\(\int \sin ^{-1-\frac {a^2}{a^2+b^2}}(c+d x) (a+b \sin (c+d x))^2 \, dx\) [219]

Optimal result

Integrand size = 36, antiderivative size = 142 \[ \int \sin ^{-1-\frac {a^2}{a^2+b^2}}(c+d x) (a+b \sin (c+d x))^2 \, dx=-\frac {\left (a^2+b^2\right ) \cos (c+d x) \sin ^{-\frac {a^2}{a^2+b^2}}(c+d x)}{d}+\frac {2 a \left (a^2+b^2\right ) \cos (c+d x) \operatorname {Hypergeometric2F1}\left (\frac {1}{2},\frac {b^2}{2 \left (a^2+b^2\right )},\frac {1}{2} \left (3-\frac {a^2}{a^2+b^2}\right ),\sin ^2(c+d x)\right ) \sin ^{\frac {b^2}{a^2+b^2}}(c+d x)}{b d \sqrt {\cos ^2(c+d x)}} \] Output:

-(a^2+b^2)*cos(d*x+c)/d/(sin(d*x+c)^(a^2/(a^2+b^2)))+2*a*(a^2+b^2)*cos(d*x 
+c)*hypergeom([1/2, b^2/(2*a^2+2*b^2)],[3/2-1/2*a^2/(a^2+b^2)],sin(d*x+c)^ 
2)*sin(d*x+c)^(b^2/(a^2+b^2))/b/d/(cos(d*x+c)^2)^(1/2)
 

Mathematica [A] (verified)

Time = 0.59 (sec) , antiderivative size = 249, normalized size of antiderivative = 1.75 \[ \int \sin ^{-1-\frac {a^2}{a^2+b^2}}(c+d x) (a+b \sin (c+d x))^2 \, dx=\frac {\left (a^2+b^2\right ) \cos (c+d x) \sin ^{-\frac {a^2}{a^2+b^2}}(c+d x) \left (-b \left (a^2+2 b^2\right ) \operatorname {Hypergeometric2F1}\left (\frac {1}{2},-\frac {a^2}{2 \left (a^2+b^2\right )},1-\frac {a^2}{2 \left (a^2+b^2\right )},\sin ^2(c+d x)\right )+\sin (c+d x) \left (2 a \left (a^2+2 b^2\right ) \operatorname {Hypergeometric2F1}\left (\frac {1}{2},\frac {b^2}{2 \left (a^2+b^2\right )},\frac {1}{2} \left (3-\frac {a^2}{a^2+b^2}\right ),\sin ^2(c+d x)\right )+b^3 \operatorname {Hypergeometric2F1}\left (\frac {1}{2},1-\frac {a^2}{2 \left (a^2+b^2\right )},2-\frac {a^2}{2 \left (a^2+b^2\right )},\sin ^2(c+d x)\right ) \sin (c+d x)\right )\right )}{b \left (a^2+2 b^2\right ) d \sqrt {\cos ^2(c+d x)}} \] Input:

Integrate[Sin[c + d*x]^(-1 - a^2/(a^2 + b^2))*(a + b*Sin[c + d*x])^2,x]
 

Output:

((a^2 + b^2)*Cos[c + d*x]*(-(b*(a^2 + 2*b^2)*Hypergeometric2F1[1/2, -1/2*a 
^2/(a^2 + b^2), 1 - a^2/(2*(a^2 + b^2)), Sin[c + d*x]^2]) + Sin[c + d*x]*( 
2*a*(a^2 + 2*b^2)*Hypergeometric2F1[1/2, b^2/(2*(a^2 + b^2)), (3 - a^2/(a^ 
2 + b^2))/2, Sin[c + d*x]^2] + b^3*Hypergeometric2F1[1/2, 1 - a^2/(2*(a^2 
+ b^2)), 2 - a^2/(2*(a^2 + b^2)), Sin[c + d*x]^2]*Sin[c + d*x])))/(b*(a^2 
+ 2*b^2)*d*Sqrt[Cos[c + d*x]^2]*Sin[c + d*x]^(a^2/(a^2 + b^2)))
 

Rubi [A] (verified)

Time = 0.48 (sec) , antiderivative size = 142, normalized size of antiderivative = 1.00, number of steps used = 5, number of rules used = 5, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.139, Rules used = {3042, 3268, 3042, 3122, 3490}

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.

\(\displaystyle \int \sin ^{-\frac {a^2}{a^2+b^2}-1}(c+d x) (a+b \sin (c+d x))^2 \, dx\)

\(\Big \downarrow \) 3042

\(\displaystyle \int \sin (c+d x)^{-\frac {a^2}{a^2+b^2}-1} (a+b \sin (c+d x))^2dx\)

\(\Big \downarrow \) 3268

\(\displaystyle 2 a b \int \sin ^{-\frac {a^2}{a^2+b^2}}(c+d x)dx+\int \sin ^{-\frac {a^2}{a^2+b^2}-1}(c+d x) \left (a^2+b^2 \sin ^2(c+d x)\right )dx\)

\(\Big \downarrow \) 3042

\(\displaystyle 2 a b \int \sin (c+d x)^{-\frac {a^2}{a^2+b^2}}dx+\int \sin (c+d x)^{-\frac {a^2}{a^2+b^2}-1} \left (a^2+b^2 \sin (c+d x)^2\right )dx\)

\(\Big \downarrow \) 3122

\(\displaystyle \int \sin (c+d x)^{-\frac {a^2}{a^2+b^2}-1} \left (a^2+b^2 \sin (c+d x)^2\right )dx+\frac {2 a \left (a^2+b^2\right ) \cos (c+d x) \sin ^{\frac {b^2}{a^2+b^2}}(c+d x) \operatorname {Hypergeometric2F1}\left (\frac {1}{2},\frac {b^2}{2 \left (a^2+b^2\right )},\frac {1}{2} \left (3-\frac {a^2}{a^2+b^2}\right ),\sin ^2(c+d x)\right )}{b d \sqrt {\cos ^2(c+d x)}}\)

\(\Big \downarrow \) 3490

\(\displaystyle \frac {2 a \left (a^2+b^2\right ) \cos (c+d x) \sin ^{\frac {b^2}{a^2+b^2}}(c+d x) \operatorname {Hypergeometric2F1}\left (\frac {1}{2},\frac {b^2}{2 \left (a^2+b^2\right )},\frac {1}{2} \left (3-\frac {a^2}{a^2+b^2}\right ),\sin ^2(c+d x)\right )}{b d \sqrt {\cos ^2(c+d x)}}-\frac {\left (a^2+b^2\right ) \cos (c+d x) \sin ^{-\frac {a^2}{a^2+b^2}}(c+d x)}{d}\)

Input:

Int[Sin[c + d*x]^(-1 - a^2/(a^2 + b^2))*(a + b*Sin[c + d*x])^2,x]
 

Output:

-(((a^2 + b^2)*Cos[c + d*x])/(d*Sin[c + d*x]^(a^2/(a^2 + b^2)))) + (2*a*(a 
^2 + b^2)*Cos[c + d*x]*Hypergeometric2F1[1/2, b^2/(2*(a^2 + b^2)), (3 - a^ 
2/(a^2 + b^2))/2, Sin[c + d*x]^2]*Sin[c + d*x]^(b^2/(a^2 + b^2)))/(b*d*Sqr 
t[Cos[c + d*x]^2])
 

Defintions of rubi rules used

rule 3042

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

rule 3122

Int[((b_.)*sin[(c_.) + (d_.)*(x_)])^(n_), x_Symbol] :> Simp[Cos[c + d*x]*(( 
b*Sin[c + d*x])^(n + 1)/(b*d*(n + 1)*Sqrt[Cos[c + d*x]^2]))*Hypergeometric2 
F1[1/2, (n + 1)/2, (n + 3)/2, Sin[c + d*x]^2], x] /; FreeQ[{b, c, d, n}, x] 
 &&  !IntegerQ[2*n]
 

rule 3268

Int[((b_.)*sin[(e_.) + (f_.)*(x_)])^(m_)*((c_) + (d_.)*sin[(e_.) + (f_.)*(x 
_)])^2, x_Symbol] :> Simp[2*c*(d/b)   Int[(b*Sin[e + f*x])^(m + 1), x], x] 
+ Int[(b*Sin[e + f*x])^m*(c^2 + d^2*Sin[e + f*x]^2), x] /; FreeQ[{b, c, d, 
e, f, m}, x]
 

rule 3490

Int[((b_.)*sin[(e_.) + (f_.)*(x_)])^(m_.)*((A_) + (C_.)*sin[(e_.) + (f_.)*( 
x_)]^2), x_Symbol] :> Simp[A*Cos[e + f*x]*((b*Sin[e + f*x])^(m + 1)/(b*f*(m 
 + 1))), x] /; FreeQ[{b, e, f, A, C, m}, x] && EqQ[A*(m + 2) + C*(m + 1), 0 
]
 

Maple [F]

\[\int \sin \left (d x +c \right )^{-1-\frac {a^{2}}{a^{2}+b^{2}}} \left (a +b \sin \left (d x +c \right )\right )^{2}d x\]

Input:

int(sin(d*x+c)^(-1-a^2/(a^2+b^2))*(a+b*sin(d*x+c))^2,x)
 

Output:

int(sin(d*x+c)^(-1-a^2/(a^2+b^2))*(a+b*sin(d*x+c))^2,x)
 

Fricas [F]

\[ \int \sin ^{-1-\frac {a^2}{a^2+b^2}}(c+d x) (a+b \sin (c+d x))^2 \, dx=\int { {\left (b \sin \left (d x + c\right ) + a\right )}^{2} \sin \left (d x + c\right )^{-\frac {a^{2}}{a^{2} + b^{2}} - 1} \,d x } \] Input:

integrate(sin(d*x+c)^(-1-a^2/(a^2+b^2))*(a+b*sin(d*x+c))^2,x, algorithm="f 
ricas")
 

Output:

integral(-(b^2*cos(d*x + c)^2 - 2*a*b*sin(d*x + c) - a^2 - b^2)*sin(d*x + 
c)^(-(2*a^2 + b^2)/(a^2 + b^2)), x)
 

Sympy [F(-1)]

Timed out. \[ \int \sin ^{-1-\frac {a^2}{a^2+b^2}}(c+d x) (a+b \sin (c+d x))^2 \, dx=\text {Timed out} \] Input:

integrate(sin(d*x+c)**(-1-a**2/(a**2+b**2))*(a+b*sin(d*x+c))**2,x)
 

Output:

Timed out
 

Maxima [F]

\[ \int \sin ^{-1-\frac {a^2}{a^2+b^2}}(c+d x) (a+b \sin (c+d x))^2 \, dx=\int { {\left (b \sin \left (d x + c\right ) + a\right )}^{2} \sin \left (d x + c\right )^{-\frac {a^{2}}{a^{2} + b^{2}} - 1} \,d x } \] Input:

integrate(sin(d*x+c)^(-1-a^2/(a^2+b^2))*(a+b*sin(d*x+c))^2,x, algorithm="m 
axima")
 

Output:

integrate((b*sin(d*x + c) + a)^2*sin(d*x + c)^(-a^2/(a^2 + b^2) - 1), x)
 

Giac [F]

\[ \int \sin ^{-1-\frac {a^2}{a^2+b^2}}(c+d x) (a+b \sin (c+d x))^2 \, dx=\int { {\left (b \sin \left (d x + c\right ) + a\right )}^{2} \sin \left (d x + c\right )^{-\frac {a^{2}}{a^{2} + b^{2}} - 1} \,d x } \] Input:

integrate(sin(d*x+c)^(-1-a^2/(a^2+b^2))*(a+b*sin(d*x+c))^2,x, algorithm="g 
iac")
 

Output:

integrate((b*sin(d*x + c) + a)^2*sin(d*x + c)^(-a^2/(a^2 + b^2) - 1), x)
 

Mupad [F(-1)]

Timed out. \[ \int \sin ^{-1-\frac {a^2}{a^2+b^2}}(c+d x) (a+b \sin (c+d x))^2 \, dx=\int \frac {{\left (a+b\,\sin \left (c+d\,x\right )\right )}^2}{{\sin \left (c+d\,x\right )}^{\frac {a^2}{a^2+b^2}+1}} \,d x \] Input:

int((a + b*sin(c + d*x))^2/sin(c + d*x)^(a^2/(a^2 + b^2) + 1),x)
 

Output:

int((a + b*sin(c + d*x))^2/sin(c + d*x)^(a^2/(a^2 + b^2) + 1), x)
 

Reduce [F]

\[ \int \sin ^{-1-\frac {a^2}{a^2+b^2}}(c+d x) (a+b \sin (c+d x))^2 \, dx=\left (\int \sin \left (d x +c \right )^{\frac {b^{2}}{a^{2}+b^{2}}}d x \right ) b^{2}+\left (\int \frac {\sin \left (d x +c \right )^{\frac {b^{2}}{a^{2}+b^{2}}}}{\sin \left (d x +c \right )^{2}}d x \right ) a^{2}+2 \left (\int \frac {\sin \left (d x +c \right )^{\frac {b^{2}}{a^{2}+b^{2}}}}{\sin \left (d x +c \right )}d x \right ) a b \] Input:

int(sin(d*x+c)^(-1-a^2/(a^2+b^2))*(a+b*sin(d*x+c))^2,x)
 

Output:

int(sin(c + d*x)**(b**2/(a**2 + b**2)),x)*b**2 + int(sin(c + d*x)**(b**2/( 
a**2 + b**2))/sin(c + d*x)**2,x)*a**2 + 2*int(sin(c + d*x)**(b**2/(a**2 + 
b**2))/sin(c + d*x),x)*a*b