\(\int \tan ^4(c+d x) (a+b \tan (c+d x))^n \, dx\) [707]

Optimal result

Integrand size = 21, antiderivative size = 297 \[ \int \tan ^4(c+d x) (a+b \tan (c+d x))^n \, dx=\frac {\left (2 a^2-b^2 (2+n) (3+n)\right ) (a+b \tan (c+d x))^{1+n}}{b^3 d (1+n) (2+n) (3+n)}-\frac {\sqrt {-b^2} \operatorname {Hypergeometric2F1}\left (1,1+n,2+n,\frac {a+b \tan (c+d x)}{a-\sqrt {-b^2}}\right ) (a+b \tan (c+d x))^{1+n}}{2 b \left (a-\sqrt {-b^2}\right ) d (1+n)}+\frac {\sqrt {-b^2} \operatorname {Hypergeometric2F1}\left (1,1+n,2+n,\frac {a+b \tan (c+d x)}{a+\sqrt {-b^2}}\right ) (a+b \tan (c+d x))^{1+n}}{2 b \left (a+\sqrt {-b^2}\right ) d (1+n)}-\frac {2 a \tan (c+d x) (a+b \tan (c+d x))^{1+n}}{b^2 d (2+n) (3+n)}+\frac {\tan ^2(c+d x) (a+b \tan (c+d x))^{1+n}}{b d (3+n)} \]

[Out]

(2*a^2-b^2*(2+n)*(3+n))*(a+b*tan(d*x+c))^(1+n)/b^3/d/(1+n)/(2+n)/(3+n)-1/2*hypergeom([1, 1+n],[2+n],(a+b*tan(d
*x+c))/(a-(-b^2)^(1/2)))*(-b^2)^(1/2)*(a+b*tan(d*x+c))^(1+n)/b/d/(1+n)/(a-(-b^2)^(1/2))+1/2*hypergeom([1, 1+n]
,[2+n],(a+b*tan(d*x+c))/(a+(-b^2)^(1/2)))*(-b^2)^(1/2)*(a+b*tan(d*x+c))^(1+n)/b/d/(1+n)/(a+(-b^2)^(1/2))-2*a*t
an(d*x+c)*(a+b*tan(d*x+c))^(1+n)/b^2/d/(2+n)/(3+n)+tan(d*x+c)^2*(a+b*tan(d*x+c))^(1+n)/b/d/(3+n)

Rubi [A] (verified)

Time = 0.72 (sec) , antiderivative size = 297, normalized size of antiderivative = 1.00, number of steps used = 8, number of rules used = 6, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.286, Rules used = {3647, 3728, 3712, 3566, 726, 70} \[ \int \tan ^4(c+d x) (a+b \tan (c+d x))^n \, dx=\frac {\left (2 a^2-b^2 (n+2) (n+3)\right ) (a+b \tan (c+d x))^{n+1}}{b^3 d (n+1) (n+2) (n+3)}-\frac {\sqrt {-b^2} (a+b \tan (c+d x))^{n+1} \operatorname {Hypergeometric2F1}\left (1,n+1,n+2,\frac {a+b \tan (c+d x)}{a-\sqrt {-b^2}}\right )}{2 b d (n+1) \left (a-\sqrt {-b^2}\right )}+\frac {\sqrt {-b^2} (a+b \tan (c+d x))^{n+1} \operatorname {Hypergeometric2F1}\left (1,n+1,n+2,\frac {a+b \tan (c+d x)}{a+\sqrt {-b^2}}\right )}{2 b d (n+1) \left (a+\sqrt {-b^2}\right )}-\frac {2 a \tan (c+d x) (a+b \tan (c+d x))^{n+1}}{b^2 d (n+2) (n+3)}+\frac {\tan ^2(c+d x) (a+b \tan (c+d x))^{n+1}}{b d (n+3)} \]

[In]

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

[Out]

((2*a^2 - b^2*(2 + n)*(3 + n))*(a + b*Tan[c + d*x])^(1 + n))/(b^3*d*(1 + n)*(2 + n)*(3 + n)) - (Sqrt[-b^2]*Hyp
ergeometric2F1[1, 1 + n, 2 + n, (a + b*Tan[c + d*x])/(a - Sqrt[-b^2])]*(a + b*Tan[c + d*x])^(1 + n))/(2*b*(a -
Sqrt[-b^2])*d*(1 + n)) + (Sqrt[-b^2]*Hypergeometric2F1[1, 1 + n, 2 + n, (a + b*Tan[c + d*x])/(a + Sqrt[-b^2])
]*(a + b*Tan[c + d*x])^(1 + n))/(2*b*(a + Sqrt[-b^2])*d*(1 + n)) - (2*a*Tan[c + d*x]*(a + b*Tan[c + d*x])^(1 +
n))/(b^2*d*(2 + n)*(3 + n)) + (Tan[c + d*x]^2*(a + b*Tan[c + d*x])^(1 + n))/(b*d*(3 + n))

Rule 70

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

Rule 726

Int[((d_) + (e_.)*(x_))^(m_)/((a_) + (c_.)*(x_)^2), x_Symbol] :> Int[ExpandIntegrand[(d + e*x)^m, 1/(a + c*x^2
), x], x] /; FreeQ[{a, c, d, e, m}, x] && NeQ[c*d^2 + a*e^2, 0] &&  !IntegerQ[m]

Rule 3566

Int[((a_) + (b_.)*tan[(c_.) + (d_.)*(x_)])^(n_), x_Symbol] :> Dist[b/d, Subst[Int[(a + x)^n/(b^2 + x^2), x], x
, b*Tan[c + d*x]], x] /; FreeQ[{a, b, c, d, n}, x] && NeQ[a^2 + b^2, 0]

Rule 3647

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 - 2)*((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 - 3)*(c + d*Tan[e + f*x])^n*Simp[a^3*d*(m + n - 1) - b^2*(b*c*(m - 2) + a*d*(
1 + n)) + b*d*(m + n - 1)*(3*a^2 - b^2)*Tan[e + f*x] - b^2*(b*c*(m - 2) - a*d*(3*m + 2*n - 4))*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]
&& IntegerQ[2*m] && GtQ[m, 2] && (GeQ[n, -1] || IntegerQ[m]) &&  !(IGtQ[n, 2] && ( !IntegerQ[m] || (EqQ[c, 0]
&& NeQ[a, 0])))

Rule 3712

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

Rule 3728

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*} \text {integral}& = \frac {\tan ^2(c+d x) (a+b \tan (c+d x))^{1+n}}{b d (3+n)}+\frac {\int \tan (c+d x) (a+b \tan (c+d x))^n \left (-2 a-b (3+n) \tan (c+d x)-2 a \tan ^2(c+d x)\right ) \, dx}{b (3+n)} \\ & = -\frac {2 a \tan (c+d x) (a+b \tan (c+d x))^{1+n}}{b^2 d (2+n) (3+n)}+\frac {\tan ^2(c+d x) (a+b \tan (c+d x))^{1+n}}{b d (3+n)}+\frac {\int (a+b \tan (c+d x))^n \left (2 a^2+\left (2 a^2-b^2 (2+n) (3+n)\right ) \tan ^2(c+d x)\right ) \, dx}{b^2 (2+n) (3+n)} \\ & = \frac {\left (2 a^2-b^2 (2+n) (3+n)\right ) (a+b \tan (c+d x))^{1+n}}{b^3 d (1+n) (2+n) (3+n)}-\frac {2 a \tan (c+d x) (a+b \tan (c+d x))^{1+n}}{b^2 d (2+n) (3+n)}+\frac {\tan ^2(c+d x) (a+b \tan (c+d x))^{1+n}}{b d (3+n)}+\int (a+b \tan (c+d x))^n \, dx \\ & = \frac {\left (2 a^2-b^2 (2+n) (3+n)\right ) (a+b \tan (c+d x))^{1+n}}{b^3 d (1+n) (2+n) (3+n)}-\frac {2 a \tan (c+d x) (a+b \tan (c+d x))^{1+n}}{b^2 d (2+n) (3+n)}+\frac {\tan ^2(c+d x) (a+b \tan (c+d x))^{1+n}}{b d (3+n)}+\frac {b \text {Subst}\left (\int \frac {(a+x)^n}{b^2+x^2} \, dx,x,b \tan (c+d x)\right )}{d} \\ & = \frac {\left (2 a^2-b^2 (2+n) (3+n)\right ) (a+b \tan (c+d x))^{1+n}}{b^3 d (1+n) (2+n) (3+n)}-\frac {2 a \tan (c+d x) (a+b \tan (c+d x))^{1+n}}{b^2 d (2+n) (3+n)}+\frac {\tan ^2(c+d x) (a+b \tan (c+d x))^{1+n}}{b d (3+n)}+\frac {b \text {Subst}\left (\int \left (\frac {\sqrt {-b^2} (a+x)^n}{2 b^2 \left (\sqrt {-b^2}-x\right )}+\frac {\sqrt {-b^2} (a+x)^n}{2 b^2 \left (\sqrt {-b^2}+x\right )}\right ) \, dx,x,b \tan (c+d x)\right )}{d} \\ & = \frac {\left (2 a^2-b^2 (2+n) (3+n)\right ) (a+b \tan (c+d x))^{1+n}}{b^3 d (1+n) (2+n) (3+n)}-\frac {2 a \tan (c+d x) (a+b \tan (c+d x))^{1+n}}{b^2 d (2+n) (3+n)}+\frac {\tan ^2(c+d x) (a+b \tan (c+d x))^{1+n}}{b d (3+n)}-\frac {b \text {Subst}\left (\int \frac {(a+x)^n}{\sqrt {-b^2}-x} \, dx,x,b \tan (c+d x)\right )}{2 \sqrt {-b^2} d}-\frac {b \text {Subst}\left (\int \frac {(a+x)^n}{\sqrt {-b^2}+x} \, dx,x,b \tan (c+d x)\right )}{2 \sqrt {-b^2} d} \\ & = \frac {\left (2 a^2-b^2 (2+n) (3+n)\right ) (a+b \tan (c+d x))^{1+n}}{b^3 d (1+n) (2+n) (3+n)}+\frac {b \operatorname {Hypergeometric2F1}\left (1,1+n,2+n,\frac {a+b \tan (c+d x)}{a-\sqrt {-b^2}}\right ) (a+b \tan (c+d x))^{1+n}}{2 \sqrt {-b^2} \left (a-\sqrt {-b^2}\right ) d (1+n)}-\frac {b \operatorname {Hypergeometric2F1}\left (1,1+n,2+n,\frac {a+b \tan (c+d x)}{a+\sqrt {-b^2}}\right ) (a+b \tan (c+d x))^{1+n}}{2 \sqrt {-b^2} \left (a+\sqrt {-b^2}\right ) d (1+n)}-\frac {2 a \tan (c+d x) (a+b \tan (c+d x))^{1+n}}{b^2 d (2+n) (3+n)}+\frac {\tan ^2(c+d x) (a+b \tan (c+d x))^{1+n}}{b d (3+n)} \\ \end{align*}

Mathematica [C] (verified)

Result contains complex when optimal does not.

Time = 2.32 (sec) , antiderivative size = 249, normalized size of antiderivative = 0.84 \[ \int \tan ^4(c+d x) (a+b \tan (c+d x))^n \, dx=-\frac {(a+b \tan (c+d x))^{1+n} \left (2 (i a-b) (i a+b) \left (2 a^2-b^2 \left (6+5 n+n^2\right )\right )+i (a+i b) b^3 (2+n) (3+n) \operatorname {Hypergeometric2F1}\left (1,1+n,2+n,\frac {a+b \tan (c+d x)}{a-i b}\right )-b^3 (i a+b) (2+n) (3+n) \operatorname {Hypergeometric2F1}\left (1,1+n,2+n,\frac {a+b \tan (c+d x)}{a+i b}\right )+4 a (a-i b) (a+i b) b (1+n) \tan (c+d x)-2 (a-i b) (a+i b) b^2 (1+n) (2+n) \tan ^2(c+d x)\right )}{2 (a-i b) (a+i b) b^3 d (1+n) (2+n) (3+n)} \]

[In]

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

[Out]

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

Maple [F]

\[\int \left (\tan ^{4}\left (d x +c \right )\right ) \left (a +b \tan \left (d x +c \right )\right )^{n}d x\]

[In]

int(tan(d*x+c)^4*(a+b*tan(d*x+c))^n,x)

[Out]

int(tan(d*x+c)^4*(a+b*tan(d*x+c))^n,x)

Fricas [F]

\[ \int \tan ^4(c+d x) (a+b \tan (c+d x))^n \, dx=\int { {\left (b \tan \left (d x + c\right ) + a\right )}^{n} \tan \left (d x + c\right )^{4} \,d x } \]

[In]

integrate(tan(d*x+c)^4*(a+b*tan(d*x+c))^n,x, algorithm="fricas")

[Out]

integral((b*tan(d*x + c) + a)^n*tan(d*x + c)^4, x)

Sympy [F]

\[ \int \tan ^4(c+d x) (a+b \tan (c+d x))^n \, dx=\int \left (a + b \tan {\left (c + d x \right )}\right )^{n} \tan ^{4}{\left (c + d x \right )}\, dx \]

[In]

integrate(tan(d*x+c)**4*(a+b*tan(d*x+c))**n,x)

[Out]

Integral((a + b*tan(c + d*x))**n*tan(c + d*x)**4, x)

Maxima [F]

\[ \int \tan ^4(c+d x) (a+b \tan (c+d x))^n \, dx=\int { {\left (b \tan \left (d x + c\right ) + a\right )}^{n} \tan \left (d x + c\right )^{4} \,d x } \]

[In]

integrate(tan(d*x+c)^4*(a+b*tan(d*x+c))^n,x, algorithm="maxima")

[Out]

integrate((b*tan(d*x + c) + a)^n*tan(d*x + c)^4, x)

Giac [F]

\[ \int \tan ^4(c+d x) (a+b \tan (c+d x))^n \, dx=\int { {\left (b \tan \left (d x + c\right ) + a\right )}^{n} \tan \left (d x + c\right )^{4} \,d x } \]

[In]

integrate(tan(d*x+c)^4*(a+b*tan(d*x+c))^n,x, algorithm="giac")

[Out]

integrate((b*tan(d*x + c) + a)^n*tan(d*x + c)^4, x)

Mupad [F(-1)]

Timed out. \[ \int \tan ^4(c+d x) (a+b \tan (c+d x))^n \, dx=\int {\mathrm {tan}\left (c+d\,x\right )}^4\,{\left (a+b\,\mathrm {tan}\left (c+d\,x\right )\right )}^n \,d x \]

[In]

int(tan(c + d*x)^4*(a + b*tan(c + d*x))^n,x)

[Out]

int(tan(c + d*x)^4*(a + b*tan(c + d*x))^n, x)