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

Optimal. Leaf size=389 \[ \frac{\sqrt{2} \cos (e+f x) (a \sin (e+f x)+a)^m (2 c (2 C m+C)-d (-A (2 m+3)+2 B m+3 B+2 C m-C)) \sqrt{\frac{c+d \sin (e+f x)}{c-d}} F_1\left (m+\frac{1}{2};\frac{1}{2},\frac{1}{2};m+\frac{3}{2};\frac{1}{2} (\sin (e+f x)+1),-\frac{d (\sin (e+f x)+1)}{c-d}\right )}{d f (2 m+1) (2 m+3) \sqrt{1-\sin (e+f x)} \sqrt{c+d \sin (e+f x)}}-\frac{\sqrt{2} \cos (e+f x) (2 c C (m+1)-d (B (2 m+3)+2 C m)) (a \sin (e+f x)+a)^{m+1} \sqrt{\frac{c+d \sin (e+f x)}{c-d}} F_1\left (m+\frac{3}{2};\frac{1}{2},\frac{1}{2};m+\frac{5}{2};\frac{1}{2} (\sin (e+f x)+1),-\frac{d (\sin (e+f x)+1)}{c-d}\right )}{a d f (2 m+3)^2 \sqrt{1-\sin (e+f x)} \sqrt{c+d \sin (e+f x)}}-\frac{2 C \cos (e+f x) (a \sin (e+f x)+a)^m \sqrt{c+d \sin (e+f x)}}{d f (2 m+3)} \]

[Out]

(-2*C*Cos[e + f*x]*(a + a*Sin[e + f*x])^m*Sqrt[c + d*Sin[e + f*x]])/(d*f*(3 + 2*m)) + (Sqrt[2]*(2*c*(C + 2*C*m
) - d*(3*B - C + 2*B*m + 2*C*m - A*(3 + 2*m)))*AppellF1[1/2 + m, 1/2, 1/2, 3/2 + m, (1 + Sin[e + f*x])/2, -((d
*(1 + Sin[e + f*x]))/(c - d))]*Cos[e + f*x]*(a + a*Sin[e + f*x])^m*Sqrt[(c + d*Sin[e + f*x])/(c - d)])/(d*f*(1
+ 2*m)*(3 + 2*m)*Sqrt[1 - Sin[e + f*x]]*Sqrt[c + d*Sin[e + f*x]]) - (Sqrt[2]*(2*c*C*(1 + m) - d*(2*C*m + B*(3
+ 2*m)))*AppellF1[3/2 + m, 1/2, 1/2, 5/2 + m, (1 + Sin[e + f*x])/2, -((d*(1 + Sin[e + f*x]))/(c - d))]*Cos[e
+ f*x]*(a + a*Sin[e + f*x])^(1 + m)*Sqrt[(c + d*Sin[e + f*x])/(c - d)])/(a*d*f*(3 + 2*m)^2*Sqrt[1 - Sin[e + f*
x]]*Sqrt[c + d*Sin[e + f*x]])

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Rubi [A]  time = 0.928285, antiderivative size = 386, normalized size of antiderivative = 0.99, number of steps used = 10, number of rules used = 6, integrand size = 47, \(\frac{\text{number of rules}}{\text{integrand size}}\) = 0.128, Rules used = {3045, 2987, 2788, 140, 139, 138} \[ \frac{\sqrt{2} \cos (e+f x) (a \sin (e+f x)+a)^m (2 c (2 C m+C)-d (-A (2 m+3)+2 B m+3 B+2 C m-C)) \sqrt{\frac{c+d \sin (e+f x)}{c-d}} F_1\left (m+\frac{1}{2};\frac{1}{2},\frac{1}{2};m+\frac{3}{2};\frac{1}{2} (\sin (e+f x)+1),-\frac{d (\sin (e+f x)+1)}{c-d}\right )}{d f (2 m+1) (2 m+3) \sqrt{1-\sin (e+f x)} \sqrt{c+d \sin (e+f x)}}+\frac{\sqrt{2} \cos (e+f x) (B d (2 m+3)-2 c C (m+1)+2 C d m) (a \sin (e+f x)+a)^{m+1} \sqrt{\frac{c+d \sin (e+f x)}{c-d}} F_1\left (m+\frac{3}{2};\frac{1}{2},\frac{1}{2};m+\frac{5}{2};\frac{1}{2} (\sin (e+f x)+1),-\frac{d (\sin (e+f x)+1)}{c-d}\right )}{a d f (2 m+3)^2 \sqrt{1-\sin (e+f x)} \sqrt{c+d \sin (e+f x)}}-\frac{2 C \cos (e+f x) (a \sin (e+f x)+a)^m \sqrt{c+d \sin (e+f x)}}{d f (2 m+3)} \]

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 + d*Sin[e + f*x]],x]

[Out]

(-2*C*Cos[e + f*x]*(a + a*Sin[e + f*x])^m*Sqrt[c + d*Sin[e + f*x]])/(d*f*(3 + 2*m)) + (Sqrt[2]*(2*c*(C + 2*C*m
) - d*(3*B - C + 2*B*m + 2*C*m - A*(3 + 2*m)))*AppellF1[1/2 + m, 1/2, 1/2, 3/2 + m, (1 + Sin[e + f*x])/2, -((d
*(1 + Sin[e + f*x]))/(c - d))]*Cos[e + f*x]*(a + a*Sin[e + f*x])^m*Sqrt[(c + d*Sin[e + f*x])/(c - d)])/(d*f*(1
+ 2*m)*(3 + 2*m)*Sqrt[1 - Sin[e + f*x]]*Sqrt[c + d*Sin[e + f*x]]) + (Sqrt[2]*(2*C*d*m - 2*c*C*(1 + m) + B*d*(
3 + 2*m))*AppellF1[3/2 + m, 1/2, 1/2, 5/2 + m, (1 + Sin[e + f*x])/2, -((d*(1 + Sin[e + f*x]))/(c - d))]*Cos[e
+ f*x]*(a + a*Sin[e + f*x])^(1 + m)*Sqrt[(c + d*Sin[e + f*x])/(c - d)])/(a*d*f*(3 + 2*m)^2*Sqrt[1 - Sin[e + f*
x]]*Sqrt[c + d*Sin[e + f*x]])

Rule 3045

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

Rule 2987

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

Rule 2788

Int[((a_) + (b_.)*sin[(e_.) + (f_.)*(x_)])^(m_)*((c_) + (d_.)*sin[(e_.) + (f_.)*(x_)])^(n_.), x_Symbol] :> Dis
t[(a^2*Cos[e + f*x])/(f*Sqrt[a + b*Sin[e + f*x]]*Sqrt[a - b*Sin[e + f*x]]), Subst[Int[((a + b*x)^(m - 1/2)*(c
+ d*x)^n)/Sqrt[a - b*x], x], x, Sin[e + f*x]], x] /; FreeQ[{a, b, c, d, e, f, m, n}, x] && NeQ[b*c - a*d, 0] &
& EqQ[a^2 - b^2, 0] && NeQ[c^2 - d^2, 0] &&  !IntegerQ[m]

Rule 140

Int[((a_) + (b_.)*(x_))^(m_)*((c_.) + (d_.)*(x_))^(n_)*((e_.) + (f_.)*(x_))^(p_), x_Symbol] :> Dist[(c + d*x)^
FracPart[n]/((b/(b*c - a*d))^IntPart[n]*((b*(c + d*x))/(b*c - a*d))^FracPart[n]), Int[(a + b*x)^m*((b*c)/(b*c
- a*d) + (b*d*x)/(b*c - a*d))^n*(e + f*x)^p, x], x] /; FreeQ[{a, b, c, d, e, f, m, n, p}, x] &&  !IntegerQ[m]
&&  !IntegerQ[n] &&  !IntegerQ[p] &&  !GtQ[b/(b*c - a*d), 0] &&  !SimplerQ[c + d*x, a + b*x] &&  !SimplerQ[e +
f*x, a + b*x]

Rule 139

Int[((a_) + (b_.)*(x_))^(m_)*((c_.) + (d_.)*(x_))^(n_)*((e_.) + (f_.)*(x_))^(p_), x_Symbol] :> Dist[(e + f*x)^
FracPart[p]/((b/(b*e - a*f))^IntPart[p]*((b*(e + f*x))/(b*e - a*f))^FracPart[p]), Int[(a + b*x)^m*(c + d*x)^n*
((b*e)/(b*e - a*f) + (b*f*x)/(b*e - a*f))^p, x], x] /; FreeQ[{a, b, c, d, e, f, m, n, p}, x] &&  !IntegerQ[m]
&&  !IntegerQ[n] &&  !IntegerQ[p] && GtQ[b/(b*c - a*d), 0] &&  !GtQ[b/(b*e - a*f), 0]

Rule 138

Int[((a_) + (b_.)*(x_))^(m_)*((c_.) + (d_.)*(x_))^(n_)*((e_.) + (f_.)*(x_))^(p_), x_Symbol] :> Simp[((a + b*x)
^(m + 1)*AppellF1[m + 1, -n, -p, m + 2, -((d*(a + b*x))/(b*c - a*d)), -((f*(a + b*x))/(b*e - a*f))])/(b*(m + 1
)*(b/(b*c - a*d))^n*(b/(b*e - a*f))^p), x] /; FreeQ[{a, b, c, d, e, f, m, n, p}, x] &&  !IntegerQ[m] &&  !Inte
gerQ[n] &&  !IntegerQ[p] && GtQ[b/(b*c - a*d), 0] && GtQ[b/(b*e - a*f), 0] &&  !(GtQ[d/(d*a - c*b), 0] && GtQ[
d/(d*e - c*f), 0] && SimplerQ[c + d*x, a + b*x]) &&  !(GtQ[f/(f*a - e*b), 0] && GtQ[f/(f*c - e*d), 0] && Simpl
erQ[e + f*x, a + b*x])

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+d \sin (e+f x)}} \, dx &=-\frac{2 C \cos (e+f x) (a+a \sin (e+f x))^m \sqrt{c+d \sin (e+f x)}}{d f (3+2 m)}+\frac{2 \int \frac{(a+a \sin (e+f x))^m \left (\frac{1}{2} a (A d (3+2 m)+C (d+2 c m))+\frac{1}{2} a (2 C d m-2 c C (1+m)+B d (3+2 m)) \sin (e+f x)\right )}{\sqrt{c+d \sin (e+f x)}} \, dx}{a d (3+2 m)}\\ &=-\frac{2 C \cos (e+f x) (a+a \sin (e+f x))^m \sqrt{c+d \sin (e+f x)}}{d f (3+2 m)}+\frac{(2 C d m-2 c C (1+m)+B d (3+2 m)) \int \frac{(a+a \sin (e+f x))^{1+m}}{\sqrt{c+d \sin (e+f x)}} \, dx}{a d (3+2 m)}+\frac{(2 c (C+2 C m)-d (3 B-C+2 B m+2 C m-A (3+2 m))) \int \frac{(a+a \sin (e+f x))^m}{\sqrt{c+d \sin (e+f x)}} \, dx}{d (3+2 m)}\\ &=-\frac{2 C \cos (e+f x) (a+a \sin (e+f x))^m \sqrt{c+d \sin (e+f x)}}{d f (3+2 m)}+\frac{(a (2 C d m-2 c C (1+m)+B d (3+2 m)) \cos (e+f x)) \operatorname{Subst}\left (\int \frac{(a+a x)^{\frac{1}{2}+m}}{\sqrt{a-a x} \sqrt{c+d x}} \, dx,x,\sin (e+f x)\right )}{d f (3+2 m) \sqrt{a-a \sin (e+f x)} \sqrt{a+a \sin (e+f x)}}+\frac{\left (a^2 (2 c (C+2 C m)-d (3 B-C+2 B m+2 C m-A (3+2 m))) \cos (e+f x)\right ) \operatorname{Subst}\left (\int \frac{(a+a x)^{-\frac{1}{2}+m}}{\sqrt{a-a x} \sqrt{c+d x}} \, dx,x,\sin (e+f x)\right )}{d f (3+2 m) \sqrt{a-a \sin (e+f x)} \sqrt{a+a \sin (e+f x)}}\\ &=-\frac{2 C \cos (e+f x) (a+a \sin (e+f x))^m \sqrt{c+d \sin (e+f x)}}{d f (3+2 m)}+\frac{\left (a (2 C d m-2 c C (1+m)+B d (3+2 m)) \cos (e+f x) \sqrt{\frac{a-a \sin (e+f x)}{a}}\right ) \operatorname{Subst}\left (\int \frac{(a+a x)^{\frac{1}{2}+m}}{\sqrt{\frac{1}{2}-\frac{x}{2}} \sqrt{c+d x}} \, dx,x,\sin (e+f x)\right )}{\sqrt{2} d f (3+2 m) (a-a \sin (e+f x)) \sqrt{a+a \sin (e+f x)}}+\frac{\left (a^2 (2 c (C+2 C m)-d (3 B-C+2 B m+2 C m-A (3+2 m))) \cos (e+f x) \sqrt{\frac{a-a \sin (e+f x)}{a}}\right ) \operatorname{Subst}\left (\int \frac{(a+a x)^{-\frac{1}{2}+m}}{\sqrt{\frac{1}{2}-\frac{x}{2}} \sqrt{c+d x}} \, dx,x,\sin (e+f x)\right )}{\sqrt{2} d f (3+2 m) (a-a \sin (e+f x)) \sqrt{a+a \sin (e+f x)}}\\ &=-\frac{2 C \cos (e+f x) (a+a \sin (e+f x))^m \sqrt{c+d \sin (e+f x)}}{d f (3+2 m)}+\frac{\left (a (2 C d m-2 c C (1+m)+B d (3+2 m)) \cos (e+f x) \sqrt{\frac{a-a \sin (e+f x)}{a}} \sqrt{\frac{a (c+d \sin (e+f x))}{a c-a d}}\right ) \operatorname{Subst}\left (\int \frac{(a+a x)^{\frac{1}{2}+m}}{\sqrt{\frac{1}{2}-\frac{x}{2}} \sqrt{\frac{a c}{a c-a d}+\frac{a d x}{a c-a d}}} \, dx,x,\sin (e+f x)\right )}{\sqrt{2} d f (3+2 m) (a-a \sin (e+f x)) \sqrt{a+a \sin (e+f x)} \sqrt{c+d \sin (e+f x)}}+\frac{\left (a^2 (2 c (C+2 C m)-d (3 B-C+2 B m+2 C m-A (3+2 m))) \cos (e+f x) \sqrt{\frac{a-a \sin (e+f x)}{a}} \sqrt{\frac{a (c+d \sin (e+f x))}{a c-a d}}\right ) \operatorname{Subst}\left (\int \frac{(a+a x)^{-\frac{1}{2}+m}}{\sqrt{\frac{1}{2}-\frac{x}{2}} \sqrt{\frac{a c}{a c-a d}+\frac{a d x}{a c-a d}}} \, dx,x,\sin (e+f x)\right )}{\sqrt{2} d f (3+2 m) (a-a \sin (e+f x)) \sqrt{a+a \sin (e+f x)} \sqrt{c+d \sin (e+f x)}}\\ &=-\frac{2 C \cos (e+f x) (a+a \sin (e+f x))^m \sqrt{c+d \sin (e+f x)}}{d f (3+2 m)}+\frac{\sqrt{2} (2 c (C+2 C m)-d (3 B-C+2 B m+2 C m-A (3+2 m))) F_1\left (\frac{1}{2}+m;\frac{1}{2},\frac{1}{2};\frac{3}{2}+m;\frac{1}{2} (1+\sin (e+f x)),-\frac{d (1+\sin (e+f x))}{c-d}\right ) \cos (e+f x) (a+a \sin (e+f x))^m \sqrt{\frac{c+d \sin (e+f x)}{c-d}}}{d f (1+2 m) (3+2 m) \sqrt{1-\sin (e+f x)} \sqrt{c+d \sin (e+f x)}}+\frac{\sqrt{2} (2 C d m-2 c C (1+m)+B d (3+2 m)) F_1\left (\frac{3}{2}+m;\frac{1}{2},\frac{1}{2};\frac{5}{2}+m;\frac{1}{2} (1+\sin (e+f x)),-\frac{d (1+\sin (e+f x))}{c-d}\right ) \cos (e+f x) \sqrt{1-\sin (e+f x)} (a+a \sin (e+f x))^{1+m} \sqrt{\frac{c+d \sin (e+f x)}{c-d}}}{d f (3+2 m)^2 (a-a \sin (e+f x)) \sqrt{c+d \sin (e+f x)}}\\ \end{align*}

Mathematica [B]  time = 34.5322, size = 9783, normalized size = 25.15 \[ \text{Result too large to show} \]

Warning: Unable to verify antiderivative.

[In]

Integrate[((a + a*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]

[Out]

Result too large to show

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Maple [F]  time = 0.748, size = 0, normalized size = 0. \begin{align*} \int{ \left ( a+a\sin \left ( fx+e \right ) \right ) ^{m} \left ( A+B\sin \left ( fx+e \right ) +C \left ( \sin \left ( fx+e \right ) \right ) ^{2} \right ){\frac{1}{\sqrt{c+d\sin \left ( fx+e \right ) }}}}\, dx \end{align*}

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+d*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+d*sin(f*x+e))^(1/2),x)

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Maxima [F]  time = 0., size = 0, normalized size = 0. \begin{align*} \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{d \sin \left (f x + e\right ) + c}}\,{d x} \end{align*}

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+d*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(d*sin(f*x + e) + c), x)

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Fricas [F]  time = 0., size = 0, normalized size = 0. \begin{align*}{\rm integral}\left (-\frac{{\left (C \cos \left (f x + e\right )^{2} - B \sin \left (f x + e\right ) - A - C\right )}{\left (a \sin \left (f x + e\right ) + a\right )}^{m}}{\sqrt{d \sin \left (f x + e\right ) + c}}, x\right ) \end{align*}

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+d*sin(f*x+e))^(1/2),x, algorithm="fricas")

[Out]

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

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Sympy [F(-1)]  time = 0., size = 0, normalized size = 0. \begin{align*} \text{Timed out} \end{align*}

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+d*sin(f*x+e))**(1/2),x)

[Out]

Timed out

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Giac [F]  time = 0., size = 0, normalized size = 0. \begin{align*} \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{d \sin \left (f x + e\right ) + c}}\,{d x} \end{align*}

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+d*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(d*sin(f*x + e) + c), x)