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

Optimal. Leaf size=274 \[ \frac{\sqrt{2} (A-B) \cos (e+f x) (a \sin (e+f x)+a)^m \sqrt{\frac{c+d \sin (e+f x)}{c-d}} F_1\left (m+\frac{1}{2};\frac{1}{2},\frac{1}{2};m+\frac{3}{2};\frac{1}{2} (\sin (e+f x)+1),-\frac{d (\sin (e+f x)+1)}{c-d}\right )}{f (2 m+1) \sqrt{1-\sin (e+f x)} \sqrt{c+d \sin (e+f x)}}+\frac{\sqrt{2} B \cos (e+f x) (a \sin (e+f x)+a)^{m+1} \sqrt{\frac{c+d \sin (e+f x)}{c-d}} F_1\left (m+\frac{3}{2};\frac{1}{2},\frac{1}{2};m+\frac{5}{2};\frac{1}{2} (\sin (e+f x)+1),-\frac{d (\sin (e+f x)+1)}{c-d}\right )}{a f (2 m+3) \sqrt{1-\sin (e+f x)} \sqrt{c+d \sin (e+f x)}} \]

[Out]

(Sqrt[2]*(A - B)*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)])/(f*(1 + 2*m)*Sqrt[1 - Sin[e + f*x]]*S
qrt[c + d*Sin[e + f*x]]) + (Sqrt[2]*B*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*f*(3 +
2*m)*Sqrt[1 - Sin[e + f*x]]*Sqrt[c + d*Sin[e + f*x]])

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Rubi [A]  time = 0.5435, antiderivative size = 274, normalized size of antiderivative = 1., number of steps used = 9, number of rules used = 5, integrand size = 37, \(\frac{\text{number of rules}}{\text{integrand size}}\) = 0.135, Rules used = {2987, 2788, 140, 139, 138} \[ \frac{\sqrt{2} (A-B) \cos (e+f x) (a \sin (e+f x)+a)^m \sqrt{\frac{c+d \sin (e+f x)}{c-d}} F_1\left (m+\frac{1}{2};\frac{1}{2},\frac{1}{2};m+\frac{3}{2};\frac{1}{2} (\sin (e+f x)+1),-\frac{d (\sin (e+f x)+1)}{c-d}\right )}{f (2 m+1) \sqrt{1-\sin (e+f x)} \sqrt{c+d \sin (e+f x)}}+\frac{\sqrt{2} B \cos (e+f x) (a \sin (e+f x)+a)^{m+1} \sqrt{\frac{c+d \sin (e+f x)}{c-d}} F_1\left (m+\frac{3}{2};\frac{1}{2},\frac{1}{2};m+\frac{5}{2};\frac{1}{2} (\sin (e+f x)+1),-\frac{d (\sin (e+f x)+1)}{c-d}\right )}{a f (2 m+3) \sqrt{1-\sin (e+f x)} \sqrt{c+d \sin (e+f x)}} \]

Antiderivative was successfully verified.

[In]

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

[Out]

(Sqrt[2]*(A - B)*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)])/(f*(1 + 2*m)*Sqrt[1 - Sin[e + f*x]]*S
qrt[c + d*Sin[e + f*x]]) + (Sqrt[2]*B*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*f*(3 +
2*m)*Sqrt[1 - Sin[e + f*x]]*Sqrt[c + d*Sin[e + f*x]])

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

Mathematica [B]  time = 6.31583, size = 672, normalized size = 2.45 \[ \frac{6 (c+d) \cot \left (\frac{1}{4} (2 e+2 f x+\pi )\right ) \sin ^2\left (\frac{1}{4} (2 e+2 f x+\pi )\right )^{\frac{1}{2}-m} \cos ^2\left (\frac{1}{4} (2 e+2 f x-\pi )\right )^{m-\frac{1}{2}} (a (\sin (e+f x)+1))^m \left (\frac{(B c-A d) F_1\left (\frac{1}{2};\frac{1}{2}-m,\frac{1}{2};\frac{3}{2};\cos ^2\left (\frac{1}{4} (2 e+2 f x+\pi )\right ),\frac{2 d \sin ^2\left (\frac{1}{4} (2 e+2 f x-\pi )\right )}{c+d}\right )}{\cos ^2\left (\frac{1}{4} (2 e+2 f x+\pi )\right ) \left (2 d F_1\left (\frac{3}{2};\frac{1}{2}-m,\frac{3}{2};\frac{5}{2};\cos ^2\left (\frac{1}{4} (2 e+2 f x+\pi )\right ),\frac{2 d \sin ^2\left (\frac{1}{4} (2 e+2 f x-\pi )\right )}{c+d}\right )-(2 m-1) (c+d) F_1\left (\frac{3}{2};\frac{3}{2}-m,\frac{1}{2};\frac{5}{2};\cos ^2\left (\frac{1}{4} (2 e+2 f x+\pi )\right ),\frac{2 d \sin ^2\left (\frac{1}{4} (2 e+2 f x-\pi )\right )}{c+d}\right )\right )+3 (c+d) F_1\left (\frac{1}{2};\frac{1}{2}-m,\frac{1}{2};\frac{3}{2};\cos ^2\left (\frac{1}{4} (2 e+2 f x+\pi )\right ),\frac{2 d \sin ^2\left (\frac{1}{4} (2 e+2 f x-\pi )\right )}{c+d}\right )}-\frac{B (c+d \sin (e+f x)) F_1\left (\frac{1}{2};\frac{1}{2}-m,-\frac{1}{2};\frac{3}{2};\cos ^2\left (\frac{1}{4} (2 e+2 f x+\pi )\right ),\frac{2 d \sin ^2\left (\frac{1}{4} (2 e+2 f x-\pi )\right )}{c+d}\right )}{\cos ^2\left (\frac{1}{4} (2 e+2 f x+\pi )\right ) \left (-2 d F_1\left (\frac{3}{2};\frac{1}{2}-m,\frac{1}{2};\frac{5}{2};\cos ^2\left (\frac{1}{4} (2 e+2 f x+\pi )\right ),\frac{2 d \sin ^2\left (\frac{1}{4} (2 e+2 f x-\pi )\right )}{c+d}\right )-(2 m-1) (c+d) F_1\left (\frac{3}{2};\frac{3}{2}-m,-\frac{1}{2};\frac{5}{2};\cos ^2\left (\frac{1}{4} (2 e+2 f x+\pi )\right ),\frac{2 d \sin ^2\left (\frac{1}{4} (2 e+2 f x-\pi )\right )}{c+d}\right )\right )+3 (c+d) F_1\left (\frac{1}{2};\frac{1}{2}-m,-\frac{1}{2};\frac{3}{2};\cos ^2\left (\frac{1}{4} (2 e+2 f x+\pi )\right ),\frac{2 d \sin ^2\left (\frac{1}{4} (2 e+2 f x-\pi )\right )}{c+d}\right )}\right )}{d f \sqrt{c+d \sin (e+f x)}} \]

Warning: Unable to verify antiderivative.

[In]

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

[Out]

(6*(c + d)*(Cos[(2*e - Pi + 2*f*x)/4]^2)^(-1/2 + m)*Cot[(2*e + Pi + 2*f*x)/4]*(a*(1 + Sin[e + f*x]))^m*(((B*c
- A*d)*AppellF1[1/2, 1/2 - m, 1/2, 3/2, Cos[(2*e + Pi + 2*f*x)/4]^2, (2*d*Sin[(2*e - Pi + 2*f*x)/4]^2)/(c + d)
])/(3*(c + d)*AppellF1[1/2, 1/2 - m, 1/2, 3/2, Cos[(2*e + Pi + 2*f*x)/4]^2, (2*d*Sin[(2*e - Pi + 2*f*x)/4]^2)/
(c + d)] + (2*d*AppellF1[3/2, 1/2 - m, 3/2, 5/2, Cos[(2*e + Pi + 2*f*x)/4]^2, (2*d*Sin[(2*e - Pi + 2*f*x)/4]^2
)/(c + d)] - (c + d)*(-1 + 2*m)*AppellF1[3/2, 3/2 - m, 1/2, 5/2, Cos[(2*e + Pi + 2*f*x)/4]^2, (2*d*Sin[(2*e -
Pi + 2*f*x)/4]^2)/(c + d)])*Cos[(2*e + Pi + 2*f*x)/4]^2) - (B*AppellF1[1/2, 1/2 - m, -1/2, 3/2, Cos[(2*e + Pi
+ 2*f*x)/4]^2, (2*d*Sin[(2*e - Pi + 2*f*x)/4]^2)/(c + d)]*(c + d*Sin[e + f*x]))/(3*(c + d)*AppellF1[1/2, 1/2 -
m, -1/2, 3/2, Cos[(2*e + Pi + 2*f*x)/4]^2, (2*d*Sin[(2*e - Pi + 2*f*x)/4]^2)/(c + d)] + (-2*d*AppellF1[3/2, 1
/2 - m, 1/2, 5/2, Cos[(2*e + Pi + 2*f*x)/4]^2, (2*d*Sin[(2*e - Pi + 2*f*x)/4]^2)/(c + d)] - (c + d)*(-1 + 2*m)
*AppellF1[3/2, 3/2 - m, -1/2, 5/2, Cos[(2*e + Pi + 2*f*x)/4]^2, (2*d*Sin[(2*e - Pi + 2*f*x)/4]^2)/(c + d)])*Co
s[(2*e + Pi + 2*f*x)/4]^2))*(Sin[(2*e + Pi + 2*f*x)/4]^2)^(1/2 - m))/(d*f*Sqrt[c + d*Sin[e + f*x]])

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Maple [F]  time = 0.343, 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 ) \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+d*sin(f*x+e))^(1/2),x)

[Out]

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

[Out]

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

[Out]

integral((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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Sympy [F]  time = 0., size = 0, normalized size = 0. \begin{align*} \int \frac{\left (a \left (\sin{\left (e + f x \right )} + 1\right )\right )^{m} \left (A + B \sin{\left (e + f x \right )}\right )}{\sqrt{c + d \sin{\left (e + f x \right )}}}\, dx \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+d*sin(f*x+e))**(1/2),x)

[Out]

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

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Giac [F]  time = 0., size = 0, normalized size = 0. \begin{align*} \int \frac{{\left (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+d*sin(f*x+e))^(1/2),x, algorithm="giac")

[Out]

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