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

Optimal. Leaf size=438 \[ -\frac{b \left (2 a^2 A b^3 \left (-m^2+3 m+1\right )-a^4 A b \left (m^2-5 m+6\right )-2 a^3 b^2 B \left (-m^2+m+3\right )+a^5 B \left (m^2-3 m+2\right )+a b^4 B m (m+1)+A b^5 (1-m) m\right ) \tan ^{m+1}(c+d x) \text{Hypergeometric2F1}\left (1,m+1,m+2,-\frac{b \tan (c+d x)}{a}\right )}{2 a^3 d (m+1) \left (a^2+b^2\right )^3}+\frac{\left (a^3 A+3 a^2 b B-3 a A b^2-b^3 B\right ) \tan ^{m+1}(c+d x) \text{Hypergeometric2F1}\left (1,\frac{m+1}{2},\frac{m+3}{2},-\tan ^2(c+d x)\right )}{d (m+1) \left (a^2+b^2\right )^3}-\frac{\left (3 a^2 A b+a^3 (-B)+3 a b^2 B-A b^3\right ) \tan ^{m+2}(c+d x) \text{Hypergeometric2F1}\left (1,\frac{m+2}{2},\frac{m+4}{2},-\tan ^2(c+d x)\right )}{d (m+2) \left (a^2+b^2\right )^3}+\frac{b \left (a^2 A b (5-m)+a^3 (-B) (3-m)+a b^2 B (m+1)+A b^3 (1-m)\right ) \tan ^{m+1}(c+d x)}{2 a^2 d \left (a^2+b^2\right )^2 (a+b \tan (c+d x))}+\frac{b (A b-a B) \tan ^{m+1}(c+d x)}{2 a d \left (a^2+b^2\right ) (a+b \tan (c+d x))^2} \]

[Out]

((a^3*A - 3*a*A*b^2 + 3*a^2*b*B - b^3*B)*Hypergeometric2F1[1, (1 + m)/2, (3 + m)/2, -Tan[c + d*x]^2]*Tan[c + d
*x]^(1 + m))/((a^2 + b^2)^3*d*(1 + m)) - (b*(A*b^5*(1 - m)*m + a*b^4*B*m*(1 + m) - 2*a^3*b^2*B*(3 + m - m^2) +
2*a^2*A*b^3*(1 + 3*m - m^2) - a^4*A*b*(6 - 5*m + m^2) + a^5*B*(2 - 3*m + m^2))*Hypergeometric2F1[1, 1 + m, 2
+ m, -((b*Tan[c + d*x])/a)]*Tan[c + d*x]^(1 + m))/(2*a^3*(a^2 + b^2)^3*d*(1 + m)) - ((3*a^2*A*b - A*b^3 - a^3*
B + 3*a*b^2*B)*Hypergeometric2F1[1, (2 + m)/2, (4 + m)/2, -Tan[c + d*x]^2]*Tan[c + d*x]^(2 + m))/((a^2 + b^2)^
3*d*(2 + m)) + (b*(A*b - a*B)*Tan[c + d*x]^(1 + m))/(2*a*(a^2 + b^2)*d*(a + b*Tan[c + d*x])^2) + (b*(A*b^3*(1
- m) - a^3*B*(3 - m) + a^2*A*b*(5 - m) + a*b^2*B*(1 + m))*Tan[c + d*x]^(1 + m))/(2*a^2*(a^2 + b^2)^2*d*(a + b*
Tan[c + d*x]))

________________________________________________________________________________________

Rubi [A]  time = 1.28485, antiderivative size = 438, normalized size of antiderivative = 1., number of steps used = 10, number of rules used = 8, integrand size = 31, \(\frac{\text{number of rules}}{\text{integrand size}}\) = 0.258, Rules used = {3609, 3649, 3653, 3538, 3476, 364, 3634, 64} \[ -\frac{b \left (2 a^2 A b^3 \left (-m^2+3 m+1\right )-a^4 A b \left (m^2-5 m+6\right )-2 a^3 b^2 B \left (-m^2+m+3\right )+a^5 B \left (m^2-3 m+2\right )+a b^4 B m (m+1)+A b^5 (1-m) m\right ) \tan ^{m+1}(c+d x) \, _2F_1\left (1,m+1;m+2;-\frac{b \tan (c+d x)}{a}\right )}{2 a^3 d (m+1) \left (a^2+b^2\right )^3}+\frac{\left (a^3 A+3 a^2 b B-3 a A b^2-b^3 B\right ) \tan ^{m+1}(c+d x) \, _2F_1\left (1,\frac{m+1}{2};\frac{m+3}{2};-\tan ^2(c+d x)\right )}{d (m+1) \left (a^2+b^2\right )^3}-\frac{\left (3 a^2 A b+a^3 (-B)+3 a b^2 B-A b^3\right ) \tan ^{m+2}(c+d x) \, _2F_1\left (1,\frac{m+2}{2};\frac{m+4}{2};-\tan ^2(c+d x)\right )}{d (m+2) \left (a^2+b^2\right )^3}+\frac{b \left (a^2 A b (5-m)+a^3 (-B) (3-m)+a b^2 B (m+1)+A b^3 (1-m)\right ) \tan ^{m+1}(c+d x)}{2 a^2 d \left (a^2+b^2\right )^2 (a+b \tan (c+d x))}+\frac{b (A b-a B) \tan ^{m+1}(c+d x)}{2 a d \left (a^2+b^2\right ) (a+b \tan (c+d x))^2} \]

Antiderivative was successfully verified.

[In]

Int[(Tan[c + d*x]^m*(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)*Hypergeometric2F1[1, (1 + m)/2, (3 + m)/2, -Tan[c + d*x]^2]*Tan[c + d
*x]^(1 + m))/((a^2 + b^2)^3*d*(1 + m)) - (b*(A*b^5*(1 - m)*m + a*b^4*B*m*(1 + m) - 2*a^3*b^2*B*(3 + m - m^2) +
2*a^2*A*b^3*(1 + 3*m - m^2) - a^4*A*b*(6 - 5*m + m^2) + a^5*B*(2 - 3*m + m^2))*Hypergeometric2F1[1, 1 + m, 2
+ m, -((b*Tan[c + d*x])/a)]*Tan[c + d*x]^(1 + m))/(2*a^3*(a^2 + b^2)^3*d*(1 + m)) - ((3*a^2*A*b - A*b^3 - a^3*
B + 3*a*b^2*B)*Hypergeometric2F1[1, (2 + m)/2, (4 + m)/2, -Tan[c + d*x]^2]*Tan[c + d*x]^(2 + m))/((a^2 + b^2)^
3*d*(2 + m)) + (b*(A*b - a*B)*Tan[c + d*x]^(1 + m))/(2*a*(a^2 + b^2)*d*(a + b*Tan[c + d*x])^2) + (b*(A*b^3*(1
- m) - a^3*B*(3 - m) + a^2*A*b*(5 - m) + a*b^2*B*(1 + m))*Tan[c + d*x]^(1 + m))/(2*a^2*(a^2 + b^2)^2*d*(a + b*
Tan[c + d*x]))

Rule 3609

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*(A*b - a*B)*(a + b*Tan[e + f*x])^(m + 1)*(c + d*Tan[e + f*x])^(n
+ 1))/(f*(m + 1)*(b*c - a*d)*(a^2 + b^2)), x] + Dist[1/((m + 1)*(b*c - a*d)*(a^2 + b^2)), Int[(a + b*Tan[e +
f*x])^(m + 1)*(c + d*Tan[e + f*x])^n*Simp[b*B*(b*c*(m + 1) + a*d*(n + 1)) + A*(a*(b*c - a*d)*(m + 1) - b^2*d*(
m + n + 2)) - (A*b - a*B)*(b*c - a*d)*(m + 1)*Tan[e + f*x] - b*d*(A*b - a*B)*(m + n + 2)*Tan[e + f*x]^2, x], x
], x] /; FreeQ[{a, b, c, d, e, f, A, B, n}, x] && NeQ[b*c - a*d, 0] && NeQ[a^2 + b^2, 0] && NeQ[c^2 + d^2, 0]
&& LtQ[m, -1] && (IntegerQ[m] || IntegersQ[2*m, 2*n]) &&  !(ILtQ[n, -1] && ( !IntegerQ[m] || (EqQ[c, 0] && NeQ
[a, 0])))

Rule 3649

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*b^2 - a*(b*B - a*C))*(a + b*T
an[e + f*x])^(m + 1)*(c + d*Tan[e + f*x])^(n + 1))/(f*(m + 1)*(b*c - a*d)*(a^2 + b^2)), x] + Dist[1/((m + 1)*(
b*c - a*d)*(a^2 + b^2)), Int[(a + b*Tan[e + f*x])^(m + 1)*(c + d*Tan[e + f*x])^n*Simp[A*(a*(b*c - a*d)*(m + 1)
- b^2*d*(m + n + 2)) + (b*B - a*C)*(b*c*(m + 1) + a*d*(n + 1)) - (m + 1)*(b*c - a*d)*(A*b - a*B - b*C)*Tan[e
+ f*x] - d*(A*b^2 - a*(b*B - a*C))*(m + n + 2)*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] && LtQ[m, -1] &&  !(ILtQ[n, -1] && ( !I
ntegerQ[m] || (EqQ[c, 0] && NeQ[a, 0])))

Rule 3653

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

Rule 3538

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

Rule 3476

Int[((b_.)*tan[(c_.) + (d_.)*(x_)])^(n_), x_Symbol] :> Dist[b/d, Subst[Int[x^n/(b^2 + x^2), x], x, b*Tan[c + d
*x]], x] /; FreeQ[{b, c, d, n}, x] &&  !IntegerQ[n]

Rule 364

Int[((c_.)*(x_))^(m_.)*((a_) + (b_.)*(x_)^(n_))^(p_), x_Symbol] :> Simp[(a^p*(c*x)^(m + 1)*Hypergeometric2F1[-
p, (m + 1)/n, (m + 1)/n + 1, -((b*x^n)/a)])/(c*(m + 1)), x] /; FreeQ[{a, b, c, m, n, p}, x] &&  !IGtQ[p, 0] &&
(ILtQ[p, 0] || GtQ[a, 0])

Rule 3634

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

Rule 64

Int[((b_.)*(x_))^(m_)*((c_) + (d_.)*(x_))^(n_), x_Symbol] :> Simp[(c^n*(b*x)^(m + 1)*Hypergeometric2F1[-n, m +
1, m + 2, -((d*x)/c)])/(b*(m + 1)), x] /; FreeQ[{b, c, d, m, n}, x] &&  !IntegerQ[m] && (IntegerQ[n] || (GtQ[
c, 0] &&  !(EqQ[n, -2^(-1)] && EqQ[c^2 - d^2, 0] && GtQ[-(d/(b*c)), 0])))

Rubi steps

\begin{align*} \int \frac{\tan ^m(c+d x) (A+B \tan (c+d x))}{(a+b \tan (c+d x))^3} \, dx &=\frac{b (A b-a B) \tan ^{1+m}(c+d x)}{2 a \left (a^2+b^2\right ) d (a+b \tan (c+d x))^2}+\frac{\int \frac{\tan ^m(c+d x) \left (2 a^2 A+A b^2 (1-m)+a b B (1+m)-2 a (A b-a B) \tan (c+d x)+b (A b-a B) (1-m) \tan ^2(c+d x)\right )}{(a+b \tan (c+d x))^2} \, dx}{2 a \left (a^2+b^2\right )}\\ &=\frac{b (A b-a B) \tan ^{1+m}(c+d x)}{2 a \left (a^2+b^2\right ) d (a+b \tan (c+d x))^2}+\frac{b \left (A b^3 (1-m)-a^3 B (3-m)+a^2 A b (5-m)+a b^2 B (1+m)\right ) \tan ^{1+m}(c+d x)}{2 a^2 \left (a^2+b^2\right )^2 d (a+b \tan (c+d x))}+\frac{\int \frac{\tan ^m(c+d x) \left (-a^2 b (A b-a B) (3-m) (1+m)+\left (a^2-b^2 m\right ) \left (2 a^2 A+A b^2 (1-m)+a b B (1+m)\right )-2 a^2 \left (2 a A b-a^2 B+b^2 B\right ) \tan (c+d x)-b m \left (A b^3 (1-m)-a^3 B (3-m)+a^2 A b (5-m)+a b^2 B (1+m)\right ) \tan ^2(c+d x)\right )}{a+b \tan (c+d x)} \, dx}{2 a^2 \left (a^2+b^2\right )^2}\\ &=\frac{b (A b-a B) \tan ^{1+m}(c+d x)}{2 a \left (a^2+b^2\right ) d (a+b \tan (c+d x))^2}+\frac{b \left (A b^3 (1-m)-a^3 B (3-m)+a^2 A b (5-m)+a b^2 B (1+m)\right ) \tan ^{1+m}(c+d x)}{2 a^2 \left (a^2+b^2\right )^2 d (a+b \tan (c+d x))}+\frac{\int \tan ^m(c+d x) \left (2 a^2 \left (a^3 A-3 a A b^2+3 a^2 b B-b^3 B\right )-2 a^2 \left (3 a^2 A b-A b^3-a^3 B+3 a b^2 B\right ) \tan (c+d x)\right ) \, dx}{2 a^2 \left (a^2+b^2\right )^3}-\frac{\left (b \left (A b^5 (1-m) m+a b^4 B m (1+m)-2 a^3 b^2 B \left (3+m-m^2\right )+2 a^2 A b^3 \left (1+3 m-m^2\right )-a^4 A b \left (6-5 m+m^2\right )+a^5 B \left (2-3 m+m^2\right )\right )\right ) \int \frac{\tan ^m(c+d x) \left (1+\tan ^2(c+d x)\right )}{a+b \tan (c+d x)} \, dx}{2 a^2 \left (a^2+b^2\right )^3}\\ &=\frac{b (A b-a B) \tan ^{1+m}(c+d x)}{2 a \left (a^2+b^2\right ) d (a+b \tan (c+d x))^2}+\frac{b \left (A b^3 (1-m)-a^3 B (3-m)+a^2 A b (5-m)+a b^2 B (1+m)\right ) \tan ^{1+m}(c+d x)}{2 a^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 ^{1+m}(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 ) \int \tan ^m(c+d x) \, dx}{\left (a^2+b^2\right )^3}-\frac{\left (b \left (A b^5 (1-m) m+a b^4 B m (1+m)-2 a^3 b^2 B \left (3+m-m^2\right )+2 a^2 A b^3 \left (1+3 m-m^2\right )-a^4 A b \left (6-5 m+m^2\right )+a^5 B \left (2-3 m+m^2\right )\right )\right ) \operatorname{Subst}\left (\int \frac{x^m}{a+b x} \, dx,x,\tan (c+d x)\right )}{2 a^2 \left (a^2+b^2\right )^3 d}\\ &=-\frac{b \left (A b^5 (1-m) m+a b^4 B m (1+m)-2 a^3 b^2 B \left (3+m-m^2\right )+2 a^2 A b^3 \left (1+3 m-m^2\right )-a^4 A b \left (6-5 m+m^2\right )+a^5 B \left (2-3 m+m^2\right )\right ) \, _2F_1\left (1,1+m;2+m;-\frac{b \tan (c+d x)}{a}\right ) \tan ^{1+m}(c+d x)}{2 a^3 \left (a^2+b^2\right )^3 d (1+m)}+\frac{b (A b-a B) \tan ^{1+m}(c+d x)}{2 a \left (a^2+b^2\right ) d (a+b \tan (c+d x))^2}+\frac{b \left (A b^3 (1-m)-a^3 B (3-m)+a^2 A b (5-m)+a b^2 B (1+m)\right ) \tan ^{1+m}(c+d x)}{2 a^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 ) \operatorname{Subst}\left (\int \frac{x^{1+m}}{1+x^2} \, dx,x,\tan (c+d x)\right )}{\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 ) \operatorname{Subst}\left (\int \frac{x^m}{1+x^2} \, dx,x,\tan (c+d x)\right )}{\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 ) \, _2F_1\left (1,\frac{1+m}{2};\frac{3+m}{2};-\tan ^2(c+d x)\right ) \tan ^{1+m}(c+d x)}{\left (a^2+b^2\right )^3 d (1+m)}-\frac{b \left (A b^5 (1-m) m+a b^4 B m (1+m)-2 a^3 b^2 B \left (3+m-m^2\right )+2 a^2 A b^3 \left (1+3 m-m^2\right )-a^4 A b \left (6-5 m+m^2\right )+a^5 B \left (2-3 m+m^2\right )\right ) \, _2F_1\left (1,1+m;2+m;-\frac{b \tan (c+d x)}{a}\right ) \tan ^{1+m}(c+d x)}{2 a^3 \left (a^2+b^2\right )^3 d (1+m)}-\frac{\left (3 a^2 A b-A b^3-a^3 B+3 a b^2 B\right ) \, _2F_1\left (1,\frac{2+m}{2};\frac{4+m}{2};-\tan ^2(c+d x)\right ) \tan ^{2+m}(c+d x)}{\left (a^2+b^2\right )^3 d (2+m)}+\frac{b (A b-a B) \tan ^{1+m}(c+d x)}{2 a \left (a^2+b^2\right ) d (a+b \tan (c+d x))^2}+\frac{b \left (A b^3 (1-m)-a^3 B (3-m)+a^2 A b (5-m)+a b^2 B (1+m)\right ) \tan ^{1+m}(c+d x)}{2 a^2 \left (a^2+b^2\right )^2 d (a+b \tan (c+d x))}\\ \end{align*}

Mathematica [A]  time = 6.24346, size = 534, normalized size = 1.22 \[ \frac{\frac{\frac{\left (-a^2 b m \left (a^2 A b (5-m)+a^3 (-B) (3-m)+a b^2 B (m+1)+A b^3 (1-m)\right )+b^2 \left (\left (a^2-b^2 m\right ) \left (2 a^2 A+a b B (m+1)+A b^2 (1-m)\right )-a^2 b (3-m) (m+1) (A b-a B)\right )+2 a^3 b \left (a^2 (-B)+2 a A b+b^2 B\right )\right ) \tan ^{m+1}(c+d x) \text{Hypergeometric2F1}\left (1,m+1,m+2,-\frac{b \tan (c+d x)}{a}\right )}{a d (m+1) \left (a^2+b^2\right )}+\frac{\frac{2 a^2 \left (a^3 A+3 a^2 b B-3 a A b^2-b^3 B\right ) \tan ^{m+1}(c+d x) \text{Hypergeometric2F1}\left (1,\frac{m+1}{2},\frac{m+1}{2}+1,-\tan ^2(c+d x)\right )}{d (m+1)}-\frac{2 a^2 \left (3 a^2 A b+a^3 (-B)+3 a b^2 B-A b^3\right ) \tan ^{m+2}(c+d x) \text{Hypergeometric2F1}\left (1,\frac{m+2}{2},\frac{m+2}{2}+1,-\tan ^2(c+d x)\right )}{d (m+2)}}{a^2+b^2}}{a \left (a^2+b^2\right )}+\frac{\left (b^2 \left (2 a^2 A+a b B (m+1)+A b^2 (1-m)\right )-a (-a b (1-m) (A b-a B)-2 a b (A b-a B))\right ) \tan ^{m+1}(c+d x)}{a d \left (a^2+b^2\right ) (a+b \tan (c+d x))}}{2 a \left (a^2+b^2\right )}+\frac{b (A b-a B) \tan ^{m+1}(c+d x)}{2 a d \left (a^2+b^2\right ) (a+b \tan (c+d x))^2} \]

Antiderivative was successfully verified.

[In]

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

[Out]

(b*(A*b - a*B)*Tan[c + d*x]^(1 + m))/(2*a*(a^2 + b^2)*d*(a + b*Tan[c + d*x])^2) + (((-(a*(-2*a*b*(A*b - a*B) -
a*b*(A*b - a*B)*(1 - m))) + b^2*(2*a^2*A + A*b^2*(1 - m) + a*b*B*(1 + m)))*Tan[c + d*x]^(1 + m))/(a*(a^2 + b^
2)*d*(a + b*Tan[c + d*x])) + (((2*a^3*b*(2*a*A*b - a^2*B + b^2*B) - a^2*b*m*(A*b^3*(1 - m) - a^3*B*(3 - m) + a
^2*A*b*(5 - m) + a*b^2*B*(1 + m)) + b^2*(-(a^2*b*(A*b - a*B)*(3 - m)*(1 + m)) + (a^2 - b^2*m)*(2*a^2*A + A*b^2
*(1 - m) + a*b*B*(1 + m))))*Hypergeometric2F1[1, 1 + m, 2 + m, -((b*Tan[c + d*x])/a)]*Tan[c + d*x]^(1 + m))/(a
*(a^2 + b^2)*d*(1 + m)) + ((2*a^2*(a^3*A - 3*a*A*b^2 + 3*a^2*b*B - b^3*B)*Hypergeometric2F1[1, (1 + m)/2, 1 +
(1 + m)/2, -Tan[c + d*x]^2]*Tan[c + d*x]^(1 + m))/(d*(1 + m)) - (2*a^2*(3*a^2*A*b - A*b^3 - a^3*B + 3*a*b^2*B)
*Hypergeometric2F1[1, (2 + m)/2, 1 + (2 + m)/2, -Tan[c + d*x]^2]*Tan[c + d*x]^(2 + m))/(d*(2 + m)))/(a^2 + b^2
))/(a*(a^2 + b^2)))/(2*a*(a^2 + b^2))

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Maple [F]  time = 0.578, size = 0, normalized size = 0. \begin{align*} \int{\frac{ \left ( \tan \left ( dx+c \right ) \right ) ^{m} \left ( A+B\tan \left ( dx+c \right ) \right ) }{ \left ( a+b\tan \left ( dx+c \right ) \right ) ^{3}}}\, dx \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

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

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

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

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

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

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

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

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

[Out]

Timed out

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

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

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