3.282 \(\int \frac{\tan ^4(c+d x) (A+B \tan (c+d x))}{(a+b \tan (c+d x))^3} \, dx\)

Optimal. Leaf size=331 \[ \frac{a (A b-a B) \tan ^3(c+d x)}{2 b d \left (a^2+b^2\right ) (a+b \tan (c+d x))^2}+\frac{a \left (a^2 A b-3 a^3 B-7 a b^2 B+5 A b^3\right ) \tan ^2(c+d x)}{2 b^2 d \left (a^2+b^2\right )^2 (a+b \tan (c+d x))}-\frac{\left (a^3 A b-6 a^2 b^2 B-3 a^4 B+3 a A b^3-b^4 B\right ) \tan (c+d x)}{b^3 d \left (a^2+b^2\right )^2}+\frac{a^2 \left (3 a^2 A b^3+a^4 A b-9 a^3 b^2 B-3 a^5 B-10 a b^4 B+6 A b^5\right ) \log (a+b \tan (c+d x))}{b^4 d \left (a^2+b^2\right )^3}+\frac{\left (3 a^2 A b+a^3 (-B)+3 a b^2 B-A b^3\right ) \log (\cos (c+d x))}{d \left (a^2+b^2\right )^3}+\frac{x \left (a^3 A+3 a^2 b B-3 a A b^2-b^3 B\right )}{\left (a^2+b^2\right )^3} \]

[Out]

((a^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[Cos[c
+ d*x]])/((a^2 + b^2)^3*d) + (a^2*(a^4*A*b + 3*a^2*A*b^3 + 6*A*b^5 - 3*a^5*B - 9*a^3*b^2*B - 10*a*b^4*B)*Log[
a + b*Tan[c + d*x]])/(b^4*(a^2 + b^2)^3*d) - ((a^3*A*b + 3*a*A*b^3 - 3*a^4*B - 6*a^2*b^2*B - b^4*B)*Tan[c + d*
x])/(b^3*(a^2 + b^2)^2*d) + (a*(A*b - a*B)*Tan[c + d*x]^3)/(2*b*(a^2 + b^2)*d*(a + b*Tan[c + d*x])^2) + (a*(a^
2*A*b + 5*A*b^3 - 3*a^3*B - 7*a*b^2*B)*Tan[c + d*x]^2)/(2*b^2*(a^2 + b^2)^2*d*(a + b*Tan[c + d*x]))

________________________________________________________________________________________

Rubi [A]  time = 0.798083, antiderivative size = 331, normalized size of antiderivative = 1., number of steps used = 7, number of rules used = 7, integrand size = 31, \(\frac{\text{number of rules}}{\text{integrand size}}\) = 0.226, Rules used = {3605, 3645, 3647, 3626, 3617, 31, 3475} \[ \frac{a (A b-a B) \tan ^3(c+d x)}{2 b d \left (a^2+b^2\right ) (a+b \tan (c+d x))^2}+\frac{a \left (a^2 A b-3 a^3 B-7 a b^2 B+5 A b^3\right ) \tan ^2(c+d x)}{2 b^2 d \left (a^2+b^2\right )^2 (a+b \tan (c+d x))}-\frac{\left (a^3 A b-6 a^2 b^2 B-3 a^4 B+3 a A b^3-b^4 B\right ) \tan (c+d x)}{b^3 d \left (a^2+b^2\right )^2}+\frac{a^2 \left (3 a^2 A b^3+a^4 A b-9 a^3 b^2 B-3 a^5 B-10 a b^4 B+6 A b^5\right ) \log (a+b \tan (c+d x))}{b^4 d \left (a^2+b^2\right )^3}+\frac{\left (3 a^2 A b+a^3 (-B)+3 a b^2 B-A b^3\right ) \log (\cos (c+d x))}{d \left (a^2+b^2\right )^3}+\frac{x \left (a^3 A+3 a^2 b B-3 a A b^2-b^3 B\right )}{\left (a^2+b^2\right )^3} \]

Antiderivative was successfully verified.

[In]

Int[(Tan[c + d*x]^4*(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[Cos[c
+ d*x]])/((a^2 + b^2)^3*d) + (a^2*(a^4*A*b + 3*a^2*A*b^3 + 6*A*b^5 - 3*a^5*B - 9*a^3*b^2*B - 10*a*b^4*B)*Log[
a + b*Tan[c + d*x]])/(b^4*(a^2 + b^2)^3*d) - ((a^3*A*b + 3*a*A*b^3 - 3*a^4*B - 6*a^2*b^2*B - b^4*B)*Tan[c + d*
x])/(b^3*(a^2 + b^2)^2*d) + (a*(A*b - a*B)*Tan[c + d*x]^3)/(2*b*(a^2 + b^2)*d*(a + b*Tan[c + d*x])^2) + (a*(a^
2*A*b + 5*A*b^3 - 3*a^3*B - 7*a*b^2*B)*Tan[c + d*x]^2)/(2*b^2*(a^2 + b^2)^2*d*(a + b*Tan[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 3645

Int[((a_.) + (b_.)*tan[(e_.) + (f_.)*(x_)])^(m_)*((c_.) + (d_.)*tan[(e_.) + (f_.)*(x_)])^(n_)*((A_.) + (B_.)*t
an[(e_.) + (f_.)*(x_)] + (C_.)*tan[(e_.) + (f_.)*(x_)]^2), x_Symbol] :> Simp[((A*d^2 + c*(c*C - B*d))*(a + b*T
an[e + f*x])^m*(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)), I
nt[(a + b*Tan[e + f*x])^(m - 1)*(c + d*Tan[e + f*x])^(n + 1)*Simp[A*d*(b*d*m - a*c*(n + 1)) + (c*C - B*d)*(b*c
*m + a*d*(n + 1)) - d*(n + 1)*((A - C)*(b*c - a*d) + B*(a*c + b*d))*Tan[e + f*x] - b*(d*(B*c - A*d)*(m + n + 1
) - C*(c^2*m - d^2*(n + 1)))*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[a^2 + b^2, 0] && NeQ[c^2 + d^2, 0] && GtQ[m, 0] && LtQ[n, -1]

Rule 3647

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

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 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 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]

Rubi steps

\begin{align*} \int \frac{\tan ^4(c+d x) (A+B \tan (c+d x))}{(a+b \tan (c+d x))^3} \, dx &=\frac{a (A b-a B) \tan ^3(c+d x)}{2 b \left (a^2+b^2\right ) d (a+b \tan (c+d x))^2}+\frac{\int \frac{\tan ^2(c+d x) \left (-3 a (A b-a B)+2 b (A b-a B) \tan (c+d x)-\left (a A b-3 a^2 B-2 b^2 B\right ) \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 ^3(c+d x)}{2 b \left (a^2+b^2\right ) d (a+b \tan (c+d x))^2}+\frac{a \left (a^2 A b+5 A b^3-3 a^3 B-7 a b^2 B\right ) \tan ^2(c+d x)}{2 b^2 \left (a^2+b^2\right )^2 d (a+b \tan (c+d x))}+\frac{\int \frac{\tan (c+d x) \left (-2 a \left (a^2 A b+5 A b^3-3 a^3 B-7 a b^2 B\right )-2 b^2 \left (a^2 A-A b^2+2 a b B\right ) \tan (c+d x)-2 \left (a^3 A b+3 a A b^3-3 a^4 B-6 a^2 b^2 B-b^4 B\right ) \tan ^2(c+d x)\right )}{a+b \tan (c+d x)} \, dx}{2 b^2 \left (a^2+b^2\right )^2}\\ &=-\frac{\left (a^3 A b+3 a A b^3-3 a^4 B-6 a^2 b^2 B-b^4 B\right ) \tan (c+d x)}{b^3 \left (a^2+b^2\right )^2 d}+\frac{a (A b-a B) \tan ^3(c+d x)}{2 b \left (a^2+b^2\right ) d (a+b \tan (c+d x))^2}+\frac{a \left (a^2 A b+5 A b^3-3 a^3 B-7 a b^2 B\right ) \tan ^2(c+d x)}{2 b^2 \left (a^2+b^2\right )^2 d (a+b \tan (c+d x))}+\frac{\int \frac{2 a \left (a^3 A b+3 a A b^3-3 a^4 B-6 a^2 b^2 B-b^4 B\right )-2 b^3 \left (2 a A b-a^2 B+b^2 B\right ) \tan (c+d x)+2 \left (a^2+b^2\right )^2 (A b-3 a B) \tan ^2(c+d x)}{a+b \tan (c+d x)} \, dx}{2 b^3 \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{\left (a^3 A b+3 a A b^3-3 a^4 B-6 a^2 b^2 B-b^4 B\right ) \tan (c+d x)}{b^3 \left (a^2+b^2\right )^2 d}+\frac{a (A b-a B) \tan ^3(c+d x)}{2 b \left (a^2+b^2\right ) d (a+b \tan (c+d x))^2}+\frac{a \left (a^2 A b+5 A b^3-3 a^3 B-7 a b^2 B\right ) \tan ^2(c+d x)}{2 b^2 \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 \tan (c+d x) \, dx}{\left (a^2+b^2\right )^3}+\frac{\left (a^2 \left (a^4 A b+3 a^2 A b^3+6 A b^5-3 a^5 B-9 a^3 b^2 B-10 a b^4 B\right )\right ) \int \frac{1+\tan ^2(c+d x)}{a+b \tan (c+d x)} \, dx}{b^3 \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 (\cos (c+d x))}{\left (a^2+b^2\right )^3 d}-\frac{\left (a^3 A b+3 a A b^3-3 a^4 B-6 a^2 b^2 B-b^4 B\right ) \tan (c+d x)}{b^3 \left (a^2+b^2\right )^2 d}+\frac{a (A b-a B) \tan ^3(c+d x)}{2 b \left (a^2+b^2\right ) d (a+b \tan (c+d x))^2}+\frac{a \left (a^2 A b+5 A b^3-3 a^3 B-7 a b^2 B\right ) \tan ^2(c+d x)}{2 b^2 \left (a^2+b^2\right )^2 d (a+b \tan (c+d x))}+\frac{\left (a^2 \left (a^4 A b+3 a^2 A b^3+6 A b^5-3 a^5 B-9 a^3 b^2 B-10 a b^4 B\right )\right ) \operatorname{Subst}\left (\int \frac{1}{a+x} \, dx,x,b \tan (c+d x)\right )}{b^4 \left (a^2+b^2\right )^3 d}\\ &=\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 (\cos (c+d x))}{\left (a^2+b^2\right )^3 d}+\frac{a^2 \left (a^4 A b+3 a^2 A b^3+6 A b^5-3 a^5 B-9 a^3 b^2 B-10 a b^4 B\right ) \log (a+b \tan (c+d x))}{b^4 \left (a^2+b^2\right )^3 d}-\frac{\left (a^3 A b+3 a A b^3-3 a^4 B-6 a^2 b^2 B-b^4 B\right ) \tan (c+d x)}{b^3 \left (a^2+b^2\right )^2 d}+\frac{a (A b-a B) \tan ^3(c+d x)}{2 b \left (a^2+b^2\right ) d (a+b \tan (c+d x))^2}+\frac{a \left (a^2 A b+5 A b^3-3 a^3 B-7 a b^2 B\right ) \tan ^2(c+d x)}{2 b^2 \left (a^2+b^2\right )^2 d (a+b \tan (c+d x))}\\ \end{align*}

Mathematica [C]  time = 6.67341, size = 1146, normalized size = 3.46 \[ \frac{(a B-A b) \sec ^2(c+d x) (a \cos (c+d x)+b \sin (c+d x)) (A+B \tan (c+d x)) a^4}{2 (a-i b)^2 (a+i b)^2 b^2 d (A \cos (c+d x)+B \sin (c+d x)) (a+b \tan (c+d x))^3}+\frac{\sec ^2(c+d x) (a \cos (c+d x)+b \sin (c+d x))^2 \left (2 B \sin (c+d x) a^5-A b \sin (c+d x) a^4+5 b^2 B \sin (c+d x) a^3-4 A b^3 \sin (c+d x) a^2\right ) (A+B \tan (c+d x))}{(a-i b)^2 (a+i b)^2 b^3 d (A \cos (c+d x)+B \sin (c+d x)) (a+b \tan (c+d x))^3}+\frac{B \sec ^2(c+d x) (a \cos (c+d x)+b \sin (c+d x))^3 \tan (c+d x) (A+B \tan (c+d x))}{b^3 d (A \cos (c+d x)+B \sin (c+d x)) (a+b \tan (c+d x))^3}+\frac{\left (A a^3+3 b B a^2-3 A b^2 a-b^3 B\right ) (c+d x) \sec ^2(c+d x) (a \cos (c+d x)+b \sin (c+d x))^3 (A+B \tan (c+d x))}{(a-i b)^3 (a+i b)^3 d (A \cos (c+d x)+B \sin (c+d x)) (a+b \tan (c+d x))^3}+\frac{\left (6 a^2 A b^{13}+6 i a^3 A b^{12}-10 a^3 B b^{12}+15 a^4 A b^{11}-10 i a^4 B b^{11}+15 i a^5 A b^{10}-29 a^5 B b^{10}+13 a^6 A b^9-29 i a^6 B b^9+13 i a^7 A b^8-31 a^7 B b^8+5 a^8 A b^7-31 i a^8 B b^7+5 i a^9 A b^6-15 a^9 B b^6+a^{10} A b^5-15 i a^{10} B b^5+i a^{11} A b^4-3 a^{11} B b^4-3 i a^{12} B b^3\right ) (c+d x) \sec ^2(c+d x) (a \cos (c+d x)+b \sin (c+d x))^3 (A+B \tan (c+d x))}{(a-i b)^6 (a+i b)^5 b^7 d (A \cos (c+d x)+B \sin (c+d x)) (a+b \tan (c+d x))^3}-\frac{i \left (-3 B a^7+A b a^6-9 b^2 B a^5+3 A b^3 a^4-10 b^4 B a^3+6 A b^5 a^2\right ) \tan ^{-1}(\tan (c+d x)) \sec ^2(c+d x) (a \cos (c+d x)+b \sin (c+d x))^3 (A+B \tan (c+d x))}{b^4 \left (a^2+b^2\right )^3 d (A \cos (c+d x)+B \sin (c+d x)) (a+b \tan (c+d x))^3}+\frac{(3 a B-A b) \log (\cos (c+d x)) \sec ^2(c+d x) (a \cos (c+d x)+b \sin (c+d x))^3 (A+B \tan (c+d x))}{b^4 d (A \cos (c+d x)+B \sin (c+d x)) (a+b \tan (c+d x))^3}+\frac{\left (-3 B a^7+A b a^6-9 b^2 B a^5+3 A b^3 a^4-10 b^4 B a^3+6 A b^5 a^2\right ) \log \left ((a \cos (c+d x)+b \sin (c+d x))^2\right ) \sec ^2(c+d x) (a \cos (c+d x)+b \sin (c+d x))^3 (A+B \tan (c+d x))}{2 b^4 \left (a^2+b^2\right )^3 d (A \cos (c+d x)+B \sin (c+d x)) (a+b \tan (c+d x))^3} \]

Antiderivative was successfully verified.

[In]

Integrate[(Tan[c + d*x]^4*(A + B*Tan[c + d*x]))/(a + b*Tan[c + d*x])^3,x]

[Out]

(a^4*(-(A*b) + a*B)*Sec[c + d*x]^2*(a*Cos[c + d*x] + b*Sin[c + d*x])*(A + B*Tan[c + d*x]))/(2*(a - I*b)^2*(a +
I*b)^2*b^2*d*(A*Cos[c + d*x] + B*Sin[c + d*x])*(a + b*Tan[c + d*x])^3) + ((a^3*A - 3*a*A*b^2 + 3*a^2*b*B - b^
3*B)*(c + d*x)*Sec[c + d*x]^2*(a*Cos[c + d*x] + b*Sin[c + d*x])^3*(A + B*Tan[c + d*x]))/((a - I*b)^3*(a + I*b)
^3*d*(A*Cos[c + d*x] + B*Sin[c + d*x])*(a + b*Tan[c + d*x])^3) + ((I*a^11*A*b^4 + a^10*A*b^5 + (5*I)*a^9*A*b^6
+ 5*a^8*A*b^7 + (13*I)*a^7*A*b^8 + 13*a^6*A*b^9 + (15*I)*a^5*A*b^10 + 15*a^4*A*b^11 + (6*I)*a^3*A*b^12 + 6*a^
2*A*b^13 - (3*I)*a^12*b^3*B - 3*a^11*b^4*B - (15*I)*a^10*b^5*B - 15*a^9*b^6*B - (31*I)*a^8*b^7*B - 31*a^7*b^8*
B - (29*I)*a^6*b^9*B - 29*a^5*b^10*B - (10*I)*a^4*b^11*B - 10*a^3*b^12*B)*(c + d*x)*Sec[c + d*x]^2*(a*Cos[c +
d*x] + b*Sin[c + d*x])^3*(A + B*Tan[c + d*x]))/((a - I*b)^6*(a + I*b)^5*b^7*d*(A*Cos[c + d*x] + B*Sin[c + d*x]
)*(a + b*Tan[c + d*x])^3) - (I*(a^6*A*b + 3*a^4*A*b^3 + 6*a^2*A*b^5 - 3*a^7*B - 9*a^5*b^2*B - 10*a^3*b^4*B)*Ar
cTan[Tan[c + d*x]]*Sec[c + d*x]^2*(a*Cos[c + d*x] + b*Sin[c + d*x])^3*(A + B*Tan[c + d*x]))/(b^4*(a^2 + b^2)^3
*d*(A*Cos[c + d*x] + B*Sin[c + d*x])*(a + b*Tan[c + d*x])^3) + ((-(A*b) + 3*a*B)*Log[Cos[c + d*x]]*Sec[c + d*x
]^2*(a*Cos[c + d*x] + b*Sin[c + d*x])^3*(A + B*Tan[c + d*x]))/(b^4*d*(A*Cos[c + d*x] + B*Sin[c + d*x])*(a + b*
Tan[c + d*x])^3) + ((a^6*A*b + 3*a^4*A*b^3 + 6*a^2*A*b^5 - 3*a^7*B - 9*a^5*b^2*B - 10*a^3*b^4*B)*Log[(a*Cos[c
+ d*x] + b*Sin[c + d*x])^2]*Sec[c + d*x]^2*(a*Cos[c + d*x] + b*Sin[c + d*x])^3*(A + B*Tan[c + d*x]))/(2*b^4*(a
^2 + b^2)^3*d*(A*Cos[c + d*x] + B*Sin[c + d*x])*(a + b*Tan[c + d*x])^3) + (Sec[c + d*x]^2*(a*Cos[c + d*x] + b*
Sin[c + d*x])^2*(-(a^4*A*b*Sin[c + d*x]) - 4*a^2*A*b^3*Sin[c + d*x] + 2*a^5*B*Sin[c + d*x] + 5*a^3*b^2*B*Sin[c
+ d*x])*(A + B*Tan[c + d*x]))/((a - I*b)^2*(a + I*b)^2*b^3*d*(A*Cos[c + d*x] + B*Sin[c + d*x])*(a + b*Tan[c +
d*x])^3) + (B*Sec[c + d*x]^2*(a*Cos[c + d*x] + b*Sin[c + d*x])^3*Tan[c + d*x]*(A + B*Tan[c + d*x]))/(b^3*d*(A
*Cos[c + d*x] + B*Sin[c + d*x])*(a + b*Tan[c + d*x])^3)

________________________________________________________________________________________

Maple [A]  time = 0.05, size = 619, normalized size = 1.9 \begin{align*} \text{result too large to display} \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

int(tan(d*x+c)^4*(A+B*tan(d*x+c))/(a+b*tan(d*x+c))^3,x)

[Out]

1/d*B/b^3*tan(d*x+c)-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)*B*a^3-3/2/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*arcta
n(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+1/d/b^3*a^6/(a^2+b^2)^3*ln(a+b*tan(d*x+c))*A+3/d/b*a^4/(a^2+b^2)^3*ln(a+b*t
an(d*x+c))*A+6/d*b*a^2/(a^2+b^2)^3*ln(a+b*tan(d*x+c))*A-3/d/b^4*a^7/(a^2+b^2)^3*ln(a+b*tan(d*x+c))*B-9/d/b^2*a
^5/(a^2+b^2)^3*ln(a+b*tan(d*x+c))*B-10/d*a^3/(a^2+b^2)^3*ln(a+b*tan(d*x+c))*B-1/2/d/b^3*a^4/(a^2+b^2)/(a+b*tan
(d*x+c))^2*A+1/2/d/b^4*a^5/(a^2+b^2)/(a+b*tan(d*x+c))^2*B+2/d/b^3*a^5/(a^2+b^2)^2/(a+b*tan(d*x+c))*A+4/d/b*a^3
/(a^2+b^2)^2/(a+b*tan(d*x+c))*A-3/d/b^4*a^6/(a^2+b^2)^2/(a+b*tan(d*x+c))*B-5/d/b^2*a^4/(a^2+b^2)^2/(a+b*tan(d*
x+c))*B

________________________________________________________________________________________

Maxima [A]  time = 1.556, size = 525, normalized size = 1.59 \begin{align*} \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 (3 \, B a^{7} - A a^{6} b + 9 \, B a^{5} b^{2} - 3 \, A a^{4} b^{3} + 10 \, B a^{3} b^{4} - 6 \, A a^{2} b^{5}\right )} \log \left (b \tan \left (d x + c\right ) + a\right )}{a^{6} b^{4} + 3 \, a^{4} b^{6} + 3 \, a^{2} b^{8} + b^{10}} + \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{5 \, B a^{7} - 3 \, A a^{6} b + 9 \, B a^{5} b^{2} - 7 \, A a^{4} b^{3} + 2 \,{\left (3 \, B a^{6} b - 2 \, A a^{5} b^{2} + 5 \, B a^{4} b^{3} - 4 \, A a^{3} b^{4}\right )} \tan \left (d x + c\right )}{a^{6} b^{4} + 2 \, a^{4} b^{6} + a^{2} b^{8} +{\left (a^{4} b^{6} + 2 \, a^{2} b^{8} + b^{10}\right )} \tan \left (d x + c\right )^{2} + 2 \,{\left (a^{5} b^{5} + 2 \, a^{3} b^{7} + a b^{9}\right )} \tan \left (d x + c\right )} + \frac{2 \, B \tan \left (d x + c\right )}{b^{3}}}{2 \, d} \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

________________________________________________________________________________________

Fricas [B]  time = 2.91801, size = 1895, normalized size = 5.73 \begin{align*} \text{result too large to display} \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate(tan(d*x+c)^4*(A+B*tan(d*x+c))/(a+b*tan(d*x+c))^3,x, algorithm="fricas")

[Out]

-1/2*(3*B*a^7*b^2 - A*a^6*b^3 + 9*B*a^5*b^4 - 7*A*a^4*b^5 - 2*(B*a^6*b^3 + 3*B*a^4*b^5 + 3*B*a^2*b^7 + B*b^9)*
tan(d*x + c)^3 - 2*(A*a^5*b^4 + 3*B*a^4*b^5 - 3*A*a^3*b^6 - B*a^2*b^7)*d*x - (9*B*a^7*b^2 - 3*A*a^6*b^3 + 23*B
*a^5*b^4 - 9*A*a^4*b^5 + 12*B*a^3*b^6 + 4*B*a*b^8 + 2*(A*a^3*b^6 + 3*B*a^2*b^7 - 3*A*a*b^8 - B*b^9)*d*x)*tan(d
*x + c)^2 + (3*B*a^9 - A*a^8*b + 9*B*a^7*b^2 - 3*A*a^6*b^3 + 10*B*a^5*b^4 - 6*A*a^4*b^5 + (3*B*a^7*b^2 - A*a^6
*b^3 + 9*B*a^5*b^4 - 3*A*a^4*b^5 + 10*B*a^3*b^6 - 6*A*a^2*b^7)*tan(d*x + c)^2 + 2*(3*B*a^8*b - A*a^7*b^2 + 9*B
*a^6*b^3 - 3*A*a^5*b^4 + 10*B*a^4*b^5 - 6*A*a^3*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)) - (3*B*a^9 - A*a^8*b + 9*B*a^7*b^2 - 3*A*a^6*b^3 + 9*B*a^5*b^4 - 3*A*a^4*b^5 +
3*B*a^3*b^6 - A*a^2*b^7 + (3*B*a^7*b^2 - A*a^6*b^3 + 9*B*a^5*b^4 - 3*A*a^4*b^5 + 9*B*a^3*b^6 - 3*A*a^2*b^7 + 3
*B*a*b^8 - A*b^9)*tan(d*x + c)^2 + 2*(3*B*a^8*b - A*a^7*b^2 + 9*B*a^6*b^3 - 3*A*a^5*b^4 + 9*B*a^4*b^5 - 3*A*a^
3*b^6 + 3*B*a^2*b^7 - A*a*b^8)*tan(d*x + c))*log(1/(tan(d*x + c)^2 + 1)) - 2*(3*B*a^8*b - A*a^7*b^2 + 6*B*a^6*
b^3 - 3*A*a^5*b^4 - 2*B*a^4*b^5 + 4*A*a^3*b^6 + B*a^2*b^7 + 2*(A*a^4*b^5 + 3*B*a^3*b^6 - 3*A*a^2*b^7 - B*a*b^8
)*d*x)*tan(d*x + c))/((a^6*b^6 + 3*a^4*b^8 + 3*a^2*b^10 + b^12)*d*tan(d*x + c)^2 + 2*(a^7*b^5 + 3*a^5*b^7 + 3*
a^3*b^9 + a*b^11)*d*tan(d*x + c) + (a^8*b^4 + 3*a^6*b^6 + 3*a^4*b^8 + a^2*b^10)*d)

________________________________________________________________________________________

Sympy [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(tan(d*x+c)**4*(A+B*tan(d*x+c))/(a+b*tan(d*x+c))**3,x)

[Out]

Timed out

________________________________________________________________________________________

Giac [A]  time = 2.45969, size = 682, normalized size = 2.06 \begin{align*} \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 (3 \, B a^{7} - A a^{6} b + 9 \, B a^{5} b^{2} - 3 \, A a^{4} b^{3} + 10 \, B a^{3} b^{4} - 6 \, A a^{2} b^{5}\right )} \log \left ({\left | b \tan \left (d x + c\right ) + a \right |}\right )}{a^{6} b^{4} + 3 \, a^{4} b^{6} + 3 \, a^{2} b^{8} + b^{10}} + \frac{2 \, B \tan \left (d x + c\right )}{b^{3}} + \frac{9 \, B a^{7} b^{2} \tan \left (d x + c\right )^{2} - 3 \, A a^{6} b^{3} \tan \left (d x + c\right )^{2} + 27 \, B a^{5} b^{4} \tan \left (d x + c\right )^{2} - 9 \, A a^{4} b^{5} \tan \left (d x + c\right )^{2} + 30 \, B a^{3} b^{6} \tan \left (d x + c\right )^{2} - 18 \, A a^{2} b^{7} \tan \left (d x + c\right )^{2} + 12 \, B a^{8} b \tan \left (d x + c\right ) - 2 \, A a^{7} b^{2} \tan \left (d x + c\right ) + 38 \, B a^{6} b^{3} \tan \left (d x + c\right ) - 6 \, A a^{5} b^{4} \tan \left (d x + c\right ) + 50 \, B a^{4} b^{5} \tan \left (d x + c\right ) - 28 \, A a^{3} b^{6} \tan \left (d x + c\right ) + 4 \, B a^{9} + 13 \, B a^{7} b^{2} + A a^{6} b^{3} + 21 \, B a^{5} b^{4} - 11 \, A a^{4} b^{5}}{{\left (a^{6} b^{4} + 3 \, a^{4} b^{6} + 3 \, a^{2} b^{8} + b^{10}\right )}{\left (b \tan \left (d x + c\right ) + a\right )}^{2}}}{2 \, d} \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate(tan(d*x+c)^4*(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*(3*B*a^7 - A*a^6*b +
9*B*a^5*b^2 - 3*A*a^4*b^3 + 10*B*a^3*b^4 - 6*A*a^2*b^5)*log(abs(b*tan(d*x + c) + a))/(a^6*b^4 + 3*a^4*b^6 + 3*
a^2*b^8 + b^10) + 2*B*tan(d*x + c)/b^3 + (9*B*a^7*b^2*tan(d*x + c)^2 - 3*A*a^6*b^3*tan(d*x + c)^2 + 27*B*a^5*b
^4*tan(d*x + c)^2 - 9*A*a^4*b^5*tan(d*x + c)^2 + 30*B*a^3*b^6*tan(d*x + c)^2 - 18*A*a^2*b^7*tan(d*x + c)^2 + 1
2*B*a^8*b*tan(d*x + c) - 2*A*a^7*b^2*tan(d*x + c) + 38*B*a^6*b^3*tan(d*x + c) - 6*A*a^5*b^4*tan(d*x + c) + 50*
B*a^4*b^5*tan(d*x + c) - 28*A*a^3*b^6*tan(d*x + c) + 4*B*a^9 + 13*B*a^7*b^2 + A*a^6*b^3 + 21*B*a^5*b^4 - 11*A*
a^4*b^5)/((a^6*b^4 + 3*a^4*b^6 + 3*a^2*b^8 + b^10)*(b*tan(d*x + c) + a)^2))/d