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

Optimal. Leaf size=250 \[ \frac {a (A b-a B) \tan ^2(c+d x)}{2 b d \left (a^2+b^2\right ) (a+b \tan (c+d x))^2}-\frac {a^2 \left (2 A b^3-a B \left (a^2+3 b^2\right )\right )}{b^3 d \left (a^2+b^2\right )^2 (a+b \tan (c+d x))}+\frac {\left (a^3 A+3 a^2 b B-3 a A b^2-b^3 B\right ) \log (\cos (c+d x))}{d \left (a^2+b^2\right )^3}-\frac {x \left (a^3 (-B)+3 a^2 A b+3 a b^2 B-A b^3\right )}{\left (a^2+b^2\right )^3}+\frac {a \left (a^5 B+3 a^3 b^2 B+a^2 A b^3+6 a b^4 B-3 A b^5\right ) \log (a+b \tan (c+d x))}{b^3 d \left (a^2+b^2\right )^3} \]

[Out]

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

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Rubi [A]  time = 0.49, antiderivative size = 250, normalized size of antiderivative = 1.00, number of steps used = 6, number of rules used = 6, integrand size = 31, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.194, Rules used = {3605, 3635, 3626, 3617, 31, 3475} \[ \frac {a (A b-a B) \tan ^2(c+d x)}{2 b d \left (a^2+b^2\right ) (a+b \tan (c+d x))^2}-\frac {a^2 \left (2 A b^3-a B \left (a^2+3 b^2\right )\right )}{b^3 d \left (a^2+b^2\right )^2 (a+b \tan (c+d x))}+\frac {a \left (a^2 A b^3+3 a^3 b^2 B+a^5 B+6 a b^4 B-3 A b^5\right ) \log (a+b \tan (c+d x))}{b^3 d \left (a^2+b^2\right )^3}+\frac {\left (a^3 A+3 a^2 b B-3 a A b^2-b^3 B\right ) \log (\cos (c+d x))}{d \left (a^2+b^2\right )^3}-\frac {x \left (3 a^2 A b+a^3 (-B)+3 a b^2 B-A b^3\right )}{\left (a^2+b^2\right )^3} \]

Antiderivative was successfully verified.

[In]

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

[Out]

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

Rule 31

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

Rule 3475

Int[tan[(c_.) + (d_.)*(x_)], x_Symbol] :> -Simp[Log[RemoveContent[Cos[c + d*x], x]]/d, x] /; FreeQ[{c, d}, x]

Rule 3605

Int[((a_.) + (b_.)*tan[(e_.) + (f_.)*(x_)])^(m_)*((A_.) + (B_.)*tan[(e_.) + (f_.)*(x_)])*((c_.) + (d_.)*tan[(e
_.) + (f_.)*(x_)])^(n_), x_Symbol] :> Simp[((b*c - a*d)*(B*c - A*d)*(a + b*Tan[e + f*x])^(m - 1)*(c + d*Tan[e
+ f*x])^(n + 1))/(d*f*(n + 1)*(c^2 + d^2)), x] - Dist[1/(d*(n + 1)*(c^2 + d^2)), Int[(a + b*Tan[e + f*x])^(m -
2)*(c + d*Tan[e + f*x])^(n + 1)*Simp[a*A*d*(b*d*(m - 1) - a*c*(n + 1)) + (b*B*c - (A*b + a*B)*d)*(b*c*(m - 1)
+ a*d*(n + 1)) - d*((a*A - b*B)*(b*c - a*d) + (A*b + a*B)*(a*c + b*d))*(n + 1)*Tan[e + f*x] - b*(d*(A*b*c + a
*B*c - a*A*d)*(m + n) - b*B*(c^2*(m - 1) - d^2*(n + 1)))*Tan[e + f*x]^2, x], x], x] /; FreeQ[{a, b, c, d, e, f
, A, B}, x] && NeQ[b*c - a*d, 0] && NeQ[a^2 + b^2, 0] && NeQ[c^2 + d^2, 0] && GtQ[m, 1] && LtQ[n, -1] && (Inte
gerQ[m] || IntegersQ[2*m, 2*n])

Rule 3617

Int[((a_.) + (b_.)*tan[(e_.) + (f_.)*(x_)])^(m_.)*((A_) + (C_.)*tan[(e_.) + (f_.)*(x_)]^2), x_Symbol] :> Dist[
A/(b*f), Subst[Int[(a + x)^m, x], x, b*Tan[e + f*x]], x] /; FreeQ[{a, b, e, f, A, C, m}, x] && EqQ[A, C]

Rule 3626

Int[((A_) + (B_.)*tan[(e_.) + (f_.)*(x_)] + (C_.)*tan[(e_.) + (f_.)*(x_)]^2)/((a_.) + (b_.)*tan[(e_.) + (f_.)*
(x_)]), x_Symbol] :> Simp[((a*A + b*B - a*C)*x)/(a^2 + b^2), x] + (Dist[(A*b^2 - a*b*B + a^2*C)/(a^2 + b^2), I
nt[(1 + Tan[e + f*x]^2)/(a + b*Tan[e + f*x]), x], x] - Dist[(A*b - a*B - b*C)/(a^2 + b^2), Int[Tan[e + f*x], x
], x]) /; FreeQ[{a, b, e, f, A, B, C}, x] && NeQ[A*b^2 - a*b*B + a^2*C, 0] && NeQ[a^2 + b^2, 0] && NeQ[A*b - a
*B - b*C, 0]

Rule 3635

Int[((a_.) + (b_.)*tan[(e_.) + (f_.)*(x_)])*((c_.) + (d_.)*tan[(e_.) + (f_.)*(x_)])^(n_)*((A_.) + (B_.)*tan[(e
_.) + (f_.)*(x_)] + (C_.)*tan[(e_.) + (f_.)*(x_)]^2), x_Symbol] :> -Simp[((b*c - a*d)*(c^2*C - B*c*d + A*d^2)*
(c + d*Tan[e + f*x])^(n + 1))/(d^2*f*(n + 1)*(c^2 + d^2)), x] + Dist[1/(d*(c^2 + d^2)), Int[(c + d*Tan[e + f*x
])^(n + 1)*Simp[a*d*(A*c - c*C + B*d) + b*(c^2*C - B*c*d + A*d^2) + d*(A*b*c + a*B*c - b*c*C - a*A*d + b*B*d +
a*C*d)*Tan[e + f*x] + b*C*(c^2 + d^2)*Tan[e + f*x]^2, x], x], x] /; FreeQ[{a, b, c, d, e, f, A, B, C}, x] &&
NeQ[b*c - a*d, 0] && NeQ[c^2 + d^2, 0] && LtQ[n, -1]

Rubi steps

\begin {align*} \int \frac {\tan ^3(c+d x) (A+B \tan (c+d x))}{(a+b \tan (c+d x))^3} \, dx &=\frac {a (A b-a B) \tan ^2(c+d x)}{2 b \left (a^2+b^2\right ) d (a+b \tan (c+d x))^2}+\frac {\int \frac {\tan (c+d x) \left (-2 a (A b-a B)+2 b (A b-a B) \tan (c+d x)+2 \left (a^2+b^2\right ) B \tan ^2(c+d x)\right )}{(a+b \tan (c+d x))^2} \, dx}{2 b \left (a^2+b^2\right )}\\ &=\frac {a (A b-a B) \tan ^2(c+d x)}{2 b \left (a^2+b^2\right ) d (a+b \tan (c+d x))^2}-\frac {a^2 \left (2 A b^3-a \left (a^2+3 b^2\right ) B\right )}{b^3 \left (a^2+b^2\right )^2 d (a+b \tan (c+d x))}+\frac {\int \frac {-2 a \left (2 A b^3-a \left (a^2+3 b^2\right ) B\right )-2 b^2 \left (a^2 A-A b^2+2 a b B\right ) \tan (c+d x)+2 \left (a^2+b^2\right )^2 B \tan ^2(c+d x)}{a+b \tan (c+d x)} \, dx}{2 b^2 \left (a^2+b^2\right )^2}\\ &=-\frac {\left (3 a^2 A b-A b^3-a^3 B+3 a b^2 B\right ) x}{\left (a^2+b^2\right )^3}+\frac {a (A b-a B) \tan ^2(c+d x)}{2 b \left (a^2+b^2\right ) d (a+b \tan (c+d x))^2}-\frac {a^2 \left (2 A b^3-a \left (a^2+3 b^2\right ) B\right )}{b^3 \left (a^2+b^2\right )^2 d (a+b \tan (c+d x))}-\frac {\left (a^3 A-3 a A b^2+3 a^2 b B-b^3 B\right ) \int \tan (c+d x) \, dx}{\left (a^2+b^2\right )^3}+\frac {\left (a \left (a^2 A b^3-3 A b^5+a^5 B+3 a^3 b^2 B+6 a b^4 B\right )\right ) \int \frac {1+\tan ^2(c+d x)}{a+b \tan (c+d x)} \, dx}{b^2 \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 ) x}{\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 ) \log (\cos (c+d x))}{\left (a^2+b^2\right )^3 d}+\frac {a (A b-a B) \tan ^2(c+d x)}{2 b \left (a^2+b^2\right ) d (a+b \tan (c+d x))^2}-\frac {a^2 \left (2 A b^3-a \left (a^2+3 b^2\right ) B\right )}{b^3 \left (a^2+b^2\right )^2 d (a+b \tan (c+d x))}+\frac {\left (a \left (a^2 A b^3-3 A b^5+a^5 B+3 a^3 b^2 B+6 a b^4 B\right )\right ) \operatorname {Subst}\left (\int \frac {1}{a+x} \, dx,x,b \tan (c+d x)\right )}{b^3 \left (a^2+b^2\right )^3 d}\\ &=-\frac {\left (3 a^2 A b-A b^3-a^3 B+3 a b^2 B\right ) x}{\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 ) \log (\cos (c+d x))}{\left (a^2+b^2\right )^3 d}+\frac {a \left (a^2 A b^3-3 A b^5+a^5 B+3 a^3 b^2 B+6 a b^4 B\right ) \log (a+b \tan (c+d x))}{b^3 \left (a^2+b^2\right )^3 d}+\frac {a (A b-a B) \tan ^2(c+d x)}{2 b \left (a^2+b^2\right ) d (a+b \tan (c+d x))^2}-\frac {a^2 \left (2 A b^3-a \left (a^2+3 b^2\right ) B\right )}{b^3 \left (a^2+b^2\right )^2 d (a+b \tan (c+d x))}\\ \end {align*}

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Mathematica [C]  time = 4.89, size = 462, normalized size = 1.85 \[ \frac {\sec ^2(c+d x) (A+B \tan (c+d x)) (a \cos (c+d x)+b \sin (c+d x)) \left (-2 a b \left (a^2+b^2\right ) \left (a B \left (a^2+4 b^2\right )-3 A b^3\right ) \sin (c+d x) (a \cos (c+d x)+b \sin (c+d x))-2 B \left (a^2+b^2\right )^3 \log (\cos (c+d x)) (a \cos (c+d x)+b \sin (c+d x))^2+a^3 b^2 \left (a^2+b^2\right ) (A b-a B)+2 b^3 (c+d x) \left (a^3 B-3 a^2 A b-3 a b^2 B+A b^3\right ) (a \cos (c+d x)+b \sin (c+d x))^2+2 i a (c+d x) \left (a^5 B+3 a^3 b^2 B+a^2 A b^3+6 a b^4 B-3 A b^5\right ) (a \cos (c+d x)+b \sin (c+d x))^2+a \left (a^5 B+3 a^3 b^2 B+a^2 A b^3+6 a b^4 B-3 A b^5\right ) (a \cos (c+d x)+b \sin (c+d x))^2 \log \left ((a \cos (c+d x)+b \sin (c+d x))^2\right )-2 i a \left (a^5 B+3 a^3 b^2 B+a^2 A b^3+6 a b^4 B-3 A b^5\right ) \tan ^{-1}(\tan (c+d x)) (a \cos (c+d x)+b \sin (c+d x))^2\right )}{2 b^3 d \left (a^2+b^2\right )^3 (a+b \tan (c+d x))^3 (A \cos (c+d x)+B \sin (c+d x))} \]

Antiderivative was successfully verified.

[In]

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

[Out]

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

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fricas [B]  time = 0.89, size = 666, normalized size = 2.66 \[ \frac {B a^{6} b^{2} + A a^{5} b^{3} + 7 \, B a^{4} b^{4} - 5 \, A a^{3} b^{5} + 2 \, {\left (B a^{5} b^{3} - 3 \, A a^{4} b^{4} - 3 \, B a^{3} b^{5} + A a^{2} b^{6}\right )} d x - {\left (3 \, B a^{6} b^{2} - A a^{5} b^{3} + 9 \, B a^{4} b^{4} - 7 \, A a^{3} b^{5} - 2 \, {\left (B a^{3} b^{5} - 3 \, A a^{2} b^{6} - 3 \, B a b^{7} + A b^{8}\right )} d x\right )} \tan \left (d x + c\right )^{2} + {\left (B a^{8} + 3 \, B a^{6} b^{2} + A a^{5} b^{3} + 6 \, B a^{4} b^{4} - 3 \, A a^{3} b^{5} + {\left (B a^{6} b^{2} + 3 \, B a^{4} b^{4} + A a^{3} b^{5} + 6 \, B a^{2} b^{6} - 3 \, A a b^{7}\right )} \tan \left (d x + c\right )^{2} + 2 \, {\left (B a^{7} b + 3 \, B a^{5} b^{3} + A a^{4} b^{4} + 6 \, B a^{3} b^{5} - 3 \, A a^{2} b^{6}\right )} \tan \left (d x + c\right )\right )} \log \left (\frac {b^{2} \tan \left (d x + c\right )^{2} + 2 \, a b \tan \left (d x + c\right ) + a^{2}}{\tan \left (d x + c\right )^{2} + 1}\right ) - {\left (B a^{8} + 3 \, B a^{6} b^{2} + 3 \, B a^{4} b^{4} + B a^{2} b^{6} + {\left (B a^{6} b^{2} + 3 \, B a^{4} b^{4} + 3 \, B a^{2} b^{6} + B b^{8}\right )} \tan \left (d x + c\right )^{2} + 2 \, {\left (B a^{7} b + 3 \, B a^{5} b^{3} + 3 \, B a^{3} b^{5} + B a b^{7}\right )} \tan \left (d x + c\right )\right )} \log \left (\frac {1}{\tan \left (d x + c\right )^{2} + 1}\right ) - 2 \, {\left (B a^{7} b + 3 \, B a^{5} b^{3} - 3 \, A a^{4} b^{4} - 4 \, B a^{3} b^{5} + 3 \, A a^{2} b^{6} - 2 \, {\left (B a^{4} b^{4} - 3 \, A a^{3} b^{5} - 3 \, B a^{2} b^{6} + A a b^{7}\right )} d x\right )} \tan \left (d x + c\right )}{2 \, {\left ({\left (a^{6} b^{5} + 3 \, a^{4} b^{7} + 3 \, a^{2} b^{9} + b^{11}\right )} d \tan \left (d x + c\right )^{2} + 2 \, {\left (a^{7} b^{4} + 3 \, a^{5} b^{6} + 3 \, a^{3} b^{8} + a b^{10}\right )} d \tan \left (d x + c\right ) + {\left (a^{8} b^{3} + 3 \, a^{6} b^{5} + 3 \, a^{4} b^{7} + a^{2} b^{9}\right )} d\right )}} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

1/2*(B*a^6*b^2 + A*a^5*b^3 + 7*B*a^4*b^4 - 5*A*a^3*b^5 + 2*(B*a^5*b^3 - 3*A*a^4*b^4 - 3*B*a^3*b^5 + A*a^2*b^6)
*d*x - (3*B*a^6*b^2 - A*a^5*b^3 + 9*B*a^4*b^4 - 7*A*a^3*b^5 - 2*(B*a^3*b^5 - 3*A*a^2*b^6 - 3*B*a*b^7 + A*b^8)*
d*x)*tan(d*x + c)^2 + (B*a^8 + 3*B*a^6*b^2 + A*a^5*b^3 + 6*B*a^4*b^4 - 3*A*a^3*b^5 + (B*a^6*b^2 + 3*B*a^4*b^4
+ A*a^3*b^5 + 6*B*a^2*b^6 - 3*A*a*b^7)*tan(d*x + c)^2 + 2*(B*a^7*b + 3*B*a^5*b^3 + A*a^4*b^4 + 6*B*a^3*b^5 - 3
*A*a^2*b^6)*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)) - (B*a^8 +
3*B*a^6*b^2 + 3*B*a^4*b^4 + B*a^2*b^6 + (B*a^6*b^2 + 3*B*a^4*b^4 + 3*B*a^2*b^6 + B*b^8)*tan(d*x + c)^2 + 2*(B
*a^7*b + 3*B*a^5*b^3 + 3*B*a^3*b^5 + B*a*b^7)*tan(d*x + c))*log(1/(tan(d*x + c)^2 + 1)) - 2*(B*a^7*b + 3*B*a^5
*b^3 - 3*A*a^4*b^4 - 4*B*a^3*b^5 + 3*A*a^2*b^6 - 2*(B*a^4*b^4 - 3*A*a^3*b^5 - 3*B*a^2*b^6 + A*a*b^7)*d*x)*tan(
d*x + c))/((a^6*b^5 + 3*a^4*b^7 + 3*a^2*b^9 + b^11)*d*tan(d*x + c)^2 + 2*(a^7*b^4 + 3*a^5*b^6 + 3*a^3*b^8 + a*
b^10)*d*tan(d*x + c) + (a^8*b^3 + 3*a^6*b^5 + 3*a^4*b^7 + a^2*b^9)*d)

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giac [A]  time = 1.36, size = 458, normalized size = 1.83 \[ \frac {\frac {2 \, {\left (B a^{3} - 3 \, A a^{2} b - 3 \, B a b^{2} + A b^{3}\right )} {\left (d x + c\right )}}{a^{6} + 3 \, a^{4} b^{2} + 3 \, a^{2} b^{4} + b^{6}} - \frac {{\left (A a^{3} + 3 \, B a^{2} b - 3 \, A a b^{2} - B b^{3}\right )} \log \left (\tan \left (d x + c\right )^{2} + 1\right )}{a^{6} + 3 \, a^{4} b^{2} + 3 \, a^{2} b^{4} + b^{6}} + \frac {2 \, {\left (B a^{6} + 3 \, B a^{4} b^{2} + A a^{3} b^{3} + 6 \, B a^{2} b^{4} - 3 \, A a b^{5}\right )} \log \left ({\left | b \tan \left (d x + c\right ) + a \right |}\right )}{a^{6} b^{3} + 3 \, a^{4} b^{5} + 3 \, a^{2} b^{7} + b^{9}} - \frac {3 \, B a^{6} b \tan \left (d x + c\right )^{2} + 9 \, B a^{4} b^{3} \tan \left (d x + c\right )^{2} + 3 \, A a^{3} b^{4} \tan \left (d x + c\right )^{2} + 18 \, B a^{2} b^{5} \tan \left (d x + c\right )^{2} - 9 \, A a b^{6} \tan \left (d x + c\right )^{2} + 2 \, B a^{7} \tan \left (d x + c\right ) + 2 \, A a^{6} b \tan \left (d x + c\right ) + 6 \, B a^{5} b^{2} \tan \left (d x + c\right ) + 14 \, A a^{4} b^{3} \tan \left (d x + c\right ) + 28 \, B a^{3} b^{4} \tan \left (d x + c\right ) - 12 \, A a^{2} b^{5} \tan \left (d x + c\right ) + A a^{7} - B a^{6} b + 9 \, A a^{5} b^{2} + 11 \, B a^{4} b^{3} - 4 \, A a^{3} b^{4}}{{\left (a^{6} b^{2} + 3 \, a^{4} b^{4} + 3 \, a^{2} b^{6} + b^{8}\right )} {\left (b \tan \left (d x + c\right ) + a\right )}^{2}}}{2 \, d} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

1/2*(2*(B*a^3 - 3*A*a^2*b - 3*B*a*b^2 + A*b^3)*(d*x + c)/(a^6 + 3*a^4*b^2 + 3*a^2*b^4 + b^6) - (A*a^3 + 3*B*a^
2*b - 3*A*a*b^2 - B*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^6 + 3*B*a^4*b^2
+ A*a^3*b^3 + 6*B*a^2*b^4 - 3*A*a*b^5)*log(abs(b*tan(d*x + c) + a))/(a^6*b^3 + 3*a^4*b^5 + 3*a^2*b^7 + b^9) -
(3*B*a^6*b*tan(d*x + c)^2 + 9*B*a^4*b^3*tan(d*x + c)^2 + 3*A*a^3*b^4*tan(d*x + c)^2 + 18*B*a^2*b^5*tan(d*x + c
)^2 - 9*A*a*b^6*tan(d*x + c)^2 + 2*B*a^7*tan(d*x + c) + 2*A*a^6*b*tan(d*x + c) + 6*B*a^5*b^2*tan(d*x + c) + 14
*A*a^4*b^3*tan(d*x + c) + 28*B*a^3*b^4*tan(d*x + c) - 12*A*a^2*b^5*tan(d*x + c) + A*a^7 - B*a^6*b + 9*A*a^5*b^
2 + 11*B*a^4*b^3 - 4*A*a^3*b^4)/((a^6*b^2 + 3*a^4*b^4 + 3*a^2*b^6 + b^8)*(b*tan(d*x + c) + a)^2))/d

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maple [B]  time = 0.27, size = 566, normalized size = 2.26 \[ \frac {a^{3} \ln \left (a +b \tan \left (d x +c \right )\right ) A}{d \left (a^{2}+b^{2}\right )^{3}}-\frac {3 a \,b^{2} \ln \left (a +b \tan \left (d x +c \right )\right ) A}{d \left (a^{2}+b^{2}\right )^{3}}+\frac {a^{6} \ln \left (a +b \tan \left (d x +c \right )\right ) B}{d \left (a^{2}+b^{2}\right )^{3} b^{3}}+\frac {3 a^{4} \ln \left (a +b \tan \left (d x +c \right )\right ) B}{d \left (a^{2}+b^{2}\right )^{3} b}+\frac {6 a^{2} b \ln \left (a +b \tan \left (d x +c \right )\right ) B}{d \left (a^{2}+b^{2}\right )^{3}}-\frac {a^{4} A}{d \,b^{2} \left (a^{2}+b^{2}\right )^{2} \left (a +b \tan \left (d x +c \right )\right )}-\frac {3 a^{2} A}{d \left (a^{2}+b^{2}\right )^{2} \left (a +b \tan \left (d x +c \right )\right )}+\frac {2 a^{5} B}{d \,b^{3} \left (a^{2}+b^{2}\right )^{2} \left (a +b \tan \left (d x +c \right )\right )}+\frac {4 a^{3} B}{d b \left (a^{2}+b^{2}\right )^{2} \left (a +b \tan \left (d x +c \right )\right )}+\frac {a^{3} A}{2 d \,b^{2} \left (a^{2}+b^{2}\right ) \left (a +b \tan \left (d x +c \right )\right )^{2}}-\frac {a^{4} B}{2 d \,b^{3} \left (a^{2}+b^{2}\right ) \left (a +b \tan \left (d x +c \right )\right )^{2}}-\frac {\ln \left (1+\tan ^{2}\left (d x +c \right )\right ) A \,a^{3}}{2 d \left (a^{2}+b^{2}\right )^{3}}+\frac {3 \ln \left (1+\tan ^{2}\left (d x +c \right )\right ) A a \,b^{2}}{2 d \left (a^{2}+b^{2}\right )^{3}}-\frac {3 \ln \left (1+\tan ^{2}\left (d x +c \right )\right ) a^{2} b B}{2 d \left (a^{2}+b^{2}\right )^{3}}+\frac {\ln \left (1+\tan ^{2}\left (d x +c \right )\right ) b^{3} B}{2 d \left (a^{2}+b^{2}\right )^{3}}-\frac {3 A \arctan \left (\tan \left (d x +c \right )\right ) a^{2} b}{d \left (a^{2}+b^{2}\right )^{3}}+\frac {A \arctan \left (\tan \left (d x +c \right )\right ) b^{3}}{d \left (a^{2}+b^{2}\right )^{3}}+\frac {B \arctan \left (\tan \left (d x +c \right )\right ) a^{3}}{d \left (a^{2}+b^{2}\right )^{3}}-\frac {3 B \arctan \left (\tan \left (d x +c \right )\right ) a \,b^{2}}{d \left (a^{2}+b^{2}\right )^{3}} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

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

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maxima [A]  time = 0.78, size = 366, normalized size = 1.46 \[ \frac {\frac {2 \, {\left (B a^{3} - 3 \, A a^{2} b - 3 \, B a b^{2} + A b^{3}\right )} {\left (d x + c\right )}}{a^{6} + 3 \, a^{4} b^{2} + 3 \, a^{2} b^{4} + b^{6}} + \frac {2 \, {\left (B a^{6} + 3 \, B a^{4} b^{2} + A a^{3} b^{3} + 6 \, B a^{2} b^{4} - 3 \, A a b^{5}\right )} \log \left (b \tan \left (d x + c\right ) + a\right )}{a^{6} b^{3} + 3 \, a^{4} b^{5} + 3 \, a^{2} b^{7} + b^{9}} - \frac {{\left (A a^{3} + 3 \, B a^{2} b - 3 \, A a b^{2} - B b^{3}\right )} \log \left (\tan \left (d x + c\right )^{2} + 1\right )}{a^{6} + 3 \, a^{4} b^{2} + 3 \, a^{2} b^{4} + b^{6}} + \frac {3 \, B a^{6} - A a^{5} b + 7 \, B a^{4} b^{2} - 5 \, A a^{3} b^{3} + 2 \, {\left (2 \, B a^{5} b - A a^{4} b^{2} + 4 \, B a^{3} b^{3} - 3 \, A a^{2} b^{4}\right )} \tan \left (d x + c\right )}{a^{6} b^{3} + 2 \, a^{4} b^{5} + a^{2} b^{7} + {\left (a^{4} b^{5} + 2 \, a^{2} b^{7} + b^{9}\right )} \tan \left (d x + c\right )^{2} + 2 \, {\left (a^{5} b^{4} + 2 \, a^{3} b^{6} + a b^{8}\right )} \tan \left (d x + c\right )}}{2 \, d} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

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

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mupad [B]  time = 6.86, size = 307, normalized size = 1.23 \[ \frac {\frac {3\,B\,a^6-A\,a^5\,b+7\,B\,a^4\,b^2-5\,A\,a^3\,b^3}{2\,b^3\,\left (a^4+2\,a^2\,b^2+b^4\right )}-\frac {a^2\,\mathrm {tan}\left (c+d\,x\right )\,\left (-2\,B\,a^3+A\,a^2\,b-4\,B\,a\,b^2+3\,A\,b^3\right )}{b^2\,\left (a^4+2\,a^2\,b^2+b^4\right )}}{d\,\left (a^2+2\,a\,b\,\mathrm {tan}\left (c+d\,x\right )+b^2\,{\mathrm {tan}\left (c+d\,x\right )}^2\right )}+\frac {\ln \left (\mathrm {tan}\left (c+d\,x\right )-\mathrm {i}\right )\,\left (-B+A\,1{}\mathrm {i}\right )}{2\,d\,\left (-a^3\,1{}\mathrm {i}+3\,a^2\,b+a\,b^2\,3{}\mathrm {i}-b^3\right )}+\frac {\ln \left (\mathrm {tan}\left (c+d\,x\right )+1{}\mathrm {i}\right )\,\left (A-B\,1{}\mathrm {i}\right )}{2\,d\,\left (-a^3+a^2\,b\,3{}\mathrm {i}+3\,a\,b^2-b^3\,1{}\mathrm {i}\right )}+\frac {a\,\ln \left (a+b\,\mathrm {tan}\left (c+d\,x\right )\right )\,\left (B\,a^5+3\,B\,a^3\,b^2+A\,a^2\,b^3+6\,B\,a\,b^4-3\,A\,b^5\right )}{b^3\,d\,{\left (a^2+b^2\right )}^3} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

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

[Out]

Exception raised: AttributeError

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