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]

-1/4*(a-I*b)^(5/3)*x-1/4*(a+I*b)^(5/3)*x+1/4*I*(a-I*b)^(5/3)*ln(cos(d*x+c))/d-1/4*I*(a+I*b)^(5/3)*ln(cos(d*x+c
))/d+3/4*I*(a-I*b)^(5/3)*ln((a-I*b)^(1/3)-(a+b*tan(d*x+c))^(1/3))/d-3/4*I*(a+I*b)^(5/3)*ln((a+I*b)^(1/3)-(a+b*
tan(d*x+c))^(1/3))/d+1/2*I*(a-I*b)^(5/3)*arctan(1/3*(1+2*(a+b*tan(d*x+c))^(1/3)/(a-I*b)^(1/3))*3^(1/2))*3^(1/2
)/d-1/2*I*(a+I*b)^(5/3)*arctan(1/3*(1+2*(a+b*tan(d*x+c))^(1/3)/(a+I*b)^(1/3))*3^(1/2))*3^(1/2)/d+3/2*b*(a+b*ta
n(d*x+c))^(2/3)/d

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Rubi [A]  time = 0.38, antiderivative size = 329, normalized size of antiderivative = 1.00, number of steps used = 12, number of rules used = 7, integrand size = 14, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.500, 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 31

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

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 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 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 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 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 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]

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*}

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Mathematica [A]  time = 0.98, 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)

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fricas [F(-1)]  time = 0.00, size = 0, normalized size = 0.00 \[ \text {Timed out} \]

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

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giac [F]  time = 0.00, size = 0, normalized size = 0.00 \[ \int {\left (b \tan \left (d x + c\right ) + a\right )}^{\frac {5}{3}}\,{d x} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

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

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maple [C]  time = 0.22, size = 96, normalized size = 0.29 \[ \frac {3 b \left (a +b \tan \left (d x +c \right )\right )^{\frac {2}{3}}}{2 d}+\frac {b \left (\munderset {\textit {\_R} =\RootOf \left (\textit {\_Z}^{6}-2 a \,\textit {\_Z}^{3}+a^{2}+b^{2}\right )}{\sum }\frac {\left (2 a \,\textit {\_R}^{4}+\left (-a^{2}-b^{2}\right ) \textit {\_R} \right ) \ln \left (\left (a +b \tan \left (d x +c \right )\right )^{\frac {1}{3}}-\textit {\_R} \right )}{\textit {\_R}^{5}-\textit {\_R}^{2} a}\right )}{2 d} \]

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))

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maxima [F]  time = 0.00, size = 0, normalized size = 0.00 \[ \int {\left (b \tan \left (d x + c\right ) + a\right )}^{\frac {5}{3}}\,{d x} \]

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)

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mupad [B]  time = 10.09, size = 1540, normalized size = 4.68 \[ \text {result too large to display} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

int((a + b*tan(c + d*x))^(5/3),x)

[Out]

log(((-(a*1i + b)^5/d^3)^(2/3)*(((-(a*1i + b)^5/d^3)^(1/3)*(1944*a*b^4*(-(a*1i + b)^5/d^3)^(2/3)*(a^2 + b^2) +
(1944*b^4*(a^2 - b^2)*(a^2 + b^2)^2*(a + b*tan(c + d*x))^(1/3))/d^2))/2 + (1944*a*b^5*(3*a^6 + 3*b^6 - 7*a^2*
b^4 - 7*a^4*b^2))/d^3))/4 + (243*b^5*(3*a^2 - b^2)*(a^2 + b^2)^4*(a + b*tan(c + d*x))^(1/3))/d^5)*(-(a*b^4*5i
+ 5*a^4*b + a^5*1i + b^5 - 10*a^2*b^3 - a^3*b^2*10i)/(8*d^3))^(1/3) + log(((((1944*a*b^4*(a^2 + b^2)*(((a + b*
1i)^5*1i)/d^3)^(2/3) + (1944*b^4*(a^2 - b^2)*(a^2 + b^2)^2*(a + b*tan(c + d*x))^(1/3))/d^2)*(((a + b*1i)^5*1i)
/d^3)^(1/3))/2 + (1944*a*b^5*(3*a^6 + 3*b^6 - 7*a^2*b^4 - 7*a^4*b^2))/d^3)*(((a + b*1i)^5*1i)/d^3)^(2/3))/4 +
(243*b^5*(3*a^2 - b^2)*(a^2 + b^2)^4*(a + b*tan(c + d*x))^(1/3))/d^5)*((a*b^4*5i - 5*a^4*b + a^5*1i - b^5 + 10
*a^2*b^3 - a^3*b^2*10i)/(8*d^3))^(1/3) - log((243*b^5*(3*a^2 - b^2)*(a^2 + b^2)^4*(a + b*tan(c + d*x))^(1/3))/
d^5 - (((3^(1/2)*1i)/2 - 1/2)*((((3^(1/2)*1i)/2 + 1/2)*((1944*b^4*(a^2 - b^2)*(a^2 + b^2)^2*(a + b*tan(c + d*x
))^(1/3))/d^2 + 1944*a*b^4*((3^(1/2)*1i)/2 - 1/2)*(a^2 + b^2)*(((a + b*1i)^5*1i)/d^3)^(2/3))*(((a + b*1i)^5*1i
)/d^3)^(1/3))/2 - (1944*a*b^5*(3*a^6 + 3*b^6 - 7*a^2*b^4 - 7*a^4*b^2))/d^3)*(((a + b*1i)^5*1i)/d^3)^(2/3))/4)*
((3^(1/2)*1i)/2 + 1/2)*((a*b^4*5i - 5*a^4*b + a^5*1i - b^5 + 10*a^2*b^3 - a^3*b^2*10i)/(8*d^3))^(1/3) + log((2
43*b^5*(3*a^2 - b^2)*(a^2 + b^2)^4*(a + b*tan(c + d*x))^(1/3))/d^5 - (((3^(1/2)*1i)/2 + 1/2)*((((3^(1/2)*1i)/2
- 1/2)*((1944*b^4*(a^2 - b^2)*(a^2 + b^2)^2*(a + b*tan(c + d*x))^(1/3))/d^2 - 1944*a*b^4*((3^(1/2)*1i)/2 + 1/
2)*(a^2 + b^2)*(((a + b*1i)^5*1i)/d^3)^(2/3))*(((a + b*1i)^5*1i)/d^3)^(1/3))/2 + (1944*a*b^5*(3*a^6 + 3*b^6 -
7*a^2*b^4 - 7*a^4*b^2))/d^3)*(((a + b*1i)^5*1i)/d^3)^(2/3))/4)*((3^(1/2)*1i)/2 - 1/2)*((a*b^4*5i - 5*a^4*b + a
^5*1i - b^5 + 10*a^2*b^3 - a^3*b^2*10i)/(8*d^3))^(1/3) - log((243*b^5*(3*a^2 - b^2)*(a^2 + b^2)^4*(a + b*tan(c
+ d*x))^(1/3))/d^5 - ((-(a*1i + b)^5/d^3)^(2/3)*((3^(1/2)*1i)/2 - 1/2)*(((-(a*1i + b)^5/d^3)^(1/3)*((3^(1/2)*
1i)/2 + 1/2)*((1944*b^4*(a^2 - b^2)*(a^2 + b^2)^2*(a + b*tan(c + d*x))^(1/3))/d^2 + 1944*a*b^4*(-(a*1i + b)^5/
d^3)^(2/3)*((3^(1/2)*1i)/2 - 1/2)*(a^2 + b^2)))/2 - (1944*a*b^5*(3*a^6 + 3*b^6 - 7*a^2*b^4 - 7*a^4*b^2))/d^3))
/4)*((3^(1/2)*1i)/2 + 1/2)*(-(a*b^4*5i + 5*a^4*b + a^5*1i + b^5 - 10*a^2*b^3 - a^3*b^2*10i)/(8*d^3))^(1/3) + l
og((243*b^5*(3*a^2 - b^2)*(a^2 + b^2)^4*(a + b*tan(c + d*x))^(1/3))/d^5 - ((-(a*1i + b)^5/d^3)^(2/3)*((3^(1/2)
*1i)/2 + 1/2)*(((-(a*1i + b)^5/d^3)^(1/3)*((3^(1/2)*1i)/2 - 1/2)*((1944*b^4*(a^2 - b^2)*(a^2 + b^2)^2*(a + b*t
an(c + d*x))^(1/3))/d^2 - 1944*a*b^4*(-(a*1i + b)^5/d^3)^(2/3)*((3^(1/2)*1i)/2 + 1/2)*(a^2 + b^2)))/2 + (1944*
a*b^5*(3*a^6 + 3*b^6 - 7*a^2*b^4 - 7*a^4*b^2))/d^3))/4)*((3^(1/2)*1i)/2 - 1/2)*(-(a*b^4*5i + 5*a^4*b + a^5*1i
+ b^5 - 10*a^2*b^3 - a^3*b^2*10i)/(8*d^3))^(1/3) + (3*b*(a + b*tan(c + d*x))^(2/3))/(2*d)

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sympy [F]  time = 0.00, size = 0, normalized size = 0.00 \[ \int \left (a + b \tan {\left (c + d x \right )}\right )^{\frac {5}{3}}\, dx \]

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)

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