3.21 \(\int \frac {(a+a \sin (e+f x))^m (A+B \sin (e+f x)+C \sin ^2(e+f x))}{\sqrt {c-c \sin (e+f x)}} \, dx\)

Optimal. Leaf size=170 \[ \frac {(A+B+C) \cos (e+f x) (a \sin (e+f x)+a)^m \, _2F_1\left (1,m+\frac {1}{2};m+\frac {3}{2};\frac {1}{2} (\sin (e+f x)+1)\right )}{f (2 m+1) \sqrt {c-c \sin (e+f x)}}-\frac {2 B \cos (e+f x) (a \sin (e+f x)+a)^m}{f (2 m+1) \sqrt {c-c \sin (e+f x)}}-\frac {2 C \cos (e+f x) (a \sin (e+f x)+a)^{m+1}}{a f (2 m+3) \sqrt {c-c \sin (e+f x)}} \]

[Out]

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

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Rubi [A]  time = 0.47, antiderivative size = 170, normalized size of antiderivative = 1.00, number of steps used = 5, number of rules used = 5, integrand size = 48, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.104, Rules used = {3037, 2973, 2745, 2667, 68} \[ \frac {(A+B+C) \cos (e+f x) (a \sin (e+f x)+a)^m \, _2F_1\left (1,m+\frac {1}{2};m+\frac {3}{2};\frac {1}{2} (\sin (e+f x)+1)\right )}{f (2 m+1) \sqrt {c-c \sin (e+f x)}}-\frac {2 B \cos (e+f x) (a \sin (e+f x)+a)^m}{f (2 m+1) \sqrt {c-c \sin (e+f x)}}-\frac {2 C \cos (e+f x) (a \sin (e+f x)+a)^{m+1}}{a f (2 m+3) \sqrt {c-c \sin (e+f x)}} \]

Antiderivative was successfully verified.

[In]

Int[((a + a*Sin[e + f*x])^m*(A + B*Sin[e + f*x] + C*Sin[e + f*x]^2))/Sqrt[c - c*Sin[e + f*x]],x]

[Out]

(-2*B*Cos[e + f*x]*(a + a*Sin[e + f*x])^m)/(f*(1 + 2*m)*Sqrt[c - c*Sin[e + f*x]]) + ((A + B + C)*Cos[e + f*x]*
Hypergeometric2F1[1, 1/2 + m, 3/2 + m, (1 + Sin[e + f*x])/2]*(a + a*Sin[e + f*x])^m)/(f*(1 + 2*m)*Sqrt[c - c*S
in[e + f*x]]) - (2*C*Cos[e + f*x]*(a + a*Sin[e + f*x])^(1 + m))/(a*f*(3 + 2*m)*Sqrt[c - c*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]

Rule 3037

Int[(((a_.) + (b_.)*sin[(e_.) + (f_.)*(x_)])^(m_.)*((A_.) + (B_.)*sin[(e_.) + (f_.)*(x_)] + (C_.)*sin[(e_.) +
(f_.)*(x_)]^2))/Sqrt[(c_.) + (d_.)*sin[(e_.) + (f_.)*(x_)]], x_Symbol] :> Simp[(-2*C*Cos[e + f*x]*(a + b*Sin[e
+ f*x])^(m + 1))/(b*f*(2*m + 3)*Sqrt[c + d*Sin[e + f*x]]), x] + Int[((a + b*Sin[e + f*x])^m*Simp[A + C + B*Si
n[e + f*x], x])/Sqrt[c + d*Sin[e + f*x]], x] /; FreeQ[{a, b, c, d, e, f, A, B, C, m}, x] && EqQ[b*c + a*d, 0]
&& EqQ[a^2 - b^2, 0] &&  !LtQ[m, -2^(-1)]

Rubi steps

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

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Mathematica [F]  time = 180.00, size = 0, normalized size = 0.00 \[ \text {\$Aborted} \]

Verification is Not applicable to the result.

[In]

Integrate[((a + a*Sin[e + f*x])^m*(A + B*Sin[e + f*x] + C*Sin[e + f*x]^2))/Sqrt[c - c*Sin[e + f*x]],x]

[Out]

$Aborted

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fricas [F]  time = 0.47, size = 0, normalized size = 0.00 \[ {\rm integral}\left (\frac {{\left (C \cos \left (f x + e\right )^{2} - B \sin \left (f x + e\right ) - A - C\right )} \sqrt {-c \sin \left (f x + e\right ) + c} {\left (a \sin \left (f x + e\right ) + a\right )}^{m}}{c \sin \left (f x + e\right ) - c}, x\right ) \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

integral((C*cos(f*x + e)^2 - B*sin(f*x + e) - A - C)*sqrt(-c*sin(f*x + e) + c)*(a*sin(f*x + e) + a)^m/(c*sin(f
*x + e) - c), x)

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giac [F]  time = 0.00, size = 0, normalized size = 0.00 \[ \int \frac {{\left (C \sin \left (f x + e\right )^{2} + B \sin \left (f x + e\right ) + A\right )} {\left (a \sin \left (f x + e\right ) + a\right )}^{m}}{\sqrt {-c \sin \left (f x + e\right ) + c}}\,{d x} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

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

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maple [F]  time = 4.33, size = 0, normalized size = 0.00 \[ \int \frac {\left (a +a \sin \left (f x +e \right )\right )^{m} \left (A +B \sin \left (f x +e \right )+C \left (\sin ^{2}\left (f x +e \right )\right )\right )}{\sqrt {c -c \sin \left (f x +e \right )}}\, dx \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

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

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maxima [F]  time = 0.00, size = 0, normalized size = 0.00 \[ \int \frac {{\left (C \sin \left (f x + e\right )^{2} + B \sin \left (f x + e\right ) + A\right )} {\left (a \sin \left (f x + e\right ) + a\right )}^{m}}{\sqrt {-c \sin \left (f x + e\right ) + c}}\,{d x} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

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

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mupad [F]  time = 0.00, size = -1, normalized size = -0.01 \[ \int \frac {{\left (a+a\,\sin \left (e+f\,x\right )\right )}^m\,\left (C\,{\sin \left (e+f\,x\right )}^2+B\,\sin \left (e+f\,x\right )+A\right )}{\sqrt {c-c\,\sin \left (e+f\,x\right )}} \,d x \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

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

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sympy [F]  time = 0.00, size = 0, normalized size = 0.00 \[ \int \frac {\left (a \left (\sin {\left (e + f x \right )} + 1\right )\right )^{m} \left (A + B \sin {\left (e + f x \right )} + C \sin ^{2}{\left (e + f x \right )}\right )}{\sqrt {- c \left (\sin {\left (e + f x \right )} - 1\right )}}\, dx \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

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

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