3.286 \(\int \frac {A+B \tan (c+d x)}{(a+b \tan (c+d x))^3} \, dx\)

Optimal. Leaf size=175 \[ -\frac {A b-a B}{2 d \left (a^2+b^2\right ) (a+b \tan (c+d x))^2}-\frac {a^2 (-B)+2 a A b+b^2 B}{d \left (a^2+b^2\right )^2 (a+b \tan (c+d x))}+\frac {\left (a^3 (-B)+3 a^2 A b+3 a b^2 B-A b^3\right ) \log (a \cos (c+d x)+b \sin (c+d x))}{d \left (a^2+b^2\right )^3}+\frac {x \left (a^3 A+3 a^2 b B-3 a A b^2-b^3 B\right )}{\left (a^2+b^2\right )^3} \]

[Out]

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

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Rubi [A]  time = 0.27, antiderivative size = 175, normalized size of antiderivative = 1.00, number of steps used = 4, number of rules used = 3, integrand size = 23, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.130, Rules used = {3529, 3531, 3530} \[ -\frac {A b-a B}{2 d \left (a^2+b^2\right ) (a+b \tan (c+d x))^2}-\frac {a^2 (-B)+2 a A b+b^2 B}{d \left (a^2+b^2\right )^2 (a+b \tan (c+d x))}+\frac {\left (3 a^2 A b+a^3 (-B)+3 a b^2 B-A b^3\right ) \log (a \cos (c+d x)+b \sin (c+d x))}{d \left (a^2+b^2\right )^3}+\frac {x \left (a^3 A+3 a^2 b B-3 a A b^2-b^3 B\right )}{\left (a^2+b^2\right )^3} \]

Antiderivative was successfully verified.

[In]

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

[Out]

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

Rule 3529

Int[((a_.) + (b_.)*tan[(e_.) + (f_.)*(x_)])^(m_)*((c_.) + (d_.)*tan[(e_.) + (f_.)*(x_)]), x_Symbol] :> Simp[((
b*c - a*d)*(a + b*Tan[e + f*x])^(m + 1))/(f*(m + 1)*(a^2 + b^2)), x] + Dist[1/(a^2 + b^2), Int[(a + b*Tan[e +
f*x])^(m + 1)*Simp[a*c + b*d - (b*c - a*d)*Tan[e + f*x], x], x], x] /; FreeQ[{a, b, c, d, e, f}, x] && NeQ[b*c
- a*d, 0] && NeQ[a^2 + b^2, 0] && LtQ[m, -1]

Rule 3530

Int[((c_) + (d_.)*tan[(e_.) + (f_.)*(x_)])/((a_) + (b_.)*tan[(e_.) + (f_.)*(x_)]), x_Symbol] :> Simp[(c*Log[Re
moveContent[a*Cos[e + f*x] + b*Sin[e + f*x], x]])/(b*f), x] /; FreeQ[{a, b, c, d, e, f}, x] && NeQ[b*c - a*d,
0] && NeQ[a^2 + b^2, 0] && EqQ[a*c + b*d, 0]

Rule 3531

Int[((c_.) + (d_.)*tan[(e_.) + (f_.)*(x_)])/((a_.) + (b_.)*tan[(e_.) + (f_.)*(x_)]), x_Symbol] :> Simp[((a*c +
b*d)*x)/(a^2 + b^2), x] + Dist[(b*c - a*d)/(a^2 + b^2), Int[(b - a*Tan[e + f*x])/(a + b*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] && NeQ[a*c + b*d, 0]

Rubi steps

\begin {align*} \int \frac {A+B \tan (c+d x)}{(a+b \tan (c+d x))^3} \, dx &=-\frac {A b-a B}{2 \left (a^2+b^2\right ) d (a+b \tan (c+d x))^2}+\frac {\int \frac {a A+b B-(A b-a B) \tan (c+d x)}{(a+b \tan (c+d x))^2} \, dx}{a^2+b^2}\\ &=-\frac {A b-a B}{2 \left (a^2+b^2\right ) d (a+b \tan (c+d x))^2}-\frac {2 a A b-a^2 B+b^2 B}{\left (a^2+b^2\right )^2 d (a+b \tan (c+d x))}+\frac {\int \frac {a^2 A-A b^2+2 a b B-\left (2 a A b-a^2 B+b^2 B\right ) \tan (c+d x)}{a+b \tan (c+d x)} \, dx}{\left (a^2+b^2\right )^2}\\ &=\frac {\left (a^3 A-3 a A b^2+3 a^2 b B-b^3 B\right ) x}{\left (a^2+b^2\right )^3}-\frac {A b-a B}{2 \left (a^2+b^2\right ) d (a+b \tan (c+d x))^2}-\frac {2 a A b-a^2 B+b^2 B}{\left (a^2+b^2\right )^2 d (a+b \tan (c+d x))}+\frac {\left (3 a^2 A b-A b^3-a^3 B+3 a b^2 B\right ) \int \frac {b-a \tan (c+d x)}{a+b \tan (c+d x)} \, dx}{\left (a^2+b^2\right )^3}\\ &=\frac {\left (a^3 A-3 a A b^2+3 a^2 b B-b^3 B\right ) x}{\left (a^2+b^2\right )^3}+\frac {\left (3 a^2 A b-A b^3-a^3 B+3 a b^2 B\right ) \log (a \cos (c+d x)+b \sin (c+d x))}{\left (a^2+b^2\right )^3 d}-\frac {A b-a B}{2 \left (a^2+b^2\right ) d (a+b \tan (c+d x))^2}-\frac {2 a A b-a^2 B+b^2 B}{\left (a^2+b^2\right )^2 d (a+b \tan (c+d x))}\\ \end {align*}

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Mathematica [C]  time = 4.77, size = 243, normalized size = 1.39 \[ -\frac {(A b-a B) \left (\frac {b \left (\frac {\left (a^2+b^2\right ) \left (5 a^2+4 a b \tan (c+d x)+b^2\right )}{(a+b \tan (c+d x))^2}+\left (2 b^2-6 a^2\right ) \log (a+b \tan (c+d x))\right )}{\left (a^2+b^2\right )^3}+\frac {i \log (-\tan (c+d x)+i)}{(a+i b)^3}-\frac {\log (\tan (c+d x)+i)}{(b+i a)^3}\right )+B \left (\frac {2 b \left (\frac {a^2+b^2}{a+b \tan (c+d x)}-2 a \log (a+b \tan (c+d x))\right )}{\left (a^2+b^2\right )^2}+\frac {i \log (-\tan (c+d x)+i)}{(a+i b)^2}-\frac {i \log (\tan (c+d x)+i)}{(a-i b)^2}\right )}{2 b d} \]

Antiderivative was successfully verified.

[In]

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

[Out]

-1/2*(B*((I*Log[I - Tan[c + d*x]])/(a + I*b)^2 - (I*Log[I + Tan[c + d*x]])/(a - I*b)^2 + (2*b*(-2*a*Log[a + b*
Tan[c + d*x]] + (a^2 + b^2)/(a + b*Tan[c + d*x])))/(a^2 + b^2)^2) + (A*b - a*B)*((I*Log[I - Tan[c + d*x]])/(a
+ I*b)^3 - Log[I + Tan[c + d*x]]/(I*a + b)^3 + (b*((-6*a^2 + 2*b^2)*Log[a + b*Tan[c + d*x]] + ((a^2 + b^2)*(5*
a^2 + b^2 + 4*a*b*Tan[c + d*x]))/(a + b*Tan[c + d*x])^2))/(a^2 + b^2)^3))/(b*d)

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fricas [B]  time = 0.62, size = 482, normalized size = 2.75 \[ \frac {5 \, B a^{3} b^{2} - 7 \, A a^{2} b^{3} - B a b^{4} - A b^{5} + 2 \, {\left (A a^{5} + 3 \, B a^{4} b - 3 \, A a^{3} b^{2} - B a^{2} b^{3}\right )} d x - {\left (3 \, B a^{3} b^{2} - 5 \, A a^{2} b^{3} - 3 \, B a b^{4} + A b^{5} - 2 \, {\left (A a^{3} b^{2} + 3 \, B a^{2} b^{3} - 3 \, A a b^{4} - B b^{5}\right )} d x\right )} \tan \left (d x + c\right )^{2} - {\left (B a^{5} - 3 \, A a^{4} b - 3 \, B a^{3} b^{2} + A a^{2} b^{3} + {\left (B a^{3} b^{2} - 3 \, A a^{2} b^{3} - 3 \, B a b^{4} + A b^{5}\right )} \tan \left (d x + c\right )^{2} + 2 \, {\left (B a^{4} b - 3 \, A a^{3} b^{2} - 3 \, B a^{2} b^{3} + A a b^{4}\right )} \tan \left (d x + c\right )\right )} \log \left (\frac {b^{2} \tan \left (d x + c\right )^{2} + 2 \, a b \tan \left (d x + c\right ) + a^{2}}{\tan \left (d x + c\right )^{2} + 1}\right ) - 2 \, {\left (2 \, B a^{4} b - 3 \, A a^{3} b^{2} - 3 \, B a^{2} b^{3} + 3 \, A a b^{4} + B b^{5} - 2 \, {\left (A a^{4} b + 3 \, B a^{3} b^{2} - 3 \, A a^{2} b^{3} - B a b^{4}\right )} d x\right )} \tan \left (d x + c\right )}{2 \, {\left ({\left (a^{6} b^{2} + 3 \, a^{4} b^{4} + 3 \, a^{2} b^{6} + b^{8}\right )} d \tan \left (d x + c\right )^{2} + 2 \, {\left (a^{7} b + 3 \, a^{5} b^{3} + 3 \, a^{3} b^{5} + a b^{7}\right )} d \tan \left (d x + c\right ) + {\left (a^{8} + 3 \, a^{6} b^{2} + 3 \, a^{4} b^{4} + a^{2} b^{6}\right )} d\right )}} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

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

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giac [B]  time = 0.69, size = 409, normalized size = 2.34 \[ \frac {\frac {2 \, {\left (A a^{3} + 3 \, B a^{2} b - 3 \, A a b^{2} - B b^{3}\right )} {\left (d x + c\right )}}{a^{6} + 3 \, a^{4} b^{2} + 3 \, a^{2} b^{4} + b^{6}} + \frac {{\left (B a^{3} - 3 \, A a^{2} b - 3 \, B a b^{2} + A b^{3}\right )} \log \left (\tan \left (d x + c\right )^{2} + 1\right )}{a^{6} + 3 \, a^{4} b^{2} + 3 \, a^{2} b^{4} + b^{6}} - \frac {2 \, {\left (B a^{3} b - 3 \, A a^{2} b^{2} - 3 \, B a b^{3} + A b^{4}\right )} \log \left ({\left | b \tan \left (d x + c\right ) + a \right |}\right )}{a^{6} b + 3 \, a^{4} b^{3} + 3 \, a^{2} b^{5} + b^{7}} + \frac {3 \, B a^{3} b^{2} \tan \left (d x + c\right )^{2} - 9 \, A a^{2} b^{3} \tan \left (d x + c\right )^{2} - 9 \, B a b^{4} \tan \left (d x + c\right )^{2} + 3 \, A b^{5} \tan \left (d x + c\right )^{2} + 8 \, B a^{4} b \tan \left (d x + c\right ) - 22 \, A a^{3} b^{2} \tan \left (d x + c\right ) - 18 \, B a^{2} b^{3} \tan \left (d x + c\right ) + 2 \, A a b^{4} \tan \left (d x + c\right ) - 2 \, B b^{5} \tan \left (d x + c\right ) + 6 \, B a^{5} - 14 \, A a^{4} b - 7 \, B a^{3} b^{2} - 3 \, A a^{2} b^{3} - B a b^{4} - A b^{5}}{{\left (a^{6} + 3 \, a^{4} b^{2} + 3 \, a^{2} b^{4} + b^{6}\right )} {\left (b \tan \left (d x + c\right ) + a\right )}^{2}}}{2 \, d} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

1/2*(2*(A*a^3 + 3*B*a^2*b - 3*A*a*b^2 - B*b^3)*(d*x + c)/(a^6 + 3*a^4*b^2 + 3*a^2*b^4 + b^6) + (B*a^3 - 3*A*a^
2*b - 3*B*a*b^2 + A*b^3)*log(tan(d*x + c)^2 + 1)/(a^6 + 3*a^4*b^2 + 3*a^2*b^4 + b^6) - 2*(B*a^3*b - 3*A*a^2*b^
2 - 3*B*a*b^3 + A*b^4)*log(abs(b*tan(d*x + c) + a))/(a^6*b + 3*a^4*b^3 + 3*a^2*b^5 + b^7) + (3*B*a^3*b^2*tan(d
*x + c)^2 - 9*A*a^2*b^3*tan(d*x + c)^2 - 9*B*a*b^4*tan(d*x + c)^2 + 3*A*b^5*tan(d*x + c)^2 + 8*B*a^4*b*tan(d*x
+ c) - 22*A*a^3*b^2*tan(d*x + c) - 18*B*a^2*b^3*tan(d*x + c) + 2*A*a*b^4*tan(d*x + c) - 2*B*b^5*tan(d*x + c)
+ 6*B*a^5 - 14*A*a^4*b - 7*B*a^3*b^2 - 3*A*a^2*b^3 - B*a*b^4 - A*b^5)/((a^6 + 3*a^4*b^2 + 3*a^2*b^4 + b^6)*(b*
tan(d*x + c) + a)^2))/d

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maple [B]  time = 0.28, size = 483, normalized size = 2.76 \[ \frac {3 b \,a^{2} \ln \left (a +b \tan \left (d x +c \right )\right ) A}{d \left (a^{2}+b^{2}\right )^{3}}-\frac {\ln \left (a +b \tan \left (d x +c \right )\right ) A \,b^{3}}{d \left (a^{2}+b^{2}\right )^{3}}-\frac {a^{3} \ln \left (a +b \tan \left (d x +c \right )\right ) B}{d \left (a^{2}+b^{2}\right )^{3}}+\frac {3 \ln \left (a +b \tan \left (d x +c \right )\right ) B a \,b^{2}}{d \left (a^{2}+b^{2}\right )^{3}}-\frac {A b}{2 d \left (a^{2}+b^{2}\right ) \left (a +b \tan \left (d x +c \right )\right )^{2}}+\frac {a B}{2 d \left (a^{2}+b^{2}\right ) \left (a +b \tan \left (d x +c \right )\right )^{2}}-\frac {2 a b A}{d \left (a^{2}+b^{2}\right )^{2} \left (a +b \tan \left (d x +c \right )\right )}+\frac {a^{2} B}{d \left (a^{2}+b^{2}\right )^{2} \left (a +b \tan \left (d x +c \right )\right )}-\frac {b^{2} B}{d \left (a^{2}+b^{2}\right )^{2} \left (a +b \tan \left (d x +c \right )\right )}-\frac {3 \ln \left (1+\tan ^{2}\left (d x +c \right )\right ) A \,a^{2} b}{2 d \left (a^{2}+b^{2}\right )^{3}}+\frac {\ln \left (1+\tan ^{2}\left (d x +c \right )\right ) A \,b^{3}}{2 d \left (a^{2}+b^{2}\right )^{3}}+\frac {\ln \left (1+\tan ^{2}\left (d x +c \right )\right ) a^{3} B}{2 d \left (a^{2}+b^{2}\right )^{3}}-\frac {3 \ln \left (1+\tan ^{2}\left (d x +c \right )\right ) B a \,b^{2}}{2 d \left (a^{2}+b^{2}\right )^{3}}+\frac {A \arctan \left (\tan \left (d x +c \right )\right ) a^{3}}{d \left (a^{2}+b^{2}\right )^{3}}-\frac {3 A \arctan \left (\tan \left (d x +c \right )\right ) a \,b^{2}}{d \left (a^{2}+b^{2}\right )^{3}}+\frac {3 B \arctan \left (\tan \left (d x +c \right )\right ) a^{2} b}{d \left (a^{2}+b^{2}\right )^{3}}-\frac {B \arctan \left (\tan \left (d x +c \right )\right ) b^{3}}{d \left (a^{2}+b^{2}\right )^{3}} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

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

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maxima [A]  time = 1.00, size = 321, normalized size = 1.83 \[ \frac {\frac {2 \, {\left (A a^{3} + 3 \, B a^{2} b - 3 \, A a b^{2} - B b^{3}\right )} {\left (d x + c\right )}}{a^{6} + 3 \, a^{4} b^{2} + 3 \, a^{2} b^{4} + b^{6}} - \frac {2 \, {\left (B a^{3} - 3 \, A a^{2} b - 3 \, B a b^{2} + A b^{3}\right )} \log \left (b \tan \left (d x + c\right ) + a\right )}{a^{6} + 3 \, a^{4} b^{2} + 3 \, a^{2} b^{4} + b^{6}} + \frac {{\left (B a^{3} - 3 \, A a^{2} b - 3 \, B a b^{2} + A b^{3}\right )} \log \left (\tan \left (d x + c\right )^{2} + 1\right )}{a^{6} + 3 \, a^{4} b^{2} + 3 \, a^{2} b^{4} + b^{6}} + \frac {3 \, B a^{3} - 5 \, A a^{2} b - B a b^{2} - A b^{3} + 2 \, {\left (B a^{2} b - 2 \, A a b^{2} - B b^{3}\right )} \tan \left (d x + c\right )}{a^{6} + 2 \, a^{4} b^{2} + a^{2} b^{4} + {\left (a^{4} b^{2} + 2 \, a^{2} b^{4} + b^{6}\right )} \tan \left (d x + c\right )^{2} + 2 \, {\left (a^{5} b + 2 \, a^{3} b^{3} + a b^{5}\right )} \tan \left (d x + c\right )}}{2 \, d} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

1/2*(2*(A*a^3 + 3*B*a^2*b - 3*A*a*b^2 - B*b^3)*(d*x + c)/(a^6 + 3*a^4*b^2 + 3*a^2*b^4 + b^6) - 2*(B*a^3 - 3*A*
a^2*b - 3*B*a*b^2 + A*b^3)*log(b*tan(d*x + c) + a)/(a^6 + 3*a^4*b^2 + 3*a^2*b^4 + b^6) + (B*a^3 - 3*A*a^2*b -
3*B*a*b^2 + A*b^3)*log(tan(d*x + c)^2 + 1)/(a^6 + 3*a^4*b^2 + 3*a^2*b^4 + b^6) + (3*B*a^3 - 5*A*a^2*b - B*a*b^
2 - A*b^3 + 2*(B*a^2*b - 2*A*a*b^2 - B*b^3)*tan(d*x + c))/(a^6 + 2*a^4*b^2 + a^2*b^4 + (a^4*b^2 + 2*a^2*b^4 +
b^6)*tan(d*x + c)^2 + 2*(a^5*b + 2*a^3*b^3 + a*b^5)*tan(d*x + c)))/d

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mupad [B]  time = 6.52, size = 279, normalized size = 1.59 \[ \frac {\ln \left (a+b\,\mathrm {tan}\left (c+d\,x\right )\right )\,\left (\frac {3\,A\,b-B\,a}{{\left (a^2+b^2\right )}^2}-\frac {4\,b^2\,\left (A\,b-B\,a\right )}{{\left (a^2+b^2\right )}^3}\right )}{d}-\frac {\frac {-3\,B\,a^3+5\,A\,a^2\,b+B\,a\,b^2+A\,b^3}{2\,\left (a^4+2\,a^2\,b^2+b^4\right )}+\frac {\mathrm {tan}\left (c+d\,x\right )\,\left (-B\,a^2\,b+2\,A\,a\,b^2+B\,b^3\right )}{a^4+2\,a^2\,b^2+b^4}}{d\,\left (a^2+2\,a\,b\,\mathrm {tan}\left (c+d\,x\right )+b^2\,{\mathrm {tan}\left (c+d\,x\right )}^2\right )}+\frac {\ln \left (\mathrm {tan}\left (c+d\,x\right )-\mathrm {i}\right )\,\left (-B+A\,1{}\mathrm {i}\right )}{2\,d\,\left (-a^3-a^2\,b\,3{}\mathrm {i}+3\,a\,b^2+b^3\,1{}\mathrm {i}\right )}+\frac {\ln \left (\mathrm {tan}\left (c+d\,x\right )+1{}\mathrm {i}\right )\,\left (A-B\,1{}\mathrm {i}\right )}{2\,d\,\left (-a^3\,1{}\mathrm {i}-3\,a^2\,b+a\,b^2\,3{}\mathrm {i}+b^3\right )} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

(log(a + b*tan(c + d*x))*((3*A*b - B*a)/(a^2 + b^2)^2 - (4*b^2*(A*b - B*a))/(a^2 + b^2)^3))/d - ((A*b^3 - 3*B*
a^3 + 5*A*a^2*b + B*a*b^2)/(2*(a^4 + b^4 + 2*a^2*b^2)) + (tan(c + d*x)*(B*b^3 + 2*A*a*b^2 - B*a^2*b))/(a^4 + b
^4 + 2*a^2*b^2))/(d*(a^2 + b^2*tan(c + d*x)^2 + 2*a*b*tan(c + d*x))) + (log(tan(c + d*x) - 1i)*(A*1i - B))/(2*
d*(3*a*b^2 - a^2*b*3i - a^3 + b^3*1i)) + (log(tan(c + d*x) + 1i)*(A - B*1i))/(2*d*(a*b^2*3i - 3*a^2*b - a^3*1i
+ b^3))

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sympy [F(-2)]  time = 0.00, size = 0, normalized size = 0.00 \[ \text {Exception raised: AttributeError} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

Exception raised: AttributeError

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