\(\int \csc ^6(e+f x) (a+b \sec ^2(e+f x))^p \, dx\) [143]

Optimal result

Integrand size = 23, antiderivative size = 192 \[ \int \csc ^6(e+f x) \left (a+b \sec ^2(e+f x)\right )^p \, dx=-\frac {(10 a+b (7+2 p)) \cot ^3(e+f x) \left (a+b+b \tan ^2(e+f x)\right )^{1+p}}{15 (a+b)^2 f}-\frac {\cot ^5(e+f x) \left (a+b+b \tan ^2(e+f x)\right )^{1+p}}{5 (a+b) f}-\frac {\left (15 a^2+20 a b (1+p)+4 b^2 \left (2+3 p+p^2\right )\right ) \cot (e+f x) \operatorname {Hypergeometric2F1}\left (-\frac {1}{2},-p,\frac {1}{2},-\frac {b \tan ^2(e+f x)}{a+b}\right ) \left (a+b+b \tan ^2(e+f x)\right )^p \left (1+\frac {b \tan ^2(e+f x)}{a+b}\right )^{-p}}{15 (a+b)^2 f} \]

[Out]

-1/15*(10*a+b*(7+2*p))*cot(f*x+e)^3*(a+b+b*tan(f*x+e)^2)^(p+1)/(a+b)^2/f-1/5*cot(f*x+e)^5*(a+b+b*tan(f*x+e)^2)
^(p+1)/(a+b)/f-1/15*(15*a^2+20*a*b*(p+1)+4*b^2*(p^2+3*p+2))*cot(f*x+e)*hypergeom([-1/2, -p],[1/2],-b*tan(f*x+e
)^2/(a+b))*(a+b+b*tan(f*x+e)^2)^p/(a+b)^2/f/((1+b*tan(f*x+e)^2/(a+b))^p)

Rubi [A] (verified)

Time = 0.20 (sec) , antiderivative size = 192, normalized size of antiderivative = 1.00, number of steps used = 5, number of rules used = 5, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.217, Rules used = {4217, 473, 464, 372, 371} \[ \int \csc ^6(e+f x) \left (a+b \sec ^2(e+f x)\right )^p \, dx=-\frac {\left (15 a^2+20 a b (p+1)+4 b^2 \left (p^2+3 p+2\right )\right ) \cot (e+f x) \left (a+b \tan ^2(e+f x)+b\right )^p \left (\frac {b \tan ^2(e+f x)}{a+b}+1\right )^{-p} \operatorname {Hypergeometric2F1}\left (-\frac {1}{2},-p,\frac {1}{2},-\frac {b \tan ^2(e+f x)}{a+b}\right )}{15 f (a+b)^2}-\frac {\cot ^5(e+f x) \left (a+b \tan ^2(e+f x)+b\right )^{p+1}}{5 f (a+b)}-\frac {(10 a+b (2 p+7)) \cot ^3(e+f x) \left (a+b \tan ^2(e+f x)+b\right )^{p+1}}{15 f (a+b)^2} \]

[In]

Int[Csc[e + f*x]^6*(a + b*Sec[e + f*x]^2)^p,x]

[Out]

-1/15*((10*a + b*(7 + 2*p))*Cot[e + f*x]^3*(a + b + b*Tan[e + f*x]^2)^(1 + p))/((a + b)^2*f) - (Cot[e + f*x]^5
*(a + b + b*Tan[e + f*x]^2)^(1 + p))/(5*(a + b)*f) - ((15*a^2 + 20*a*b*(1 + p) + 4*b^2*(2 + 3*p + p^2))*Cot[e
+ f*x]*Hypergeometric2F1[-1/2, -p, 1/2, -((b*Tan[e + f*x]^2)/(a + b))]*(a + b + b*Tan[e + f*x]^2)^p)/(15*(a +
b)^2*f*(1 + (b*Tan[e + f*x]^2)/(a + b))^p)

Rule 371

Int[((c_.)*(x_))^(m_.)*((a_) + (b_.)*(x_)^(n_))^(p_), x_Symbol] :> Simp[a^p*((c*x)^(m + 1)/(c*(m + 1)))*Hyperg
eometric2F1[-p, (m + 1)/n, (m + 1)/n + 1, (-b)*(x^n/a)], x] /; FreeQ[{a, b, c, m, n, p}, x] &&  !IGtQ[p, 0] &&
(ILtQ[p, 0] || GtQ[a, 0])

Rule 372

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

Rule 464

Int[((e_.)*(x_))^(m_.)*((a_) + (b_.)*(x_)^(n_))^(p_.)*((c_) + (d_.)*(x_)^(n_)), x_Symbol] :> Simp[c*(e*x)^(m +
1)*((a + b*x^n)^(p + 1)/(a*e*(m + 1))), x] + Dist[(a*d*(m + 1) - b*c*(m + n*(p + 1) + 1))/(a*e^n*(m + 1)), In
t[(e*x)^(m + n)*(a + b*x^n)^p, x], x] /; FreeQ[{a, b, c, d, e, p}, x] && NeQ[b*c - a*d, 0] && (IntegerQ[n] ||
GtQ[e, 0]) && ((GtQ[n, 0] && LtQ[m, -1]) || (LtQ[n, 0] && GtQ[m + n, -1])) &&  !ILtQ[p, -1]

Rule 473

Int[((e_.)*(x_))^(m_)*((a_) + (b_.)*(x_)^(n_))^(p_)*((c_) + (d_.)*(x_)^(n_))^2, x_Symbol] :> Simp[c^2*(e*x)^(m
+ 1)*((a + b*x^n)^(p + 1)/(a*e*(m + 1))), x] - Dist[1/(a*e^n*(m + 1)), Int[(e*x)^(m + n)*(a + b*x^n)^p*Simp[b
*c^2*n*(p + 1) + c*(b*c - 2*a*d)*(m + 1) - a*(m + 1)*d^2*x^n, x], x], x] /; FreeQ[{a, b, c, d, e, p}, x] && Ne
Q[b*c - a*d, 0] && IGtQ[n, 0] && LtQ[m, -1] && GtQ[n, 0]

Rule 4217

Int[((a_) + (b_.)*sec[(e_.) + (f_.)*(x_)]^(n_))^(p_.)*sin[(e_.) + (f_.)*(x_)]^(m_), x_Symbol] :> With[{ff = Fr
eeFactors[Tan[e + f*x], x]}, Dist[ff^(m + 1)/f, Subst[Int[x^m*(ExpandToSum[a + b*(1 + ff^2*x^2)^(n/2), x]^p/(1
+ ff^2*x^2)^(m/2 + 1)), x], x, Tan[e + f*x]/ff], x]] /; FreeQ[{a, b, e, f, p}, x] && IntegerQ[m/2] && Integer
Q[n/2]

Rubi steps \begin{align*} \text {integral}& = \frac {\text {Subst}\left (\int \frac {\left (1+x^2\right )^2 \left (a+b+b x^2\right )^p}{x^6} \, dx,x,\tan (e+f x)\right )}{f} \\ & = -\frac {\cot ^5(e+f x) \left (a+b+b \tan ^2(e+f x)\right )^{1+p}}{5 (a+b) f}+\frac {\text {Subst}\left (\int \frac {\left (a+b+b x^2\right )^p \left (10 a+b (7+2 p)+5 (a+b) x^2\right )}{x^4} \, dx,x,\tan (e+f x)\right )}{5 (a+b) f} \\ & = -\frac {(10 a+b (7+2 p)) \cot ^3(e+f x) \left (a+b+b \tan ^2(e+f x)\right )^{1+p}}{15 (a+b)^2 f}-\frac {\cot ^5(e+f x) \left (a+b+b \tan ^2(e+f x)\right )^{1+p}}{5 (a+b) f}+\frac {\left (15 a^2+20 a b (1+p)+4 b^2 \left (2+3 p+p^2\right )\right ) \text {Subst}\left (\int \frac {\left (a+b+b x^2\right )^p}{x^2} \, dx,x,\tan (e+f x)\right )}{15 (a+b)^2 f} \\ & = -\frac {(10 a+b (7+2 p)) \cot ^3(e+f x) \left (a+b+b \tan ^2(e+f x)\right )^{1+p}}{15 (a+b)^2 f}-\frac {\cot ^5(e+f x) \left (a+b+b \tan ^2(e+f x)\right )^{1+p}}{5 (a+b) f}+\frac {\left (\left (15 a^2+20 a b (1+p)+4 b^2 \left (2+3 p+p^2\right )\right ) \left (a+b+b \tan ^2(e+f x)\right )^p \left (1+\frac {b \tan ^2(e+f x)}{a+b}\right )^{-p}\right ) \text {Subst}\left (\int \frac {\left (1+\frac {b x^2}{a+b}\right )^p}{x^2} \, dx,x,\tan (e+f x)\right )}{15 (a+b)^2 f} \\ & = -\frac {(10 a+b (7+2 p)) \cot ^3(e+f x) \left (a+b+b \tan ^2(e+f x)\right )^{1+p}}{15 (a+b)^2 f}-\frac {\cot ^5(e+f x) \left (a+b+b \tan ^2(e+f x)\right )^{1+p}}{5 (a+b) f}-\frac {\left (15 a^2+20 a b (1+p)+4 b^2 \left (2+3 p+p^2\right )\right ) \cot (e+f x) \operatorname {Hypergeometric2F1}\left (-\frac {1}{2},-p,\frac {1}{2},-\frac {b \tan ^2(e+f x)}{a+b}\right ) \left (a+b+b \tan ^2(e+f x)\right )^p \left (1+\frac {b \tan ^2(e+f x)}{a+b}\right )^{-p}}{15 (a+b)^2 f} \\ \end{align*}

Mathematica [A] (verified)

Time = 1.44 (sec) , antiderivative size = 149, normalized size of antiderivative = 0.78 \[ \int \csc ^6(e+f x) \left (a+b \sec ^2(e+f x)\right )^p \, dx=-\frac {\cot (e+f x) \left (3 \cot ^4(e+f x) \operatorname {Hypergeometric2F1}\left (-\frac {5}{2},-p,-\frac {3}{2},-\frac {b \tan ^2(e+f x)}{a+b}\right )+10 \cot ^2(e+f x) \operatorname {Hypergeometric2F1}\left (-\frac {3}{2},-p,-\frac {1}{2},-\frac {b \tan ^2(e+f x)}{a+b}\right )+15 \operatorname {Hypergeometric2F1}\left (-\frac {1}{2},-p,\frac {1}{2},-\frac {b \tan ^2(e+f x)}{a+b}\right )\right ) \left (a+b \sec ^2(e+f x)\right )^p \left (1+\frac {b \tan ^2(e+f x)}{a+b}\right )^{-p}}{15 f} \]

[In]

Integrate[Csc[e + f*x]^6*(a + b*Sec[e + f*x]^2)^p,x]

[Out]

-1/15*(Cot[e + f*x]*(3*Cot[e + f*x]^4*Hypergeometric2F1[-5/2, -p, -3/2, -((b*Tan[e + f*x]^2)/(a + b))] + 10*Co
t[e + f*x]^2*Hypergeometric2F1[-3/2, -p, -1/2, -((b*Tan[e + f*x]^2)/(a + b))] + 15*Hypergeometric2F1[-1/2, -p,
1/2, -((b*Tan[e + f*x]^2)/(a + b))])*(a + b*Sec[e + f*x]^2)^p)/(f*(1 + (b*Tan[e + f*x]^2)/(a + b))^p)

Maple [F]

\[\int \csc \left (f x +e \right )^{6} \left (a +b \sec \left (f x +e \right )^{2}\right )^{p}d x\]

[In]

int(csc(f*x+e)^6*(a+b*sec(f*x+e)^2)^p,x)

[Out]

int(csc(f*x+e)^6*(a+b*sec(f*x+e)^2)^p,x)

Fricas [F]

\[ \int \csc ^6(e+f x) \left (a+b \sec ^2(e+f x)\right )^p \, dx=\int { {\left (b \sec \left (f x + e\right )^{2} + a\right )}^{p} \csc \left (f x + e\right )^{6} \,d x } \]

[In]

integrate(csc(f*x+e)^6*(a+b*sec(f*x+e)^2)^p,x, algorithm="fricas")

[Out]

integral((b*sec(f*x + e)^2 + a)^p*csc(f*x + e)^6, x)

Sympy [F(-1)]

Timed out. \[ \int \csc ^6(e+f x) \left (a+b \sec ^2(e+f x)\right )^p \, dx=\text {Timed out} \]

[In]

integrate(csc(f*x+e)**6*(a+b*sec(f*x+e)**2)**p,x)

[Out]

Timed out

Maxima [F]

\[ \int \csc ^6(e+f x) \left (a+b \sec ^2(e+f x)\right )^p \, dx=\int { {\left (b \sec \left (f x + e\right )^{2} + a\right )}^{p} \csc \left (f x + e\right )^{6} \,d x } \]

[In]

integrate(csc(f*x+e)^6*(a+b*sec(f*x+e)^2)^p,x, algorithm="maxima")

[Out]

integrate((b*sec(f*x + e)^2 + a)^p*csc(f*x + e)^6, x)

Giac [F]

\[ \int \csc ^6(e+f x) \left (a+b \sec ^2(e+f x)\right )^p \, dx=\int { {\left (b \sec \left (f x + e\right )^{2} + a\right )}^{p} \csc \left (f x + e\right )^{6} \,d x } \]

[In]

integrate(csc(f*x+e)^6*(a+b*sec(f*x+e)^2)^p,x, algorithm="giac")

[Out]

integrate((b*sec(f*x + e)^2 + a)^p*csc(f*x + e)^6, x)

Mupad [F(-1)]

Timed out. \[ \int \csc ^6(e+f x) \left (a+b \sec ^2(e+f x)\right )^p \, dx=\int \frac {{\left (a+\frac {b}{{\cos \left (e+f\,x\right )}^2}\right )}^p}{{\sin \left (e+f\,x\right )}^6} \,d x \]

[In]

int((a + b/cos(e + f*x)^2)^p/sin(e + f*x)^6,x)

[Out]

int((a + b/cos(e + f*x)^2)^p/sin(e + f*x)^6, x)