3.3.18 \(\int \frac {(d \sin (e+f x))^m}{(a+b \sin (e+f x))^3} \, dx\) [218]

3.3.18.1 Optimal result

Integrand size = 23, antiderivative size = 406 \[ \int \frac {(d \sin (e+f x))^m}{(a+b \sin (e+f x))^3} \, dx=-\frac {3 a b^2 \operatorname {AppellF1}\left (\frac {1}{2},\frac {1}{2} (-1-m),3,\frac {3}{2},\cos ^2(e+f x),-\frac {b^2 \cos ^2(e+f x)}{a^2-b^2}\right ) \cos (e+f x) (d \sin (e+f x))^{1+m} \sin ^2(e+f x)^{\frac {1}{2} (-1-m)}}{\left (a^2-b^2\right )^3 d f}-\frac {a^3 d \operatorname {AppellF1}\left (\frac {1}{2},\frac {1-m}{2},3,\frac {3}{2},\cos ^2(e+f x),-\frac {b^2 \cos ^2(e+f x)}{a^2-b^2}\right ) \cos (e+f x) (d \sin (e+f x))^{-1+m} \sin ^2(e+f x)^{\frac {1-m}{2}}}{\left (a^2-b^2\right )^3 f}+\frac {b^3 \operatorname {AppellF1}\left (\frac {1}{2},\frac {1}{2} (-2-m),3,\frac {3}{2},\cos ^2(e+f x),-\frac {b^2 \cos ^2(e+f x)}{a^2-b^2}\right ) \cos (e+f x) (d \sin (e+f x))^m \sin ^2(e+f x)^{-m/2}}{\left (a^2-b^2\right )^3 f}+\frac {3 a^2 b \operatorname {AppellF1}\left (\frac {1}{2},-\frac {m}{2},3,\frac {3}{2},\cos ^2(e+f x),-\frac {b^2 \cos ^2(e+f x)}{a^2-b^2}\right ) \cos (e+f x) (d \sin (e+f x))^m \sin ^2(e+f x)^{-m/2}}{\left (a^2-b^2\right )^3 f} \]

output

-3*a*b^2*AppellF1(1/2,-1/2-1/2*m,3,3/2,cos(f*x+e)^2,-b^2*cos(f*x+e)^2/(a^2 
-b^2))*cos(f*x+e)*(d*sin(f*x+e))^(1+m)*(sin(f*x+e)^2)^(-1/2-1/2*m)/(a^2-b^ 
2)^3/d/f-a^3*d*AppellF1(1/2,-1/2*m+1/2,3,3/2,cos(f*x+e)^2,-b^2*cos(f*x+e)^ 
2/(a^2-b^2))*cos(f*x+e)*(d*sin(f*x+e))^(-1+m)*(sin(f*x+e)^2)^(-1/2*m+1/2)/ 
(a^2-b^2)^3/f+b^3*AppellF1(1/2,-1-1/2*m,3,3/2,cos(f*x+e)^2,-b^2*cos(f*x+e) 
^2/(a^2-b^2))*cos(f*x+e)*(d*sin(f*x+e))^m/(a^2-b^2)^3/f/((sin(f*x+e)^2)^(1 
/2*m))+3*a^2*b*AppellF1(1/2,-1/2*m,3,3/2,cos(f*x+e)^2,-b^2*cos(f*x+e)^2/(a 
^2-b^2))*cos(f*x+e)*(d*sin(f*x+e))^m/(a^2-b^2)^3/f/((sin(f*x+e)^2)^(1/2*m) 
)
 

3.3.18.2 Mathematica [B] (warning: unable to verify)

Leaf count is larger than twice the leaf count of optimal. \(2298\) vs. \(2(406)=812\).

Time = 19.19 (sec) , antiderivative size = 2298, normalized size of antiderivative = 5.66 \[ \int \frac {(d \sin (e+f x))^m}{(a+b \sin (e+f x))^3} \, dx=\text {Result too large to show} \]

input

Integrate[(d*Sin[e + f*x])^m/(a + b*Sin[e + f*x])^3,x]
 

output

-(((Sec[e + f*x]^2)^(m/2)*(d*Sin[e + f*x])^m*Tan[e + f*x]*(Tan[e + f*x]/Sq 
rt[Sec[e + f*x]^2])^m*(-(a*(a^2 + 3*b^2)*(2 + m)*AppellF1[(1 + m)/2, (-2 + 
 m)/2, 2, (3 + m)/2, -Tan[e + f*x]^2, (-1 + b^2/a^2)*Tan[e + f*x]^2]) + b* 
(4*a*b*(2 + m)*AppellF1[(1 + m)/2, (-2 + m)/2, 3, (3 + m)/2, -Tan[e + f*x] 
^2, (-1 + b^2/a^2)*Tan[e + f*x]^2] + (1 + m)*((3*a^2 + b^2)*AppellF1[(2 + 
m)/2, (-1 + m)/2, 2, (4 + m)/2, -Tan[e + f*x]^2, (-1 + b^2/a^2)*Tan[e + f* 
x]^2] - 4*b^2*AppellF1[(2 + m)/2, (-1 + m)/2, 3, (4 + m)/2, -Tan[e + f*x]^ 
2, (-1 + b^2/a^2)*Tan[e + f*x]^2])*Tan[e + f*x])))/(a^4*(a^2 - b^2)*f*(1 + 
 m)*(2 + m)*(a + b*Sin[e + f*x])^3*(-(((Sec[e + f*x]^2)^(1 + m/2)*(Tan[e + 
 f*x]/Sqrt[Sec[e + f*x]^2])^m*(-(a*(a^2 + 3*b^2)*(2 + m)*AppellF1[(1 + m)/ 
2, (-2 + m)/2, 2, (3 + m)/2, -Tan[e + f*x]^2, (-1 + b^2/a^2)*Tan[e + f*x]^ 
2]) + b*(4*a*b*(2 + m)*AppellF1[(1 + m)/2, (-2 + m)/2, 3, (3 + m)/2, -Tan[ 
e + f*x]^2, (-1 + b^2/a^2)*Tan[e + f*x]^2] + (1 + m)*((3*a^2 + b^2)*Appell 
F1[(2 + m)/2, (-1 + m)/2, 2, (4 + m)/2, -Tan[e + f*x]^2, (-1 + b^2/a^2)*Ta 
n[e + f*x]^2] - 4*b^2*AppellF1[(2 + m)/2, (-1 + m)/2, 3, (4 + m)/2, -Tan[e 
 + f*x]^2, (-1 + b^2/a^2)*Tan[e + f*x]^2])*Tan[e + f*x])))/(a^4*(a^2 - b^2 
)*(1 + m)*(2 + m))) - (m*(Sec[e + f*x]^2)^(m/2)*Tan[e + f*x]^2*(Tan[e + f* 
x]/Sqrt[Sec[e + f*x]^2])^m*(-(a*(a^2 + 3*b^2)*(2 + m)*AppellF1[(1 + m)/2, 
(-2 + m)/2, 2, (3 + m)/2, -Tan[e + f*x]^2, (-1 + b^2/a^2)*Tan[e + f*x]^2]) 
 + b*(4*a*b*(2 + m)*AppellF1[(1 + m)/2, (-2 + m)/2, 3, (3 + m)/2, -Tan[...
 

3.3.18.3 Rubi [A] (verified)

Time = 0.88 (sec) , antiderivative size = 406, 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 = {3042, 3303, 2009}

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.

\(\displaystyle \int \frac {(d \sin (e+f x))^m}{(a+b \sin (e+f x))^3} \, dx\)

\(\Big \downarrow \) 3042

\(\displaystyle \int \frac {(d \sin (e+f x))^m}{(a+b \sin (e+f x))^3}dx\)

\(\Big \downarrow \) 3303

\(\displaystyle \int \left (\frac {3 a b^2 \sin ^2(e+f x) (d \sin (e+f x))^m}{\left (a^2-b^2 \sin ^2(e+f x)\right )^3}-\frac {3 a^2 b \sin (e+f x) (d \sin (e+f x))^m}{\left (a^2-b^2 \sin ^2(e+f x)\right )^3}+\frac {b^3 \sin ^3(e+f x) (d \sin (e+f x))^m}{\left (b^2 \sin ^2(e+f x)-a^2\right )^3}+\frac {a^3 (d \sin (e+f x))^m}{\left (a^2-b^2 \sin ^2(e+f x)\right )^3}\right )dx\)

\(\Big \downarrow \) 2009

\(\displaystyle -\frac {3 a b^2 \cos (e+f x) \sin ^2(e+f x)^{\frac {1}{2} (-m-1)} (d \sin (e+f x))^{m+1} \operatorname {AppellF1}\left (\frac {1}{2},\frac {1}{2} (-m-1),3,\frac {3}{2},\cos ^2(e+f x),-\frac {b^2 \cos ^2(e+f x)}{a^2-b^2}\right )}{d f \left (a^2-b^2\right )^3}+\frac {3 a^2 b \cos (e+f x) \sin ^2(e+f x)^{-m/2} (d \sin (e+f x))^m \operatorname {AppellF1}\left (\frac {1}{2},-\frac {m}{2},3,\frac {3}{2},\cos ^2(e+f x),-\frac {b^2 \cos ^2(e+f x)}{a^2-b^2}\right )}{f \left (a^2-b^2\right )^3}+\frac {b^3 \cos (e+f x) \sin ^2(e+f x)^{-m/2} (d \sin (e+f x))^m \operatorname {AppellF1}\left (\frac {1}{2},\frac {1}{2} (-m-2),3,\frac {3}{2},\cos ^2(e+f x),-\frac {b^2 \cos ^2(e+f x)}{a^2-b^2}\right )}{f \left (a^2-b^2\right )^3}-\frac {a^3 d \cos (e+f x) \sin ^2(e+f x)^{\frac {1-m}{2}} (d \sin (e+f x))^{m-1} \operatorname {AppellF1}\left (\frac {1}{2},\frac {1-m}{2},3,\frac {3}{2},\cos ^2(e+f x),-\frac {b^2 \cos ^2(e+f x)}{a^2-b^2}\right )}{f \left (a^2-b^2\right )^3}\)

input

Int[(d*Sin[e + f*x])^m/(a + b*Sin[e + f*x])^3,x]
 

output

(-3*a*b^2*AppellF1[1/2, (-1 - m)/2, 3, 3/2, Cos[e + f*x]^2, -((b^2*Cos[e + 
 f*x]^2)/(a^2 - b^2))]*Cos[e + f*x]*(d*Sin[e + f*x])^(1 + m)*(Sin[e + f*x] 
^2)^((-1 - m)/2))/((a^2 - b^2)^3*d*f) - (a^3*d*AppellF1[1/2, (1 - m)/2, 3, 
 3/2, Cos[e + f*x]^2, -((b^2*Cos[e + f*x]^2)/(a^2 - b^2))]*Cos[e + f*x]*(d 
*Sin[e + f*x])^(-1 + m)*(Sin[e + f*x]^2)^((1 - m)/2))/((a^2 - b^2)^3*f) + 
(b^3*AppellF1[1/2, (-2 - m)/2, 3, 3/2, Cos[e + f*x]^2, -((b^2*Cos[e + f*x] 
^2)/(a^2 - b^2))]*Cos[e + f*x]*(d*Sin[e + f*x])^m)/((a^2 - b^2)^3*f*(Sin[e 
 + f*x]^2)^(m/2)) + (3*a^2*b*AppellF1[1/2, -1/2*m, 3, 3/2, Cos[e + f*x]^2, 
 -((b^2*Cos[e + f*x]^2)/(a^2 - b^2))]*Cos[e + f*x]*(d*Sin[e + f*x])^m)/((a 
^2 - b^2)^3*f*(Sin[e + f*x]^2)^(m/2))
 

3.3.18.3.1 Defintions of rubi rules used

rule 2009

Int[u_, x_Symbol] :> Simp[IntSum[u, x], x] /; SumQ[u]
 

rule 3042

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

rule 3303

Int[((d_.)*sin[(e_.) + (f_.)*(x_)])^(n_.)*((a_) + (b_.)*sin[(e_.) + (f_.)*( 
x_)])^(m_.), x_Symbol] :> Int[ExpandTrig[(d*sin[e + f*x])^n*(1/((a - b*sin[ 
e + f*x])^m/(a^2 - b^2*sin[e + f*x]^2)^m)), x], x] /; FreeQ[{a, b, d, e, f, 
 n}, x] && NeQ[a^2 - b^2, 0] && ILtQ[m, -1]
 

3.3.18.4 Maple [F]

\[\int \frac {\left (d \sin \left (f x +e \right )\right )^{m}}{\left (a +b \sin \left (f x +e \right )\right )^{3}}d x\]

input

int((d*sin(f*x+e))^m/(a+b*sin(f*x+e))^3,x)
 

output

int((d*sin(f*x+e))^m/(a+b*sin(f*x+e))^3,x)
 

3.3.18.5 Fricas [F]

\[ \int \frac {(d \sin (e+f x))^m}{(a+b \sin (e+f x))^3} \, dx=\int { \frac {\left (d \sin \left (f x + e\right )\right )^{m}}{{\left (b \sin \left (f x + e\right ) + a\right )}^{3}} \,d x } \]

input

integrate((d*sin(f*x+e))^m/(a+b*sin(f*x+e))^3,x, algorithm="fricas")
 

output

integral(-(d*sin(f*x + e))^m/(3*a*b^2*cos(f*x + e)^2 - a^3 - 3*a*b^2 + (b^ 
3*cos(f*x + e)^2 - 3*a^2*b - b^3)*sin(f*x + e)), x)
 

3.3.18.6 Sympy [F(-1)]

Timed out. \[ \int \frac {(d \sin (e+f x))^m}{(a+b \sin (e+f x))^3} \, dx=\text {Timed out} \]

input

integrate((d*sin(f*x+e))**m/(a+b*sin(f*x+e))**3,x)
 

output

Timed out
 

3.3.18.7 Maxima [F]

\[ \int \frac {(d \sin (e+f x))^m}{(a+b \sin (e+f x))^3} \, dx=\int { \frac {\left (d \sin \left (f x + e\right )\right )^{m}}{{\left (b \sin \left (f x + e\right ) + a\right )}^{3}} \,d x } \]

input

integrate((d*sin(f*x+e))^m/(a+b*sin(f*x+e))^3,x, algorithm="maxima")
 

output

integrate((d*sin(f*x + e))^m/(b*sin(f*x + e) + a)^3, x)
 

3.3.18.8 Giac [F]

\[ \int \frac {(d \sin (e+f x))^m}{(a+b \sin (e+f x))^3} \, dx=\int { \frac {\left (d \sin \left (f x + e\right )\right )^{m}}{{\left (b \sin \left (f x + e\right ) + a\right )}^{3}} \,d x } \]

input

integrate((d*sin(f*x+e))^m/(a+b*sin(f*x+e))^3,x, algorithm="giac")
 

output

integrate((d*sin(f*x + e))^m/(b*sin(f*x + e) + a)^3, x)
 

3.3.18.9 Mupad [F(-1)]

Timed out. \[ \int \frac {(d \sin (e+f x))^m}{(a+b \sin (e+f x))^3} \, dx=\int \frac {{\left (d\,\sin \left (e+f\,x\right )\right )}^m}{{\left (a+b\,\sin \left (e+f\,x\right )\right )}^3} \,d x \]

input

int((d*sin(e + f*x))^m/(a + b*sin(e + f*x))^3,x)
 

output

int((d*sin(e + f*x))^m/(a + b*sin(e + f*x))^3, x)