\(\int \csc ^4(e+f x) (a+b \sin ^2(e+f x))^p \, dx\) [181]

Optimal result

Integrand size = 23, antiderivative size = 101 \[ \int \csc ^4(e+f x) \left (a+b \sin ^2(e+f x)\right )^p \, dx=-\frac {\operatorname {AppellF1}\left (-\frac {3}{2},\frac {1}{2},-p,-\frac {1}{2},\sin ^2(e+f x),-\frac {b \sin ^2(e+f x)}{a}\right ) \sqrt {\cos ^2(e+f x)} \csc ^3(e+f x) \sec (e+f x) \left (a+b \sin ^2(e+f x)\right )^p \left (1+\frac {b \sin ^2(e+f x)}{a}\right )^{-p}}{3 f} \]

[Out]

-1/3*AppellF1(-3/2,1/2,-p,-1/2,sin(f*x+e)^2,-b*sin(f*x+e)^2/a)*csc(f*x+e)^3*sec(f*x+e)*(a+b*sin(f*x+e)^2)^p*(c
os(f*x+e)^2)^(1/2)/f/((1+b*sin(f*x+e)^2/a)^p)

Rubi [A] (verified)

Time = 0.07 (sec) , antiderivative size = 101, normalized size of antiderivative = 1.00, number of steps used = 3, number of rules used = 3, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.130, Rules used = {3267, 525, 524} \[ \int \csc ^4(e+f x) \left (a+b \sin ^2(e+f x)\right )^p \, dx=-\frac {\sqrt {\cos ^2(e+f x)} \csc ^3(e+f x) \sec (e+f x) \left (a+b \sin ^2(e+f x)\right )^p \left (\frac {b \sin ^2(e+f x)}{a}+1\right )^{-p} \operatorname {AppellF1}\left (-\frac {3}{2},\frac {1}{2},-p,-\frac {1}{2},\sin ^2(e+f x),-\frac {b \sin ^2(e+f x)}{a}\right )}{3 f} \]

[In]

Int[Csc[e + f*x]^4*(a + b*Sin[e + f*x]^2)^p,x]

[Out]

-1/3*(AppellF1[-3/2, 1/2, -p, -1/2, Sin[e + f*x]^2, -((b*Sin[e + f*x]^2)/a)]*Sqrt[Cos[e + f*x]^2]*Csc[e + f*x]
^3*Sec[e + f*x]*(a + b*Sin[e + f*x]^2)^p)/(f*(1 + (b*Sin[e + f*x]^2)/a)^p)

Rule 524

Int[((e_.)*(x_))^(m_.)*((a_) + (b_.)*(x_)^(n_))^(p_)*((c_) + (d_.)*(x_)^(n_))^(q_), x_Symbol] :> Simp[a^p*c^q*
((e*x)^(m + 1)/(e*(m + 1)))*AppellF1[(m + 1)/n, -p, -q, 1 + (m + 1)/n, (-b)*(x^n/a), (-d)*(x^n/c)], x] /; Free
Q[{a, b, c, d, e, m, n, p, q}, x] && NeQ[b*c - a*d, 0] && NeQ[m, -1] && NeQ[m, n - 1] && (IntegerQ[p] || GtQ[a
, 0]) && (IntegerQ[q] || GtQ[c, 0])

Rule 525

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

Rule 3267

Int[sin[(e_.) + (f_.)*(x_)]^(m_)*((a_) + (b_.)*sin[(e_.) + (f_.)*(x_)]^2)^(p_.), x_Symbol] :> With[{ff = FreeF
actors[Sin[e + f*x], x]}, Dist[ff^(m + 1)*(Sqrt[Cos[e + f*x]^2]/(f*Cos[e + f*x])), Subst[Int[x^m*((a + b*ff^2*
x^2)^p/Sqrt[1 - ff^2*x^2]), x], x, Sin[e + f*x]/ff], x]] /; FreeQ[{a, b, e, f, p}, x] && IntegerQ[m/2] &&  !In
tegerQ[p]

Rubi steps \begin{align*} \text {integral}& = \frac {\left (\sqrt {\cos ^2(e+f x)} \sec (e+f x)\right ) \text {Subst}\left (\int \frac {\left (a+b x^2\right )^p}{x^4 \sqrt {1-x^2}} \, dx,x,\sin (e+f x)\right )}{f} \\ & = \frac {\left (\sqrt {\cos ^2(e+f x)} \sec (e+f x) \left (a+b \sin ^2(e+f x)\right )^p \left (1+\frac {b \sin ^2(e+f x)}{a}\right )^{-p}\right ) \text {Subst}\left (\int \frac {\left (1+\frac {b x^2}{a}\right )^p}{x^4 \sqrt {1-x^2}} \, dx,x,\sin (e+f x)\right )}{f} \\ & = -\frac {\operatorname {AppellF1}\left (-\frac {3}{2},\frac {1}{2},-p,-\frac {1}{2},\sin ^2(e+f x),-\frac {b \sin ^2(e+f x)}{a}\right ) \sqrt {\cos ^2(e+f x)} \csc ^3(e+f x) \sec (e+f x) \left (a+b \sin ^2(e+f x)\right )^p \left (1+\frac {b \sin ^2(e+f x)}{a}\right )^{-p}}{3 f} \\ \end{align*}

Mathematica [F]

\[ \int \csc ^4(e+f x) \left (a+b \sin ^2(e+f x)\right )^p \, dx=\int \csc ^4(e+f x) \left (a+b \sin ^2(e+f x)\right )^p \, dx \]

[In]

Integrate[Csc[e + f*x]^4*(a + b*Sin[e + f*x]^2)^p,x]

[Out]

Integrate[Csc[e + f*x]^4*(a + b*Sin[e + f*x]^2)^p, x]

Maple [F]

\[\int \left (\csc ^{4}\left (f x +e \right )\right ) {\left (a +b \left (\sin ^{2}\left (f x +e \right )\right )\right )}^{p}d x\]

[In]

int(csc(f*x+e)^4*(a+b*sin(f*x+e)^2)^p,x)

[Out]

int(csc(f*x+e)^4*(a+b*sin(f*x+e)^2)^p,x)

Fricas [F]

\[ \int \csc ^4(e+f x) \left (a+b \sin ^2(e+f x)\right )^p \, dx=\int { {\left (b \sin \left (f x + e\right )^{2} + a\right )}^{p} \csc \left (f x + e\right )^{4} \,d x } \]

[In]

integrate(csc(f*x+e)^4*(a+b*sin(f*x+e)^2)^p,x, algorithm="fricas")

[Out]

integral((-b*cos(f*x + e)^2 + a + b)^p*csc(f*x + e)^4, x)

Sympy [F(-1)]

Timed out. \[ \int \csc ^4(e+f x) \left (a+b \sin ^2(e+f x)\right )^p \, dx=\text {Timed out} \]

[In]

integrate(csc(f*x+e)**4*(a+b*sin(f*x+e)**2)**p,x)

[Out]

Timed out

Maxima [F]

\[ \int \csc ^4(e+f x) \left (a+b \sin ^2(e+f x)\right )^p \, dx=\int { {\left (b \sin \left (f x + e\right )^{2} + a\right )}^{p} \csc \left (f x + e\right )^{4} \,d x } \]

[In]

integrate(csc(f*x+e)^4*(a+b*sin(f*x+e)^2)^p,x, algorithm="maxima")

[Out]

integrate((b*sin(f*x + e)^2 + a)^p*csc(f*x + e)^4, x)

Giac [F]

\[ \int \csc ^4(e+f x) \left (a+b \sin ^2(e+f x)\right )^p \, dx=\int { {\left (b \sin \left (f x + e\right )^{2} + a\right )}^{p} \csc \left (f x + e\right )^{4} \,d x } \]

[In]

integrate(csc(f*x+e)^4*(a+b*sin(f*x+e)^2)^p,x, algorithm="giac")

[Out]

integrate((b*sin(f*x + e)^2 + a)^p*csc(f*x + e)^4, x)

Mupad [F(-1)]

Timed out. \[ \int \csc ^4(e+f x) \left (a+b \sin ^2(e+f x)\right )^p \, dx=\int \frac {{\left (b\,{\sin \left (e+f\,x\right )}^2+a\right )}^p}{{\sin \left (e+f\,x\right )}^4} \,d x \]

[In]

int((a + b*sin(e + f*x)^2)^p/sin(e + f*x)^4,x)

[Out]

int((a + b*sin(e + f*x)^2)^p/sin(e + f*x)^4, x)