3.479 \(\int \tan ^m(c+d x) (a+b \tan (c+d x))^4 (A+B \tan (c+d x)) \, dx\)

Optimal. Leaf size=403 \[ \frac{\left (-6 a^2 A b^2+a^4 A-4 a^3 b B+4 a b^3 B+A b^4\right ) \tan ^{m+1}(c+d x) \text{Hypergeometric2F1}\left (1,\frac{m+1}{2},\frac{m+3}{2},-\tan ^2(c+d x)\right )}{d (m+1)}+\frac{\left (4 a^3 A b-6 a^2 b^2 B+a^4 B-4 a A b^3+b^4 B\right ) \tan ^{m+2}(c+d x) \text{Hypergeometric2F1}\left (1,\frac{m+2}{2},\frac{m+4}{2},-\tan ^2(c+d x)\right )}{d (m+2)}-\frac{b \left (-a^2 A b \left (5 m^2+37 m+68\right )-2 a^3 B \left (m^2+8 m+19\right )+4 a b^2 B \left (m^2+7 m+12\right )+A b^3 \left (m^2+7 m+12\right )\right ) \tan ^{m+1}(c+d x)}{d (m+1) (m+3) (m+4)}+\frac{b^2 \left (a^2 B \left (m^2+9 m+26\right )+2 a A b (m+4)^2-b^2 B \left (m^2+7 m+12\right )\right ) \tan ^{m+2}(c+d x)}{d (m+2) (m+3) (m+4)}+\frac{b (a B (m+7)+A b (m+4)) \tan ^{m+1}(c+d x) (a+b \tan (c+d x))^2}{d (m+3) (m+4)}+\frac{b B \tan ^{m+1}(c+d x) (a+b \tan (c+d x))^3}{d (m+4)} \]

[Out]

-((b*(A*b^3*(12 + 7*m + m^2) + 4*a*b^2*B*(12 + 7*m + m^2) - 2*a^3*B*(19 + 8*m + m^2) - a^2*A*b*(68 + 37*m + 5*
m^2))*Tan[c + d*x]^(1 + m))/(d*(1 + m)*(3 + m)*(4 + m))) + ((a^4*A - 6*a^2*A*b^2 + A*b^4 - 4*a^3*b*B + 4*a*b^3
*B)*Hypergeometric2F1[1, (1 + m)/2, (3 + m)/2, -Tan[c + d*x]^2]*Tan[c + d*x]^(1 + m))/(d*(1 + m)) + (b^2*(2*a*
A*b*(4 + m)^2 - b^2*B*(12 + 7*m + m^2) + a^2*B*(26 + 9*m + m^2))*Tan[c + d*x]^(2 + m))/(d*(2 + m)*(3 + m)*(4 +
m)) + ((4*a^3*A*b - 4*a*A*b^3 + a^4*B - 6*a^2*b^2*B + b^4*B)*Hypergeometric2F1[1, (2 + m)/2, (4 + m)/2, -Tan[
c + d*x]^2]*Tan[c + d*x]^(2 + m))/(d*(2 + m)) + (b*(A*b*(4 + m) + a*B*(7 + m))*Tan[c + d*x]^(1 + m)*(a + b*Tan
[c + d*x])^2)/(d*(3 + m)*(4 + m)) + (b*B*Tan[c + d*x]^(1 + m)*(a + b*Tan[c + d*x])^3)/(d*(4 + m))

________________________________________________________________________________________

Rubi [A]  time = 1.32834, antiderivative size = 403, normalized size of antiderivative = 1., number of steps used = 9, number of rules used = 7, integrand size = 31, \(\frac{\text{number of rules}}{\text{integrand size}}\) = 0.226, Rules used = {3607, 3647, 3637, 3630, 3538, 3476, 364} \[ \frac{\left (-6 a^2 A b^2+a^4 A-4 a^3 b B+4 a b^3 B+A b^4\right ) \tan ^{m+1}(c+d x) \, _2F_1\left (1,\frac{m+1}{2};\frac{m+3}{2};-\tan ^2(c+d x)\right )}{d (m+1)}+\frac{\left (4 a^3 A b-6 a^2 b^2 B+a^4 B-4 a A b^3+b^4 B\right ) \tan ^{m+2}(c+d x) \, _2F_1\left (1,\frac{m+2}{2};\frac{m+4}{2};-\tan ^2(c+d x)\right )}{d (m+2)}-\frac{b \left (-a^2 A b \left (5 m^2+37 m+68\right )-2 a^3 B \left (m^2+8 m+19\right )+4 a b^2 B \left (m^2+7 m+12\right )+A b^3 \left (m^2+7 m+12\right )\right ) \tan ^{m+1}(c+d x)}{d (m+1) (m+3) (m+4)}+\frac{b^2 \left (a^2 B \left (m^2+9 m+26\right )+2 a A b (m+4)^2-b^2 B \left (m^2+7 m+12\right )\right ) \tan ^{m+2}(c+d x)}{d (m+2) (m+3) (m+4)}+\frac{b (a B (m+7)+A b (m+4)) \tan ^{m+1}(c+d x) (a+b \tan (c+d x))^2}{d (m+3) (m+4)}+\frac{b B \tan ^{m+1}(c+d x) (a+b \tan (c+d x))^3}{d (m+4)} \]

Antiderivative was successfully verified.

[In]

Int[Tan[c + d*x]^m*(a + b*Tan[c + d*x])^4*(A + B*Tan[c + d*x]),x]

[Out]

-((b*(A*b^3*(12 + 7*m + m^2) + 4*a*b^2*B*(12 + 7*m + m^2) - 2*a^3*B*(19 + 8*m + m^2) - a^2*A*b*(68 + 37*m + 5*
m^2))*Tan[c + d*x]^(1 + m))/(d*(1 + m)*(3 + m)*(4 + m))) + ((a^4*A - 6*a^2*A*b^2 + A*b^4 - 4*a^3*b*B + 4*a*b^3
*B)*Hypergeometric2F1[1, (1 + m)/2, (3 + m)/2, -Tan[c + d*x]^2]*Tan[c + d*x]^(1 + m))/(d*(1 + m)) + (b^2*(2*a*
A*b*(4 + m)^2 - b^2*B*(12 + 7*m + m^2) + a^2*B*(26 + 9*m + m^2))*Tan[c + d*x]^(2 + m))/(d*(2 + m)*(3 + m)*(4 +
m)) + ((4*a^3*A*b - 4*a*A*b^3 + a^4*B - 6*a^2*b^2*B + b^4*B)*Hypergeometric2F1[1, (2 + m)/2, (4 + m)/2, -Tan[
c + d*x]^2]*Tan[c + d*x]^(2 + m))/(d*(2 + m)) + (b*(A*b*(4 + m) + a*B*(7 + m))*Tan[c + d*x]^(1 + m)*(a + b*Tan
[c + d*x])^2)/(d*(3 + m)*(4 + m)) + (b*B*Tan[c + d*x]^(1 + m)*(a + b*Tan[c + d*x])^3)/(d*(4 + m))

Rule 3607

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

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 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 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 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])

Rubi steps

\begin{align*} \int \tan ^m(c+d x) (a+b \tan (c+d x))^4 (A+B \tan (c+d x)) \, dx &=\frac{b B \tan ^{1+m}(c+d x) (a+b \tan (c+d x))^3}{d (4+m)}+\frac{\int \tan ^m(c+d x) (a+b \tan (c+d x))^2 \left (-a (b B (1+m)-a A (4+m))+\left (2 a A b+a^2 B-b^2 B\right ) (4+m) \tan (c+d x)+b (A b (4+m)+a B (7+m)) \tan ^2(c+d x)\right ) \, dx}{4+m}\\ &=\frac{b (A b (4+m)+a B (7+m)) \tan ^{1+m}(c+d x) (a+b \tan (c+d x))^2}{d (3+m) (4+m)}+\frac{b B \tan ^{1+m}(c+d x) (a+b \tan (c+d x))^3}{d (4+m)}+\frac{\int \tan ^m(c+d x) (a+b \tan (c+d x)) \left (-a (a (3+m) (b B (1+m)-a A (4+m))+b (1+m) (A b (4+m)+a B (7+m)))+\left (3 a^2 A b-A b^3+a^3 B-3 a b^2 B\right ) (3+m) (4+m) \tan (c+d x)+b \left (2 a A b (4+m)^2-b^2 B \left (12+7 m+m^2\right )+a^2 B \left (26+9 m+m^2\right )\right ) \tan ^2(c+d x)\right ) \, dx}{12+7 m+m^2}\\ &=\frac{b^2 \left (2 a A b (4+m)^2-b^2 B \left (12+7 m+m^2\right )+a^2 B \left (26+9 m+m^2\right )\right ) \tan ^{2+m}(c+d x)}{d (2+m) \left (12+7 m+m^2\right )}+\frac{b (A b (4+m)+a B (7+m)) \tan ^{1+m}(c+d x) (a+b \tan (c+d x))^2}{d (3+m) (4+m)}+\frac{b B \tan ^{1+m}(c+d x) (a+b \tan (c+d x))^3}{d (4+m)}-\frac{\int \tan ^m(c+d x) \left (a^2 (2+m) (a (3+m) (b B (1+m)-a A (4+m))+b (1+m) (A b (4+m)+a B (7+m)))-\left (4 a^3 A b-4 a A b^3+a^4 B-6 a^2 b^2 B+b^4 B\right ) (2+m) (3+m) (4+m) \tan (c+d x)+b (2+m) \left (A b^3 \left (12+7 m+m^2\right )+4 a b^2 B \left (12+7 m+m^2\right )-2 a^3 B \left (19+8 m+m^2\right )-a^2 A b \left (68+37 m+5 m^2\right )\right ) \tan ^2(c+d x)\right ) \, dx}{24+26 m+9 m^2+m^3}\\ &=-\frac{b \left (A b^3 \left (12+7 m+m^2\right )+4 a b^2 B \left (12+7 m+m^2\right )-2 a^3 B \left (19+8 m+m^2\right )-a^2 A b \left (68+37 m+5 m^2\right )\right ) \tan ^{1+m}(c+d x)}{d (1+m) \left (12+7 m+m^2\right )}+\frac{b^2 \left (2 a A b (4+m)^2-b^2 B \left (12+7 m+m^2\right )+a^2 B \left (26+9 m+m^2\right )\right ) \tan ^{2+m}(c+d x)}{d (2+m) \left (12+7 m+m^2\right )}+\frac{b (A b (4+m)+a B (7+m)) \tan ^{1+m}(c+d x) (a+b \tan (c+d x))^2}{d (3+m) (4+m)}+\frac{b B \tan ^{1+m}(c+d x) (a+b \tan (c+d x))^3}{d (4+m)}-\frac{\int \tan ^m(c+d x) \left (-\left (a^4 A-6 a^2 A b^2+A b^4-4 a^3 b B+4 a b^3 B\right ) (2+m) (3+m) (4+m)-\left (4 a^3 A b-4 a A b^3+a^4 B-6 a^2 b^2 B+b^4 B\right ) (2+m) (3+m) (4+m) \tan (c+d x)\right ) \, dx}{24+26 m+9 m^2+m^3}\\ &=-\frac{b \left (A b^3 \left (12+7 m+m^2\right )+4 a b^2 B \left (12+7 m+m^2\right )-2 a^3 B \left (19+8 m+m^2\right )-a^2 A b \left (68+37 m+5 m^2\right )\right ) \tan ^{1+m}(c+d x)}{d (1+m) \left (12+7 m+m^2\right )}+\frac{b^2 \left (2 a A b (4+m)^2-b^2 B \left (12+7 m+m^2\right )+a^2 B \left (26+9 m+m^2\right )\right ) \tan ^{2+m}(c+d x)}{d (2+m) \left (12+7 m+m^2\right )}+\frac{b (A b (4+m)+a B (7+m)) \tan ^{1+m}(c+d x) (a+b \tan (c+d x))^2}{d (3+m) (4+m)}+\frac{b B \tan ^{1+m}(c+d x) (a+b \tan (c+d x))^3}{d (4+m)}+\left (a^4 A-6 a^2 A b^2+A b^4-4 a^3 b B+4 a b^3 B\right ) \int \tan ^m(c+d x) \, dx+\left (4 a^3 A b-4 a A b^3+a^4 B-6 a^2 b^2 B+b^4 B\right ) \int \tan ^{1+m}(c+d x) \, dx\\ &=-\frac{b \left (A b^3 \left (12+7 m+m^2\right )+4 a b^2 B \left (12+7 m+m^2\right )-2 a^3 B \left (19+8 m+m^2\right )-a^2 A b \left (68+37 m+5 m^2\right )\right ) \tan ^{1+m}(c+d x)}{d (1+m) \left (12+7 m+m^2\right )}+\frac{b^2 \left (2 a A b (4+m)^2-b^2 B \left (12+7 m+m^2\right )+a^2 B \left (26+9 m+m^2\right )\right ) \tan ^{2+m}(c+d x)}{d (2+m) \left (12+7 m+m^2\right )}+\frac{b (A b (4+m)+a B (7+m)) \tan ^{1+m}(c+d x) (a+b \tan (c+d x))^2}{d (3+m) (4+m)}+\frac{b B \tan ^{1+m}(c+d x) (a+b \tan (c+d x))^3}{d (4+m)}+\frac{\left (a^4 A-6 a^2 A b^2+A b^4-4 a^3 b B+4 a b^3 B\right ) \operatorname{Subst}\left (\int \frac{x^m}{1+x^2} \, dx,x,\tan (c+d x)\right )}{d}+\frac{\left (4 a^3 A b-4 a A b^3+a^4 B-6 a^2 b^2 B+b^4 B\right ) \operatorname{Subst}\left (\int \frac{x^{1+m}}{1+x^2} \, dx,x,\tan (c+d x)\right )}{d}\\ &=-\frac{b \left (A b^3 \left (12+7 m+m^2\right )+4 a b^2 B \left (12+7 m+m^2\right )-2 a^3 B \left (19+8 m+m^2\right )-a^2 A b \left (68+37 m+5 m^2\right )\right ) \tan ^{1+m}(c+d x)}{d (1+m) \left (12+7 m+m^2\right )}+\frac{\left (a^4 A-6 a^2 A b^2+A b^4-4 a^3 b B+4 a 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)}{d (1+m)}+\frac{b^2 \left (2 a A b (4+m)^2-b^2 B \left (12+7 m+m^2\right )+a^2 B \left (26+9 m+m^2\right )\right ) \tan ^{2+m}(c+d x)}{d (2+m) \left (12+7 m+m^2\right )}+\frac{\left (4 a^3 A b-4 a A b^3+a^4 B-6 a^2 b^2 B+b^4 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)}{d (2+m)}+\frac{b (A b (4+m)+a B (7+m)) \tan ^{1+m}(c+d x) (a+b \tan (c+d x))^2}{d (3+m) (4+m)}+\frac{b B \tan ^{1+m}(c+d x) (a+b \tan (c+d x))^3}{d (4+m)}\\ \end{align*}

Mathematica [A]  time = 5.47464, size = 355, normalized size = 0.88 \[ \frac{\tan ^{m+1}(c+d x) \left ((m+2) (m+3) (m+4) \left (-6 a^2 A b^2+a^4 A-4 a^3 b B+4 a b^3 B+A b^4\right ) \text{Hypergeometric2F1}\left (1,\frac{m+1}{2},\frac{m+3}{2},-\tan ^2(c+d x)\right )+(m+1) (m+3) (m+4) \left (4 a^3 A b-6 a^2 b^2 B+a^4 B-4 a A b^3+b^4 B\right ) \tan (c+d x) \text{Hypergeometric2F1}\left (1,\frac{m+2}{2},\frac{m+4}{2},-\tan ^2(c+d x)\right )+b^2 (m+1) \tan (c+d x) \left (a^2 B \left (m^2+9 m+26\right )+2 a A b (m+4)^2-b^2 B \left (m^2+7 m+12\right )\right )-b (m+2) \left (-a^2 A b \left (5 m^2+37 m+68\right )-2 a^3 B \left (m^2+8 m+19\right )+4 a b^2 B \left (m^2+7 m+12\right )+A b^3 \left (m^2+7 m+12\right )\right )+b (m+1) (m+2) (a B (m+7)+A b (m+4)) (a+b \tan (c+d x))^2+b B (m+1) (m+2) (m+3) (a+b \tan (c+d x))^3\right )}{d (m+1) (m+2) (m+3) (m+4)} \]

Antiderivative was successfully verified.

[In]

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

[Out]

(Tan[c + d*x]^(1 + m)*(-(b*(2 + m)*(A*b^3*(12 + 7*m + m^2) + 4*a*b^2*B*(12 + 7*m + m^2) - 2*a^3*B*(19 + 8*m +
m^2) - a^2*A*b*(68 + 37*m + 5*m^2))) + (a^4*A - 6*a^2*A*b^2 + A*b^4 - 4*a^3*b*B + 4*a*b^3*B)*(2 + m)*(3 + m)*(
4 + m)*Hypergeometric2F1[1, (1 + m)/2, (3 + m)/2, -Tan[c + d*x]^2] + b^2*(1 + m)*(2*a*A*b*(4 + m)^2 - b^2*B*(1
2 + 7*m + m^2) + a^2*B*(26 + 9*m + m^2))*Tan[c + d*x] + (4*a^3*A*b - 4*a*A*b^3 + a^4*B - 6*a^2*b^2*B + b^4*B)*
(1 + m)*(3 + m)*(4 + m)*Hypergeometric2F1[1, (2 + m)/2, (4 + m)/2, -Tan[c + d*x]^2]*Tan[c + d*x] + b*(1 + m)*(
2 + m)*(A*b*(4 + m) + a*B*(7 + m))*(a + b*Tan[c + d*x])^2 + b*B*(1 + m)*(2 + m)*(3 + m)*(a + b*Tan[c + d*x])^3
))/(d*(1 + m)*(2 + m)*(3 + m)*(4 + m))

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Maple [F]  time = 0.598, size = 0, normalized size = 0. \begin{align*} \int \left ( \tan \left ( dx+c \right ) \right ) ^{m} \left ( a+b\tan \left ( dx+c \right ) \right ) ^{4} \left ( A+B\tan \left ( dx+c \right ) \right ) \, 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))^4*(A+B*tan(d*x+c)),x)

[Out]

int(tan(d*x+c)^m*(a+b*tan(d*x+c))^4*(A+B*tan(d*x+c)),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(tan(d*x+c)^m*(a+b*tan(d*x+c))^4*(A+B*tan(d*x+c)),x, algorithm="maxima")

[Out]

Timed out

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Fricas [F]  time = 0., size = 0, normalized size = 0. \begin{align*}{\rm integral}\left ({\left (B b^{4} \tan \left (d x + c\right )^{5} + A a^{4} +{\left (4 \, B a b^{3} + A b^{4}\right )} \tan \left (d x + c\right )^{4} + 2 \,{\left (3 \, B a^{2} b^{2} + 2 \, A a b^{3}\right )} \tan \left (d x + c\right )^{3} + 2 \,{\left (2 \, B a^{3} b + 3 \, A a^{2} b^{2}\right )} \tan \left (d x + c\right )^{2} +{\left (B a^{4} + 4 \, A a^{3} b\right )} \tan \left (d x + c\right )\right )} \tan \left (d x + c\right )^{m}, 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))^4*(A+B*tan(d*x+c)),x, algorithm="fricas")

[Out]

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

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Sympy [F]  time = 0., size = 0, normalized size = 0. \begin{align*} \int \left (A + B \tan{\left (c + d x \right )}\right ) \left (a + b \tan{\left (c + d x \right )}\right )^{4} \tan ^{m}{\left (c + d x \right )}\, dx \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))**4*(A+B*tan(d*x+c)),x)

[Out]

Integral((A + B*tan(c + d*x))*(a + b*tan(c + d*x))**4*tan(c + d*x)**m, x)

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Giac [F]  time = 0., size = 0, normalized size = 0. \begin{align*} \int{\left (B \tan \left (d x + c\right ) + A\right )}{\left (b \tan \left (d x + c\right ) + a\right )}^{4} \tan \left (d x + c\right )^{m}\,{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))^4*(A+B*tan(d*x+c)),x, algorithm="giac")

[Out]

integrate((B*tan(d*x + c) + A)*(b*tan(d*x + c) + a)^4*tan(d*x + c)^m, x)