3.204 \(\int \frac {(A+B \sin (e+f x)) (c+c \sin (e+f x))^m}{\sqrt {a-a \sin (e+f x)}} \, dx\)

Optimal. Leaf size=118 \[ \frac {(A+B) \cos (e+f x) (c \sin (e+f x)+c)^m \, _2F_1\left (1,m+\frac {1}{2};m+\frac {3}{2};\frac {1}{2} (\sin (e+f x)+1)\right )}{f (2 m+1) \sqrt {a-a \sin (e+f x)}}-\frac {2 B \cos (e+f x) (c \sin (e+f x)+c)^m}{f (2 m+1) \sqrt {a-a \sin (e+f x)}} \]

[Out]

-2*B*cos(f*x+e)*(c+c*sin(f*x+e))^m/f/(1+2*m)/(a-a*sin(f*x+e))^(1/2)+(A+B)*cos(f*x+e)*hypergeom([1, 1/2+m],[3/2
+m],1/2+1/2*sin(f*x+e))*(c+c*sin(f*x+e))^m/f/(1+2*m)/(a-a*sin(f*x+e))^(1/2)

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Rubi [A]  time = 0.29, antiderivative size = 118, normalized size of antiderivative = 1.00, number of steps used = 4, number of rules used = 4, integrand size = 38, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.105, Rules used = {2973, 2745, 2667, 68} \[ \frac {(A+B) \cos (e+f x) (c \sin (e+f x)+c)^m \, _2F_1\left (1,m+\frac {1}{2};m+\frac {3}{2};\frac {1}{2} (\sin (e+f x)+1)\right )}{f (2 m+1) \sqrt {a-a \sin (e+f x)}}-\frac {2 B \cos (e+f x) (c \sin (e+f x)+c)^m}{f (2 m+1) \sqrt {a-a \sin (e+f x)}} \]

Antiderivative was successfully verified.

[In]

Int[((A + B*Sin[e + f*x])*(c + c*Sin[e + f*x])^m)/Sqrt[a - a*Sin[e + f*x]],x]

[Out]

(-2*B*Cos[e + f*x]*(c + c*Sin[e + f*x])^m)/(f*(1 + 2*m)*Sqrt[a - a*Sin[e + f*x]]) + ((A + B)*Cos[e + f*x]*Hype
rgeometric2F1[1, 1/2 + m, 3/2 + m, (1 + Sin[e + f*x])/2]*(c + c*Sin[e + f*x])^m)/(f*(1 + 2*m)*Sqrt[a - a*Sin[e
+ f*x]])

Rule 68

Int[((a_) + (b_.)*(x_))^(m_)*((c_) + (d_.)*(x_))^(n_), x_Symbol] :> Simp[((b*c - a*d)^n*(a + b*x)^(m + 1)*Hype
rgeometric2F1[-n, m + 1, m + 2, -((d*(a + b*x))/(b*c - a*d))])/(b^(n + 1)*(m + 1)), x] /; FreeQ[{a, b, c, d, m
}, x] && NeQ[b*c - a*d, 0] &&  !IntegerQ[m] && IntegerQ[n]

Rule 2667

Int[cos[(e_.) + (f_.)*(x_)]^(p_.)*((a_) + (b_.)*sin[(e_.) + (f_.)*(x_)])^(m_.), x_Symbol] :> Dist[1/(b^p*f), S
ubst[Int[(a + x)^(m + (p - 1)/2)*(a - x)^((p - 1)/2), x], x, b*Sin[e + f*x]], x] /; FreeQ[{a, b, e, f, m}, x]
&& IntegerQ[(p - 1)/2] && EqQ[a^2 - b^2, 0] && (GeQ[p, -1] ||  !IntegerQ[m + 1/2])

Rule 2745

Int[((a_) + (b_.)*sin[(e_.) + (f_.)*(x_)])^(m_)*((c_) + (d_.)*sin[(e_.) + (f_.)*(x_)])^(n_), x_Symbol] :> Dist
[(a^IntPart[m]*c^IntPart[m]*(a + b*Sin[e + f*x])^FracPart[m]*(c + d*Sin[e + f*x])^FracPart[m])/Cos[e + f*x]^(2
*FracPart[m]), Int[Cos[e + f*x]^(2*m)*(c + d*Sin[e + f*x])^(n - m), x], x] /; FreeQ[{a, b, c, d, e, f, m, n},
x] && EqQ[b*c + a*d, 0] && EqQ[a^2 - b^2, 0] && (FractionQ[m] ||  !FractionQ[n])

Rule 2973

Int[((a_) + (b_.)*sin[(e_.) + (f_.)*(x_)])^(m_.)*((A_.) + (B_.)*sin[(e_.) + (f_.)*(x_)])*((c_) + (d_.)*sin[(e_
.) + (f_.)*(x_)])^(n_), x_Symbol] :> -Simp[(B*Cos[e + f*x]*(a + b*Sin[e + f*x])^m*(c + d*Sin[e + f*x])^n)/(f*(
m + n + 1)), x] - Dist[(B*c*(m - n) - A*d*(m + n + 1))/(d*(m + n + 1)), Int[(a + b*Sin[e + f*x])^m*(c + d*Sin[
e + f*x])^n, x], x] /; FreeQ[{a, b, c, d, e, f, A, B, m, n}, x] && EqQ[b*c + a*d, 0] && EqQ[a^2 - b^2, 0] &&
!LtQ[m, -2^(-1)] && NeQ[m + n + 1, 0]

Rubi steps

\begin {align*} \int \frac {(A+B \sin (e+f x)) (c+c \sin (e+f x))^m}{\sqrt {a-a \sin (e+f x)}} \, dx &=-\frac {2 B \cos (e+f x) (c+c \sin (e+f x))^m}{f (1+2 m) \sqrt {a-a \sin (e+f x)}}-(-A-B) \int \frac {(c+c \sin (e+f x))^m}{\sqrt {a-a \sin (e+f x)}} \, dx\\ &=-\frac {2 B \cos (e+f x) (c+c \sin (e+f x))^m}{f (1+2 m) \sqrt {a-a \sin (e+f x)}}-\frac {((-A-B) \cos (e+f x)) \int \sec (e+f x) (c+c \sin (e+f x))^{\frac {1}{2}+m} \, dx}{\sqrt {a-a \sin (e+f x)} \sqrt {c+c \sin (e+f x)}}\\ &=-\frac {2 B \cos (e+f x) (c+c \sin (e+f x))^m}{f (1+2 m) \sqrt {a-a \sin (e+f x)}}-\frac {((-A-B) c \cos (e+f x)) \operatorname {Subst}\left (\int \frac {(c+x)^{-\frac {1}{2}+m}}{c-x} \, dx,x,c \sin (e+f x)\right )}{f \sqrt {a-a \sin (e+f x)} \sqrt {c+c \sin (e+f x)}}\\ &=-\frac {2 B \cos (e+f x) (c+c \sin (e+f x))^m}{f (1+2 m) \sqrt {a-a \sin (e+f x)}}+\frac {(A+B) \cos (e+f x) \, _2F_1\left (1,\frac {1}{2}+m;\frac {3}{2}+m;\frac {1}{2} (1+\sin (e+f x))\right ) (c+c \sin (e+f x))^m}{f (1+2 m) \sqrt {a-a \sin (e+f x)}}\\ \end {align*}

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Mathematica [A]  time = 2.38, size = 200, normalized size = 1.69 \[ \frac {2^{-2 m-\frac {3}{2}} \sin \left (\frac {1}{4} (2 e+2 f x+\pi )\right ) \left (\cos \left (\frac {1}{2} (e+f x)\right )-\sin \left (\frac {1}{2} (e+f x)\right )\right ) (c (\sin (e+f x)+1))^m \left (2^{2 m+1} (A+B) \, _2F_1\left (1,2 m+1;2 (m+1);\sin \left (\frac {1}{4} (2 e+2 f x+\pi )\right )\right )+(A+B) \sec ^2\left (\frac {1}{8} (2 e+2 f x-\pi )\right )^{2 m+1} \, _2F_1\left (2 m+1,2 m+1;2 (m+1);\frac {1}{2} \left (1-\tan ^2\left (\frac {1}{8} (2 e+2 f x-\pi )\right )\right )\right )-B 2^{2 m+3}\right )}{(2 f m+f) \sqrt {a-a \sin (e+f x)}} \]

Antiderivative was successfully verified.

[In]

Integrate[((A + B*Sin[e + f*x])*(c + c*Sin[e + f*x])^m)/Sqrt[a - a*Sin[e + f*x]],x]

[Out]

(2^(-3/2 - 2*m)*(-(2^(3 + 2*m)*B) + 2^(1 + 2*m)*(A + B)*Hypergeometric2F1[1, 1 + 2*m, 2*(1 + m), Sin[(2*e + Pi
+ 2*f*x)/4]] + (A + B)*Hypergeometric2F1[1 + 2*m, 1 + 2*m, 2*(1 + m), (1 - Tan[(2*e - Pi + 2*f*x)/8]^2)/2]*(S
ec[(2*e - Pi + 2*f*x)/8]^2)^(1 + 2*m))*(Cos[(e + f*x)/2] - Sin[(e + f*x)/2])*(c*(1 + Sin[e + f*x]))^m*Sin[(2*e
+ Pi + 2*f*x)/4])/((f + 2*f*m)*Sqrt[a - a*Sin[e + f*x]])

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fricas [F]  time = 0.46, size = 0, normalized size = 0.00 \[ {\rm integral}\left (-\frac {{\left (B \sin \left (f x + e\right ) + A\right )} \sqrt {-a \sin \left (f x + e\right ) + a} {\left (c \sin \left (f x + e\right ) + c\right )}^{m}}{a \sin \left (f x + e\right ) - a}, x\right ) \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate((A+B*sin(f*x+e))*(c+c*sin(f*x+e))^m/(a-a*sin(f*x+e))^(1/2),x, algorithm="fricas")

[Out]

integral(-(B*sin(f*x + e) + A)*sqrt(-a*sin(f*x + e) + a)*(c*sin(f*x + e) + c)^m/(a*sin(f*x + e) - a), x)

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giac [F]  time = 0.00, size = 0, normalized size = 0.00 \[ \int \frac {{\left (B \sin \left (f x + e\right ) + A\right )} {\left (c \sin \left (f x + e\right ) + c\right )}^{m}}{\sqrt {-a \sin \left (f x + e\right ) + a}}\,{d x} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate((A+B*sin(f*x+e))*(c+c*sin(f*x+e))^m/(a-a*sin(f*x+e))^(1/2),x, algorithm="giac")

[Out]

integrate((B*sin(f*x + e) + A)*(c*sin(f*x + e) + c)^m/sqrt(-a*sin(f*x + e) + a), x)

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maple [F]  time = 1.12, size = 0, normalized size = 0.00 \[ \int \frac {\left (A +B \sin \left (f x +e \right )\right ) \left (c +c \sin \left (f x +e \right )\right )^{m}}{\sqrt {a -a \sin \left (f x +e \right )}}\, dx \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

int((A+B*sin(f*x+e))*(c+c*sin(f*x+e))^m/(a-a*sin(f*x+e))^(1/2),x)

[Out]

int((A+B*sin(f*x+e))*(c+c*sin(f*x+e))^m/(a-a*sin(f*x+e))^(1/2),x)

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maxima [F]  time = 0.00, size = 0, normalized size = 0.00 \[ \int \frac {{\left (B \sin \left (f x + e\right ) + A\right )} {\left (c \sin \left (f x + e\right ) + c\right )}^{m}}{\sqrt {-a \sin \left (f x + e\right ) + a}}\,{d x} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate((A+B*sin(f*x+e))*(c+c*sin(f*x+e))^m/(a-a*sin(f*x+e))^(1/2),x, algorithm="maxima")

[Out]

integrate((B*sin(f*x + e) + A)*(c*sin(f*x + e) + c)^m/sqrt(-a*sin(f*x + e) + a), x)

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mupad [F]  time = 0.00, size = -1, normalized size = -0.01 \[ \int \frac {\left (A+B\,\sin \left (e+f\,x\right )\right )\,{\left (c+c\,\sin \left (e+f\,x\right )\right )}^m}{\sqrt {a-a\,\sin \left (e+f\,x\right )}} \,d x \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

int(((A + B*sin(e + f*x))*(c + c*sin(e + f*x))^m)/(a - a*sin(e + f*x))^(1/2),x)

[Out]

int(((A + B*sin(e + f*x))*(c + c*sin(e + f*x))^m)/(a - a*sin(e + f*x))^(1/2), x)

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sympy [F]  time = 0.00, size = 0, normalized size = 0.00 \[ \int \frac {\left (c \left (\sin {\left (e + f x \right )} + 1\right )\right )^{m} \left (A + B \sin {\left (e + f x \right )}\right )}{\sqrt {- a \left (\sin {\left (e + f x \right )} - 1\right )}}\, dx \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate((A+B*sin(f*x+e))*(c+c*sin(f*x+e))**m/(a-a*sin(f*x+e))**(1/2),x)

[Out]

Integral((c*(sin(e + f*x) + 1))**m*(A + B*sin(e + f*x))/sqrt(-a*(sin(e + f*x) - 1)), x)

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