3.1311 \(\int \frac{(a+b \tan (e+f x))^m}{(c+d \tan (e+f x))^3} \, dx\)

Optimal. Leaf size=455 \[ \frac{d^2 \left (2 a^2 d^2 \left (3 c^2-d^2\right )-4 a b c d \left (c^2 (3-m)-d^2 (m+1)\right )+b^2 \left (-\left (2 c^2 d^2 \left (-m^2+3 m+1\right )+c^4 \left (-\left (m^2-5 m+6\right )\right )+d^4 (1-m) m\right )\right )\right ) (a+b \tan (e+f x))^{m+1} \, _2F_1\left (1,m+1;m+2;-\frac{d (a+b \tan (e+f x))}{b c-a d}\right )}{2 f (m+1) \left (c^2+d^2\right )^3 (b c-a d)^3}-\frac{d^2 \left (4 a c d-b \left (c^2 (5-m)+d^2 (1-m)\right )\right ) (a+b \tan (e+f x))^{m+1}}{2 f \left (c^2+d^2\right )^2 (b c-a d)^2 (c+d \tan (e+f x))}+\frac{d^2 (a+b \tan (e+f x))^{m+1}}{2 f \left (c^2+d^2\right ) (b c-a d) (c+d \tan (e+f x))^2}+\frac{(a+b \tan (e+f x))^{m+1} \, _2F_1\left (1,m+1;m+2;\frac{a+b \tan (e+f x)}{a-i b}\right )}{2 f (m+1) (b+i a) (c-i d)^3}+\frac{(a+b \tan (e+f x))^{m+1} \, _2F_1\left (1,m+1;m+2;\frac{a+b \tan (e+f x)}{a+i b}\right )}{2 f (m+1) (a+i b) (-d+i c)^3} \]

[Out]

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

________________________________________________________________________________________

Rubi [A]  time = 1.49052, antiderivative size = 455, normalized size of antiderivative = 1., number of steps used = 10, number of rules used = 7, integrand size = 25, \(\frac{\text{number of rules}}{\text{integrand size}}\) = 0.28, Rules used = {3569, 3649, 3653, 3539, 3537, 68, 3634} \[ \frac{d^2 \left (2 a^2 d^2 \left (3 c^2-d^2\right )-4 a b c d \left (c^2 (3-m)-d^2 (m+1)\right )+b^2 \left (-\left (2 c^2 d^2 \left (-m^2+3 m+1\right )+c^4 \left (-\left (m^2-5 m+6\right )\right )+d^4 (1-m) m\right )\right )\right ) (a+b \tan (e+f x))^{m+1} \, _2F_1\left (1,m+1;m+2;-\frac{d (a+b \tan (e+f x))}{b c-a d}\right )}{2 f (m+1) \left (c^2+d^2\right )^3 (b c-a d)^3}-\frac{d^2 \left (4 a c d-b c^2 (5-m)-b d^2 (1-m)\right ) (a+b \tan (e+f x))^{m+1}}{2 f \left (c^2+d^2\right )^2 (b c-a d)^2 (c+d \tan (e+f x))}+\frac{d^2 (a+b \tan (e+f x))^{m+1}}{2 f \left (c^2+d^2\right ) (b c-a d) (c+d \tan (e+f x))^2}+\frac{(a+b \tan (e+f x))^{m+1} \, _2F_1\left (1,m+1;m+2;\frac{a+b \tan (e+f x)}{a-i b}\right )}{2 f (m+1) (b+i a) (c-i d)^3}+\frac{(a+b \tan (e+f x))^{m+1} \, _2F_1\left (1,m+1;m+2;\frac{a+b \tan (e+f x)}{a+i b}\right )}{2 f (m+1) (a+i b) (-d+i c)^3} \]

Antiderivative was successfully verified.

[In]

Int[(a + b*Tan[e + f*x])^m/(c + d*Tan[e + f*x])^3,x]

[Out]

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

Rule 3569

Int[((a_.) + (b_.)*tan[(e_.) + (f_.)*(x_)])^(m_)*((c_.) + (d_.)*tan[(e_.) + (f_.)*(x_)])^(n_), x_Symbol] :> Si
mp[(b^2*(a + b*Tan[e + f*x])^(m + 1)*(c + d*Tan[e + f*x])^(n + 1))/(f*(m + 1)*(a^2 + b^2)*(b*c - a*d)), x] + D
ist[1/((m + 1)*(a^2 + b^2)*(b*c - a*d)), Int[(a + b*Tan[e + f*x])^(m + 1)*(c + d*Tan[e + f*x])^n*Simp[a*(b*c -
a*d)*(m + 1) - b^2*d*(m + n + 2) - b*(b*c - a*d)*(m + 1)*Tan[e + f*x] - b^2*d*(m + n + 2)*Tan[e + f*x]^2, x],
x], x] /; FreeQ[{a, b, c, d, e, f, n}, x] && NeQ[b*c - a*d, 0] && NeQ[a^2 + b^2, 0] && NeQ[c^2 + d^2, 0] && I
ntegerQ[2*m] && LtQ[m, -1] && (LtQ[n, 0] || IntegerQ[m]) &&  !(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 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 68

Int[((a_) + (b_.)*(x_))^(m_)*((c_) + (d_.)*(x_))^(n_), x_Symbol] :> Simp[((b*c - a*d)^n*(a + b*x)^(m + 1)*Hype
rgeometric2F1[-n, m + 1, m + 2, -((d*(a + b*x))/(b*c - a*d))])/(b^(n + 1)*(m + 1)), x] /; FreeQ[{a, b, c, d, m
}, x] && NeQ[b*c - a*d, 0] &&  !IntegerQ[m] && IntegerQ[n]

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]

Rubi steps

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

Mathematica [A]  time = 6.26372, size = 670, normalized size = 1.47 \[ -\frac{d^2 (a+b \tan (e+f x))^{m+1}}{2 f \left (c^2+d^2\right ) (a d-b c) (c+d \tan (e+f x))^2}-\frac{-\frac{\left (d^2 \left (2 c (b c-a d)+b d^2 (1-m)\right )-c \left (-2 d^2 (b c-a d)-b c d^2 (1-m)\right )\right ) (a+b \tan (e+f x))^{m+1}}{f \left (c^2+d^2\right ) (a d-b c) (c+d \tan (e+f x))}-\frac{-\frac{\left (b c^2 d^2 m \left (4 a c d-b c^2 (5-m)-b d^2 (1-m)\right )+d^2 \left (d^2 (-(2 a d-b c (3-m))) (a d-b c (m+1))-\left (-2 a c d+2 b c^2+b d^2 (1-m)\right ) \left (a c d-b \left (c^2-d^2 m\right )\right )\right )+4 c^2 d^2 (b c-a d)^2\right ) (a+b \tan (e+f x))^{m+1} \, _2F_1\left (1,m+1;m+2;\frac{d (a+b \tan (e+f x))}{a d-b c}\right )}{f (m+1) \left (c^2+d^2\right ) (a d-b c)}+\frac{\frac{i \left (2 c \left (c^2-3 d^2\right ) (b c-a d)^2-2 i d \left (3 c^2-d^2\right ) (b c-a d)^2\right ) (a+b \tan (e+f x))^{m+1} \, _2F_1\left (1,m+1;m+2;\frac{-i a-i b \tan (e+f x)}{b-i a}\right )}{2 f (m+1) (a+i b)}-\frac{i \left (2 c \left (c^2-3 d^2\right ) (b c-a d)^2+2 i d \left (3 c^2-d^2\right ) (b c-a d)^2\right ) (a+b \tan (e+f x))^{m+1} \, _2F_1\left (1,m+1;m+2;-\frac{i a+i b \tan (e+f x)}{-i a-b}\right )}{2 f (m+1) (a-i b)}}{c^2+d^2}}{\left (c^2+d^2\right ) (a d-b c)}}{2 \left (c^2+d^2\right ) (a d-b c)} \]

Antiderivative was successfully verified.

[In]

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

[Out]

-(d^2*(a + b*Tan[e + f*x])^(1 + m))/(2*(-(b*c) + a*d)*(c^2 + d^2)*f*(c + d*Tan[e + f*x])^2) - (-(((d^2*(2*c*(b
*c - a*d) + b*d^2*(1 - m)) - c*(-2*d^2*(b*c - a*d) - b*c*d^2*(1 - m)))*(a + b*Tan[e + f*x])^(1 + m))/((-(b*c)
+ a*d)*(c^2 + d^2)*f*(c + d*Tan[e + f*x]))) - (-(((4*c^2*d^2*(b*c - a*d)^2 + b*c^2*d^2*(4*a*c*d - b*d^2*(1 - m
) - b*c^2*(5 - m))*m + d^2*(-(d^2*(2*a*d - b*c*(3 - m))*(a*d - b*c*(1 + m))) - (2*b*c^2 - 2*a*c*d + b*d^2*(1 -
m))*(a*c*d - b*(c^2 - d^2*m))))*Hypergeometric2F1[1, 1 + m, 2 + m, (d*(a + b*Tan[e + f*x]))/(-(b*c) + a*d)]*(
a + b*Tan[e + f*x])^(1 + m))/((-(b*c) + a*d)*(c^2 + d^2)*f*(1 + m))) + (((I/2)*(2*c*(b*c - a*d)^2*(c^2 - 3*d^2
) - (2*I)*d*(b*c - a*d)^2*(3*c^2 - d^2))*Hypergeometric2F1[1, 1 + m, 2 + m, ((-I)*a - I*b*Tan[e + f*x])/((-I)*
a + b)]*(a + b*Tan[e + f*x])^(1 + m))/((a + I*b)*f*(1 + m)) - ((I/2)*(2*c*(b*c - a*d)^2*(c^2 - 3*d^2) + (2*I)*
d*(b*c - a*d)^2*(3*c^2 - d^2))*Hypergeometric2F1[1, 1 + m, 2 + m, -((I*a + I*b*Tan[e + f*x])/((-I)*a - b))]*(a
+ b*Tan[e + f*x])^(1 + m))/((a - I*b)*f*(1 + m)))/(c^2 + d^2))/((-(b*c) + a*d)*(c^2 + d^2)))/(2*(-(b*c) + a*d
)*(c^2 + d^2))

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Maple [F]  time = 0.483, size = 0, normalized size = 0. \begin{align*} \int{\frac{ \left ( a+b\tan \left ( fx+e \right ) \right ) ^{m}}{ \left ( c+d\tan \left ( fx+e \right ) \right ) ^{3}}}\, dx \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

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

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Maxima [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(f*x+e))^m/(c+d*tan(f*x+e))^3,x, algorithm="maxima")

[Out]

Timed out

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

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

integral((b*tan(f*x + e) + a)^m/(d^3*tan(f*x + e)^3 + 3*c*d^2*tan(f*x + e)^2 + 3*c^2*d*tan(f*x + e) + c^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((a+b*tan(f*x+e))**m/(c+d*tan(f*x+e))**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 (f x + e\right ) + a\right )}^{m}}{{\left (d \tan \left (f x + e\right ) + c\right )}^{3}}\,{d x} \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

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