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

Optimal. Leaf size=377 \[ -\frac {\sqrt {3} \sqrt [3]{a-i b} (B+i A) \tan ^{-1}\left (\frac {1+\frac {2 \sqrt [3]{a+b \tan (c+d x)}}{\sqrt [3]{a-i b}}}{\sqrt {3}}\right )}{2 d}+\frac {\sqrt {3} \sqrt [3]{a+i b} (-B+i A) \tan ^{-1}\left (\frac {1+\frac {2 \sqrt [3]{a+b \tan (c+d x)}}{\sqrt [3]{a+i b}}}{\sqrt {3}}\right )}{2 d}+\frac {3 \sqrt [3]{a-i b} (B+i A) \log \left (-\sqrt [3]{a+b \tan (c+d x)}+\sqrt [3]{a-i b}\right )}{4 d}-\frac {3 \sqrt [3]{a+i b} (-B+i A) \log \left (-\sqrt [3]{a+b \tan (c+d x)}+\sqrt [3]{a+i b}\right )}{4 d}-\frac {\sqrt [3]{a+i b} (-B+i A) \log (\cos (c+d x))}{4 d}+\frac {\sqrt [3]{a-i b} (B+i A) \log (\cos (c+d x))}{4 d}-\frac {1}{4} x \sqrt [3]{a-i b} (A-i B)-\frac {1}{4} x \sqrt [3]{a+i b} (A+i B)+\frac {3 B \sqrt [3]{a+b \tan (c+d x)}}{d} \]

[Out]

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

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Rubi [A]  time = 0.41, antiderivative size = 377, normalized size of antiderivative = 1.00, number of steps used = 12, number of rules used = 7, integrand size = 25, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.280, Rules used = {3528, 3539, 3537, 57, 617, 204, 31} \[ -\frac {\sqrt {3} \sqrt [3]{a-i b} (B+i A) \tan ^{-1}\left (\frac {1+\frac {2 \sqrt [3]{a+b \tan (c+d x)}}{\sqrt [3]{a-i b}}}{\sqrt {3}}\right )}{2 d}+\frac {\sqrt {3} \sqrt [3]{a+i b} (-B+i A) \tan ^{-1}\left (\frac {1+\frac {2 \sqrt [3]{a+b \tan (c+d x)}}{\sqrt [3]{a+i b}}}{\sqrt {3}}\right )}{2 d}+\frac {3 \sqrt [3]{a-i b} (B+i A) \log \left (-\sqrt [3]{a+b \tan (c+d x)}+\sqrt [3]{a-i b}\right )}{4 d}-\frac {3 \sqrt [3]{a+i b} (-B+i A) \log \left (-\sqrt [3]{a+b \tan (c+d x)}+\sqrt [3]{a+i b}\right )}{4 d}-\frac {\sqrt [3]{a+i b} (-B+i A) \log (\cos (c+d x))}{4 d}+\frac {\sqrt [3]{a-i b} (B+i A) \log (\cos (c+d x))}{4 d}-\frac {1}{4} x \sqrt [3]{a-i b} (A-i B)-\frac {1}{4} x \sqrt [3]{a+i b} (A+i B)+\frac {3 B \sqrt [3]{a+b \tan (c+d x)}}{d} \]

Antiderivative was successfully verified.

[In]

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

[Out]

-((a - I*b)^(1/3)*(A - I*B)*x)/4 - ((a + I*b)^(1/3)*(A + I*B)*x)/4 - (Sqrt[3]*(a - I*b)^(1/3)*(I*A + B)*ArcTan
[(1 + (2*(a + b*Tan[c + d*x])^(1/3))/(a - I*b)^(1/3))/Sqrt[3]])/(2*d) + (Sqrt[3]*(a + I*b)^(1/3)*(I*A - B)*Arc
Tan[(1 + (2*(a + b*Tan[c + d*x])^(1/3))/(a + I*b)^(1/3))/Sqrt[3]])/(2*d) - ((a + I*b)^(1/3)*(I*A - B)*Log[Cos[
c + d*x]])/(4*d) + ((a - I*b)^(1/3)*(I*A + B)*Log[Cos[c + d*x]])/(4*d) + (3*(a - I*b)^(1/3)*(I*A + B)*Log[(a -
I*b)^(1/3) - (a + b*Tan[c + d*x])^(1/3)])/(4*d) - (3*(a + I*b)^(1/3)*(I*A - B)*Log[(a + I*b)^(1/3) - (a + b*T
an[c + d*x])^(1/3)])/(4*d) + (3*B*(a + b*Tan[c + d*x])^(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 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 3528

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

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Mathematica [A]  time = 0.94, size = 347, normalized size = 0.92 \[ \frac {i \left ((A-i B) \left (3 \sqrt [3]{a+b \tan (c+d x)}-\frac {1}{2} \sqrt [3]{a-i b} \left (2 \sqrt {3} \tan ^{-1}\left (\frac {1+\frac {2 \sqrt [3]{a+b \tan (c+d x)}}{\sqrt [3]{a-i b}}}{\sqrt {3}}\right )-2 \log \left (-\sqrt [3]{a+b \tan (c+d x)}+\sqrt [3]{a-i b}\right )+\log \left (\sqrt [3]{a-i b} \sqrt [3]{a+b \tan (c+d x)}+(a+b \tan (c+d x))^{2/3}+(a-i b)^{2/3}\right )\right )\right )-(A+i B) \left (3 \sqrt [3]{a+b \tan (c+d x)}-\frac {1}{2} \sqrt [3]{a+i b} \left (2 \sqrt {3} \tan ^{-1}\left (\frac {1+\frac {2 \sqrt [3]{a+b \tan (c+d x)}}{\sqrt [3]{a+i b}}}{\sqrt {3}}\right )-2 \log \left (-\sqrt [3]{a+b \tan (c+d x)}+\sqrt [3]{a+i b}\right )+\log \left (\sqrt [3]{a+i b} \sqrt [3]{a+b \tan (c+d x)}+(a+b \tan (c+d x))^{2/3}+(a+i b)^{2/3}\right )\right )\right )\right )}{2 d} \]

Antiderivative was successfully verified.

[In]

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

[Out]

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

[Out]

Timed out

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

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate((a+b*tan(d*x+c))^(1/3)*(A+B*tan(d*x+c)),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 = 0.33, size = 99, normalized size = 0.26 \[ \frac {3 B \left (a +b \tan \left (d x +c \right )\right )^{\frac {1}{3}}}{d}+\frac {\munderset {\textit {\_R} =\RootOf \left (\textit {\_Z}^{6}-2 \textit {\_Z}^{3} a +a^{2}+b^{2}\right )}{\sum }\frac {\left (\left (A b +a B \right ) \textit {\_R}^{3}-a^{2} B -b^{2} B \right ) \ln \left (\left (a +b \tan \left (d x +c \right )\right )^{\frac {1}{3}}-\textit {\_R} \right )}{\textit {\_R}^{5}-\textit {\_R}^{2} a}}{2 d} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

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

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate((a+b*tan(d*x+c))^(1/3)*(A+B*tan(d*x+c)),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 = 17.68, size = 2537, normalized size = 6.73 \[ \text {result too large to display} \]

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(a*d^7*(((-A^6*a^2*d^6)^(1/2) - A^3*b*d^3)/d^6)^(4/3) + A*b*(a + b*tan(c + d*x))^(1/3)*(-A^6*a^2*d^6)^(1/2)
- A^4*a^2*d^3*(a + b*tan(c + d*x))^(1/3) + 2*A^3*a*b*d^4*(((-A^6*a^2*d^6)^(1/2) - A^3*b*d^3)/d^6)^(1/3))*(((-
A^6*a^2*d^6)^(1/2) - A^3*b*d^3)/(8*d^6))^(1/3) + log(A*b*(a + b*tan(c + d*x))^(1/3)*(-A^6*a^2*d^6)^(1/2) + A^4
*a^2*d^3*(a + b*tan(c + d*x))^(1/3) + a*d*(-((-A^6*a^2*d^6)^(1/2) + A^3*b*d^3)/d^6)^(1/3)*(-A^6*a^2*d^6)^(1/2)
- A^3*a*b*d^4*(-((-A^6*a^2*d^6)^(1/2) + A^3*b*d^3)/d^6)^(1/3))*(-((-A^6*a^2*d^6)^(1/2) + A^3*b*d^3)/(8*d^6))^
(1/3) + log(d*(((-B^6*b^2*d^6)^(1/2) + B^3*a*d^3)/d^6)^(1/3) - B*(a + b*tan(c + d*x))^(1/3))*(((-B^6*b^2*d^6)^
(1/2) + B^3*a*d^3)/(8*d^6))^(1/3) + log(d*(-((-B^6*b^2*d^6)^(1/2) - B^3*a*d^3)/d^6)^(1/3) - B*(a + b*tan(c + d
*x))^(1/3))*(-((-B^6*b^2*d^6)^(1/2) - B^3*a*d^3)/(8*d^6))^(1/3) + (log(- ((3^(1/2)*1i - 1)*(((3^(1/2)*1i - 1)^
2*(1944*a*b^4*(3^(1/2)*1i - 1)*(((-B^6*b^2*d^6)^(1/2) + B^3*a*d^3)/d^6)^(1/3)*(a^2 + b^2) - (3888*B*a*b^4*(a^2
+ b^2)*(a + b*tan(c + d*x))^(1/3))/d)*(((-B^6*b^2*d^6)^(1/2) + B^3*a*d^3)/d^6)^(2/3))/16 - (972*B^3*b^4*(a^4
- b^4))/d^3)*(((-B^6*b^2*d^6)^(1/2) + B^3*a*d^3)/d^6)^(1/3))/4 - (486*B^4*b^4*(a^4 - b^4)*(a + b*tan(c + d*x))
^(1/3))/d^4)*(3^(1/2)*1i - 1)*((-B^6*b^2*d^6)^(1/2)/(8*d^6) + (B^3*a)/(8*d^3))^(1/3))/2 - (log(- ((3^(1/2)*1i
+ 1)*(((3^(1/2)*1i + 1)^2*(1944*a*b^4*(3^(1/2)*1i + 1)*(((-B^6*b^2*d^6)^(1/2) + B^3*a*d^3)/d^6)^(1/3)*(a^2 + b
^2) + (3888*B*a*b^4*(a^2 + b^2)*(a + b*tan(c + d*x))^(1/3))/d)*(((-B^6*b^2*d^6)^(1/2) + B^3*a*d^3)/d^6)^(2/3))
/16 + (972*B^3*b^4*(a^4 - b^4))/d^3)*(((-B^6*b^2*d^6)^(1/2) + B^3*a*d^3)/d^6)^(1/3))/4 - (486*B^4*b^4*(a^4 - b
^4)*(a + b*tan(c + d*x))^(1/3))/d^4)*(3^(1/2)*1i + 1)*((-B^6*b^2*d^6)^(1/2)/(8*d^6) + (B^3*a)/(8*d^3))^(1/3))/
2 + (log(- ((3^(1/2)*1i - 1)*(((3^(1/2)*1i - 1)^2*(1944*a*b^4*(3^(1/2)*1i - 1)*(-((-B^6*b^2*d^6)^(1/2) - B^3*a
*d^3)/d^6)^(1/3)*(a^2 + b^2) - (3888*B*a*b^4*(a^2 + b^2)*(a + b*tan(c + d*x))^(1/3))/d)*(-((-B^6*b^2*d^6)^(1/2
) - B^3*a*d^3)/d^6)^(2/3))/16 - (972*B^3*b^4*(a^4 - b^4))/d^3)*(-((-B^6*b^2*d^6)^(1/2) - B^3*a*d^3)/d^6)^(1/3)
)/4 - (486*B^4*b^4*(a^4 - b^4)*(a + b*tan(c + d*x))^(1/3))/d^4)*(3^(1/2)*1i - 1)*((B^3*a)/(8*d^3) - (-B^6*b^2*
d^6)^(1/2)/(8*d^6))^(1/3))/2 - (log(- ((3^(1/2)*1i + 1)*(((3^(1/2)*1i + 1)^2*(1944*a*b^4*(3^(1/2)*1i + 1)*(-((
-B^6*b^2*d^6)^(1/2) - B^3*a*d^3)/d^6)^(1/3)*(a^2 + b^2) + (3888*B*a*b^4*(a^2 + b^2)*(a + b*tan(c + d*x))^(1/3)
)/d)*(-((-B^6*b^2*d^6)^(1/2) - B^3*a*d^3)/d^6)^(2/3))/16 + (972*B^3*b^4*(a^4 - b^4))/d^3)*(-((-B^6*b^2*d^6)^(1
/2) - B^3*a*d^3)/d^6)^(1/3))/4 - (486*B^4*b^4*(a^4 - b^4)*(a + b*tan(c + d*x))^(1/3))/d^4)*(3^(1/2)*1i + 1)*((
B^3*a)/(8*d^3) - (-B^6*b^2*d^6)^(1/2)/(8*d^6))^(1/3))/2 + (3*B*(a + b*tan(c + d*x))^(1/3))/d + (log(((((3^(1/2
)*1i - 1)^2*((3888*A*b^5*(a^2 + b^2)*(a + b*tan(c + d*x))^(1/3))/d + 1944*a*b^4*(3^(1/2)*1i - 1)*(((-A^6*a^2*d
^6)^(1/2) - A^3*b*d^3)/d^6)^(1/3)*(a^2 + b^2))*(((-A^6*a^2*d^6)^(1/2) - A^3*b*d^3)/d^6)^(2/3))/16 + (1944*A^3*
a*b^5*(a^2 + b^2))/d^3)*(3^(1/2)*1i - 1)*(((-A^6*a^2*d^6)^(1/2) - A^3*b*d^3)/d^6)^(1/3))/4 - (486*A^4*b^4*(a^4
- b^4)*(a + b*tan(c + d*x))^(1/3))/d^4)*(3^(1/2)*1i - 1)*((-A^6*a^2*d^6)^(1/2)/(8*d^6) - (A^3*b)/(8*d^3))^(1/
3))/2 + (log(((((3^(1/2)*1i - 1)^2*((3888*A*b^5*(a^2 + b^2)*(a + b*tan(c + d*x))^(1/3))/d + 1944*a*b^4*(3^(1/2
)*1i - 1)*(-((-A^6*a^2*d^6)^(1/2) + A^3*b*d^3)/d^6)^(1/3)*(a^2 + b^2))*(-((-A^6*a^2*d^6)^(1/2) + A^3*b*d^3)/d^
6)^(2/3))/16 + (1944*A^3*a*b^5*(a^2 + b^2))/d^3)*(3^(1/2)*1i - 1)*(-((-A^6*a^2*d^6)^(1/2) + A^3*b*d^3)/d^6)^(1
/3))/4 - (486*A^4*b^4*(a^4 - b^4)*(a + b*tan(c + d*x))^(1/3))/d^4)*(3^(1/2)*1i - 1)*(- (-A^6*a^2*d^6)^(1/2)/(8
*d^6) - (A^3*b)/(8*d^3))^(1/3))/2 - (log(- ((((3^(1/2)*1i + 1)^2*((3888*A*b^5*(a^2 + b^2)*(a + b*tan(c + d*x))
^(1/3))/d - 1944*a*b^4*(3^(1/2)*1i + 1)*(((-A^6*a^2*d^6)^(1/2) - A^3*b*d^3)/d^6)^(1/3)*(a^2 + b^2))*(((-A^6*a^
2*d^6)^(1/2) - A^3*b*d^3)/d^6)^(2/3))/16 + (1944*A^3*a*b^5*(a^2 + b^2))/d^3)*(3^(1/2)*1i + 1)*(((-A^6*a^2*d^6)
^(1/2) - A^3*b*d^3)/d^6)^(1/3))/4 - (486*A^4*b^4*(a^4 - b^4)*(a + b*tan(c + d*x))^(1/3))/d^4)*(3^(1/2)*1i + 1)
*((-A^6*a^2*d^6)^(1/2)/(8*d^6) - (A^3*b)/(8*d^3))^(1/3))/2 - (log(- ((((3^(1/2)*1i + 1)^2*((3888*A*b^5*(a^2 +
b^2)*(a + b*tan(c + d*x))^(1/3))/d - 1944*a*b^4*(3^(1/2)*1i + 1)*(-((-A^6*a^2*d^6)^(1/2) + A^3*b*d^3)/d^6)^(1/
3)*(a^2 + b^2))*(-((-A^6*a^2*d^6)^(1/2) + A^3*b*d^3)/d^6)^(2/3))/16 + (1944*A^3*a*b^5*(a^2 + b^2))/d^3)*(3^(1/
2)*1i + 1)*(-((-A^6*a^2*d^6)^(1/2) + A^3*b*d^3)/d^6)^(1/3))/4 - (486*A^4*b^4*(a^4 - b^4)*(a + b*tan(c + d*x))^
(1/3))/d^4)*(3^(1/2)*1i + 1)*(- (-A^6*a^2*d^6)^(1/2)/(8*d^6) - (A^3*b)/(8*d^3))^(1/3))/2

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

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

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

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