3.335 \(\int \frac{(A+B \sin (e+f x)) (c+d \sin (e+f x))^n}{(a+a \sin (e+f x))^{3/2}} \, dx\)

Optimal. Leaf size=269 \[ \frac{d (A-B) \cos (e+f x) \sqrt{\frac{d (1-\sin (e+f x))}{c+d}} (c+d \sin (e+f x))^{n+1} F_1\left (n+1;\frac{1}{2},2;n+2;\frac{c+d \sin (e+f x)}{c+d},\frac{c+d \sin (e+f x)}{c-d}\right )}{f (n+1) (c-d)^2 (a-a \sin (e+f x)) \sqrt{a \sin (e+f x)+a}}-\frac{B \cos (e+f x) \sqrt{\frac{d (1-\sin (e+f x))}{c+d}} (c+d \sin (e+f x))^{n+1} F_1\left (n+1;\frac{1}{2},1;n+2;\frac{c+d \sin (e+f x)}{c+d},\frac{c+d \sin (e+f x)}{c-d}\right )}{a f (n+1) (c-d) (1-\sin (e+f x)) \sqrt{a \sin (e+f x)+a}} \]

[Out]

-((B*AppellF1[1 + n, 1/2, 1, 2 + n, (c + d*Sin[e + f*x])/(c + d), (c + d*Sin[e + f*x])/(c - d)]*Cos[e + f*x]*S
qrt[(d*(1 - Sin[e + f*x]))/(c + d)]*(c + d*Sin[e + f*x])^(1 + n))/(a*(c - d)*f*(1 + n)*(1 - Sin[e + f*x])*Sqrt
[a + a*Sin[e + f*x]])) + ((A - B)*d*AppellF1[1 + n, 1/2, 2, 2 + n, (c + d*Sin[e + f*x])/(c + d), (c + d*Sin[e
+ f*x])/(c - d)]*Cos[e + f*x]*Sqrt[(d*(1 - Sin[e + f*x]))/(c + d)]*(c + d*Sin[e + f*x])^(1 + n))/((c - d)^2*f*
(1 + n)*(a - a*Sin[e + f*x])*Sqrt[a + a*Sin[e + f*x]])

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Rubi [A]  time = 0.469883, antiderivative size = 269, normalized size of antiderivative = 1., number of steps used = 7, number of rules used = 4, integrand size = 37, \(\frac{\text{number of rules}}{\text{integrand size}}\) = 0.108, Rules used = {2987, 2788, 137, 136} \[ \frac{d (A-B) \cos (e+f x) \sqrt{\frac{d (1-\sin (e+f x))}{c+d}} (c+d \sin (e+f x))^{n+1} F_1\left (n+1;\frac{1}{2},2;n+2;\frac{c+d \sin (e+f x)}{c+d},\frac{c+d \sin (e+f x)}{c-d}\right )}{f (n+1) (c-d)^2 (a-a \sin (e+f x)) \sqrt{a \sin (e+f x)+a}}-\frac{B \cos (e+f x) \sqrt{\frac{d (1-\sin (e+f x))}{c+d}} (c+d \sin (e+f x))^{n+1} F_1\left (n+1;\frac{1}{2},1;n+2;\frac{c+d \sin (e+f x)}{c+d},\frac{c+d \sin (e+f x)}{c-d}\right )}{a f (n+1) (c-d) (1-\sin (e+f x)) \sqrt{a \sin (e+f x)+a}} \]

Antiderivative was successfully verified.

[In]

Int[((A + B*Sin[e + f*x])*(c + d*Sin[e + f*x])^n)/(a + a*Sin[e + f*x])^(3/2),x]

[Out]

-((B*AppellF1[1 + n, 1/2, 1, 2 + n, (c + d*Sin[e + f*x])/(c + d), (c + d*Sin[e + f*x])/(c - d)]*Cos[e + f*x]*S
qrt[(d*(1 - Sin[e + f*x]))/(c + d)]*(c + d*Sin[e + f*x])^(1 + n))/(a*(c - d)*f*(1 + n)*(1 - Sin[e + f*x])*Sqrt
[a + a*Sin[e + f*x]])) + ((A - B)*d*AppellF1[1 + n, 1/2, 2, 2 + n, (c + d*Sin[e + f*x])/(c + d), (c + d*Sin[e
+ f*x])/(c - d)]*Cos[e + f*x]*Sqrt[(d*(1 - Sin[e + f*x]))/(c + d)]*(c + d*Sin[e + f*x])^(1 + n))/((c - d)^2*f*
(1 + n)*(a - a*Sin[e + f*x])*Sqrt[a + a*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 137

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}, x] &&  !IntegerQ[m] &&
!IntegerQ[n] && IntegerQ[p] &&  !GtQ[b/(b*c - a*d), 0] &&  !SimplerQ[c + d*x, a + b*x]

Rule 136

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

Rubi steps

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

Mathematica [B]  time = 10.1795, size = 603, normalized size = 2.24 \[ \frac{\sec (e+f x) (c+d \sin (e+f x))^n \left (a A (\sin (e+f x)+1) \left (a \sqrt{2-2 \sin (e+f x)} (\sin (e+f x)+1) \left (\frac{c+d \sin (e+f x)}{c-d}\right )^{-n} F_1\left (1;\frac{1}{2},-n;2;\frac{1}{2} (\sin (e+f x)+1),\frac{d (\sin (e+f x)+1)}{d-c}\right )-\frac{4 \sqrt{\frac{\sin (e+f x)-1}{\sin (e+f x)+1}} \left (\frac{c-d}{d \sin (e+f x)+d}+1\right )^{-n} \left (2 a (2 n+1) F_1\left (\frac{1}{2}-n;-\frac{1}{2},-n;\frac{3}{2}-n;\frac{2}{\sin (e+f x)+1},\frac{d-c}{\sin (e+f x) d+d}\right )+a (2 n-1) (\sin (e+f x)+1) F_1\left (-n-\frac{1}{2};-\frac{1}{2},-n;\frac{1}{2}-n;\frac{2}{\sin (e+f x)+1},\frac{d-c}{\sin (e+f x) d+d}\right )\right )}{4 n^2-1}\right )+a B (\sin (e+f x)+1) \left (a \sqrt{2-2 \sin (e+f x)} (\sin (e+f x)+1) \left (\frac{c+d \sin (e+f x)}{c-d}\right )^{-n} F_1\left (1;\frac{1}{2},-n;2;\frac{1}{2} (\sin (e+f x)+1),\frac{d (\sin (e+f x)+1)}{d-c}\right )-\frac{4 \sqrt{\frac{\sin (e+f x)-1}{\sin (e+f x)+1}} \left (\frac{c-d}{d \sin (e+f x)+d}+1\right )^{-n} \left (a (2 n-1) (\sin (e+f x)+1) F_1\left (-n-\frac{1}{2};-\frac{1}{2},-n;\frac{1}{2}-n;\frac{2}{\sin (e+f x)+1},\frac{d-c}{\sin (e+f x) d+d}\right )-2 a (2 n+1) F_1\left (\frac{1}{2}-n;-\frac{1}{2},-n;\frac{3}{2}-n;\frac{2}{\sin (e+f x)+1},\frac{d-c}{\sin (e+f x) d+d}\right )\right )}{4 n^2-1}\right )\right )}{8 a^3 f \sqrt{a (\sin (e+f x)+1)}} \]

Warning: Unable to verify antiderivative.

[In]

Integrate[((A + B*Sin[e + f*x])*(c + d*Sin[e + f*x])^n)/(a + a*Sin[e + f*x])^(3/2),x]

[Out]

(Sec[e + f*x]*(c + d*Sin[e + f*x])^n*(a*B*(1 + Sin[e + f*x])*((a*AppellF1[1, 1/2, -n, 2, (1 + Sin[e + f*x])/2,
(d*(1 + Sin[e + f*x]))/(-c + d)]*Sqrt[2 - 2*Sin[e + f*x]]*(1 + Sin[e + f*x]))/((c + d*Sin[e + f*x])/(c - d))^
n - (4*Sqrt[(-1 + Sin[e + f*x])/(1 + Sin[e + f*x])]*(-2*a*(1 + 2*n)*AppellF1[1/2 - n, -1/2, -n, 3/2 - n, 2/(1
+ Sin[e + f*x]), (-c + d)/(d + d*Sin[e + f*x])] + a*(-1 + 2*n)*AppellF1[-1/2 - n, -1/2, -n, 1/2 - n, 2/(1 + Si
n[e + f*x]), (-c + d)/(d + d*Sin[e + f*x])]*(1 + Sin[e + f*x])))/((-1 + 4*n^2)*(1 + (c - d)/(d + d*Sin[e + f*x
]))^n)) + a*A*(1 + Sin[e + f*x])*((a*AppellF1[1, 1/2, -n, 2, (1 + Sin[e + f*x])/2, (d*(1 + Sin[e + f*x]))/(-c
+ d)]*Sqrt[2 - 2*Sin[e + f*x]]*(1 + Sin[e + f*x]))/((c + d*Sin[e + f*x])/(c - d))^n - (4*Sqrt[(-1 + Sin[e + f*
x])/(1 + Sin[e + f*x])]*(2*a*(1 + 2*n)*AppellF1[1/2 - n, -1/2, -n, 3/2 - n, 2/(1 + Sin[e + f*x]), (-c + d)/(d
+ d*Sin[e + f*x])] + a*(-1 + 2*n)*AppellF1[-1/2 - n, -1/2, -n, 1/2 - n, 2/(1 + Sin[e + f*x]), (-c + d)/(d + d*
Sin[e + f*x])]*(1 + Sin[e + f*x])))/((-1 + 4*n^2)*(1 + (c - d)/(d + d*Sin[e + f*x]))^n))))/(8*a^3*f*Sqrt[a*(1
+ Sin[e + f*x])])

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Maple [F]  time = 0.345, size = 0, normalized size = 0. \begin{align*} \int{ \left ( A+B\sin \left ( fx+e \right ) \right ) \left ( c+d\sin \left ( fx+e \right ) \right ) ^{n} \left ( a+a\sin \left ( fx+e \right ) \right ) ^{-{\frac{3}{2}}}}\, dx \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

int((A+B*sin(f*x+e))*(c+d*sin(f*x+e))^n/(a+a*sin(f*x+e))^(3/2),x)

[Out]

int((A+B*sin(f*x+e))*(c+d*sin(f*x+e))^n/(a+a*sin(f*x+e))^(3/2),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 (d \sin \left (f x + e\right ) + c\right )}^{n}}{{\left (a \sin \left (f x + e\right ) + a\right )}^{\frac{3}{2}}}\,{d x} \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate((A+B*sin(f*x+e))*(c+d*sin(f*x+e))^n/(a+a*sin(f*x+e))^(3/2),x, algorithm="maxima")

[Out]

integrate((B*sin(f*x + e) + A)*(d*sin(f*x + e) + c)^n/(a*sin(f*x + e) + a)^(3/2), 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 )} \sqrt{a \sin \left (f x + e\right ) + a}{\left (d \sin \left (f x + e\right ) + c\right )}^{n}}{a^{2} \cos \left (f x + e\right )^{2} - 2 \, a^{2} \sin \left (f x + e\right ) - 2 \, a^{2}}, x\right ) \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate((A+B*sin(f*x+e))*(c+d*sin(f*x+e))^n/(a+a*sin(f*x+e))^(3/2),x, algorithm="fricas")

[Out]

integral(-(B*sin(f*x + e) + A)*sqrt(a*sin(f*x + e) + a)*(d*sin(f*x + e) + c)^n/(a^2*cos(f*x + e)^2 - 2*a^2*sin
(f*x + e) - 2*a^2), 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+B*sin(f*x+e))*(c+d*sin(f*x+e))**n/(a+a*sin(f*x+e))**(3/2),x)

[Out]

Timed out

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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 (d \sin \left (f x + e\right ) + c\right )}^{n}}{{\left (a \sin \left (f x + e\right ) + a\right )}^{\frac{3}{2}}}\,{d x} \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate((A+B*sin(f*x+e))*(c+d*sin(f*x+e))^n/(a+a*sin(f*x+e))^(3/2),x, algorithm="giac")

[Out]

integrate((B*sin(f*x + e) + A)*(d*sin(f*x + e) + c)^n/(a*sin(f*x + e) + a)^(3/2), x)