3.165 \(\int (a+b \tan (e+f x))^m (c+d \tan (e+f x))^3 (A+B \tan (e+f x)+C \tan ^2(e+f x)) \, dx\)

Optimal. Leaf size=560 \[ \frac {d \tan (e+f x) (a+b \tan (e+f x))^{m+1} \left (b^2 d (m+3) (m+4) (d (A-C)+B c)-2 (b c-a d) (3 a C d-b (B d (m+4)+3 c C))\right )}{b^3 f (m+2) (m+3) (m+4)}+\frac {(a+b \tan (e+f x))^{m+1} \left (b c (m+2) \left (b^2 d (m+3) (m+4) (d (A-C)+B c)-2 (b c-a d) (3 a C d-b (B d (m+4)+3 c C))\right )+d \left (b^3 (m+2) (m+3) (m+4) \left (2 c d (A-C)+B \left (c^2-d^2\right )\right )-a \left (b^2 d (m+3) (m+4) (d (A-C)+B c)-2 (b c-a d) (3 a C d-b (B d (m+4)+3 c C))\right )\right )\right )}{b^4 f (m+1) (m+2) (m+3) (m+4)}+\frac {(c-i d)^3 (A-i B-C) (a+b \tan (e+f x))^{m+1} \, _2F_1\left (1,m+1;m+2;\frac {a+b \tan (e+f x)}{a-i b}\right )}{2 f (m+1) (b+i a)}-\frac {(c+i d)^3 (A+i B-C) (a+b \tan (e+f x))^{m+1} \, _2F_1\left (1,m+1;m+2;\frac {a+b \tan (e+f x)}{a+i b}\right )}{2 f (m+1) (-b+i a)}-\frac {(c+d \tan (e+f x))^2 (3 a C d-b (B d (m+4)+3 c C)) (a+b \tan (e+f x))^{m+1}}{b^2 f (m+3) (m+4)}+\frac {C (c+d \tan (e+f x))^3 (a+b \tan (e+f x))^{m+1}}{b f (m+4)} \]

[Out]

(b*c*(2+m)*(b^2*d*(B*c+(A-C)*d)*(3+m)*(4+m)-2*(-a*d+b*c)*(3*a*C*d-b*(3*c*C+B*d*(4+m))))+d*(b^3*(2*c*(A-C)*d+B*
(c^2-d^2))*(2+m)*(3+m)*(4+m)-a*(b^2*d*(B*c+(A-C)*d)*(3+m)*(4+m)-2*(-a*d+b*c)*(3*a*C*d-b*(3*c*C+B*d*(4+m))))))*
(a+b*tan(f*x+e))^(1+m)/b^4/f/(1+m)/(2+m)/(3+m)/(4+m)+1/2*(A-I*B-C)*(c-I*d)^3*hypergeom([1, 1+m],[2+m],(a+b*tan
(f*x+e))/(a-I*b))*(a+b*tan(f*x+e))^(1+m)/(I*a+b)/f/(1+m)-1/2*(A+I*B-C)*(c+I*d)^3*hypergeom([1, 1+m],[2+m],(a+b
*tan(f*x+e))/(a+I*b))*(a+b*tan(f*x+e))^(1+m)/(I*a-b)/f/(1+m)+d*(b^2*d*(B*c+(A-C)*d)*(3+m)*(4+m)-2*(-a*d+b*c)*(
3*a*C*d-b*(3*c*C+B*d*(4+m))))*tan(f*x+e)*(a+b*tan(f*x+e))^(1+m)/b^3/f/(2+m)/(3+m)/(4+m)-(3*a*C*d-b*(3*c*C+B*d*
(4+m)))*(a+b*tan(f*x+e))^(1+m)*(c+d*tan(f*x+e))^2/b^2/f/(3+m)/(4+m)+C*(a+b*tan(f*x+e))^(1+m)*(c+d*tan(f*x+e))^
3/b/f/(4+m)

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Rubi [A]  time = 2.38, antiderivative size = 551, normalized size of antiderivative = 0.98, number of steps used = 9, number of rules used = 6, integrand size = 45, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.133, Rules used = {3647, 3637, 3630, 3539, 3537, 68} \[ \frac {(a+b \tan (e+f x))^{m+1} \left (d \left (b^3 (m+2) (m+3) (m+4) \left (2 c d (A-C)+B \left (c^2-d^2\right )\right )-a \left (2 (b c-a d) (-3 a C d+b B d (m+4)+3 b c C)+b^2 d (m+3) (m+4) (d (A-C)+B c)\right )\right )+b c (m+2) \left (2 (b c-a d) (-3 a C d+b B d (m+4)+3 b c C)+b^2 d (m+3) (m+4) (d (A-C)+B c)\right )\right )}{b^4 f (m+1) (m+2) (m+3) (m+4)}+\frac {d \tan (e+f x) (a+b \tan (e+f x))^{m+1} \left (2 (b c-a d) (-3 a C d+b B d (m+4)+3 b c C)+b^2 d (m+3) (m+4) (d (A-C)+B c)\right )}{b^3 f (m+2) (m+3) (m+4)}+\frac {(c-i d)^3 (A-i B-C) (a+b \tan (e+f x))^{m+1} \, _2F_1\left (1,m+1;m+2;\frac {a+b \tan (e+f x)}{a-i b}\right )}{2 f (m+1) (b+i a)}-\frac {(c+i d)^3 (A+i B-C) (a+b \tan (e+f x))^{m+1} \, _2F_1\left (1,m+1;m+2;\frac {a+b \tan (e+f x)}{a+i b}\right )}{2 f (m+1) (-b+i a)}+\frac {(c+d \tan (e+f x))^2 (-3 a C d+b B d (m+4)+3 b c C) (a+b \tan (e+f x))^{m+1}}{b^2 f (m+3) (m+4)}+\frac {C (c+d \tan (e+f x))^3 (a+b \tan (e+f x))^{m+1}}{b f (m+4)} \]

Antiderivative was successfully verified.

[In]

Int[(a + b*Tan[e + f*x])^m*(c + d*Tan[e + f*x])^3*(A + B*Tan[e + f*x] + C*Tan[e + f*x]^2),x]

[Out]

((b*c*(2 + m)*(b^2*d*(B*c + (A - C)*d)*(3 + m)*(4 + m) + 2*(b*c - a*d)*(3*b*c*C - 3*a*C*d + b*B*d*(4 + m))) +
d*(b^3*(2*c*(A - C)*d + B*(c^2 - d^2))*(2 + m)*(3 + m)*(4 + m) - a*(b^2*d*(B*c + (A - C)*d)*(3 + m)*(4 + m) +
2*(b*c - a*d)*(3*b*c*C - 3*a*C*d + b*B*d*(4 + m)))))*(a + b*Tan[e + f*x])^(1 + m))/(b^4*f*(1 + m)*(2 + m)*(3 +
m)*(4 + m)) + ((A - I*B - C)*(c - I*d)^3*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)*f*(1 + m)) - ((A + I*B - C)*(c + I*d)^3*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)*f*(1 + m)) + (d*(b^2*d*(B*c
+ (A - C)*d)*(3 + m)*(4 + m) + 2*(b*c - a*d)*(3*b*c*C - 3*a*C*d + b*B*d*(4 + m)))*Tan[e + f*x]*(a + b*Tan[e +
f*x])^(1 + m))/(b^3*f*(2 + m)*(3 + m)*(4 + m)) + ((3*b*c*C - 3*a*C*d + b*B*d*(4 + m))*(a + b*Tan[e + f*x])^(1
+ m)*(c + d*Tan[e + f*x])^2)/(b^2*f*(3 + m)*(4 + m)) + (C*(a + b*Tan[e + f*x])^(1 + m)*(c + d*Tan[e + f*x])^3)
/(b*f*(4 + m))

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 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 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 3630

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

Rule 3637

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*Tan[e + f*x]*(c + d*Tan[e + f*x])
^(n + 1))/(d*f*(n + 2)), x] - Dist[1/(d*(n + 2)), Int[(c + d*Tan[e + f*x])^n*Simp[b*c*C - a*A*d*(n + 2) - (A*b
+ a*B - b*C)*d*(n + 2)*Tan[e + f*x] - (a*C*d*(n + 2) - b*(c*C - B*d*(n + 2)))*Tan[e + f*x]^2, x], x], x] /; F
reeQ[{a, b, c, d, e, f, A, B, C, n}, x] && NeQ[b*c - a*d, 0] && NeQ[c^2 + d^2, 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])))

Rubi steps

\begin {align*} \int (a+b \tan (e+f x))^m (c+d \tan (e+f x))^3 \left (A+B \tan (e+f x)+C \tan ^2(e+f x)\right ) \, dx &=\frac {C (a+b \tan (e+f x))^{1+m} (c+d \tan (e+f x))^3}{b f (4+m)}+\frac {\int (a+b \tan (e+f x))^m (c+d \tan (e+f x))^2 \left (A b c (4+m)-C (3 a d+b c (1+m))+b (B c+(A-C) d) (4+m) \tan (e+f x)+(3 b c C-3 a C d+b B d (4+m)) \tan ^2(e+f x)\right ) \, dx}{b (4+m)}\\ &=\frac {(3 b c C-3 a C d+b B d (4+m)) (a+b \tan (e+f x))^{1+m} (c+d \tan (e+f x))^2}{b^2 f (3+m) (4+m)}+\frac {C (a+b \tan (e+f x))^{1+m} (c+d \tan (e+f x))^3}{b f (4+m)}+\frac {\int (a+b \tan (e+f x))^m (c+d \tan (e+f x)) \left (-(2 a d+b c (1+m)) (3 b c C-3 a C d+b B d (4+m))+b c (3+m) (A b c (4+m)-C (3 a d+b c (1+m)))+b^2 \left (2 c (A-C) d+B \left (c^2-d^2\right )\right ) (3+m) (4+m) \tan (e+f x)+\left (b^2 d (B c+(A-C) d) (3+m) (4+m)+2 (b c-a d) (3 b c C-3 a C d+b B d (4+m))\right ) \tan ^2(e+f x)\right ) \, dx}{b^2 (3+m) (4+m)}\\ &=\frac {d \left (b^2 d (B c+(A-C) d) (3+m) (4+m)+2 (b c-a d) (3 b c C-3 a C d+b B d (4+m))\right ) \tan (e+f x) (a+b \tan (e+f x))^{1+m}}{b^3 f (2+m) (3+m) (4+m)}+\frac {(3 b c C-3 a C d+b B d (4+m)) (a+b \tan (e+f x))^{1+m} (c+d \tan (e+f x))^2}{b^2 f (3+m) (4+m)}+\frac {C (a+b \tan (e+f x))^{1+m} (c+d \tan (e+f x))^3}{b f (4+m)}-\frac {\int (a+b \tan (e+f x))^m \left (a d \left (b^2 d (B c+(A-C) d) (3+m) (4+m)+2 (b c-a d) (3 b c C-3 a C d+b B d (4+m))\right )+b c (2+m) ((2 a d+b c (1+m)) (3 b c C-3 a C d+b B d (4+m))-b c (3+m) (A b c (4+m)-C (3 a d+b c (1+m))))-b^3 \left ((A-C) d \left (3 c^2-d^2\right )+B \left (c^3-3 c d^2\right )\right ) (2+m) \left (12+7 m+m^2\right ) \tan (e+f x)-\left (b c (2+m) \left (b^2 d (B c+(A-C) d) (3+m) (4+m)+2 (b c-a d) (3 b c C-3 a C d+b B d (4+m))\right )+d \left (b^3 \left (2 c (A-C) d+B \left (c^2-d^2\right )\right ) (2+m) (3+m) (4+m)-a \left (b^2 d (B c+(A-C) d) (3+m) (4+m)+2 (b c-a d) (3 b c C-3 a C d+b B d (4+m))\right )\right )\right ) \tan ^2(e+f x)\right ) \, dx}{b^3 (2+m) (3+m) (4+m)}\\ &=\frac {\left (b c (2+m) \left (b^2 d (B c+(A-C) d) (3+m) (4+m)+2 (b c-a d) (3 b c C-3 a C d+b B d (4+m))\right )+d \left (b^3 \left (2 c (A-C) d+B \left (c^2-d^2\right )\right ) (2+m) (3+m) (4+m)-a \left (b^2 d (B c+(A-C) d) (3+m) (4+m)+2 (b c-a d) (3 b c C-3 a C d+b B d (4+m))\right )\right )\right ) (a+b \tan (e+f x))^{1+m}}{b^4 f (1+m) (2+m) (3+m) (4+m)}+\frac {d \left (b^2 d (B c+(A-C) d) (3+m) (4+m)+2 (b c-a d) (3 b c C-3 a C d+b B d (4+m))\right ) \tan (e+f x) (a+b \tan (e+f x))^{1+m}}{b^3 f (2+m) (3+m) (4+m)}+\frac {(3 b c C-3 a C d+b B d (4+m)) (a+b \tan (e+f x))^{1+m} (c+d \tan (e+f x))^2}{b^2 f (3+m) (4+m)}+\frac {C (a+b \tan (e+f x))^{1+m} (c+d \tan (e+f x))^3}{b f (4+m)}-\frac {\int (a+b \tan (e+f x))^m \left (-b^3 \left (A c^3-c^3 C-3 B c^2 d-3 A c d^2+3 c C d^2+B d^3\right ) (2+m) (3+m) (4+m)-b^3 \left ((A-C) d \left (3 c^2-d^2\right )+B \left (c^3-3 c d^2\right )\right ) (2+m) \left (12+7 m+m^2\right ) \tan (e+f x)\right ) \, dx}{b^3 (2+m) (3+m) (4+m)}\\ &=\frac {\left (b c (2+m) \left (b^2 d (B c+(A-C) d) (3+m) (4+m)+2 (b c-a d) (3 b c C-3 a C d+b B d (4+m))\right )+d \left (b^3 \left (2 c (A-C) d+B \left (c^2-d^2\right )\right ) (2+m) (3+m) (4+m)-a \left (b^2 d (B c+(A-C) d) (3+m) (4+m)+2 (b c-a d) (3 b c C-3 a C d+b B d (4+m))\right )\right )\right ) (a+b \tan (e+f x))^{1+m}}{b^4 f (1+m) (2+m) (3+m) (4+m)}+\frac {d \left (b^2 d (B c+(A-C) d) (3+m) (4+m)+2 (b c-a d) (3 b c C-3 a C d+b B d (4+m))\right ) \tan (e+f x) (a+b \tan (e+f x))^{1+m}}{b^3 f (2+m) (3+m) (4+m)}+\frac {(3 b c C-3 a C d+b B d (4+m)) (a+b \tan (e+f x))^{1+m} (c+d \tan (e+f x))^2}{b^2 f (3+m) (4+m)}+\frac {C (a+b \tan (e+f x))^{1+m} (c+d \tan (e+f x))^3}{b f (4+m)}+\frac {1}{2} \left ((A-i B-C) (c-i d)^3\right ) \int (1+i \tan (e+f x)) (a+b \tan (e+f x))^m \, dx+\frac {1}{2} \left ((A+i B-C) (c+i d)^3\right ) \int (1-i \tan (e+f x)) (a+b \tan (e+f x))^m \, dx\\ &=\frac {\left (b c (2+m) \left (b^2 d (B c+(A-C) d) (3+m) (4+m)+2 (b c-a d) (3 b c C-3 a C d+b B d (4+m))\right )+d \left (b^3 \left (2 c (A-C) d+B \left (c^2-d^2\right )\right ) (2+m) (3+m) (4+m)-a \left (b^2 d (B c+(A-C) d) (3+m) (4+m)+2 (b c-a d) (3 b c C-3 a C d+b B d (4+m))\right )\right )\right ) (a+b \tan (e+f x))^{1+m}}{b^4 f (1+m) (2+m) (3+m) (4+m)}+\frac {d \left (b^2 d (B c+(A-C) d) (3+m) (4+m)+2 (b c-a d) (3 b c C-3 a C d+b B d (4+m))\right ) \tan (e+f x) (a+b \tan (e+f x))^{1+m}}{b^3 f (2+m) (3+m) (4+m)}+\frac {(3 b c C-3 a C d+b B d (4+m)) (a+b \tan (e+f x))^{1+m} (c+d \tan (e+f x))^2}{b^2 f (3+m) (4+m)}+\frac {C (a+b \tan (e+f x))^{1+m} (c+d \tan (e+f x))^3}{b f (4+m)}+\frac {\left ((i A+B-i C) (c-i d)^3\right ) \operatorname {Subst}\left (\int \frac {(a-i b x)^m}{-1+x} \, dx,x,i \tan (e+f x)\right )}{2 f}-\frac {\left (i (A+i B-C) (c+i d)^3\right ) \operatorname {Subst}\left (\int \frac {(a+i b x)^m}{-1+x} \, dx,x,-i \tan (e+f x)\right )}{2 f}\\ &=\frac {\left (b c (2+m) \left (b^2 d (B c+(A-C) d) (3+m) (4+m)+2 (b c-a d) (3 b c C-3 a C d+b B d (4+m))\right )+d \left (b^3 \left (2 c (A-C) d+B \left (c^2-d^2\right )\right ) (2+m) (3+m) (4+m)-a \left (b^2 d (B c+(A-C) d) (3+m) (4+m)+2 (b c-a d) (3 b c C-3 a C d+b B d (4+m))\right )\right )\right ) (a+b \tan (e+f x))^{1+m}}{b^4 f (1+m) (2+m) (3+m) (4+m)}-\frac {(i A+B-i C) (c-i d)^3 \, _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) f (1+m)}-\frac {(A+i B-C) (c+i d)^3 \, _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) f (1+m)}+\frac {d \left (b^2 d (B c+(A-C) d) (3+m) (4+m)+2 (b c-a d) (3 b c C-3 a C d+b B d (4+m))\right ) \tan (e+f x) (a+b \tan (e+f x))^{1+m}}{b^3 f (2+m) (3+m) (4+m)}+\frac {(3 b c C-3 a C d+b B d (4+m)) (a+b \tan (e+f x))^{1+m} (c+d \tan (e+f x))^2}{b^2 f (3+m) (4+m)}+\frac {C (a+b \tan (e+f x))^{1+m} (c+d \tan (e+f x))^3}{b f (4+m)}\\ \end {align*}

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Mathematica [B]  time = 6.41, size = 1390, normalized size = 2.48 \[ \frac {C (c+d \tan (e+f x))^3 (a+b \tan (e+f x))^{m+1}}{b f (m+4)}+\frac {\frac {(3 b c C-3 a d C+b B d (m+4)) (c+d \tan (e+f x))^2 (a+b \tan (e+f x))^{m+1}}{b f (m+3)}+\frac {\frac {d \left (d (B c+(A-C) d) (m+3) (m+4) b^2+2 (b c-a d) (3 b c C-3 a d C+b B d (m+4))\right ) \tan (e+f x) (a+b \tan (e+f x))^{m+1}}{b f (m+2)}-\frac {\frac {\left (d \left (a \left (d (B c+(A-C) d) (m+3) (m+4) b^2+2 (b c-a d) (3 b c C-3 a d C+b B d (m+4))\right )-b^3 \left (2 c (A-C) d+B \left (c^2-d^2\right )\right ) (m+2) (m+3) (m+4)\right )-b c (m+2) \left (d (B c+(A-C) d) (m+3) (m+4) b^2+2 (b c-a d) (3 b c C-3 a d C+b B d (m+4))\right )\right ) (a+b \tan (e+f x))^{m+1}}{b f (m+1)}+\frac {i \left (a d \left (d (B c+(A-C) d) (m+3) (m+4) b^2+2 (b c-a d) (3 b c C-3 a d C+b B d (m+4))\right )+b c (m+2) \left (d (B c+(A-C) d) (m+3) (m+4) b^2+2 (b c-a d) (3 b c C-3 a d C+b B d (m+4))\right )-b c (m+2) (b c (m+3) (A b c (m+4)-C (3 a d+b c (m+1)))-(2 a d+b c (m+1)) (3 b c C-3 a d C+b B d (m+4)))-d \left (a \left (d (B c+(A-C) d) (m+3) (m+4) b^2+2 (b c-a d) (3 b c C-3 a d C+b B d (m+4))\right )-b^3 \left (2 c (A-C) d+B \left (c^2-d^2\right )\right ) (m+2) (m+3) (m+4)\right )-i b (m+2) \left (c \left (2 c (A-C) d+B \left (c^2-d^2\right )\right ) (m+3) (m+4) b^2-d \left (d (B c+(A-C) d) (m+3) (m+4) b^2+2 (b c-a d) (3 b c C-3 a d C+b B d (m+4))\right )+d (b c (m+3) (A b c (m+4)-C (3 a d+b c (m+1)))-(2 a d+b c (m+1)) (3 b c C-3 a d C+b B d (m+4)))\right )\right ) \, _2F_1\left (1,m+1;m+2;\frac {-i a-i b \tan (e+f x)}{b-i a}\right ) (a+b \tan (e+f x))^{m+1}}{2 (a+i b) f (m+1)}-\frac {i \left (a d \left (d (B c+(A-C) d) (m+3) (m+4) b^2+2 (b c-a d) (3 b c C-3 a d C+b B d (m+4))\right )+b c (m+2) \left (d (B c+(A-C) d) (m+3) (m+4) b^2+2 (b c-a d) (3 b c C-3 a d C+b B d (m+4))\right )-b c (m+2) (b c (m+3) (A b c (m+4)-C (3 a d+b c (m+1)))-(2 a d+b c (m+1)) (3 b c C-3 a d C+b B d (m+4)))-d \left (a \left (d (B c+(A-C) d) (m+3) (m+4) b^2+2 (b c-a d) (3 b c C-3 a d C+b B d (m+4))\right )-b^3 \left (2 c (A-C) d+B \left (c^2-d^2\right )\right ) (m+2) (m+3) (m+4)\right )+i b (m+2) \left (c \left (2 c (A-C) d+B \left (c^2-d^2\right )\right ) (m+3) (m+4) b^2-d \left (d (B c+(A-C) d) (m+3) (m+4) b^2+2 (b c-a d) (3 b c C-3 a d C+b B d (m+4))\right )+d (b c (m+3) (A b c (m+4)-C (3 a d+b c (m+1)))-(2 a d+b c (m+1)) (3 b c C-3 a d C+b B d (m+4)))\right )\right ) \, _2F_1\left (1,m+1;m+2;-\frac {i a+i b \tan (e+f x)}{-i a-b}\right ) (a+b \tan (e+f x))^{m+1}}{2 (a-i b) f (m+1)}}{b (m+2)}}{b (m+3)}}{b (m+4)} \]

Antiderivative was successfully verified.

[In]

Integrate[(a + b*Tan[e + f*x])^m*(c + d*Tan[e + f*x])^3*(A + B*Tan[e + f*x] + C*Tan[e + f*x]^2),x]

[Out]

(C*(a + b*Tan[e + f*x])^(1 + m)*(c + d*Tan[e + f*x])^3)/(b*f*(4 + m)) + (((3*b*c*C - 3*a*C*d + b*B*d*(4 + m))*
(a + b*Tan[e + f*x])^(1 + m)*(c + d*Tan[e + f*x])^2)/(b*f*(3 + m)) + ((d*(b^2*d*(B*c + (A - C)*d)*(3 + m)*(4 +
m) + 2*(b*c - a*d)*(3*b*c*C - 3*a*C*d + b*B*d*(4 + m)))*Tan[e + f*x]*(a + b*Tan[e + f*x])^(1 + m))/(b*f*(2 +
m)) - (((-(b*c*(2 + m)*(b^2*d*(B*c + (A - C)*d)*(3 + m)*(4 + m) + 2*(b*c - a*d)*(3*b*c*C - 3*a*C*d + b*B*d*(4
+ m)))) + d*(-(b^3*(2*c*(A - C)*d + B*(c^2 - d^2))*(2 + m)*(3 + m)*(4 + m)) + a*(b^2*d*(B*c + (A - C)*d)*(3 +
m)*(4 + m) + 2*(b*c - a*d)*(3*b*c*C - 3*a*C*d + b*B*d*(4 + m)))))*(a + b*Tan[e + f*x])^(1 + m))/(b*f*(1 + m))
+ ((I/2)*(a*d*(b^2*d*(B*c + (A - C)*d)*(3 + m)*(4 + m) + 2*(b*c - a*d)*(3*b*c*C - 3*a*C*d + b*B*d*(4 + m))) +
b*c*(2 + m)*(b^2*d*(B*c + (A - C)*d)*(3 + m)*(4 + m) + 2*(b*c - a*d)*(3*b*c*C - 3*a*C*d + b*B*d*(4 + m))) - b*
c*(2 + m)*(-((2*a*d + b*c*(1 + m))*(3*b*c*C - 3*a*C*d + b*B*d*(4 + m))) + b*c*(3 + m)*(A*b*c*(4 + m) - C*(3*a*
d + b*c*(1 + m)))) - d*(-(b^3*(2*c*(A - C)*d + B*(c^2 - d^2))*(2 + m)*(3 + m)*(4 + m)) + a*(b^2*d*(B*c + (A -
C)*d)*(3 + m)*(4 + m) + 2*(b*c - a*d)*(3*b*c*C - 3*a*C*d + b*B*d*(4 + m)))) - I*b*(2 + m)*(b^2*c*(2*c*(A - C)*
d + B*(c^2 - d^2))*(3 + m)*(4 + m) - d*(b^2*d*(B*c + (A - C)*d)*(3 + m)*(4 + m) + 2*(b*c - a*d)*(3*b*c*C - 3*a
*C*d + b*B*d*(4 + m))) + d*(-((2*a*d + b*c*(1 + m))*(3*b*c*C - 3*a*C*d + b*B*d*(4 + m))) + b*c*(3 + m)*(A*b*c*
(4 + m) - C*(3*a*d + b*c*(1 + m))))))*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)*(a*d*(b^2*d*(B*c + (A - C)*d)*(3 + m)*(4 + m
) + 2*(b*c - a*d)*(3*b*c*C - 3*a*C*d + b*B*d*(4 + m))) + b*c*(2 + m)*(b^2*d*(B*c + (A - C)*d)*(3 + m)*(4 + m)
+ 2*(b*c - a*d)*(3*b*c*C - 3*a*C*d + b*B*d*(4 + m))) - b*c*(2 + m)*(-((2*a*d + b*c*(1 + m))*(3*b*c*C - 3*a*C*d
+ b*B*d*(4 + m))) + b*c*(3 + m)*(A*b*c*(4 + m) - C*(3*a*d + b*c*(1 + m)))) - d*(-(b^3*(2*c*(A - C)*d + B*(c^2
- d^2))*(2 + m)*(3 + m)*(4 + m)) + a*(b^2*d*(B*c + (A - C)*d)*(3 + m)*(4 + m) + 2*(b*c - a*d)*(3*b*c*C - 3*a*
C*d + b*B*d*(4 + m)))) + I*b*(2 + m)*(b^2*c*(2*c*(A - C)*d + B*(c^2 - d^2))*(3 + m)*(4 + m) - d*(b^2*d*(B*c +
(A - C)*d)*(3 + m)*(4 + m) + 2*(b*c - a*d)*(3*b*c*C - 3*a*C*d + b*B*d*(4 + m))) + d*(-((2*a*d + b*c*(1 + m))*(
3*b*c*C - 3*a*C*d + b*B*d*(4 + m))) + b*c*(3 + m)*(A*b*c*(4 + m) - C*(3*a*d + b*c*(1 + m))))))*Hypergeometric2
F1[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)))/(b*(2 + m)))/(b*(3 + m)))/(b*(4 + m))

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fricas [F]  time = 0.91, size = 0, normalized size = 0.00 \[ {\rm integral}\left ({\left (C d^{3} \tan \left (f x + e\right )^{5} + {\left (3 \, C c d^{2} + B d^{3}\right )} \tan \left (f x + e\right )^{4} + A c^{3} + {\left (3 \, C c^{2} d + 3 \, B c d^{2} + A d^{3}\right )} \tan \left (f x + e\right )^{3} + {\left (C c^{3} + 3 \, B c^{2} d + 3 \, A c d^{2}\right )} \tan \left (f x + e\right )^{2} + {\left (B c^{3} + 3 \, A c^{2} d\right )} \tan \left (f x + e\right )\right )} {\left (b \tan \left (f x + e\right ) + a\right )}^{m}, x\right ) \]

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*(A+B*tan(f*x+e)+C*tan(f*x+e)^2),x, algorithm="fricas")

[Out]

integral((C*d^3*tan(f*x + e)^5 + (3*C*c*d^2 + B*d^3)*tan(f*x + e)^4 + A*c^3 + (3*C*c^2*d + 3*B*c*d^2 + A*d^3)*
tan(f*x + e)^3 + (C*c^3 + 3*B*c^2*d + 3*A*c*d^2)*tan(f*x + e)^2 + (B*c^3 + 3*A*c^2*d)*tan(f*x + e))*(b*tan(f*x
+ e) + a)^m, x)

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giac [F]  time = 0.00, size = 0, normalized size = 0.00 \[ \int {\left (C \tan \left (f x + e\right )^{2} + B \tan \left (f x + e\right ) + A\right )} {\left (d \tan \left (f x + e\right ) + c\right )}^{3} {\left (b \tan \left (f x + e\right ) + a\right )}^{m}\,{d x} \]

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*(A+B*tan(f*x+e)+C*tan(f*x+e)^2),x, algorithm="giac")

[Out]

integrate((C*tan(f*x + e)^2 + B*tan(f*x + e) + A)*(d*tan(f*x + e) + c)^3*(b*tan(f*x + e) + a)^m, x)

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maple [F]  time = 2.62, size = 0, normalized size = 0.00 \[ \int \left (a +b \tan \left (f x +e \right )\right )^{m} \left (c +d \tan \left (f x +e \right )\right )^{3} \left (A +B \tan \left (f x +e \right )+C \left (\tan ^{2}\left (f x +e \right )\right )\right )\, dx \]

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*(A+B*tan(f*x+e)+C*tan(f*x+e)^2),x)

[Out]

int((a+b*tan(f*x+e))^m*(c+d*tan(f*x+e))^3*(A+B*tan(f*x+e)+C*tan(f*x+e)^2),x)

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maxima [F(-1)]  time = 0.00, size = 0, normalized size = 0.00 \[ \text {Timed out} \]

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*(A+B*tan(f*x+e)+C*tan(f*x+e)^2),x, algorithm="maxima")

[Out]

Timed out

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mupad [F]  time = 0.00, size = -1, normalized size = -0.00 \[ \int {\left (a+b\,\mathrm {tan}\left (e+f\,x\right )\right )}^m\,{\left (c+d\,\mathrm {tan}\left (e+f\,x\right )\right )}^3\,\left (C\,{\mathrm {tan}\left (e+f\,x\right )}^2+B\,\mathrm {tan}\left (e+f\,x\right )+A\right ) \,d x \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

int((a + b*tan(e + f*x))^m*(c + d*tan(e + f*x))^3*(A + B*tan(e + f*x) + C*tan(e + f*x)^2),x)

[Out]

int((a + b*tan(e + f*x))^m*(c + d*tan(e + f*x))^3*(A + B*tan(e + f*x) + C*tan(e + f*x)^2), x)

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sympy [F]  time = 0.00, size = 0, normalized size = 0.00 \[ \int \left (a + b \tan {\left (e + f x \right )}\right )^{m} \left (c + d \tan {\left (e + f x \right )}\right )^{3} \left (A + B \tan {\left (e + f x \right )} + C \tan ^{2}{\left (e + f x \right )}\right )\, dx \]

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*(A+B*tan(f*x+e)+C*tan(f*x+e)**2),x)

[Out]

Integral((a + b*tan(e + f*x))**m*(c + d*tan(e + f*x))**3*(A + B*tan(e + f*x) + C*tan(e + f*x)**2), x)

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