3.688 \(\int (a+b \tan (c+d x))^{5/3} \, dx\)

Optimal. Leaf size=329 \[ \frac{i \sqrt{3} (a-i b)^{5/3} \tan ^{-1}\left (\frac{1+\frac{2 \sqrt [3]{a+b \tan (c+d x)}}{\sqrt [3]{a-i b}}}{\sqrt{3}}\right )}{2 d}-\frac{i \sqrt{3} (a+i b)^{5/3} \tan ^{-1}\left (\frac{1+\frac{2 \sqrt [3]{a+b \tan (c+d x)}}{\sqrt [3]{a+i b}}}{\sqrt{3}}\right )}{2 d}+\frac{3 b (a+b \tan (c+d x))^{2/3}}{2 d}+\frac{3 i (a-i b)^{5/3} \log \left (-\sqrt [3]{a+b \tan (c+d x)}+\sqrt [3]{a-i b}\right )}{4 d}-\frac{3 i (a+i b)^{5/3} \log \left (-\sqrt [3]{a+b \tan (c+d x)}+\sqrt [3]{a+i b}\right )}{4 d}+\frac{i (a-i b)^{5/3} \log (\cos (c+d x))}{4 d}-\frac{i (a+i b)^{5/3} \log (\cos (c+d x))}{4 d}-\frac{1}{4} x (a-i b)^{5/3}-\frac{1}{4} x (a+i b)^{5/3} \]

[Out]

-((a - I*b)^(5/3)*x)/4 - ((a + I*b)^(5/3)*x)/4 + ((I/2)*Sqrt[3]*(a - I*b)^(5/3)*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)^(5/3)*ArcTan[(1 + (2*(a + b*Tan[c + d*x])
^(1/3))/(a + I*b)^(1/3))/Sqrt[3]])/d + ((I/4)*(a - I*b)^(5/3)*Log[Cos[c + d*x]])/d - ((I/4)*(a + I*b)^(5/3)*Lo
g[Cos[c + d*x]])/d + (((3*I)/4)*(a - I*b)^(5/3)*Log[(a - I*b)^(1/3) - (a + b*Tan[c + d*x])^(1/3)])/d - (((3*I)
/4)*(a + I*b)^(5/3)*Log[(a + I*b)^(1/3) - (a + b*Tan[c + d*x])^(1/3)])/d + (3*b*(a + b*Tan[c + d*x])^(2/3))/(2
*d)

________________________________________________________________________________________

Rubi [A]  time = 0.376091, antiderivative size = 329, 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, 55, 617, 204, 31} \[ \frac{i \sqrt{3} (a-i b)^{5/3} \tan ^{-1}\left (\frac{1+\frac{2 \sqrt [3]{a+b \tan (c+d x)}}{\sqrt [3]{a-i b}}}{\sqrt{3}}\right )}{2 d}-\frac{i \sqrt{3} (a+i b)^{5/3} \tan ^{-1}\left (\frac{1+\frac{2 \sqrt [3]{a+b \tan (c+d x)}}{\sqrt [3]{a+i b}}}{\sqrt{3}}\right )}{2 d}+\frac{3 b (a+b \tan (c+d x))^{2/3}}{2 d}+\frac{3 i (a-i b)^{5/3} \log \left (-\sqrt [3]{a+b \tan (c+d x)}+\sqrt [3]{a-i b}\right )}{4 d}-\frac{3 i (a+i b)^{5/3} \log \left (-\sqrt [3]{a+b \tan (c+d x)}+\sqrt [3]{a+i b}\right )}{4 d}+\frac{i (a-i b)^{5/3} \log (\cos (c+d x))}{4 d}-\frac{i (a+i b)^{5/3} \log (\cos (c+d x))}{4 d}-\frac{1}{4} x (a-i b)^{5/3}-\frac{1}{4} x (a+i b)^{5/3} \]

Antiderivative was successfully verified.

[In]

Int[(a + b*Tan[c + d*x])^(5/3),x]

[Out]

-((a - I*b)^(5/3)*x)/4 - ((a + I*b)^(5/3)*x)/4 + ((I/2)*Sqrt[3]*(a - I*b)^(5/3)*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)^(5/3)*ArcTan[(1 + (2*(a + b*Tan[c + d*x])
^(1/3))/(a + I*b)^(1/3))/Sqrt[3]])/d + ((I/4)*(a - I*b)^(5/3)*Log[Cos[c + d*x]])/d - ((I/4)*(a + I*b)^(5/3)*Lo
g[Cos[c + d*x]])/d + (((3*I)/4)*(a - I*b)^(5/3)*Log[(a - I*b)^(1/3) - (a + b*Tan[c + d*x])^(1/3)])/d - (((3*I)
/4)*(a + I*b)^(5/3)*Log[(a + I*b)^(1/3) - (a + b*Tan[c + d*x])^(1/3)])/d + (3*b*(a + b*Tan[c + d*x])^(2/3))/(2
*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 55

Int[1/(((a_.) + (b_.)*(x_))*((c_.) + (d_.)*(x_))^(1/3)), x_Symbol] :> With[{q = Rt[(b*c - a*d)/b, 3]}, -Simp[L
og[RemoveContent[a + b*x, x]]/(2*b*q), x] + (Dist[3/(2*b), Subst[Int[1/(q^2 + q*x + x^2), x], x, (c + d*x)^(1/
3)], x] - Dist[3/(2*b*q), 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))^{5/3} \, dx &=\frac{3 b (a+b \tan (c+d x))^{2/3}}{2 d}+\int \frac{a^2-b^2+2 a b \tan (c+d x)}{\sqrt [3]{a+b \tan (c+d x)}} \, dx\\ &=\frac{3 b (a+b \tan (c+d x))^{2/3}}{2 d}+\frac{1}{2} (a-i b)^2 \int \frac{1+i \tan (c+d x)}{\sqrt [3]{a+b \tan (c+d x)}} \, dx+\frac{1}{2} (a+i b)^2 \int \frac{1-i \tan (c+d x)}{\sqrt [3]{a+b \tan (c+d x)}} \, dx\\ &=\frac{3 b (a+b \tan (c+d x))^{2/3}}{2 d}+\frac{\left (i (a-i b)^2\right ) \operatorname{Subst}\left (\int \frac{1}{(-1+x) \sqrt [3]{a-i b x}} \, 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) \sqrt [3]{a+i b x}} \, dx,x,-i \tan (c+d x)\right )}{2 d}\\ &=-\frac{1}{4} (a-i b)^{5/3} x-\frac{1}{4} (a+i b)^{5/3} x+\frac{i (a-i b)^{5/3} \log (\cos (c+d x))}{4 d}-\frac{i (a+i b)^{5/3} \log (\cos (c+d x))}{4 d}+\frac{3 b (a+b \tan (c+d x))^{2/3}}{2 d}-\frac{\left (3 i (a-i b)^{5/3}\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)^2\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)^{5/3}\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)^2\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)^{5/3} x-\frac{1}{4} (a+i b)^{5/3} x+\frac{i (a-i b)^{5/3} \log (\cos (c+d x))}{4 d}-\frac{i (a+i b)^{5/3} \log (\cos (c+d x))}{4 d}+\frac{3 i (a-i b)^{5/3} \log \left (\sqrt [3]{a-i b}-\sqrt [3]{a+b \tan (c+d x)}\right )}{4 d}-\frac{3 i (a+i b)^{5/3} \log \left (\sqrt [3]{a+i b}-\sqrt [3]{a+b \tan (c+d x)}\right )}{4 d}+\frac{3 b (a+b \tan (c+d x))^{2/3}}{2 d}-\frac{\left (3 i (a-i b)^{5/3}\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)^{5/3}\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)^{5/3} x-\frac{1}{4} (a+i b)^{5/3} x+\frac{i \sqrt{3} (a-i b)^{5/3} \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)^{5/3} \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)^{5/3} \log (\cos (c+d x))}{4 d}-\frac{i (a+i b)^{5/3} \log (\cos (c+d x))}{4 d}+\frac{3 i (a-i b)^{5/3} \log \left (\sqrt [3]{a-i b}-\sqrt [3]{a+b \tan (c+d x)}\right )}{4 d}-\frac{3 i (a+i b)^{5/3} \log \left (\sqrt [3]{a+i b}-\sqrt [3]{a+b \tan (c+d x)}\right )}{4 d}+\frac{3 b (a+b \tan (c+d x))^{2/3}}{2 d}\\ \end{align*}

Mathematica [A]  time = 0.943564, size = 300, normalized size = 0.91 \[ \frac{(b+i a) \left (2 \sqrt{3} (a-i b)^{2/3} \tan ^{-1}\left (\frac{1+\frac{2 \sqrt [3]{a+b \tan (c+d x)}}{\sqrt [3]{a-i b}}}{\sqrt{3}}\right )-(a-i b)^{2/3} \log (\tan (c+d x)+i)+3 \left ((a+b \tan (c+d x))^{2/3}+(a-i b)^{2/3} \log \left (-\sqrt [3]{a+b \tan (c+d x)}+\sqrt [3]{a-i b}\right )\right )\right )+(b-i a) \left (2 \sqrt{3} (a+i b)^{2/3} \tan ^{-1}\left (\frac{1+\frac{2 \sqrt [3]{a+b \tan (c+d x)}}{\sqrt [3]{a+i b}}}{\sqrt{3}}\right )-(a+i b)^{2/3} \log (-\tan (c+d x)+i)+3 \left ((a+b \tan (c+d x))^{2/3}+(a+i b)^{2/3} \log \left (-\sqrt [3]{a+b \tan (c+d x)}+\sqrt [3]{a+i b}\right )\right )\right )}{4 d} \]

Antiderivative was successfully verified.

[In]

Integrate[(a + b*Tan[c + d*x])^(5/3),x]

[Out]

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

________________________________________________________________________________________

Maple [C]  time = 0.031, size = 96, normalized size = 0.3 \begin{align*}{\frac{3\,b}{2\,d} \left ( a+b\tan \left ( dx+c \right ) \right ) ^{{\frac{2}{3}}}}+{\frac{b}{2\,d}\sum _{{\it \_R}={\it RootOf} \left ({{\it \_Z}}^{6}-2\,a{{\it \_Z}}^{3}+{a}^{2}+{b}^{2} \right ) }{\frac{2\,a{{\it \_R}}^{4}+ \left ( -{a}^{2}-{b}^{2} \right ){\it \_R}}{{{\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))^(5/3),x)

[Out]

3/2*b*(a+b*tan(d*x+c))^(2/3)/d+1/2/d*b*sum((2*a*_R^4+(-a^2-b^2)*_R)/(_R^5-_R^2*a)*ln((a+b*tan(d*x+c))^(1/3)-_R
),_R=RootOf(_Z^6-2*_Z^3*a+a^2+b^2))

________________________________________________________________________________________

Maxima [F]  time = 0., size = 0, normalized size = 0. \begin{align*} \int{\left (b \tan \left (d x + c\right ) + a\right )}^{\frac{5}{3}}\,{d x} \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate((a+b*tan(d*x+c))^(5/3),x, algorithm="maxima")

[Out]

integrate((b*tan(d*x + c) + a)^(5/3), x)

________________________________________________________________________________________

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))^(5/3),x, algorithm="fricas")

[Out]

Timed out

________________________________________________________________________________________

Sympy [F]  time = 0., size = 0, normalized size = 0. \begin{align*} \int \left (a + b \tan{\left (c + d x \right )}\right )^{\frac{5}{3}}\, dx \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate((a+b*tan(d*x+c))**(5/3),x)

[Out]

Integral((a + b*tan(c + d*x))**(5/3), x)

________________________________________________________________________________________

Giac [B]  time = 11.6869, size = 933, normalized size = 2.84 \begin{align*} \text{result too large to display} \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate((a+b*tan(d*x+c))^(5/3),x, algorithm="giac")

[Out]

-1/24*((-I*sqrt(3) + 1)*((216*I*a^5 - 1080*a^4*b - 2160*I*a^3*b^2 + 2160*a^2*b^3 + 1080*I*a*b^4 - 216*b^5)/(b^
3*d^3))^(1/3)*log(b*d^2*(sqrt(3) + I) + a*d^2*(-I*sqrt(3) + 1) + (I*a^2 - 2*a*b - I*b^2)^(1/3)*(b*tan(d*x + c)
+ a)^(1/3)*d^2*(sqrt(3) - I)) + (-I*sqrt(3) + 1)*((-216*I*a^5 - 1080*a^4*b + 2160*I*a^3*b^2 + 2160*a^2*b^3 -
1080*I*a*b^4 - 216*b^5)/(b^3*d^3))^(1/3)*log(-b*d^2*(sqrt(3) + I) + a*d^2*(-I*sqrt(3) + 1) - (-I*a^2 - 2*a*b +
I*b^2)^(1/3)*(b*tan(d*x + c) + a)^(1/3)*d^2*(sqrt(3) - I)) + (I*sqrt(3) + 1)*((-216*I*a^5 - 1080*a^4*b + 2160
*I*a^3*b^2 + 2160*a^2*b^3 - 1080*I*a*b^4 - 216*b^5)/(b^3*d^3))^(1/3)*log(b*d^2*(sqrt(3) - I) + a*d^2*(I*sqrt(3
) + 1) + (-I*a^2 - 2*a*b + I*b^2)^(1/3)*(b*tan(d*x + c) + a)^(1/3)*d^2*(sqrt(3) + I)) + (I*sqrt(3) + 1)*((216*
I*a^5 - 1080*a^4*b - 2160*I*a^3*b^2 + 2160*a^2*b^3 + 1080*I*a*b^4 - 216*b^5)/(b^3*d^3))^(1/3)*log(-b*d^2*(sqrt
(3) - I) + a*d^2*(I*sqrt(3) + 1) - (I*a^2 - 2*a*b - I*b^2)^(1/3)*(b*tan(d*x + c) + a)^(1/3)*d^2*(sqrt(3) + I))
- 2*((216*I*a^5 - 1080*a^4*b - 2160*I*a^3*b^2 + 2160*a^2*b^3 + 1080*I*a*b^4 - 216*b^5)/(b^3*d^3))^(1/3)*log(I
*a*d^2 - b*d^2 + (I*a^2 - 2*a*b - I*b^2)^(1/3)*(b*tan(d*x + c) + a)^(1/3)*d^2) - 2*((-216*I*a^5 - 1080*a^4*b +
2160*I*a^3*b^2 + 2160*a^2*b^3 - 1080*I*a*b^4 - 216*b^5)/(b^3*d^3))^(1/3)*log(-I*a*d^2 - b*d^2 + (-I*a^2 - 2*a
*b + I*b^2)^(1/3)*(b*tan(d*x + c) + a)^(1/3)*d^2) - 36*(b*tan(d*x + c) + a)^(2/3)/d)*b