3.695 \(\int \frac {1}{(a+b \tan (c+d x))^{5/3}} \, dx\)

Optimal. Leaf size=338 \[ -\frac {3 b}{2 d \left (a^2+b^2\right ) (a+b \tan (c+d x))^{2/3}}-\frac {i \sqrt {3} \tan ^{-1}\left (\frac {1+\frac {2 \sqrt [3]{a+b \tan (c+d x)}}{\sqrt [3]{a-i b}}}{\sqrt {3}}\right )}{2 d (a-i b)^{5/3}}+\frac {i \sqrt {3} \tan ^{-1}\left (\frac {1+\frac {2 \sqrt [3]{a+b \tan (c+d x)}}{\sqrt [3]{a+i b}}}{\sqrt {3}}\right )}{2 d (a+i b)^{5/3}}+\frac {3 i \log \left (-\sqrt [3]{a+b \tan (c+d x)}+\sqrt [3]{a-i b}\right )}{4 d (a-i b)^{5/3}}-\frac {3 i \log \left (-\sqrt [3]{a+b \tan (c+d x)}+\sqrt [3]{a+i b}\right )}{4 d (a+i b)^{5/3}}+\frac {i \log (\cos (c+d x))}{4 d (a-i b)^{5/3}}-\frac {i \log (\cos (c+d x))}{4 d (a+i b)^{5/3}}-\frac {x}{4 (a-i b)^{5/3}}-\frac {x}{4 (a+i b)^{5/3}} \]

[Out]

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

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Rubi [A]  time = 0.36, antiderivative size = 338, 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 = {3483, 3539, 3537, 57, 617, 204, 31} \[ -\frac {3 b}{2 d \left (a^2+b^2\right ) (a+b \tan (c+d x))^{2/3}}-\frac {i \sqrt {3} \tan ^{-1}\left (\frac {1+\frac {2 \sqrt [3]{a+b \tan (c+d x)}}{\sqrt [3]{a-i b}}}{\sqrt {3}}\right )}{2 d (a-i b)^{5/3}}+\frac {i \sqrt {3} \tan ^{-1}\left (\frac {1+\frac {2 \sqrt [3]{a+b \tan (c+d x)}}{\sqrt [3]{a+i b}}}{\sqrt {3}}\right )}{2 d (a+i b)^{5/3}}+\frac {3 i \log \left (-\sqrt [3]{a+b \tan (c+d x)}+\sqrt [3]{a-i b}\right )}{4 d (a-i b)^{5/3}}-\frac {3 i \log \left (-\sqrt [3]{a+b \tan (c+d x)}+\sqrt [3]{a+i b}\right )}{4 d (a+i b)^{5/3}}+\frac {i \log (\cos (c+d x))}{4 d (a-i b)^{5/3}}-\frac {i \log (\cos (c+d x))}{4 d (a+i b)^{5/3}}-\frac {x}{4 (a-i b)^{5/3}}-\frac {x}{4 (a+i b)^{5/3}} \]

Antiderivative was successfully verified.

[In]

Int[(a + b*Tan[c + d*x])^(-5/3),x]

[Out]

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

Rule 31

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

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 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 3483

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

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Mathematica [C]  time = 0.16, size = 106, normalized size = 0.31 \[ \frac {3 i \left ((a+i b) \, _2F_1\left (-\frac {2}{3},1;\frac {1}{3};\frac {a+b \tan (c+d x)}{a-i b}\right )-(a-i b) \, _2F_1\left (-\frac {2}{3},1;\frac {1}{3};\frac {a+b \tan (c+d x)}{a+i b}\right )\right )}{4 d \left (a^2+b^2\right ) (a+b \tan (c+d x))^{2/3}} \]

Antiderivative was successfully verified.

[In]

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

[Out]

(((3*I)/4)*((a + I*b)*Hypergeometric2F1[-2/3, 1, 1/3, (a + b*Tan[c + d*x])/(a - I*b)] - (a - I*b)*Hypergeometr
ic2F1[-2/3, 1, 1/3, (a + b*Tan[c + d*x])/(a + I*b)]))/((a^2 + b^2)*d*(a + b*Tan[c + d*x])^(2/3))

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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(1/(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 \[ \int \frac {1}{{\left (b \tan \left (d x + c\right ) + a\right )}^{\frac {5}{3}}}\,{d x} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate(1/(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 = 0.21, size = 103, normalized size = 0.30 \[ -\frac {3 b}{2 \left (a^{2}+b^{2}\right ) d \left (a +b \tan \left (d x +c \right )\right )^{\frac {2}{3}}}+\frac {b \left (\munderset {\textit {\_R} =\RootOf \left (\textit {\_Z}^{6}-2 a \,\textit {\_Z}^{3}+a^{2}+b^{2}\right )}{\sum }\frac {\left (-\textit {\_R}^{3}+2 a \right ) \ln \left (\left (a +b \tan \left (d x +c \right )\right )^{\frac {1}{3}}-\textit {\_R} \right )}{\textit {\_R}^{5}-\textit {\_R}^{2} a}\right )}{2 d \left (a^{2}+b^{2}\right )} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

int(1/(a+b*tan(d*x+c))^(5/3),x)

[Out]

-3/2*b/(a^2+b^2)/d/(a+b*tan(d*x+c))^(2/3)+1/2/d*b/(a^2+b^2)*sum((-_R^3+2*a)/(_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 \[ \int \frac {1}{{\left (b \tan \left (d x + c\right ) + a\right )}^{\frac {5}{3}}}\,{d x} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate(1/(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 = 6.44, size = 4348, normalized size = 12.86 \[ \text {result too large to display} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

int(1/(a + b*tan(c + d*x))^(5/3),x)

[Out]

(log((a + b*tan(c + d*x))^(1/3)*(486*b^18*d^5 + 2430*a^2*b^16*d^5 + 4374*a^4*b^14*d^5 + 2430*a^6*b^12*d^5 - 24
30*a^8*b^10*d^5 - 4374*a^10*b^8*d^5 - 2430*a^12*b^6*d^5 - 486*a^14*b^4*d^5) + ((1/(a^5*d^3*1i + b^5*d^3 + a*b^
4*d^3*5i + 5*a^4*b*d^3 - 10*a^2*b^3*d^3 - a^3*b^2*d^3*10i))^(1/3)*(((((1/(a^5*d^3*1i + b^5*d^3 + a*b^4*d^3*5i
+ 5*a^4*b*d^3 - 10*a^2*b^3*d^3 - a^3*b^2*d^3*10i))^(1/3)*(7776*a*b^24*d^9 + 77760*a^3*b^22*d^9 + 349920*a^5*b^
20*d^9 + 933120*a^7*b^18*d^9 + 1632960*a^9*b^16*d^9 + 1959552*a^11*b^14*d^9 + 1632960*a^13*b^12*d^9 + 933120*a
^15*b^10*d^9 + 349920*a^17*b^8*d^9 + 77760*a^19*b^6*d^9 + 7776*a^21*b^4*d^9))/2 + (a + b*tan(c + d*x))^(1/3)*(
108864*a^6*b^17*d^8 - 19440*a^2*b^21*d^8 - 15552*a^4*b^19*d^8 - 3888*b^23*d^8 + 381024*a^8*b^15*d^8 + 598752*a
^10*b^13*d^8 + 544320*a^12*b^11*d^8 + 295488*a^14*b^9*d^8 + 89424*a^16*b^7*d^8 + 11664*a^18*b^5*d^8))*(1/(a^5*
d^3*1i + b^5*d^3 + a*b^4*d^3*5i + 5*a^4*b*d^3 - 10*a^2*b^3*d^3 - a^3*b^2*d^3*10i))^(2/3))/4 + 3888*a*b^19*d^6
+ 19440*a^3*b^17*d^6 + 34992*a^5*b^15*d^6 + 19440*a^7*b^13*d^6 - 19440*a^9*b^11*d^6 - 34992*a^11*b^9*d^6 - 194
40*a^13*b^7*d^6 - 3888*a^15*b^5*d^6))/2)*(1/(a^5*d^3*1i + b^5*d^3 + a*b^4*d^3*5i + 5*a^4*b*d^3 - 10*a^2*b^3*d^
3 - a^3*b^2*d^3*10i))^(1/3))/2 + log((a + b*tan(c + d*x))^(1/3)*(486*b^18*d^5 + 2430*a^2*b^16*d^5 + 4374*a^4*b
^14*d^5 + 2430*a^6*b^12*d^5 - 2430*a^8*b^10*d^5 - 4374*a^10*b^8*d^5 - 2430*a^12*b^6*d^5 - 486*a^14*b^4*d^5) +
(1i/(8*(a^5*d^3 + b^5*d^3*1i + 5*a*b^4*d^3 + a^4*b*d^3*5i - a^2*b^3*d^3*10i - 10*a^3*b^2*d^3)))^(1/3)*(((1i/(8
*(a^5*d^3 + b^5*d^3*1i + 5*a*b^4*d^3 + a^4*b*d^3*5i - a^2*b^3*d^3*10i - 10*a^3*b^2*d^3)))^(1/3)*(7776*a*b^24*d
^9 + 77760*a^3*b^22*d^9 + 349920*a^5*b^20*d^9 + 933120*a^7*b^18*d^9 + 1632960*a^9*b^16*d^9 + 1959552*a^11*b^14
*d^9 + 1632960*a^13*b^12*d^9 + 933120*a^15*b^10*d^9 + 349920*a^17*b^8*d^9 + 77760*a^19*b^6*d^9 + 7776*a^21*b^4
*d^9) + (a + b*tan(c + d*x))^(1/3)*(108864*a^6*b^17*d^8 - 19440*a^2*b^21*d^8 - 15552*a^4*b^19*d^8 - 3888*b^23*
d^8 + 381024*a^8*b^15*d^8 + 598752*a^10*b^13*d^8 + 544320*a^12*b^11*d^8 + 295488*a^14*b^9*d^8 + 89424*a^16*b^7
*d^8 + 11664*a^18*b^5*d^8))*(1i/(8*(a^5*d^3 + b^5*d^3*1i + 5*a*b^4*d^3 + a^4*b*d^3*5i - a^2*b^3*d^3*10i - 10*a
^3*b^2*d^3)))^(2/3) + 3888*a*b^19*d^6 + 19440*a^3*b^17*d^6 + 34992*a^5*b^15*d^6 + 19440*a^7*b^13*d^6 - 19440*a
^9*b^11*d^6 - 34992*a^11*b^9*d^6 - 19440*a^13*b^7*d^6 - 3888*a^15*b^5*d^6))*(1i/(8*(a^5*d^3 + b^5*d^3*1i + 5*a
*b^4*d^3 + a^4*b*d^3*5i - a^2*b^3*d^3*10i - 10*a^3*b^2*d^3)))^(1/3) + (log((a + b*tan(c + d*x))^(1/3)*(486*b^1
8*d^5 + 2430*a^2*b^16*d^5 + 4374*a^4*b^14*d^5 + 2430*a^6*b^12*d^5 - 2430*a^8*b^10*d^5 - 4374*a^10*b^8*d^5 - 24
30*a^12*b^6*d^5 - 486*a^14*b^4*d^5) + ((3^(1/2)*1i - 1)*(1i/(8*(a^5*d^3 + b^5*d^3*1i + 5*a*b^4*d^3 + a^4*b*d^3
*5i - a^2*b^3*d^3*10i - 10*a^3*b^2*d^3)))^(1/3)*(3888*a*b^19*d^6 + 19440*a^3*b^17*d^6 + 34992*a^5*b^15*d^6 + 1
9440*a^7*b^13*d^6 - 19440*a^9*b^11*d^6 - 34992*a^11*b^9*d^6 - 19440*a^13*b^7*d^6 - 3888*a^15*b^5*d^6 + ((3^(1/
2)*1i - 1)^2*(1i/(8*(a^5*d^3 + b^5*d^3*1i + 5*a*b^4*d^3 + a^4*b*d^3*5i - a^2*b^3*d^3*10i - 10*a^3*b^2*d^3)))^(
2/3)*((a + b*tan(c + d*x))^(1/3)*(108864*a^6*b^17*d^8 - 19440*a^2*b^21*d^8 - 15552*a^4*b^19*d^8 - 3888*b^23*d^
8 + 381024*a^8*b^15*d^8 + 598752*a^10*b^13*d^8 + 544320*a^12*b^11*d^8 + 295488*a^14*b^9*d^8 + 89424*a^16*b^7*d
^8 + 11664*a^18*b^5*d^8) + ((3^(1/2)*1i - 1)*(1i/(8*(a^5*d^3 + b^5*d^3*1i + 5*a*b^4*d^3 + a^4*b*d^3*5i - a^2*b
^3*d^3*10i - 10*a^3*b^2*d^3)))^(1/3)*(7776*a*b^24*d^9 + 77760*a^3*b^22*d^9 + 349920*a^5*b^20*d^9 + 933120*a^7*
b^18*d^9 + 1632960*a^9*b^16*d^9 + 1959552*a^11*b^14*d^9 + 1632960*a^13*b^12*d^9 + 933120*a^15*b^10*d^9 + 34992
0*a^17*b^8*d^9 + 77760*a^19*b^6*d^9 + 7776*a^21*b^4*d^9))/2))/4))/2)*(3^(1/2)*1i - 1)*(1i/(8*(a^5*d^3 + b^5*d^
3*1i + 5*a*b^4*d^3 + a^4*b*d^3*5i - a^2*b^3*d^3*10i - 10*a^3*b^2*d^3)))^(1/3))/2 - (log((a + b*tan(c + d*x))^(
1/3)*(486*b^18*d^5 + 2430*a^2*b^16*d^5 + 4374*a^4*b^14*d^5 + 2430*a^6*b^12*d^5 - 2430*a^8*b^10*d^5 - 4374*a^10
*b^8*d^5 - 2430*a^12*b^6*d^5 - 486*a^14*b^4*d^5) - ((3^(1/2)*1i + 1)*(1i/(8*(a^5*d^3 + b^5*d^3*1i + 5*a*b^4*d^
3 + a^4*b*d^3*5i - a^2*b^3*d^3*10i - 10*a^3*b^2*d^3)))^(1/3)*(3888*a*b^19*d^6 + 19440*a^3*b^17*d^6 + 34992*a^5
*b^15*d^6 + 19440*a^7*b^13*d^6 - 19440*a^9*b^11*d^6 - 34992*a^11*b^9*d^6 - 19440*a^13*b^7*d^6 - 3888*a^15*b^5*
d^6 + ((3^(1/2)*1i + 1)^2*(1i/(8*(a^5*d^3 + b^5*d^3*1i + 5*a*b^4*d^3 + a^4*b*d^3*5i - a^2*b^3*d^3*10i - 10*a^3
*b^2*d^3)))^(2/3)*((a + b*tan(c + d*x))^(1/3)*(108864*a^6*b^17*d^8 - 19440*a^2*b^21*d^8 - 15552*a^4*b^19*d^8 -
3888*b^23*d^8 + 381024*a^8*b^15*d^8 + 598752*a^10*b^13*d^8 + 544320*a^12*b^11*d^8 + 295488*a^14*b^9*d^8 + 894
24*a^16*b^7*d^8 + 11664*a^18*b^5*d^8) - ((3^(1/2)*1i + 1)*(1i/(8*(a^5*d^3 + b^5*d^3*1i + 5*a*b^4*d^3 + a^4*b*d
^3*5i - a^2*b^3*d^3*10i - 10*a^3*b^2*d^3)))^(1/3)*(7776*a*b^24*d^9 + 77760*a^3*b^22*d^9 + 349920*a^5*b^20*d^9
+ 933120*a^7*b^18*d^9 + 1632960*a^9*b^16*d^9 + 1959552*a^11*b^14*d^9 + 1632960*a^13*b^12*d^9 + 933120*a^15*b^1
0*d^9 + 349920*a^17*b^8*d^9 + 77760*a^19*b^6*d^9 + 7776*a^21*b^4*d^9))/2))/4))/2)*(3^(1/2)*1i + 1)*(1i/(8*(a^5
*d^3 + b^5*d^3*1i + 5*a*b^4*d^3 + a^4*b*d^3*5i - a^2*b^3*d^3*10i - 10*a^3*b^2*d^3)))^(1/3))/2 + (log((a + b*ta
n(c + d*x))^(1/3)*(486*b^18*d^5 + 2430*a^2*b^16*d^5 + 4374*a^4*b^14*d^5 + 2430*a^6*b^12*d^5 - 2430*a^8*b^10*d^
5 - 4374*a^10*b^8*d^5 - 2430*a^12*b^6*d^5 - 486*a^14*b^4*d^5) + ((3^(1/2)*1i - 1)*(1/(a^5*d^3*1i + b^5*d^3 + a
*b^4*d^3*5i + 5*a^4*b*d^3 - 10*a^2*b^3*d^3 - a^3*b^2*d^3*10i))^(1/3)*(((3^(1/2)*1i - 1)^2*(1/(a^5*d^3*1i + b^5
*d^3 + a*b^4*d^3*5i + 5*a^4*b*d^3 - 10*a^2*b^3*d^3 - a^3*b^2*d^3*10i))^(2/3)*((a + b*tan(c + d*x))^(1/3)*(1088
64*a^6*b^17*d^8 - 19440*a^2*b^21*d^8 - 15552*a^4*b^19*d^8 - 3888*b^23*d^8 + 381024*a^8*b^15*d^8 + 598752*a^10*
b^13*d^8 + 544320*a^12*b^11*d^8 + 295488*a^14*b^9*d^8 + 89424*a^16*b^7*d^8 + 11664*a^18*b^5*d^8) + ((3^(1/2)*1
i - 1)*(1/(a^5*d^3*1i + b^5*d^3 + a*b^4*d^3*5i + 5*a^4*b*d^3 - 10*a^2*b^3*d^3 - a^3*b^2*d^3*10i))^(1/3)*(7776*
a*b^24*d^9 + 77760*a^3*b^22*d^9 + 349920*a^5*b^20*d^9 + 933120*a^7*b^18*d^9 + 1632960*a^9*b^16*d^9 + 1959552*a
^11*b^14*d^9 + 1632960*a^13*b^12*d^9 + 933120*a^15*b^10*d^9 + 349920*a^17*b^8*d^9 + 77760*a^19*b^6*d^9 + 7776*
a^21*b^4*d^9))/4))/16 + 3888*a*b^19*d^6 + 19440*a^3*b^17*d^6 + 34992*a^5*b^15*d^6 + 19440*a^7*b^13*d^6 - 19440
*a^9*b^11*d^6 - 34992*a^11*b^9*d^6 - 19440*a^13*b^7*d^6 - 3888*a^15*b^5*d^6))/4)*(3^(1/2)*1i - 1)*(1/(a^5*d^3*
1i + b^5*d^3 + a*b^4*d^3*5i + 5*a^4*b*d^3 - 10*a^2*b^3*d^3 - a^3*b^2*d^3*10i))^(1/3))/4 - (log((a + b*tan(c +
d*x))^(1/3)*(486*b^18*d^5 + 2430*a^2*b^16*d^5 + 4374*a^4*b^14*d^5 + 2430*a^6*b^12*d^5 - 2430*a^8*b^10*d^5 - 43
74*a^10*b^8*d^5 - 2430*a^12*b^6*d^5 - 486*a^14*b^4*d^5) - ((3^(1/2)*1i + 1)*(1/(a^5*d^3*1i + b^5*d^3 + a*b^4*d
^3*5i + 5*a^4*b*d^3 - 10*a^2*b^3*d^3 - a^3*b^2*d^3*10i))^(1/3)*(((3^(1/2)*1i + 1)^2*(1/(a^5*d^3*1i + b^5*d^3 +
a*b^4*d^3*5i + 5*a^4*b*d^3 - 10*a^2*b^3*d^3 - a^3*b^2*d^3*10i))^(2/3)*((a + b*tan(c + d*x))^(1/3)*(108864*a^6
*b^17*d^8 - 19440*a^2*b^21*d^8 - 15552*a^4*b^19*d^8 - 3888*b^23*d^8 + 381024*a^8*b^15*d^8 + 598752*a^10*b^13*d
^8 + 544320*a^12*b^11*d^8 + 295488*a^14*b^9*d^8 + 89424*a^16*b^7*d^8 + 11664*a^18*b^5*d^8) - ((3^(1/2)*1i + 1)
*(1/(a^5*d^3*1i + b^5*d^3 + a*b^4*d^3*5i + 5*a^4*b*d^3 - 10*a^2*b^3*d^3 - a^3*b^2*d^3*10i))^(1/3)*(7776*a*b^24
*d^9 + 77760*a^3*b^22*d^9 + 349920*a^5*b^20*d^9 + 933120*a^7*b^18*d^9 + 1632960*a^9*b^16*d^9 + 1959552*a^11*b^
14*d^9 + 1632960*a^13*b^12*d^9 + 933120*a^15*b^10*d^9 + 349920*a^17*b^8*d^9 + 77760*a^19*b^6*d^9 + 7776*a^21*b
^4*d^9))/4))/16 + 3888*a*b^19*d^6 + 19440*a^3*b^17*d^6 + 34992*a^5*b^15*d^6 + 19440*a^7*b^13*d^6 - 19440*a^9*b
^11*d^6 - 34992*a^11*b^9*d^6 - 19440*a^13*b^7*d^6 - 3888*a^15*b^5*d^6))/4)*(3^(1/2)*1i + 1)*(1/(a^5*d^3*1i + b
^5*d^3 + a*b^4*d^3*5i + 5*a^4*b*d^3 - 10*a^2*b^3*d^3 - a^3*b^2*d^3*10i))^(1/3))/4 - (3*b)/(2*d*(a^2 + b^2)*(a
+ b*tan(c + d*x))^(2/3))

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sympy [F]  time = 0.00, size = 0, normalized size = 0.00 \[ \int \frac {1}{\left (a + b \tan {\left (c + d x \right )}\right )^{\frac {5}{3}}}\, dx \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate(1/(a+b*tan(d*x+c))**(5/3),x)

[Out]

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

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