3.475 \(\int \frac {A+B \tan (c+d x)}{\sqrt [3]{a+b \tan (c+d x)}} \, dx\)

Optimal. Leaf size=357 \[ \frac {\sqrt {3} (B+i A) \tan ^{-1}\left (\frac {1+\frac {2 \sqrt [3]{a+b \tan (c+d x)}}{\sqrt [3]{a-i b}}}{\sqrt {3}}\right )}{2 d \sqrt [3]{a-i b}}-\frac {\sqrt {3} (-B+i A) \tan ^{-1}\left (\frac {1+\frac {2 \sqrt [3]{a+b \tan (c+d x)}}{\sqrt [3]{a+i b}}}{\sqrt {3}}\right )}{2 d \sqrt [3]{a+i b}}+\frac {3 (B+i A) \log \left (-\sqrt [3]{a+b \tan (c+d x)}+\sqrt [3]{a-i b}\right )}{4 d \sqrt [3]{a-i b}}-\frac {3 (-B+i A) \log \left (-\sqrt [3]{a+b \tan (c+d x)}+\sqrt [3]{a+i b}\right )}{4 d \sqrt [3]{a+i b}}-\frac {(-B+i A) \log (\cos (c+d x))}{4 d \sqrt [3]{a+i b}}+\frac {(B+i A) \log (\cos (c+d x))}{4 d \sqrt [3]{a-i b}}-\frac {x (A-i B)}{4 \sqrt [3]{a-i b}}-\frac {x (A+i B)}{4 \sqrt [3]{a+i b}} \]

[Out]

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

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Rubi [A]  time = 0.28, antiderivative size = 357, normalized size of antiderivative = 1.00, number of steps used = 11, number of rules used = 6, integrand size = 25, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.240, Rules used = {3539, 3537, 55, 617, 204, 31} \[ \frac {\sqrt {3} (B+i A) \tan ^{-1}\left (\frac {1+\frac {2 \sqrt [3]{a+b \tan (c+d x)}}{\sqrt [3]{a-i b}}}{\sqrt {3}}\right )}{2 d \sqrt [3]{a-i b}}-\frac {\sqrt {3} (-B+i A) \tan ^{-1}\left (\frac {1+\frac {2 \sqrt [3]{a+b \tan (c+d x)}}{\sqrt [3]{a+i b}}}{\sqrt {3}}\right )}{2 d \sqrt [3]{a+i b}}+\frac {3 (B+i A) \log \left (-\sqrt [3]{a+b \tan (c+d x)}+\sqrt [3]{a-i b}\right )}{4 d \sqrt [3]{a-i b}}-\frac {3 (-B+i A) \log \left (-\sqrt [3]{a+b \tan (c+d x)}+\sqrt [3]{a+i b}\right )}{4 d \sqrt [3]{a+i b}}-\frac {(-B+i A) \log (\cos (c+d x))}{4 d \sqrt [3]{a+i b}}+\frac {(B+i A) \log (\cos (c+d x))}{4 d \sqrt [3]{a-i b}}-\frac {x (A-i B)}{4 \sqrt [3]{a-i b}}-\frac {x (A+i B)}{4 \sqrt [3]{a+i b}} \]

Antiderivative was successfully verified.

[In]

Int[(A + B*Tan[c + d*x])/(a + b*Tan[c + d*x])^(1/3),x]

[Out]

-((A - I*B)*x)/(4*(a - I*b)^(1/3)) - ((A + I*B)*x)/(4*(a + I*b)^(1/3)) + (Sqrt[3]*(I*A + B)*ArcTan[(1 + (2*(a
+ b*Tan[c + d*x])^(1/3))/(a - I*b)^(1/3))/Sqrt[3]])/(2*(a - I*b)^(1/3)*d) - (Sqrt[3]*(I*A - B)*ArcTan[(1 + (2*
(a + b*Tan[c + d*x])^(1/3))/(a + I*b)^(1/3))/Sqrt[3]])/(2*(a + I*b)^(1/3)*d) - ((I*A - B)*Log[Cos[c + d*x]])/(
4*(a + I*b)^(1/3)*d) + ((I*A + B)*Log[Cos[c + d*x]])/(4*(a - I*b)^(1/3)*d) + (3*(I*A + B)*Log[(a - I*b)^(1/3)
- (a + b*Tan[c + d*x])^(1/3)])/(4*(a - I*b)^(1/3)*d) - (3*(I*A - B)*Log[(a + I*b)^(1/3) - (a + b*Tan[c + d*x])
^(1/3)])/(4*(a + I*b)^(1/3)*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 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 {A+B \tan (c+d x)}{\sqrt [3]{a+b \tan (c+d x)}} \, dx &=\frac {1}{2} (A-i B) \int \frac {1+i \tan (c+d x)}{\sqrt [3]{a+b \tan (c+d x)}} \, dx+\frac {1}{2} (A+i B) \int \frac {1-i \tan (c+d x)}{\sqrt [3]{a+b \tan (c+d x)}} \, dx\\ &=-\frac {(i A-B) \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 {(i A+B) \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 {(A-i B) x}{4 \sqrt [3]{a-i b}}-\frac {(A+i B) x}{4 \sqrt [3]{a+i b}}-\frac {(i A-B) \log (\cos (c+d x))}{4 \sqrt [3]{a+i b} d}+\frac {(i A+B) \log (\cos (c+d x))}{4 \sqrt [3]{a-i b} d}-\frac {(3 (i A-B)) \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 {(3 (i A-B)) \operatorname {Subst}\left (\int \frac {1}{\sqrt [3]{a+i b}-x} \, dx,x,\sqrt [3]{a+b \tan (c+d x)}\right )}{4 \sqrt [3]{a+i b} d}+\frac {(3 (i A+B)) \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 {(3 (i A+B)) \operatorname {Subst}\left (\int \frac {1}{\sqrt [3]{a-i b}-x} \, dx,x,\sqrt [3]{a+b \tan (c+d x)}\right )}{4 \sqrt [3]{a-i b} d}\\ &=-\frac {(A-i B) x}{4 \sqrt [3]{a-i b}}-\frac {(A+i B) x}{4 \sqrt [3]{a+i b}}-\frac {(i A-B) \log (\cos (c+d x))}{4 \sqrt [3]{a+i b} d}+\frac {(i A+B) \log (\cos (c+d x))}{4 \sqrt [3]{a-i b} d}+\frac {3 (i A+B) \log \left (\sqrt [3]{a-i b}-\sqrt [3]{a+b \tan (c+d x)}\right )}{4 \sqrt [3]{a-i b} d}-\frac {3 (i A-B) \log \left (\sqrt [3]{a+i b}-\sqrt [3]{a+b \tan (c+d x)}\right )}{4 \sqrt [3]{a+i b} d}+\frac {(3 (i A-B)) \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 \sqrt [3]{a+i b} d}-\frac {(3 (i A+B)) \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 \sqrt [3]{a-i b} d}\\ &=-\frac {(A-i B) x}{4 \sqrt [3]{a-i b}}-\frac {(A+i B) x}{4 \sqrt [3]{a+i b}}+\frac {\sqrt {3} (i A+B) \tan ^{-1}\left (\frac {1+\frac {2 \sqrt [3]{a+b \tan (c+d x)}}{\sqrt [3]{a-i b}}}{\sqrt {3}}\right )}{2 \sqrt [3]{a-i b} d}-\frac {\sqrt {3} (i A-B) \tan ^{-1}\left (\frac {1+\frac {2 \sqrt [3]{a+b \tan (c+d x)}}{\sqrt [3]{a+i b}}}{\sqrt {3}}\right )}{2 \sqrt [3]{a+i b} d}-\frac {(i A-B) \log (\cos (c+d x))}{4 \sqrt [3]{a+i b} d}+\frac {(i A+B) \log (\cos (c+d x))}{4 \sqrt [3]{a-i b} d}+\frac {3 (i A+B) \log \left (\sqrt [3]{a-i b}-\sqrt [3]{a+b \tan (c+d x)}\right )}{4 \sqrt [3]{a-i b} d}-\frac {3 (i A-B) \log \left (\sqrt [3]{a+i b}-\sqrt [3]{a+b \tan (c+d x)}\right )}{4 \sqrt [3]{a+i b} d}\\ \end {align*}

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Mathematica [A]  time = 0.49, size = 227, normalized size = 0.64 \[ \frac {i \left (\frac {(A-i B) \left (2 \sqrt {3} \tan ^{-1}\left (\frac {1+\frac {2 \sqrt [3]{a+b \tan (c+d x)}}{\sqrt [3]{a-i b}}}{\sqrt {3}}\right )+3 \log \left (-\sqrt [3]{a+b \tan (c+d x)}+\sqrt [3]{a-i b}\right )-\log (\tan (c+d x)+i)\right )}{\sqrt [3]{a-i b}}-\frac {(A+i B) \left (2 \sqrt {3} \tan ^{-1}\left (\frac {1+\frac {2 \sqrt [3]{a+b \tan (c+d x)}}{\sqrt [3]{a+i b}}}{\sqrt {3}}\right )+3 \log \left (-\sqrt [3]{a+b \tan (c+d x)}+\sqrt [3]{a+i b}\right )-\log (-\tan (c+d x)+i)\right )}{\sqrt [3]{a+i b}}\right )}{4 d} \]

Antiderivative was successfully verified.

[In]

Integrate[(A + B*Tan[c + d*x])/(a + b*Tan[c + d*x])^(1/3),x]

[Out]

((I/4)*(((A - I*B)*(2*Sqrt[3]*ArcTan[(1 + (2*(a + b*Tan[c + d*x])^(1/3))/(a - I*b)^(1/3))/Sqrt[3]] - Log[I + T
an[c + d*x]] + 3*Log[(a - I*b)^(1/3) - (a + b*Tan[c + d*x])^(1/3)]))/(a - I*b)^(1/3) - ((A + I*B)*(2*Sqrt[3]*A
rcTan[(1 + (2*(a + b*Tan[c + d*x])^(1/3))/(a + I*b)^(1/3))/Sqrt[3]] - Log[I - Tan[c + d*x]] + 3*Log[(a + I*b)^
(1/3) - (a + b*Tan[c + d*x])^(1/3)]))/(a + I*b)^(1/3)))/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))/(a+b*tan(d*x+c))^(1/3),x, algorithm="fricas")

[Out]

Timed out

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

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate((A+B*tan(d*x+c))/(a+b*tan(d*x+c))^(1/3),x, algorithm="giac")

[Out]

integrate((B*tan(d*x + c) + A)/(b*tan(d*x + c) + a)^(1/3), x)

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maple [C]  time = 0.32, size = 72, normalized size = 0.20 \[ \frac {\munderset {\textit {\_R} =\RootOf \left (\textit {\_Z}^{6}-2 \textit {\_Z}^{3} a +a^{2}+b^{2}\right )}{\sum }\frac {\left (B \,\textit {\_R}^{4}+\left (A b -a B \right ) \textit {\_R} \right ) \ln \left (\left (a +b \tan \left (d x +c \right )\right )^{\frac {1}{3}}-\textit {\_R} \right )}{\textit {\_R}^{5}-\textit {\_R}^{2} a}}{2 d} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

1/2/d*sum((B*_R^4+(A*b-B*a)*_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 \[ \int \frac {B \tan \left (d x + c\right ) + A}{{\left (b \tan \left (d x + c\right ) + a\right )}^{\frac {1}{3}}}\,{d x} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate((A+B*tan(d*x+c))/(a+b*tan(d*x+c))^(1/3),x, algorithm="maxima")

[Out]

integrate((B*tan(d*x + c) + A)/(b*tan(d*x + c) + a)^(1/3), x)

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mupad [B]  time = 18.25, size = 3228, normalized size = 9.04 \[ \text {result too large to display} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

log(((((-B^6*b^2*d^6)^(1/2) + B^3*a*d^3)/(d^6*(a^2 + b^2)))^(2/3)*(972*a*b^4*(-B^6*b^2*d^6)^(1/2) - 972*B^3*b^
6*d^3 + 972*B^2*b^6*d^4*(a + b*tan(c + d*x))^(1/3)*(((-B^6*b^2*d^6)^(1/2) + B^3*a*d^3)/(d^6*(a^2 + b^2)))^(1/3
) - 972*B^2*a^2*b^4*d^4*(a + b*tan(c + d*x))^(1/3)*(((-B^6*b^2*d^6)^(1/2) + B^3*a*d^3)/(d^6*(a^2 + b^2)))^(1/3
)))/(4*d^6) + (243*B^5*a*b^4*(a + b*tan(c + d*x))^(1/3))/d^5)*((-64*B^6*b^2*d^6)^(1/2)/(64*(a^2*d^6 + b^2*d^6)
) + (B^3*a*d^3)/(8*(a^2*d^6 + b^2*d^6)))^(1/3) + log(((1944*a*b^4*(a^2 + b^2)*(((-A^6*a^2*d^6)^(1/2) + A^3*b*d
^3)/(d^6*(a^2 + b^2)))^(2/3) + (1944*A^2*b^4*(a^2 - b^2)*(a + b*tan(c + d*x))^(1/3))/d^2)*((-A^6*a^2*d^6)^(1/2
) + A^3*b*d^3))/(8*d^6*(a^2 + b^2)) + (243*A^5*b^5*(a + b*tan(c + d*x))^(1/3))/d^5)*((-64*A^6*a^2*d^6)^(1/2)/(
64*(a^2*d^6 + b^2*d^6)) + (A^3*b*d^3)/(8*(a^2*d^6 + b^2*d^6)))^(1/3) + log((243*B^5*a*b^4*(a + b*tan(c + d*x))
^(1/3))/d^5 - ((-(8*(-B^6*b^2*d^6)^(1/2) - 8*B^3*a*d^3)/(d^6*(a^2 + b^2)))^(2/3)*(972*a*b^4*(-B^6*b^2*d^6)^(1/
2) + 972*B^3*b^6*d^3 - 486*B^2*b^6*d^4*(a + b*tan(c + d*x))^(1/3)*(-(8*(-B^6*b^2*d^6)^(1/2) - 8*B^3*a*d^3)/(d^
6*(a^2 + b^2)))^(1/3) + 486*B^2*a^2*b^4*d^4*(a + b*tan(c + d*x))^(1/3)*(-(8*(-B^6*b^2*d^6)^(1/2) - 8*B^3*a*d^3
)/(d^6*(a^2 + b^2)))^(1/3)))/(16*d^6))*((B^3*a*d^3)/(8*(a^2*d^6 + b^2*d^6)) - (-64*B^6*b^2*d^6)^(1/2)/(64*(a^2
*d^6 + b^2*d^6)))^(1/3) + log((243*A^5*b^5*(a + b*tan(c + d*x))^(1/3))/d^5 - ((486*a*b^4*(a^2 + b^2)*(-(8*(-A^
6*a^2*d^6)^(1/2) - 8*A^3*b*d^3)/(d^6*(a^2 + b^2)))^(2/3) + (1944*A^2*b^4*(a^2 - b^2)*(a + b*tan(c + d*x))^(1/3
))/d^2)*((-A^6*a^2*d^6)^(1/2) - A^3*b*d^3))/(8*d^6*(a^2 + b^2)))*((A^3*b*d^3)/(8*(a^2*d^6 + b^2*d^6)) - (-64*A
^6*a^2*d^6)^(1/2)/(64*(a^2*d^6 + b^2*d^6)))^(1/3) - log(((486*a*b^4*((3^(1/2)*1i)/2 - 1/2)*(a^2 + b^2)*((8*(-A
^6*a^2*d^6)^(1/2) + 8*A^3*b*d^3)/(d^6*(a^2 + b^2)))^(2/3) + (1944*A^2*b^4*(a^2 - b^2)*(a + b*tan(c + d*x))^(1/
3))/d^2)*(8*(-A^6*a^2*d^6)^(1/2) + 8*A^3*b*d^3))/(64*d^6*(a^2 + b^2)) + (243*A^5*b^5*(a + b*tan(c + d*x))^(1/3
))/d^5)*((3^(1/2)*1i)/2 + 1/2)*(((64*A^6*b^2*d^6 - A^6*(64*a^2*d^6 + 64*b^2*d^6))^(1/2) + 8*A^3*b*d^3)/(64*(a^
2*d^6 + b^2*d^6)))^(1/3) + log((243*A^5*b^5*(a + b*tan(c + d*x))^(1/3))/d^5 - ((486*a*b^4*((3^(1/2)*1i)/2 + 1/
2)*(a^2 + b^2)*((8*(-A^6*a^2*d^6)^(1/2) + 8*A^3*b*d^3)/(d^6*(a^2 + b^2)))^(2/3) - (1944*A^2*b^4*(a^2 - b^2)*(a
+ b*tan(c + d*x))^(1/3))/d^2)*(8*(-A^6*a^2*d^6)^(1/2) + 8*A^3*b*d^3))/(64*d^6*(a^2 + b^2)))*((3^(1/2)*1i)/2 -
1/2)*(((64*A^6*b^2*d^6 - A^6*(64*a^2*d^6 + 64*b^2*d^6))^(1/2) + 8*A^3*b*d^3)/(64*(a^2*d^6 + b^2*d^6)))^(1/3)
- log((243*A^5*b^5*(a + b*tan(c + d*x))^(1/3))/d^5 - ((486*a*b^4*((3^(1/2)*1i)/2 - 1/2)*(a^2 + b^2)*(-(8*(-A^6
*a^2*d^6)^(1/2) - 8*A^3*b*d^3)/(d^6*(a^2 + b^2)))^(2/3) + (1944*A^2*b^4*(a^2 - b^2)*(a + b*tan(c + d*x))^(1/3)
)/d^2)*(8*(-A^6*a^2*d^6)^(1/2) - 8*A^3*b*d^3))/(64*d^6*(a^2 + b^2)))*((3^(1/2)*1i)/2 + 1/2)*(-((64*A^6*b^2*d^6
- A^6*(64*a^2*d^6 + 64*b^2*d^6))^(1/2) - 8*A^3*b*d^3)/(64*(a^2*d^6 + b^2*d^6)))^(1/3) + log(((486*a*b^4*((3^(
1/2)*1i)/2 + 1/2)*(a^2 + b^2)*(-(8*(-A^6*a^2*d^6)^(1/2) - 8*A^3*b*d^3)/(d^6*(a^2 + b^2)))^(2/3) - (1944*A^2*b^
4*(a^2 - b^2)*(a + b*tan(c + d*x))^(1/3))/d^2)*(8*(-A^6*a^2*d^6)^(1/2) - 8*A^3*b*d^3))/(64*d^6*(a^2 + b^2)) +
(243*A^5*b^5*(a + b*tan(c + d*x))^(1/3))/d^5)*((3^(1/2)*1i)/2 - 1/2)*(-((64*A^6*b^2*d^6 - A^6*(64*a^2*d^6 + 64
*b^2*d^6))^(1/2) - 8*A^3*b*d^3)/(64*(a^2*d^6 + b^2*d^6)))^(1/3) - log((243*B^5*a*b^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)*(486*a*b^4*((3^(1/2)*1i)/2 - 1/2)*(a^2 + b^2)*((8*(-
B^6*b^2*d^6)^(1/2) + 8*B^3*a*d^3)/(d^6*(a^2 + b^2)))^(2/3) - (1944*B^2*b^4*(a^2 - b^2)*(a + b*tan(c + d*x))^(1
/3))/d^2)*((8*(-B^6*b^2*d^6)^(1/2) + 8*B^3*a*d^3)/(d^6*(a^2 + b^2)))^(1/3))/4 + (972*B^3*b^4*(a^2 + b^2))/d^3)
*((8*(-B^6*b^2*d^6)^(1/2) + 8*B^3*a*d^3)/(d^6*(a^2 + b^2)))^(2/3))/16)*((3^(1/2)*1i)/2 + 1/2)*(((64*B^6*a^2*d^
6 - B^6*(64*a^2*d^6 + 64*b^2*d^6))^(1/2) + 8*B^3*a*d^3)/(64*(a^2*d^6 + b^2*d^6)))^(1/3) + log((((3^(1/2)*1i)/2
+ 1/2)*((((3^(1/2)*1i)/2 - 1/2)*(486*a*b^4*((3^(1/2)*1i)/2 + 1/2)*(a^2 + b^2)*((8*(-B^6*b^2*d^6)^(1/2) + 8*B^
3*a*d^3)/(d^6*(a^2 + b^2)))^(2/3) + (1944*B^2*b^4*(a^2 - b^2)*(a + b*tan(c + d*x))^(1/3))/d^2)*((8*(-B^6*b^2*d
^6)^(1/2) + 8*B^3*a*d^3)/(d^6*(a^2 + b^2)))^(1/3))/4 + (972*B^3*b^4*(a^2 + b^2))/d^3)*((8*(-B^6*b^2*d^6)^(1/2)
+ 8*B^3*a*d^3)/(d^6*(a^2 + b^2)))^(2/3))/16 + (243*B^5*a*b^4*(a + b*tan(c + d*x))^(1/3))/d^5)*((3^(1/2)*1i)/2
- 1/2)*(((64*B^6*a^2*d^6 - B^6*(64*a^2*d^6 + 64*b^2*d^6))^(1/2) + 8*B^3*a*d^3)/(64*(a^2*d^6 + b^2*d^6)))^(1/3
) - log((243*B^5*a*b^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)*(486
*a*b^4*((3^(1/2)*1i)/2 - 1/2)*(a^2 + b^2)*(-(8*(-B^6*b^2*d^6)^(1/2) - 8*B^3*a*d^3)/(d^6*(a^2 + b^2)))^(2/3) -
(1944*B^2*b^4*(a^2 - b^2)*(a + b*tan(c + d*x))^(1/3))/d^2)*(-(8*(-B^6*b^2*d^6)^(1/2) - 8*B^3*a*d^3)/(d^6*(a^2
+ b^2)))^(1/3))/4 + (972*B^3*b^4*(a^2 + b^2))/d^3)*(-(8*(-B^6*b^2*d^6)^(1/2) - 8*B^3*a*d^3)/(d^6*(a^2 + b^2)))
^(2/3))/16)*((3^(1/2)*1i)/2 + 1/2)*(-((64*B^6*a^2*d^6 - B^6*(64*a^2*d^6 + 64*b^2*d^6))^(1/2) - 8*B^3*a*d^3)/(6
4*(a^2*d^6 + b^2*d^6)))^(1/3) + log((((3^(1/2)*1i)/2 + 1/2)*((((3^(1/2)*1i)/2 - 1/2)*(486*a*b^4*((3^(1/2)*1i)/
2 + 1/2)*(a^2 + b^2)*(-(8*(-B^6*b^2*d^6)^(1/2) - 8*B^3*a*d^3)/(d^6*(a^2 + b^2)))^(2/3) + (1944*B^2*b^4*(a^2 -
b^2)*(a + b*tan(c + d*x))^(1/3))/d^2)*(-(8*(-B^6*b^2*d^6)^(1/2) - 8*B^3*a*d^3)/(d^6*(a^2 + b^2)))^(1/3))/4 + (
972*B^3*b^4*(a^2 + b^2))/d^3)*(-(8*(-B^6*b^2*d^6)^(1/2) - 8*B^3*a*d^3)/(d^6*(a^2 + b^2)))^(2/3))/16 + (243*B^5
*a*b^4*(a + b*tan(c + d*x))^(1/3))/d^5)*((3^(1/2)*1i)/2 - 1/2)*(-((64*B^6*a^2*d^6 - B^6*(64*a^2*d^6 + 64*b^2*d
^6))^(1/2) - 8*B^3*a*d^3)/(64*(a^2*d^6 + b^2*d^6)))^(1/3)

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

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

Integral((A + B*tan(c + d*x))/(a + b*tan(c + d*x))**(1/3), x)

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