\right )}{\sqrt [3]{2} d}\\ &=\frac{i a^{4/3} x}{2^{2/3}}+\frac{a^{4/3} \log (\cos (c+d x))}{2^{2/3} d}+\frac{3 a^{4/3} \log \left (\sqrt [3]{2} \sqrt [3]{a}-\sqrt [3]{a+i a \tan (c+d x)}\right )}{2^{2/3} d}+\frac{3 a \sqrt [3]{a+i a \tan (c+d x)}}{d}+\frac{3 (a+i a \tan (c+d x))^{4/3}}{4 d}+\frac{\left (3 \sqrt [3]{2} a^{4/3}\right ) \operatorname{Subst}\left (\int \frac{1}{-3-x^2} \, dx,x,1+\frac{2^{2/3} \sqrt [3]{a+i a \tan (c+d x)}}{\sqrt [3]{a}}\right )}{d}\\ &=\frac{i a^{4/3} x}{2^{2/3}}-\frac{\sqrt [3]{2} \sqrt{3} a^{4/3} \tan ^{-1}\left (\frac{1+\frac{2^{2/3} \sqrt [3]{a+i a \tan (c+d x)}}{\sqrt [3]{a}}}{\sqrt{3}}\right )}{d}+\frac{a^{4/3} \log (\cos (c+d x))}{2^{2/3} d}+\frac{3 a^{4/3} \log \left (\sqrt [3]{2} \sqrt [3]{a}-\sqrt [3]{a+i a \tan (c+d x)}\right )}{2^{2/3} d}+\frac{3 a \sqrt [3]{a+i a \tan (c+d x)}}{d}+\frac{3 (a+i a \tan (c+d x))^{4/3}}{4 d}\\ \end{align*}

Mathematica [F]  time = 180.005, size = 0, normalized size = 0. \[ \text{\$Aborted} \]

Verification is Not applicable to the result.

[In]

Integrate[Tan[c + d*x]*(a + I*a*Tan[c + d*x])^(4/3),x]

[Out]

$Aborted

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Maple [A]  time = 0.017, size = 173, normalized size = 0.9 \begin{align*}{\frac{3}{4\,d} \left ( a+ia\tan \left ( dx+c \right ) \right ) ^{{\frac{4}{3}}}}+3\,{\frac{a\sqrt [3]{a+ia\tan \left ( dx+c \right ) }}{d}}+{\frac{\sqrt [3]{2}}{d}{a}^{{\frac{4}{3}}}\ln \left ( \sqrt [3]{a+ia\tan \left ( dx+c \right ) }-\sqrt [3]{2}\sqrt [3]{a} \right ) }-{\frac{\sqrt [3]{2}}{2\,d}{a}^{{\frac{4}{3}}}\ln \left ( \left ( a+ia\tan \left ( dx+c \right ) \right ) ^{{\frac{2}{3}}}+\sqrt [3]{2}\sqrt [3]{a}\sqrt [3]{a+ia\tan \left ( dx+c \right ) }+{2}^{{\frac{2}{3}}}{a}^{{\frac{2}{3}}} \right ) }-{\frac{\sqrt [3]{2}\sqrt{3}}{d}{a}^{{\frac{4}{3}}}\arctan \left ({\frac{\sqrt{3}}{3} \left ({{2}^{{\frac{2}{3}}}\sqrt [3]{a+ia\tan \left ( dx+c \right ) }{\frac{1}{\sqrt [3]{a}}}}+1 \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))^(4/3),x)

[Out]

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

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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)*(a+I*a*tan(d*x+c))^(4/3),x, algorithm="maxima")

[Out]

Exception raised: ValueError

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Fricas [B]  time = 1.80053, size = 983, normalized size = 5.12 \begin{align*} \frac{3 \cdot 2^{\frac{1}{3}}{\left (3 \, a e^{\left (2 i \, d x + 2 i \, c\right )} + 2 \, a\right )} \left (\frac{a}{e^{\left (2 i \, d x + 2 i \, c\right )} + 1}\right )^{\frac{1}{3}} e^{\left (\frac{2}{3} i \, d x + \frac{2}{3} i \, c\right )} + 2^{\frac{1}{3}}{\left ({\left (-i \, \sqrt{3} d - d\right )} e^{\left (2 i \, d x + 2 i \, c\right )} - i \, \sqrt{3} d - d\right )} \left (\frac{a^{4}}{d^{3}}\right )^{\frac{1}{3}} \log \left (\frac{2 \cdot 2^{\frac{1}{3}} a \left (\frac{a}{e^{\left (2 i \, d x + 2 i \, c\right )} + 1}\right )^{\frac{1}{3}} e^{\left (\frac{2}{3} i \, d x + \frac{2}{3} i \, c\right )} + 2^{\frac{1}{3}}{\left (i \, \sqrt{3} d + d\right )} \left (\frac{a^{4}}{d^{3}}\right )^{\frac{1}{3}}}{2 \, a}\right ) + 2^{\frac{1}{3}}{\left ({\left (i \, \sqrt{3} d - d\right )} e^{\left (2 i \, d x + 2 i \, c\right )} + i \, \sqrt{3} d - d\right )} \left (\frac{a^{4}}{d^{3}}\right )^{\frac{1}{3}} \log \left (\frac{2 \cdot 2^{\frac{1}{3}} a \left (\frac{a}{e^{\left (2 i \, d x + 2 i \, c\right )} + 1}\right )^{\frac{1}{3}} e^{\left (\frac{2}{3} i \, d x + \frac{2}{3} i \, c\right )} + 2^{\frac{1}{3}}{\left (-i \, \sqrt{3} d + d\right )} \left (\frac{a^{4}}{d^{3}}\right )^{\frac{1}{3}}}{2 \, a}\right ) + 2 \cdot 2^{\frac{1}{3}}{\left (d e^{\left (2 i \, d x + 2 i \, c\right )} + d\right )} \left (\frac{a^{4}}{d^{3}}\right )^{\frac{1}{3}} \log \left (\frac{2^{\frac{1}{3}} a \left (\frac{a}{e^{\left (2 i \, d x + 2 i \, c\right )} + 1}\right )^{\frac{1}{3}} e^{\left (\frac{2}{3} i \, d x + \frac{2}{3} i \, c\right )} - 2^{\frac{1}{3}} \left (\frac{a^{4}}{d^{3}}\right )^{\frac{1}{3}} d}{a}\right )}{2 \,{\left (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))^(4/3),x, algorithm="fricas")

[Out]

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

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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{4}{3}} \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))**(4/3),x)

[Out]

Integral((a*(I*tan(c + d*x) + 1))**(4/3)*tan(c + d*x), x)

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Giac [F]  time = 0., size = 0, normalized size = 0. \begin{align*} \int{\left (i \, a \tan \left (d x + c\right ) + a\right )}^{\frac{4}{3}} \tan \left (d x + c\right )\,{d x} \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))^(4/3),x, algorithm="giac")

[Out]

integrate((I*a*tan(d*x + c) + a)^(4/3)*tan(d*x + c), x)