3.473 \(\int (a+b \tan (c+d x))^{2/3} (A+B \tan (c+d x)) \, dx\)

Optimal. Leaf size=379 \[ \frac{\sqrt{3} (a-i b)^{2/3} (B+i A) \tan ^{-1}\left (\frac{1+\frac{2 \sqrt [3]{a+b \tan (c+d x)}}{\sqrt [3]{a-i b}}}{\sqrt{3}}\right )}{2 d}-\frac{\sqrt{3} (a+i b)^{2/3} (-B+i A) \tan ^{-1}\left (\frac{1+\frac{2 \sqrt [3]{a+b \tan (c+d x)}}{\sqrt [3]{a+i b}}}{\sqrt{3}}\right )}{2 d}+\frac{3 (a-i b)^{2/3} (B+i A) \log \left (-\sqrt [3]{a+b \tan (c+d x)}+\sqrt [3]{a-i b}\right )}{4 d}-\frac{3 (a+i b)^{2/3} (-B+i A) \log \left (-\sqrt [3]{a+b \tan (c+d x)}+\sqrt [3]{a+i b}\right )}{4 d}-\frac{(a+i b)^{2/3} (-B+i A) \log (\cos (c+d x))}{4 d}+\frac{(a-i b)^{2/3} (B+i A) \log (\cos (c+d x))}{4 d}-\frac{1}{4} x (a-i b)^{2/3} (A-i B)-\frac{1}{4} x (a+i b)^{2/3} (A+i B)+\frac{3 B (a+b \tan (c+d x))^{2/3}}{2 d} \]

[Out]

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

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Rubi [A]  time = 0.440236, antiderivative size = 379, normalized size of antiderivative = 1., number of steps used = 12, number of rules used = 7, integrand size = 25, \(\frac{\text{number of rules}}{\text{integrand size}}\) = 0.28, Rules used = {3528, 3539, 3537, 55, 617, 204, 31} \[ \frac{\sqrt{3} (a-i b)^{2/3} (B+i A) \tan ^{-1}\left (\frac{1+\frac{2 \sqrt [3]{a+b \tan (c+d x)}}{\sqrt [3]{a-i b}}}{\sqrt{3}}\right )}{2 d}-\frac{\sqrt{3} (a+i b)^{2/3} (-B+i A) \tan ^{-1}\left (\frac{1+\frac{2 \sqrt [3]{a+b \tan (c+d x)}}{\sqrt [3]{a+i b}}}{\sqrt{3}}\right )}{2 d}+\frac{3 (a-i b)^{2/3} (B+i A) \log \left (-\sqrt [3]{a+b \tan (c+d x)}+\sqrt [3]{a-i b}\right )}{4 d}-\frac{3 (a+i b)^{2/3} (-B+i A) \log \left (-\sqrt [3]{a+b \tan (c+d x)}+\sqrt [3]{a+i b}\right )}{4 d}-\frac{(a+i b)^{2/3} (-B+i A) \log (\cos (c+d x))}{4 d}+\frac{(a-i b)^{2/3} (B+i A) \log (\cos (c+d x))}{4 d}-\frac{1}{4} x (a-i b)^{2/3} (A-i B)-\frac{1}{4} x (a+i b)^{2/3} (A+i B)+\frac{3 B (a+b \tan (c+d x))^{2/3}}{2 d} \]

Antiderivative was successfully verified.

[In]

Int[(a + b*Tan[c + d*x])^(2/3)*(A + B*Tan[c + d*x]),x]

[Out]

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

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 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))^{2/3} (A+B \tan (c+d x)) \, dx &=\frac{3 B (a+b \tan (c+d x))^{2/3}}{2 d}+\int \frac{a A-b B+(A b+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) (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) (A+i B)) \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{(i (a-i b) (A-i 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) (A+i 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{1}{4} (a-i b)^{2/3} (A-i B) x-\frac{1}{4} (a+i b)^{2/3} (A+i B) x-\frac{(a+i b)^{2/3} (i A-B) \log (\cos (c+d x))}{4 d}+\frac{(a-i b)^{2/3} (i A+B) \log (\cos (c+d x))}{4 d}+\frac{3 B (a+b \tan (c+d x))^{2/3}}{2 d}+\frac{\left (3 (a+i b)^{2/3} (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{(3 i (a-i b) (A-i 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) (A+i 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{\left (3 (a-i b)^{2/3} (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{1}{4} (a-i b)^{2/3} (A-i B) x-\frac{1}{4} (a+i b)^{2/3} (A+i B) x-\frac{(a+i b)^{2/3} (i A-B) \log (\cos (c+d x))}{4 d}+\frac{(a-i b)^{2/3} (i A+B) \log (\cos (c+d x))}{4 d}+\frac{3 (a-i b)^{2/3} (i A+B) \log \left (\sqrt [3]{a-i b}-\sqrt [3]{a+b \tan (c+d x)}\right )}{4 d}-\frac{3 (a+i b)^{2/3} (i A-B) \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 (a+i b)^{2/3} (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 (a-i b)^{2/3} (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} (a-i b)^{2/3} (A-i B) x-\frac{1}{4} (a+i b)^{2/3} (A+i B) x+\frac{\sqrt{3} (a-i b)^{2/3} (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 d}-\frac{\sqrt{3} (a+i b)^{2/3} (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 d}-\frac{(a+i b)^{2/3} (i A-B) \log (\cos (c+d x))}{4 d}+\frac{(a-i b)^{2/3} (i A+B) \log (\cos (c+d x))}{4 d}+\frac{3 (a-i b)^{2/3} (i A+B) \log \left (\sqrt [3]{a-i b}-\sqrt [3]{a+b \tan (c+d x)}\right )}{4 d}-\frac{3 (a+i b)^{2/3} (i A-B) \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 = 0.873852, size = 263, normalized size = 0.69 \[ \frac{i \left ((A-i B) \left (3 (a+b \tan (c+d x))^{2/3}+(a-i b)^{2/3} \left (2 \sqrt{3} \tan ^{-1}\left (\frac{1+\frac{2 \sqrt [3]{a+b \tan (c+d x)}}{\sqrt [3]{a-i b}}}{\sqrt{3}}\right )+3 \log \left (-\sqrt [3]{a+b \tan (c+d x)}+\sqrt [3]{a-i b}\right )-\log (\tan (c+d x)+i)\right )\right )-(A+i B) \left (3 (a+b \tan (c+d x))^{2/3}+(a+i b)^{2/3} \left (2 \sqrt{3} \tan ^{-1}\left (\frac{1+\frac{2 \sqrt [3]{a+b \tan (c+d x)}}{\sqrt [3]{a+i b}}}{\sqrt{3}}\right )+3 \log \left (-\sqrt [3]{a+b \tan (c+d x)}+\sqrt [3]{a+i b}\right )-\log (-\tan (c+d x)+i)\right )\right )\right )}{4 d} \]

Antiderivative was successfully verified.

[In]

Integrate[(a + b*Tan[c + d*x])^(2/3)*(A + B*Tan[c + d*x]),x]

[Out]

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

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Maple [C]  time = 0.194, size = 101, normalized size = 0.3 \begin{align*}{\frac{3\,B}{2\,d} \left ( a+b\tan \left ( dx+c \right ) \right ) ^{{\frac{2}{3}}}}+{\frac{1}{2\,d}\sum _{{\it \_R}={\it RootOf} \left ({{\it \_Z}}^{6}-2\,{{\it \_Z}}^{3}a+{a}^{2}+{b}^{2} \right ) }{\frac{ \left ( Ab+aB \right ){{\it \_R}}^{4}+B \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))^(2/3)*(A+B*tan(d*x+c)),x)

[Out]

3/2*B*(a+b*tan(d*x+c))^(2/3)/d+1/2/d*sum(((A*b+B*a)*_R^4+B*(-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., size = 0, normalized size = 0. \begin{align*} \int{\left (B \tan \left (d x + c\right ) + A\right )}{\left (b \tan \left (d x + c\right ) + a\right )}^{\frac{2}{3}}\,{d x} \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate((a+b*tan(d*x+c))^(2/3)*(A+B*tan(d*x+c)),x, algorithm="maxima")

[Out]

integrate((B*tan(d*x + c) + A)*(b*tan(d*x + c) + a)^(2/3), x)

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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))^(2/3)*(A+B*tan(d*x+c)),x, algorithm="fricas")

[Out]

Timed out

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Sympy [F]  time = 0., size = 0, normalized size = 0. \begin{align*} \int \left (A + B \tan{\left (c + d x \right )}\right ) \left (a + b \tan{\left (c + d x \right )}\right )^{\frac{2}{3}}\, dx \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate((a+b*tan(d*x+c))**(2/3)*(A+B*tan(d*x+c)),x)

[Out]

Integral((A + B*tan(c + d*x))*(a + b*tan(c + d*x))**(2/3), x)

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Giac [B]  time = 15.4942, size = 1397, normalized size = 3.69 \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))^(2/3)*(A+B*tan(d*x+c)),x, algorithm="giac")

[Out]

-1/8*(-I*sqrt(3) + 1)*((8*I*A^3*a^2 - 24*A^2*B*a^2 - 24*I*A*B^2*a^2 + 8*B^3*a^2 - 16*A^3*a*b - 48*I*A^2*B*a*b
+ 48*A*B^2*a*b + 16*I*B^3*a*b - 8*I*A^3*b^2 + 24*A^2*B*b^2 + 24*I*A*B^2*b^2 - 8*B^3*b^2)/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)) - 1/8*(-I*sqrt(3) + 1)*((-8*I*A^3*a^2 - 24*A^2*B*a^2 + 24*I*A*B^2*a^2 + 8*B^3*a^2 - 16*A^3*a*b + 48*I*A
^2*B*a*b + 48*A*B^2*a*b - 16*I*B^3*a*b + 8*I*A^3*b^2 + 24*A^2*B*b^2 - 24*I*A*B^2*b^2 - 8*B^3*b^2)/d^3)^(1/3)*l
og(-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)) - 1/8*(I*sqrt(3) + 1)*((-8*I*A^3*a^2 - 24*A^2*B*a^2 + 24*I*A*B^2*a^2 + 8*B^3*a^2 - 16*A^3*a*
b + 48*I*A^2*B*a*b + 48*A*B^2*a*b - 16*I*B^3*a*b + 8*I*A^3*b^2 + 24*A^2*B*b^2 - 24*I*A*B^2*b^2 - 8*B^3*b^2)/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)) - 1/8*(I*sqrt(3) + 1)*((8*I*A^3*a^2 - 24*A^2*B*a^2 - 24*I*A*B^2*a^2 + 8*B^3*a^2 - 16
*A^3*a*b - 48*I*A^2*B*a*b + 48*A*B^2*a*b + 16*I*B^3*a*b - 8*I*A^3*b^2 + 24*A^2*B*b^2 + 24*I*A*B^2*b^2 - 8*B^3*
b^2)/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)) + 1/4*((8*I*A^3*a^2 - 24*A^2*B*a^2 - 24*I*A*B^2*a^2 + 8*B^3*a^2 - 16*A^3*a*b
- 48*I*A^2*B*a*b + 48*A*B^2*a*b + 16*I*B^3*a*b - 8*I*A^3*b^2 + 24*A^2*B*b^2 + 24*I*A*B^2*b^2 - 8*B^3*b^2)/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) + 1/4*((-8*I*A^3*a^
2 - 24*A^2*B*a^2 + 24*I*A*B^2*a^2 + 8*B^3*a^2 - 16*A^3*a*b + 48*I*A^2*B*a*b + 48*A*B^2*a*b - 16*I*B^3*a*b + 8*
I*A^3*b^2 + 24*A^2*B*b^2 - 24*I*A*B^2*b^2 - 8*B^3*b^2)/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) + 3/2*(b*tan(d*x + c) + a)^(2/3)*B/d