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

Optimal. Leaf size=327 \[ -\frac{i \sqrt{3} (a-i b)^{4/3} \tan ^{-1}\left (\frac{1+\frac{2 \sqrt [3]{a+b \tan (c+d x)}}{\sqrt [3]{a-i b}}}{\sqrt{3}}\right )}{2 d}+\frac{i \sqrt{3} (a+i b)^{4/3} \tan ^{-1}\left (\frac{1+\frac{2 \sqrt [3]{a+b \tan (c+d x)}}{\sqrt [3]{a+i b}}}{\sqrt{3}}\right )}{2 d}+\frac{3 b \sqrt [3]{a+b \tan (c+d x)}}{d}+\frac{3 i (a-i b)^{4/3} \log \left (-\sqrt [3]{a+b \tan (c+d x)}+\sqrt [3]{a-i b}\right )}{4 d}-\frac{3 i (a+i b)^{4/3} \log \left (-\sqrt [3]{a+b \tan (c+d x)}+\sqrt [3]{a+i b}\right )}{4 d}+\frac{i (a-i b)^{4/3} \log (\cos (c+d x))}{4 d}-\frac{i (a+i b)^{4/3} \log (\cos (c+d x))}{4 d}-\frac{1}{4} x (a-i b)^{4/3}-\frac{1}{4} x (a+i b)^{4/3} \]

[Out]

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

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Rubi [A]  time = 0.360445, antiderivative size = 327, normalized size of antiderivative = 1., number of steps used = 12, number of rules used = 7, integrand size = 14, \(\frac{\text{number of rules}}{\text{integrand size}}\) = 0.5, Rules used = {3482, 3539, 3537, 57, 617, 204, 31} \[ -\frac{i \sqrt{3} (a-i b)^{4/3} \tan ^{-1}\left (\frac{1+\frac{2 \sqrt [3]{a+b \tan (c+d x)}}{\sqrt [3]{a-i b}}}{\sqrt{3}}\right )}{2 d}+\frac{i \sqrt{3} (a+i b)^{4/3} \tan ^{-1}\left (\frac{1+\frac{2 \sqrt [3]{a+b \tan (c+d x)}}{\sqrt [3]{a+i b}}}{\sqrt{3}}\right )}{2 d}+\frac{3 b \sqrt [3]{a+b \tan (c+d x)}}{d}+\frac{3 i (a-i b)^{4/3} \log \left (-\sqrt [3]{a+b \tan (c+d x)}+\sqrt [3]{a-i b}\right )}{4 d}-\frac{3 i (a+i b)^{4/3} \log \left (-\sqrt [3]{a+b \tan (c+d x)}+\sqrt [3]{a+i b}\right )}{4 d}+\frac{i (a-i b)^{4/3} \log (\cos (c+d x))}{4 d}-\frac{i (a+i b)^{4/3} \log (\cos (c+d x))}{4 d}-\frac{1}{4} x (a-i b)^{4/3}-\frac{1}{4} x (a+i b)^{4/3} \]

Antiderivative was successfully verified.

[In]

Int[(a + b*Tan[c + d*x])^(4/3),x]

[Out]

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

Rule 3482

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

Rule 3539

Int[((a_.) + (b_.)*tan[(e_.) + (f_.)*(x_)])^(m_)*((c_.) + (d_.)*tan[(e_.) + (f_.)*(x_)]), x_Symbol] :> Dist[(c
+ I*d)/2, Int[(a + b*Tan[e + f*x])^m*(1 - I*Tan[e + f*x]), x], x] + Dist[(c - I*d)/2, Int[(a + b*Tan[e + f*x]
)^m*(1 + I*Tan[e + f*x]), x], x] /; FreeQ[{a, b, c, d, e, f, m}, x] && NeQ[b*c - a*d, 0] && NeQ[a^2 + b^2, 0]
&& NeQ[c^2 + d^2, 0] &&  !IntegerQ[m]

Rule 3537

Int[((a_.) + (b_.)*tan[(e_.) + (f_.)*(x_)])^(m_)*((c_) + (d_.)*tan[(e_.) + (f_.)*(x_)]), x_Symbol] :> Dist[(c*
d)/f, Subst[Int[(a + (b*x)/d)^m/(d^2 + c*x), x], x, d*Tan[e + f*x]], x] /; FreeQ[{a, b, c, d, e, f, m}, x] &&
NeQ[b*c - a*d, 0] && NeQ[a^2 + b^2, 0] && EqQ[c^2 + d^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 (a+b \tan (c+d x))^{4/3} \, dx &=\frac{3 b \sqrt [3]{a+b \tan (c+d x)}}{d}+\int \frac{a^2-b^2+2 a b \tan (c+d x)}{(a+b \tan (c+d x))^{2/3}} \, dx\\ &=\frac{3 b \sqrt [3]{a+b \tan (c+d x)}}{d}+\frac{1}{2} (a-i b)^2 \int \frac{1+i \tan (c+d x)}{(a+b \tan (c+d x))^{2/3}} \, dx+\frac{1}{2} (a+i b)^2 \int \frac{1-i \tan (c+d x)}{(a+b \tan (c+d x))^{2/3}} \, dx\\ &=\frac{3 b \sqrt [3]{a+b \tan (c+d x)}}{d}+\frac{\left (i (a-i b)^2\right ) \operatorname{Subst}\left (\int \frac{1}{(-1+x) (a-i b x)^{2/3}} \, dx,x,i \tan (c+d x)\right )}{2 d}-\frac{\left (i (a+i b)^2\right ) \operatorname{Subst}\left (\int \frac{1}{(-1+x) (a+i b x)^{2/3}} \, dx,x,-i \tan (c+d x)\right )}{2 d}\\ &=-\frac{1}{4} (a-i b)^{4/3} x-\frac{1}{4} (a+i b)^{4/3} x+\frac{i (a-i b)^{4/3} \log (\cos (c+d x))}{4 d}-\frac{i (a+i b)^{4/3} \log (\cos (c+d x))}{4 d}+\frac{3 b \sqrt [3]{a+b \tan (c+d x)}}{d}-\frac{\left (3 i (a-i b)^{4/3}\right ) \operatorname{Subst}\left (\int \frac{1}{\sqrt [3]{a-i b}-x} \, dx,x,\sqrt [3]{a+b \tan (c+d x)}\right )}{4 d}-\frac{\left (3 i (a-i b)^{5/3}\right ) \operatorname{Subst}\left (\int \frac{1}{(a-i b)^{2/3}+\sqrt [3]{a-i b} x+x^2} \, dx,x,\sqrt [3]{a+b \tan (c+d x)}\right )}{4 d}+\frac{\left (3 i (a+i b)^{4/3}\right ) \operatorname{Subst}\left (\int \frac{1}{\sqrt [3]{a+i b}-x} \, dx,x,\sqrt [3]{a+b \tan (c+d x)}\right )}{4 d}+\frac{\left (3 i (a+i b)^{5/3}\right ) \operatorname{Subst}\left (\int \frac{1}{(a+i b)^{2/3}+\sqrt [3]{a+i b} x+x^2} \, dx,x,\sqrt [3]{a+b \tan (c+d x)}\right )}{4 d}\\ &=-\frac{1}{4} (a-i b)^{4/3} x-\frac{1}{4} (a+i b)^{4/3} x+\frac{i (a-i b)^{4/3} \log (\cos (c+d x))}{4 d}-\frac{i (a+i b)^{4/3} \log (\cos (c+d x))}{4 d}+\frac{3 i (a-i b)^{4/3} \log \left (\sqrt [3]{a-i b}-\sqrt [3]{a+b \tan (c+d x)}\right )}{4 d}-\frac{3 i (a+i b)^{4/3} \log \left (\sqrt [3]{a+i b}-\sqrt [3]{a+b \tan (c+d x)}\right )}{4 d}+\frac{3 b \sqrt [3]{a+b \tan (c+d x)}}{d}+\frac{\left (3 i (a-i b)^{4/3}\right ) \operatorname{Subst}\left (\int \frac{1}{-3-x^2} \, dx,x,1+\frac{2 \sqrt [3]{a+b \tan (c+d x)}}{\sqrt [3]{a-i b}}\right )}{2 d}-\frac{\left (3 i (a+i b)^{4/3}\right ) \operatorname{Subst}\left (\int \frac{1}{-3-x^2} \, dx,x,1+\frac{2 \sqrt [3]{a+b \tan (c+d x)}}{\sqrt [3]{a+i b}}\right )}{2 d}\\ &=-\frac{1}{4} (a-i b)^{4/3} x-\frac{1}{4} (a+i b)^{4/3} x-\frac{i \sqrt{3} (a-i b)^{4/3} \tan ^{-1}\left (\frac{1+\frac{2 \sqrt [3]{a+b \tan (c+d x)}}{\sqrt [3]{a-i b}}}{\sqrt{3}}\right )}{2 d}+\frac{i \sqrt{3} (a+i b)^{4/3} \tan ^{-1}\left (\frac{1+\frac{2 \sqrt [3]{a+b \tan (c+d x)}}{\sqrt [3]{a+i b}}}{\sqrt{3}}\right )}{2 d}+\frac{i (a-i b)^{4/3} \log (\cos (c+d x))}{4 d}-\frac{i (a+i b)^{4/3} \log (\cos (c+d x))}{4 d}+\frac{3 i (a-i b)^{4/3} \log \left (\sqrt [3]{a-i b}-\sqrt [3]{a+b \tan (c+d x)}\right )}{4 d}-\frac{3 i (a+i b)^{4/3} \log \left (\sqrt [3]{a+i b}-\sqrt [3]{a+b \tan (c+d x)}\right )}{4 d}+\frac{3 b \sqrt [3]{a+b \tan (c+d x)}}{d}\\ \end{align*}

Mathematica [A]  time = 0.728044, size = 365, normalized size = 1.12 \[ \frac{(b+i a) \left (6 \sqrt [3]{a+b \tan (c+d x)}-\sqrt [3]{a-i b} \left (2 \sqrt{3} \tan ^{-1}\left (\frac{1+\frac{2 \sqrt [3]{a+b \tan (c+d x)}}{\sqrt [3]{a-i b}}}{\sqrt{3}}\right )+\log \left (\sqrt [3]{a-i b} \sqrt [3]{a+b \tan (c+d x)}+(a+b \tan (c+d x))^{2/3}+(a-i b)^{2/3}\right )\right )+2 \sqrt [3]{a-i b} \log \left (-\sqrt [3]{a+b \tan (c+d x)}+\sqrt [3]{a-i b}\right )\right )-(-b+i a) \left (6 \sqrt [3]{a+b \tan (c+d x)}-\sqrt [3]{a+i b} \left (2 \sqrt{3} \tan ^{-1}\left (\frac{1+\frac{2 \sqrt [3]{a+b \tan (c+d x)}}{\sqrt [3]{a+i b}}}{\sqrt{3}}\right )+\log \left (\sqrt [3]{a+i b} \sqrt [3]{a+b \tan (c+d x)}+(a+b \tan (c+d x))^{2/3}+(a+i b)^{2/3}\right )\right )+2 \sqrt [3]{a+i b} \log \left (-\sqrt [3]{a+b \tan (c+d x)}+\sqrt [3]{a+i b}\right )\right )}{4 d} \]

Antiderivative was successfully verified.

[In]

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

[Out]

((I*a + b)*(2*(a - I*b)^(1/3)*Log[(a - I*b)^(1/3) - (a + b*Tan[c + d*x])^(1/3)] - (a - I*b)^(1/3)*(2*Sqrt[3]*A
rcTan[(1 + (2*(a + b*Tan[c + d*x])^(1/3))/(a - I*b)^(1/3))/Sqrt[3]] + Log[(a - I*b)^(2/3) + (a - I*b)^(1/3)*(a
+ b*Tan[c + d*x])^(1/3) + (a + b*Tan[c + d*x])^(2/3)]) + 6*(a + b*Tan[c + d*x])^(1/3)) - (I*a - b)*(2*(a + I*
b)^(1/3)*Log[(a + I*b)^(1/3) - (a + b*Tan[c + d*x])^(1/3)] - (a + I*b)^(1/3)*(2*Sqrt[3]*ArcTan[(1 + (2*(a + b*
Tan[c + d*x])^(1/3))/(a + I*b)^(1/3))/Sqrt[3]] + Log[(a + I*b)^(2/3) + (a + I*b)^(1/3)*(a + b*Tan[c + d*x])^(1
/3) + (a + b*Tan[c + d*x])^(2/3)]) + 6*(a + b*Tan[c + d*x])^(1/3)))/(4*d)

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Maple [C]  time = 0.015, size = 93, normalized size = 0.3 \begin{align*} 3\,{\frac{b\sqrt [3]{a+b\tan \left ( dx+c \right ) }}{d}}+{\frac{b}{2\,d}\sum _{{\it \_R}={\it RootOf} \left ({{\it \_Z}}^{6}-2\,{{\it \_Z}}^{3}a+{a}^{2}+{b}^{2} \right ) }{\frac{2\,{{\it \_R}}^{3}a-{a}^{2}-{b}^{2}}{{{\it \_R}}^{5}-{{\it \_R}}^{2}a}\ln \left ( \sqrt [3]{a+b\tan \left ( dx+c \right ) }-{\it \_R} \right ) }} \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

int((a+b*tan(d*x+c))^(4/3),x)

[Out]

3*b*(a+b*tan(d*x+c))^(1/3)/d+1/2/d*b*sum((2*_R^3*a-a^2-b^2)/(_R^5-_R^2*a)*ln((a+b*tan(d*x+c))^(1/3)-_R),_R=Roo
tOf(_Z^6-2*_Z^3*a+a^2+b^2))

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Maxima [F]  time = 0., size = 0, normalized size = 0. \begin{align*} \int{\left (b \tan \left (d x + c\right ) + a\right )}^{\frac{4}{3}}\,{d x} \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

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

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Fricas [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((a+b*tan(d*x+c))^(4/3),x, algorithm="fricas")

[Out]

Timed out

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Sympy [F]  time = 0., size = 0, normalized size = 0. \begin{align*} \int \left (a + b \tan{\left (c + d x \right )}\right )^{\frac{4}{3}}\, dx \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

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

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Giac [A]  time = 6.79133, size = 579, normalized size = 1.77 \begin{align*} -\frac{1}{24} \,{\left ({\left (i \, \sqrt{3} + 1\right )} \left (\frac{216 i \, a^{4} - 864 \, a^{3} b - 1296 i \, a^{2} b^{2} + 864 \, a b^{3} + 216 i \, b^{4}}{b^{3} d^{3}}\right )^{\frac{1}{3}} \log \left ({\left (b \tan \left (d x + c\right ) + a\right )}^{\frac{2}{3}} d^{2}\right ) +{\left (-i \, \sqrt{3} + 1\right )} \left (\frac{216 i \, a^{4} - 864 \, a^{3} b - 1296 i \, a^{2} b^{2} + 864 \, a b^{3} + 216 i \, b^{4}}{b^{3} d^{3}}\right )^{\frac{1}{3}} \log \left ({\left (b \tan \left (d x + c\right ) + a\right )}^{\frac{2}{3}} d^{2}\right ) +{\left (i \, \sqrt{3} + 1\right )} \left (\frac{-216 i \, a^{4} - 864 \, a^{3} b + 1296 i \, a^{2} b^{2} + 864 \, a b^{3} - 216 i \, b^{4}}{b^{3} d^{3}}\right )^{\frac{1}{3}} \log \left ({\left (b \tan \left (d x + c\right ) + a\right )}^{\frac{2}{3}} d^{2}\right ) +{\left (-i \, \sqrt{3} + 1\right )} \left (\frac{-216 i \, a^{4} - 864 \, a^{3} b + 1296 i \, a^{2} b^{2} + 864 \, a b^{3} - 216 i \, b^{4}}{b^{3} d^{3}}\right )^{\frac{1}{3}} \log \left ({\left (b \tan \left (d x + c\right ) + a\right )}^{\frac{2}{3}} d^{2}\right ) - 2 \, \left (\frac{216 i \, a^{4} - 864 \, a^{3} b - 1296 i \, a^{2} b^{2} + 864 \, a b^{3} + 216 i \, b^{4}}{b^{3} d^{3}}\right )^{\frac{1}{3}} \log \left (i \,{\left (b \tan \left (d x + c\right ) + a\right )}^{\frac{1}{3}} d^{2} +{\left (i \, a - b\right )}^{\frac{1}{3}} d^{2}\right ) - 2 \, \left (\frac{-216 i \, a^{4} - 864 \, a^{3} b + 1296 i \, a^{2} b^{2} + 864 \, a b^{3} - 216 i \, b^{4}}{b^{3} d^{3}}\right )^{\frac{1}{3}} \log \left (-i \,{\left (b \tan \left (d x + c\right ) + a\right )}^{\frac{1}{3}} d^{2} +{\left (-i \, a - b\right )}^{\frac{1}{3}} d^{2}\right ) - \frac{72 \,{\left (b \tan \left (d x + c\right ) + a\right )}^{\frac{1}{3}}}{d}\right )} b \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate((a+b*tan(d*x+c))^(4/3),x, algorithm="giac")

[Out]

-1/24*((I*sqrt(3) + 1)*((216*I*a^4 - 864*a^3*b - 1296*I*a^2*b^2 + 864*a*b^3 + 216*I*b^4)/(b^3*d^3))^(1/3)*log(
(b*tan(d*x + c) + a)^(2/3)*d^2) + (-I*sqrt(3) + 1)*((216*I*a^4 - 864*a^3*b - 1296*I*a^2*b^2 + 864*a*b^3 + 216*
I*b^4)/(b^3*d^3))^(1/3)*log((b*tan(d*x + c) + a)^(2/3)*d^2) + (I*sqrt(3) + 1)*((-216*I*a^4 - 864*a^3*b + 1296*
I*a^2*b^2 + 864*a*b^3 - 216*I*b^4)/(b^3*d^3))^(1/3)*log((b*tan(d*x + c) + a)^(2/3)*d^2) + (-I*sqrt(3) + 1)*((-
216*I*a^4 - 864*a^3*b + 1296*I*a^2*b^2 + 864*a*b^3 - 216*I*b^4)/(b^3*d^3))^(1/3)*log((b*tan(d*x + c) + a)^(2/3
)*d^2) - 2*((216*I*a^4 - 864*a^3*b - 1296*I*a^2*b^2 + 864*a*b^3 + 216*I*b^4)/(b^3*d^3))^(1/3)*log(I*(b*tan(d*x
+ c) + a)^(1/3)*d^2 + (I*a - b)^(1/3)*d^2) - 2*((-216*I*a^4 - 864*a^3*b + 1296*I*a^2*b^2 + 864*a*b^3 - 216*I*
b^4)/(b^3*d^3))^(1/3)*log(-I*(b*tan(d*x + c) + a)^(1/3)*d^2 + (-I*a - b)^(1/3)*d^2) - 72*(b*tan(d*x + c) + a)^
(1/3)/d)*b