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

Optimal. Leaf size=279 \[ \frac {(n+1) (2 A (1-n)+2 B n+B) \cos (e+f x) (d \sin (e+f x))^{n+2} \, _2F_1\left (\frac {1}{2},\frac {n+2}{2};\frac {n+4}{2};\sin ^2(e+f x)\right )}{3 a^2 d^2 f (n+2) \sqrt {\cos ^2(e+f x)}}-\frac {n (-2 A n+A+2 B (n+1)) \cos (e+f x) (d \sin (e+f x))^{n+1} \, _2F_1\left (\frac {1}{2},\frac {n+1}{2};\frac {n+3}{2};\sin ^2(e+f x)\right )}{3 a^2 d f (n+1) \sqrt {\cos ^2(e+f x)}}+\frac {(2 A (1-n)+2 B n+B) \cos (e+f x) (d \sin (e+f x))^{n+1}}{3 a^2 d f (\sin (e+f x)+1)}+\frac {(A-B) \cos (e+f x) (d \sin (e+f x))^{n+1}}{3 d f (a \sin (e+f x)+a)^2} \]

[Out]

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

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Rubi [A]  time = 0.49, antiderivative size = 279, normalized size of antiderivative = 1.00, number of steps used = 5, number of rules used = 3, integrand size = 33, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.091, Rules used = {2978, 2748, 2643} \[ \frac {(n+1) (2 A (1-n)+2 B n+B) \cos (e+f x) (d \sin (e+f x))^{n+2} \, _2F_1\left (\frac {1}{2},\frac {n+2}{2};\frac {n+4}{2};\sin ^2(e+f x)\right )}{3 a^2 d^2 f (n+2) \sqrt {\cos ^2(e+f x)}}-\frac {n (-2 A n+A+2 B (n+1)) \cos (e+f x) (d \sin (e+f x))^{n+1} \, _2F_1\left (\frac {1}{2},\frac {n+1}{2};\frac {n+3}{2};\sin ^2(e+f x)\right )}{3 a^2 d f (n+1) \sqrt {\cos ^2(e+f x)}}+\frac {(2 A (1-n)+2 B n+B) \cos (e+f x) (d \sin (e+f x))^{n+1}}{3 a^2 d f (\sin (e+f x)+1)}+\frac {(A-B) \cos (e+f x) (d \sin (e+f x))^{n+1}}{3 d f (a \sin (e+f x)+a)^2} \]

Antiderivative was successfully verified.

[In]

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

[Out]

-(n*(A - 2*A*n + 2*B*(1 + n))*Cos[e + f*x]*Hypergeometric2F1[1/2, (1 + n)/2, (3 + n)/2, Sin[e + f*x]^2]*(d*Sin
[e + f*x])^(1 + n))/(3*a^2*d*f*(1 + n)*Sqrt[Cos[e + f*x]^2]) + ((1 + n)*(B + 2*A*(1 - n) + 2*B*n)*Cos[e + f*x]
*Hypergeometric2F1[1/2, (2 + n)/2, (4 + n)/2, Sin[e + f*x]^2]*(d*Sin[e + f*x])^(2 + n))/(3*a^2*d^2*f*(2 + n)*S
qrt[Cos[e + f*x]^2]) + ((B + 2*A*(1 - n) + 2*B*n)*Cos[e + f*x]*(d*Sin[e + f*x])^(1 + n))/(3*a^2*d*f*(1 + Sin[e
+ f*x])) + ((A - B)*Cos[e + f*x]*(d*Sin[e + f*x])^(1 + n))/(3*d*f*(a + a*Sin[e + f*x])^2)

Rule 2643

Int[((b_.)*sin[(c_.) + (d_.)*(x_)])^(n_), x_Symbol] :> Simp[(Cos[c + d*x]*(b*Sin[c + d*x])^(n + 1)*Hypergeomet
ric2F1[1/2, (n + 1)/2, (n + 3)/2, Sin[c + d*x]^2])/(b*d*(n + 1)*Sqrt[Cos[c + d*x]^2]), x] /; FreeQ[{b, c, d, n
}, x] &&  !IntegerQ[2*n]

Rule 2748

Int[((b_.)*sin[(e_.) + (f_.)*(x_)])^(m_)*((c_) + (d_.)*sin[(e_.) + (f_.)*(x_)]), x_Symbol] :> Dist[c, Int[(b*S
in[e + f*x])^m, x], x] + Dist[d/b, Int[(b*Sin[e + f*x])^(m + 1), x], x] /; FreeQ[{b, c, d, e, f, m}, x]

Rule 2978

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*(A*b - a*B)*Cos[e + f*x]*(a + b*Sin[e + f*x])^m*(c + d*Sin[e + f*
x])^(n + 1))/(a*f*(2*m + 1)*(b*c - a*d)), x] + Dist[1/(a*(2*m + 1)*(b*c - a*d)), Int[(a + b*Sin[e + f*x])^(m +
1)*(c + d*Sin[e + f*x])^n*Simp[B*(a*c*m + b*d*(n + 1)) + A*(b*c*(m + 1) - a*d*(2*m + n + 2)) + d*(A*b - a*B)*
(m + n + 2)*Sin[e + f*x], x], x], x] /; FreeQ[{a, b, c, d, e, f, A, B, n}, x] && NeQ[b*c - a*d, 0] && EqQ[a^2
- b^2, 0] && NeQ[c^2 - d^2, 0] && LtQ[m, -2^(-1)] &&  !GtQ[n, 0] && IntegerQ[2*m] && (IntegerQ[2*n] || EqQ[c,
0])

Rubi steps

\begin {align*} \int \frac {(d \sin (e+f x))^n (A+B \sin (e+f x))}{(a+a \sin (e+f x))^2} \, dx &=\frac {(A-B) \cos (e+f x) (d \sin (e+f x))^{1+n}}{3 d f (a+a \sin (e+f x))^2}+\frac {\int \frac {(d \sin (e+f x))^n (a d (2 A+B-A n+B n)+a (A-B) d n \sin (e+f x))}{a+a \sin (e+f x)} \, dx}{3 a^2 d}\\ &=\frac {(B+2 A (1-n)+2 B n) \cos (e+f x) (d \sin (e+f x))^{1+n}}{3 a^2 d f (1+\sin (e+f x))}+\frac {(A-B) \cos (e+f x) (d \sin (e+f x))^{1+n}}{3 d f (a+a \sin (e+f x))^2}+\frac {\int (d \sin (e+f x))^n \left (-a^2 d^2 n (A-2 A n+2 B (1+n))+a^2 d^2 (1+n) (2 A (1-n)+B (1+2 n)) \sin (e+f x)\right ) \, dx}{3 a^4 d^2}\\ &=\frac {(B+2 A (1-n)+2 B n) \cos (e+f x) (d \sin (e+f x))^{1+n}}{3 a^2 d f (1+\sin (e+f x))}+\frac {(A-B) \cos (e+f x) (d \sin (e+f x))^{1+n}}{3 d f (a+a \sin (e+f x))^2}+\frac {((1+n) (B+2 A (1-n)+2 B n)) \int (d \sin (e+f x))^{1+n} \, dx}{3 a^2 d}-\frac {(n (A-2 A n+2 B (1+n))) \int (d \sin (e+f x))^n \, dx}{3 a^2}\\ &=-\frac {n (A-2 A n+2 B (1+n)) \cos (e+f x) \, _2F_1\left (\frac {1}{2},\frac {1+n}{2};\frac {3+n}{2};\sin ^2(e+f x)\right ) (d \sin (e+f x))^{1+n}}{3 a^2 d f (1+n) \sqrt {\cos ^2(e+f x)}}+\frac {(1+n) (B+2 A (1-n)+2 B n) \cos (e+f x) \, _2F_1\left (\frac {1}{2},\frac {2+n}{2};\frac {4+n}{2};\sin ^2(e+f x)\right ) (d \sin (e+f x))^{2+n}}{3 a^2 d^2 f (2+n) \sqrt {\cos ^2(e+f x)}}+\frac {(B+2 A (1-n)+2 B n) \cos (e+f x) (d \sin (e+f x))^{1+n}}{3 a^2 d f (1+\sin (e+f x))}+\frac {(A-B) \cos (e+f x) (d \sin (e+f x))^{1+n}}{3 d f (a+a \sin (e+f x))^2}\\ \end {align*}

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Mathematica [A]  time = 1.33, size = 212, normalized size = 0.76 \[ \frac {\sin (e+f x) \cos (e+f x) (d \sin (e+f x))^n \left (-\frac {n (-2 A n+A+2 B (n+1)) \, _2F_1\left (\frac {1}{2},\frac {n+1}{2};\frac {n+3}{2};\sin ^2(e+f x)\right )}{(n+1) \sqrt {\cos ^2(e+f x)}}+\frac {(n+1) (-2 A (n-1)+2 B n+B) \sin (e+f x) \, _2F_1\left (\frac {1}{2},\frac {n+2}{2};\frac {n+4}{2};\sin ^2(e+f x)\right )}{(n+2) \sqrt {\cos ^2(e+f x)}}+\frac {n (B-A)}{\sin (e+f x)+1}+\frac {A (-n)+2 A+B n+B}{\sin (e+f x)+1}+\frac {A-B}{(\sin (e+f x)+1)^2}\right )}{3 a^2 f} \]

Antiderivative was successfully verified.

[In]

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

[Out]

(Cos[e + f*x]*Sin[e + f*x]*(d*Sin[e + f*x])^n*(-((n*(A - 2*A*n + 2*B*(1 + n))*Hypergeometric2F1[1/2, (1 + n)/2
, (3 + n)/2, Sin[e + f*x]^2])/((1 + n)*Sqrt[Cos[e + f*x]^2])) + ((1 + n)*(B - 2*A*(-1 + n) + 2*B*n)*Hypergeome
tric2F1[1/2, (2 + n)/2, (4 + n)/2, Sin[e + f*x]^2]*Sin[e + f*x])/((2 + n)*Sqrt[Cos[e + f*x]^2]) + (A - B)/(1 +
Sin[e + f*x])^2 + ((-A + B)*n)/(1 + Sin[e + f*x]) + (2*A + B - A*n + B*n)/(1 + Sin[e + f*x])))/(3*a^2*f)

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fricas [F]  time = 0.45, size = 0, normalized size = 0.00 \[ {\rm integral}\left (-\frac {{\left (B \sin \left (f x + e\right ) + A\right )} \left (d \sin \left (f x + e\right )\right )^{n}}{a^{2} \cos \left (f x + e\right )^{2} - 2 \, a^{2} \sin \left (f x + e\right ) - 2 \, a^{2}}, x\right ) \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

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

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

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

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

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maple [F]  time = 11.46, size = 0, normalized size = 0.00 \[ \int \frac {\left (d \sin \left (f x +e \right )\right )^{n} \left (A +B \sin \left (f x +e \right )\right )}{\left (a +a \sin \left (f x +e \right )\right )^{2}}\, dx \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

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

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

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

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

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

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

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

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sympy [F(-1)]  time = 0.00, size = 0, normalized size = 0.00 \[ \text {Timed out} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

Timed out

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