\(\int \sin ^4(e+f x) (a+b \sin ^2(e+f x))^p \, dx\) [127]

Optimal result

Integrand size = 23, antiderivative size = 101 \[ \int \sin ^4(e+f x) \left (a+b \sin ^2(e+f x)\right )^p \, dx=\frac {\operatorname {AppellF1}\left (\frac {5}{2},\frac {1}{2},-p,\frac {7}{2},\sin ^2(e+f x),-\frac {b \sin ^2(e+f x)}{a}\right ) \sqrt {\cos ^2(e+f x)} \sin ^4(e+f x) \left (a+b \sin ^2(e+f x)\right )^p \left (1+\frac {b \sin ^2(e+f x)}{a}\right )^{-p} \tan (e+f x)}{5 f} \] Output:

1/5*AppellF1(5/2,1/2,-p,7/2,sin(f*x+e)^2,-b*sin(f*x+e)^2/a)*(cos(f*x+e)^2) 
^(1/2)*sin(f*x+e)^4*(a+b*sin(f*x+e)^2)^p*tan(f*x+e)/f/((1+b*sin(f*x+e)^2/a 
)^p)
 

Mathematica [A] (verified)

Time = 1.03 (sec) , antiderivative size = 102, normalized size of antiderivative = 1.01 \[ \int \sin ^4(e+f x) \left (a+b \sin ^2(e+f x)\right )^p \, dx=\frac {\operatorname {AppellF1}\left (\frac {5}{2},\frac {1}{2},-p,\frac {7}{2},\sin ^2(e+f x),-\frac {b \sin ^2(e+f x)}{a}\right ) \sqrt {\cos ^2(e+f x)} \sin ^4(e+f x) \left (a+b \sin ^2(e+f x)\right )^p \left (\frac {a+b \sin ^2(e+f x)}{a}\right )^{-p} \tan (e+f x)}{5 f} \] Input:

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

Output:

(AppellF1[5/2, 1/2, -p, 7/2, Sin[e + f*x]^2, -((b*Sin[e + f*x]^2)/a)]*Sqrt 
[Cos[e + f*x]^2]*Sin[e + f*x]^4*(a + b*Sin[e + f*x]^2)^p*Tan[e + f*x])/(5* 
f*((a + b*Sin[e + f*x]^2)/a)^p)
 

Rubi [A] (verified)

Time = 0.30 (sec) , antiderivative size = 101, normalized size of antiderivative = 1.00, number of steps used = 5, number of rules used = 4, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.174, Rules used = {3042, 3667, 395, 394}

Below are the steps used by Rubi to obtain the solution. The rule number used for the transformation is given above next to the arrow. The rules definitions used are listed below.

\(\displaystyle \int \sin ^4(e+f x) \left (a+b \sin ^2(e+f x)\right )^p \, dx\)

\(\Big \downarrow \) 3042

\(\displaystyle \int \sin (e+f x)^4 \left (a+b \sin (e+f x)^2\right )^pdx\)

\(\Big \downarrow \) 3667

\(\displaystyle \frac {\sqrt {\cos ^2(e+f x)} \sec (e+f x) \int \frac {\sin ^4(e+f x) \left (b \sin ^2(e+f x)+a\right )^p}{\sqrt {1-\sin ^2(e+f x)}}d\sin (e+f x)}{f}\)

\(\Big \downarrow \) 395

\(\displaystyle \frac {\sqrt {\cos ^2(e+f x)} \sec (e+f x) \left (a+b \sin ^2(e+f x)\right )^p \left (\frac {b \sin ^2(e+f x)}{a}+1\right )^{-p} \int \frac {\sin ^4(e+f x) \left (\frac {b \sin ^2(e+f x)}{a}+1\right )^p}{\sqrt {1-\sin ^2(e+f x)}}d\sin (e+f x)}{f}\)

\(\Big \downarrow \) 394

\(\displaystyle \frac {\sin ^4(e+f x) \sqrt {\cos ^2(e+f x)} \tan (e+f x) \left (a+b \sin ^2(e+f x)\right )^p \left (\frac {b \sin ^2(e+f x)}{a}+1\right )^{-p} \operatorname {AppellF1}\left (\frac {5}{2},\frac {1}{2},-p,\frac {7}{2},\sin ^2(e+f x),-\frac {b \sin ^2(e+f x)}{a}\right )}{5 f}\)

Input:

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

Output:

(AppellF1[5/2, 1/2, -p, 7/2, Sin[e + f*x]^2, -((b*Sin[e + f*x]^2)/a)]*Sqrt 
[Cos[e + f*x]^2]*Sin[e + f*x]^4*(a + b*Sin[e + f*x]^2)^p*Tan[e + f*x])/(5* 
f*(1 + (b*Sin[e + f*x]^2)/a)^p)
 

Defintions of rubi rules used

rule 394

Int[((e_.)*(x_))^(m_.)*((a_) + (b_.)*(x_)^2)^(p_)*((c_) + (d_.)*(x_)^2)^(q_ 
), x_Symbol] :> Simp[a^p*c^q*((e*x)^(m + 1)/(e*(m + 1)))*AppellF1[(m + 1)/2 
, -p, -q, 1 + (m + 1)/2, (-b)*(x^2/a), (-d)*(x^2/c)], x] /; FreeQ[{a, b, c, 
 d, e, m, p, q}, x] && NeQ[b*c - a*d, 0] && NeQ[m, -1] && NeQ[m, 1] && (Int 
egerQ[p] || GtQ[a, 0]) && (IntegerQ[q] || GtQ[c, 0])
 

rule 395

Int[((e_.)*(x_))^(m_.)*((a_) + (b_.)*(x_)^2)^(p_)*((c_) + (d_.)*(x_)^2)^(q_ 
), x_Symbol] :> Simp[a^IntPart[p]*((a + b*x^2)^FracPart[p]/(1 + b*(x^2/a))^ 
FracPart[p])   Int[(e*x)^m*(1 + b*(x^2/a))^p*(c + d*x^2)^q, x], x] /; FreeQ 
[{a, b, c, d, e, m, p, q}, x] && NeQ[b*c - a*d, 0] && NeQ[m, -1] && NeQ[m, 
1] &&  !(IntegerQ[p] || GtQ[a, 0])
 

rule 3042

Int[u_, x_Symbol] :> Int[DeactivateTrig[u, x], x] /; FunctionOfTrigOfLinear 
Q[u, x]
 

rule 3667

Int[sin[(e_.) + (f_.)*(x_)]^(m_)*((a_) + (b_.)*sin[(e_.) + (f_.)*(x_)]^2)^( 
p_.), x_Symbol] :> With[{ff = FreeFactors[Sin[e + f*x], x]}, Simp[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] &&  !IntegerQ[p]
 

Maple [F]

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

Input:

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

Output:

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

Fricas [F]

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

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

Output:

integral((cos(f*x + e)^4 - 2*cos(f*x + e)^2 + 1)*(-b*cos(f*x + e)^2 + a + 
b)^p, x)
 

Sympy [F(-1)]

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

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

Output:

Timed out
 

Maxima [F]

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

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

Output:

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

Giac [F]

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

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

Output:

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

Mupad [F(-1)]

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

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

Output:

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

Reduce [F]

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

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

Output:

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