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

Optimal. Leaf size=251 \[ \frac{\sqrt [3]{2} \sqrt{3} a^{4/3} \tan ^{-1}\left (\frac{\sqrt [3]{a}+2^{2/3} \sqrt [3]{a+i a \tan (c+d x)}}{\sqrt{3} \sqrt [3]{a}}\right )}{d}-\frac{3 a^{4/3} \log \left (\sqrt [3]{2} \sqrt [3]{a}-\sqrt [3]{a+i a \tan (c+d x)}\right )}{2^{2/3} d}-\frac{a^{4/3} \log (\cos (c+d x))}{2^{2/3} d}-\frac{i a^{4/3} x}{2^{2/3}}+\frac{3 \tan ^2(c+d x) (a+i a \tan (c+d x))^{4/3}}{10 d}-\frac{6 (a+i a \tan (c+d x))^{7/3}}{35 a d}-\frac{9 (a+i a \tan (c+d x))^{4/3}}{20 d}-\frac{3 a \sqrt [3]{a+i a \tan (c+d x)}}{d} \]

[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))/(S
qrt[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 - (9*(a + I*a*Tan[c + d*x])^(4/3))/(20*d)
+ (3*Tan[c + d*x]^2*(a + I*a*Tan[c + d*x])^(4/3))/(10*d) - (6*(a + I*a*Tan[c + d*x])^(7/3))/(35*a*d)

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Rubi [A]  time = 0.34185, antiderivative size = 251, normalized size of antiderivative = 1., number of steps used = 9, number of rules used = 9, integrand size = 26, \(\frac{\text{number of rules}}{\text{integrand size}}\) = 0.346, Rules used = {3560, 3592, 3527, 3478, 3481, 57, 617, 204, 31} \[ \frac{\sqrt [3]{2} \sqrt{3} a^{4/3} \tan ^{-1}\left (\frac{\sqrt [3]{a}+2^{2/3} \sqrt [3]{a+i a \tan (c+d x)}}{\sqrt{3} \sqrt [3]{a}}\right )}{d}-\frac{3 a^{4/3} \log \left (\sqrt [3]{2} \sqrt [3]{a}-\sqrt [3]{a+i a \tan (c+d x)}\right )}{2^{2/3} d}-\frac{a^{4/3} \log (\cos (c+d x))}{2^{2/3} d}-\frac{i a^{4/3} x}{2^{2/3}}+\frac{3 \tan ^2(c+d x) (a+i a \tan (c+d x))^{4/3}}{10 d}-\frac{6 (a+i a \tan (c+d x))^{7/3}}{35 a d}-\frac{9 (a+i a \tan (c+d x))^{4/3}}{20 d}-\frac{3 a \sqrt [3]{a+i a \tan (c+d x)}}{d} \]

Antiderivative was successfully verified.

[In]

Int[Tan[c + d*x]^3*(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))/(S
qrt[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 - (9*(a + I*a*Tan[c + d*x])^(4/3))/(20*d)
+ (3*Tan[c + d*x]^2*(a + I*a*Tan[c + d*x])^(4/3))/(10*d) - (6*(a + I*a*Tan[c + d*x])^(7/3))/(35*a*d)

Rule 3560

Int[((a_) + (b_.)*tan[(e_.) + (f_.)*(x_)])^(m_)*((c_.) + (d_.)*tan[(e_.) + (f_.)*(x_)])^(n_), x_Symbol] :> Sim
p[(d*(a + b*Tan[e + f*x])^m*(c + d*Tan[e + f*x])^(n - 1))/(f*(m + n - 1)), x] - Dist[1/(a*(m + n - 1)), Int[(a
+ b*Tan[e + f*x])^m*(c + d*Tan[e + f*x])^(n - 2)*Simp[d*(b*c*m + a*d*(-1 + n)) - a*c^2*(m + n - 1) + d*(b*d*m
- a*c*(m + 2*n - 2))*Tan[e + f*x], x], x], x] /; FreeQ[{a, b, c, d, e, f, m}, x] && NeQ[b*c - a*d, 0] && EqQ[
a^2 + b^2, 0] && NeQ[c^2 + d^2, 0] && GtQ[n, 1] && NeQ[m + n - 1, 0] && (IntegerQ[n] || 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 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 ^3(c+d x) (a+i a \tan (c+d x))^{4/3} \, dx &=\frac{3 \tan ^2(c+d x) (a+i a \tan (c+d x))^{4/3}}{10 d}-\frac{3 \int \tan (c+d x) (a+i a \tan (c+d x))^{4/3} \left (2 a+\frac{4}{3} i a \tan (c+d x)\right ) \, dx}{10 a}\\ &=\frac{3 \tan ^2(c+d x) (a+i a \tan (c+d x))^{4/3}}{10 d}-\frac{6 (a+i a \tan (c+d x))^{7/3}}{35 a d}-\frac{3 \int (a+i a \tan (c+d x))^{4/3} \left (-\frac{4 i a}{3}+2 a \tan (c+d x)\right ) \, dx}{10 a}\\ &=-\frac{9 (a+i a \tan (c+d x))^{4/3}}{20 d}+\frac{3 \tan ^2(c+d x) (a+i a \tan (c+d x))^{4/3}}{10 d}-\frac{6 (a+i a \tan (c+d x))^{7/3}}{35 a 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{9 (a+i a \tan (c+d x))^{4/3}}{20 d}+\frac{3 \tan ^2(c+d x) (a+i a \tan (c+d x))^{4/3}}{10 d}-\frac{6 (a+i a \tan (c+d x))^{7/3}}{35 a 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{9 (a+i a \tan (c+d x))^{4/3}}{20 d}+\frac{3 \tan ^2(c+d x) (a+i a \tan (c+d x))^{4/3}}{10 d}-\frac{6 (a+i a \tan (c+d x))^{7/3}}{35 a 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{9 (a+i a \tan (c+d x))^{4/3}}{20 d}+\frac{3 \tan ^2(c+d x) (a+i a \tan (c+d x))^{4/3}}{10 d}-\frac{6 (a+i a \tan (c+d x))^{7/3}}{35 a 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{9 (a+i a \tan (c+d x))^{4/3}}{20 d}+\frac{3 \tan ^2(c+d x) (a+i a \tan (c+d x))^{4/3}}{10 d}-\frac{6 (a+i a \tan (c+d x))^{7/3}}{35 a 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{9 (a+i a \tan (c+d x))^{4/3}}{20 d}+\frac{3 \tan ^2(c+d x) (a+i a \tan (c+d x))^{4/3}}{10 d}-\frac{6 (a+i a \tan (c+d x))^{7/3}}{35 a d}\\ \end{align*}

Mathematica [F]  time = 180.004, size = 0, normalized size = 0. \[ \text{\$Aborted} \]

Verification is Not applicable to the result.

[In]

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

[Out]

$Aborted

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Maple [A]  time = 0.018, size = 217, normalized size = 0.9 \begin{align*} -{\frac{3}{10\,{a}^{2}d} \left ( a+ia\tan \left ( dx+c \right ) \right ) ^{{\frac{10}{3}}}}+{\frac{3}{7\,ad} \left ( a+ia\tan \left ( dx+c \right ) \right ) ^{{\frac{7}{3}}}}-{\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)^3*(a+I*a*tan(d*x+c))^(4/3),x)

[Out]

-3/10/d/a^2*(a+I*a*tan(d*x+c))^(10/3)+3/7*(a+I*a*tan(d*x+c))^(7/3)/a/d-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)^3*(a+I*a*tan(d*x+c))^(4/3),x, algorithm="maxima")

[Out]

Exception raised: ValueError

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Fricas [B]  time = 1.82993, size = 1466, normalized size = 5.84 \begin{align*} \text{result too large to display} \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))^(4/3),x, algorithm="fricas")

[Out]

-1/70*(3*2^(1/3)*(121*a*e^(6*I*d*x + 6*I*c) + 240*a*e^(4*I*d*x + 4*I*c) + 245*a*e^(2*I*d*x + 2*I*c) + 70*a)*(a
/(e^(2*I*d*x + 2*I*c) + 1))^(1/3)*e^(2/3*I*d*x + 2/3*I*c) + 35*2^(1/3)*((-I*sqrt(3)*d + d)*e^(6*I*d*x + 6*I*c)
+ 3*(-I*sqrt(3)*d + d)*e^(4*I*d*x + 4*I*c) + 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*sqr
t(3)*d - d)*(-a^4/d^3)^(1/3))/a) + 2^(1/3)*(35*(I*sqrt(3)*d + d)*e^(6*I*d*x + 6*I*c) + 105*(I*sqrt(3)*d + d)*e
^(4*I*d*x + 4*I*c) + 105*(I*sqrt(3)*d + d)*e^(2*I*d*x + 2*I*c) + 35*I*sqrt(3)*d + 35*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) - 70*2^(1/3)*(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)*(
-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^(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)

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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))**(4/3),x)

[Out]

Timed out

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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 )^{3}\,{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))^(4/3),x, algorithm="giac")

[Out]

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