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

Optimal. Leaf size=362 \[ \frac {(1-n) (n+1) (-4 A n+7 A+4 B n+3 B) \cos (e+f x) (d \sin (e+f x))^{n+2} \, _2F_1\left (\frac {1}{2},\frac {n+2}{2};\frac {n+4}{2};\sin ^2(e+f x)\right )}{15 a^3 d^2 f (n+2) \sqrt {\cos ^2(e+f x)}}-\frac {n \left (A \left (4 n^2-9 n+2\right )+B \left (-4 n^2-n+3\right )\right ) \cos (e+f x) (d \sin (e+f x))^{n+1} \, _2F_1\left (\frac {1}{2},\frac {n+1}{2};\frac {n+3}{2};\sin ^2(e+f x)\right )}{15 a^3 d f (n+1) \sqrt {\cos ^2(e+f x)}}+\frac {(1-n) (-4 A n+7 A+4 B n+3 B) \cos (e+f x) (d \sin (e+f x))^{n+1}}{15 d f \left (a^3 \sin (e+f x)+a^3\right )}+\frac {(A (5-2 n)+2 B n) \cos (e+f x) (d \sin (e+f x))^{n+1}}{15 a d f (a \sin (e+f x)+a)^2}+\frac {(A-B) \cos (e+f x) (d \sin (e+f x))^{n+1}}{5 d f (a \sin (e+f x)+a)^3} \]

[Out]

1/5*(A-B)*cos(f*x+e)*(d*sin(f*x+e))^(1+n)/d/f/(a+a*sin(f*x+e))^3+1/15*(A*(5-2*n)+2*B*n)*cos(f*x+e)*(d*sin(f*x+
e))^(1+n)/a/d/f/(a+a*sin(f*x+e))^2+1/15*(1-n)*(-4*A*n+4*B*n+7*A+3*B)*cos(f*x+e)*(d*sin(f*x+e))^(1+n)/d/f/(a^3+
a^3*sin(f*x+e))-1/15*n*(B*(-4*n^2-n+3)+A*(4*n^2-9*n+2))*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^3/d/f/(1+n)/(cos(f*x+e)^2)^(1/2)+1/15*(1-n)*(1+n)*(-4*A*n+4*B*n+7*A+3*B)*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^3/d^2/f/(2+n)/(cos(f*x+e)^2)^(1
/2)

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Rubi [A]  time = 0.85, antiderivative size = 362, normalized size of antiderivative = 1.00, number of steps used = 6, number of rules used = 3, integrand size = 33, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.091, Rules used = {2978, 2748, 2643} \[ \frac {(1-n) (n+1) (-4 A n+7 A+4 B n+3 B) \cos (e+f x) (d \sin (e+f x))^{n+2} \, _2F_1\left (\frac {1}{2},\frac {n+2}{2};\frac {n+4}{2};\sin ^2(e+f x)\right )}{15 a^3 d^2 f (n+2) \sqrt {\cos ^2(e+f x)}}-\frac {n \left (A \left (4 n^2-9 n+2\right )+B \left (-4 n^2-n+3\right )\right ) \cos (e+f x) (d \sin (e+f x))^{n+1} \, _2F_1\left (\frac {1}{2},\frac {n+1}{2};\frac {n+3}{2};\sin ^2(e+f x)\right )}{15 a^3 d f (n+1) \sqrt {\cos ^2(e+f x)}}+\frac {(1-n) (-4 A n+7 A+4 B n+3 B) \cos (e+f x) (d \sin (e+f x))^{n+1}}{15 d f \left (a^3 \sin (e+f x)+a^3\right )}+\frac {(A (5-2 n)+2 B n) \cos (e+f x) (d \sin (e+f x))^{n+1}}{15 a d f (a \sin (e+f x)+a)^2}+\frac {(A-B) \cos (e+f x) (d \sin (e+f x))^{n+1}}{5 d f (a \sin (e+f x)+a)^3} \]

Antiderivative was successfully verified.

[In]

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

[Out]

-(n*(B*(3 - n - 4*n^2) + A*(2 - 9*n + 4*n^2))*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))/(15*a^3*d*f*(1 + n)*Sqrt[Cos[e + f*x]^2]) + ((1 - n)*(1 + n)*(7*A + 3*B -
4*A*n + 4*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))/(15*a^3*d^2*f*(2 + n)*Sqrt[Cos[e + f*x]^2]) + ((A - B)*Cos[e + f*x]*(d*Sin[e + f*x])^(1 + n))/(5*d*f*(a +
a*Sin[e + f*x])^3) + ((A*(5 - 2*n) + 2*B*n)*Cos[e + f*x]*(d*Sin[e + f*x])^(1 + n))/(15*a*d*f*(a + a*Sin[e + f
*x])^2) + ((1 - n)*(7*A + 3*B - 4*A*n + 4*B*n)*Cos[e + f*x]*(d*Sin[e + f*x])^(1 + n))/(15*d*f*(a^3 + a^3*Sin[e
+ f*x]))

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))^3} \, dx &=\frac {(A-B) \cos (e+f x) (d \sin (e+f x))^{1+n}}{5 d f (a+a \sin (e+f x))^3}+\frac {\int \frac {(d \sin (e+f x))^n (a d (4 A+B-A n+B n)-a (A-B) d (1-n) \sin (e+f x))}{(a+a \sin (e+f x))^2} \, dx}{5 a^2 d}\\ &=\frac {(A-B) \cos (e+f x) (d \sin (e+f x))^{1+n}}{5 d f (a+a \sin (e+f x))^3}+\frac {(A (5-2 n)+2 B n) \cos (e+f x) (d \sin (e+f x))^{1+n}}{15 a d f (a+a \sin (e+f x))^2}+\frac {\int \frac {(d \sin (e+f x))^n \left (a^2 d^2 \left (B \left (3+n-2 n^2\right )+A \left (7-6 n+2 n^2\right )\right )+a^2 d^2 n (A (5-2 n)+2 B n) \sin (e+f x)\right )}{a+a \sin (e+f x)} \, dx}{15 a^4 d^2}\\ &=\frac {(A-B) \cos (e+f x) (d \sin (e+f x))^{1+n}}{5 d f (a+a \sin (e+f x))^3}+\frac {(A (5-2 n)+2 B n) \cos (e+f x) (d \sin (e+f x))^{1+n}}{15 a d f (a+a \sin (e+f x))^2}+\frac {(1-n) (7 A+3 B-4 A n+4 B n) \cos (e+f x) (d \sin (e+f x))^{1+n}}{15 d f \left (a^3+a^3 \sin (e+f x)\right )}+\frac {\int (d \sin (e+f x))^n \left (-a^3 d^3 n \left (B \left (3-n-4 n^2\right )+A \left (2-9 n+4 n^2\right )\right )+a^3 d^3 (1-n) (1+n) (7 A+3 B-4 A n+4 B n) \sin (e+f x)\right ) \, dx}{15 a^6 d^3}\\ &=\frac {(A-B) \cos (e+f x) (d \sin (e+f x))^{1+n}}{5 d f (a+a \sin (e+f x))^3}+\frac {(A (5-2 n)+2 B n) \cos (e+f x) (d \sin (e+f x))^{1+n}}{15 a d f (a+a \sin (e+f x))^2}+\frac {(1-n) (7 A+3 B-4 A n+4 B n) \cos (e+f x) (d \sin (e+f x))^{1+n}}{15 d f \left (a^3+a^3 \sin (e+f x)\right )}+\frac {((1-n) (1+n) (7 A+3 B-4 A n+4 B n)) \int (d \sin (e+f x))^{1+n} \, dx}{15 a^3 d}-\frac {\left (n \left (B \left (3-n-4 n^2\right )+A \left (2-9 n+4 n^2\right )\right )\right ) \int (d \sin (e+f x))^n \, dx}{15 a^3}\\ &=-\frac {n \left (B \left (3-n-4 n^2\right )+A \left (2-9 n+4 n^2\right )\right ) \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}}{15 a^3 d f (1+n) \sqrt {\cos ^2(e+f x)}}+\frac {(1-n) (1+n) (7 A+3 B-4 A n+4 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}}{15 a^3 d^2 f (2+n) \sqrt {\cos ^2(e+f x)}}+\frac {(A-B) \cos (e+f x) (d \sin (e+f x))^{1+n}}{5 d f (a+a \sin (e+f x))^3}+\frac {(A (5-2 n)+2 B n) \cos (e+f x) (d \sin (e+f x))^{1+n}}{15 a d f (a+a \sin (e+f x))^2}+\frac {(1-n) (7 A+3 B-4 A n+4 B n) \cos (e+f x) (d \sin (e+f x))^{1+n}}{15 d f \left (a^3+a^3 \sin (e+f x)\right )}\\ \end {align*}

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Mathematica [A]  time = 4.57, size = 260, normalized size = 0.72 \[ \frac {(d \sin (e+f x))^n \left (\frac {2 \sin (e+f x) \cos (e+f x) \left (n \left (A \left (-4 n^2+9 n-2\right )+B \left (4 n^2+n-3\right )\right ) \, _2F_1\left (\frac {1}{2},\frac {n+1}{2};\frac {n+3}{2};\sin ^2(e+f x)\right )+\frac {(n-1) (n+1)^2 (A (4 n-7)-B (4 n+3)) \sin (e+f x) \, _2F_1\left (\frac {1}{2},\frac {n+2}{2};\frac {n+4}{2};\sin ^2(e+f x)\right )}{n+2}\right )}{(n+1) \sqrt {\cos ^2(e+f x)}}+\frac {(n-1) (A (4 n-7)-B (4 n+3)) \sin (2 (e+f x))}{\sin (e+f x)+1}+\frac {(A (5-2 n)+2 B n) \sin (2 (e+f x))}{(\sin (e+f x)+1)^2}+\frac {3 (A-B) \sin (2 (e+f x))}{(\sin (e+f x)+1)^3}\right )}{30 a^3 f} \]

Antiderivative was successfully verified.

[In]

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

[Out]

((d*Sin[e + f*x])^n*((2*Cos[e + f*x]*Sin[e + f*x]*(n*(A*(-2 + 9*n - 4*n^2) + B*(-3 + n + 4*n^2))*Hypergeometri
c2F1[1/2, (1 + n)/2, (3 + n)/2, Sin[e + f*x]^2] + ((-1 + n)*(1 + n)^2*(A*(-7 + 4*n) - B*(3 + 4*n))*Hypergeomet
ric2F1[1/2, (2 + n)/2, (4 + n)/2, Sin[e + f*x]^2]*Sin[e + f*x])/(2 + n)))/((1 + n)*Sqrt[Cos[e + f*x]^2]) + (3*
(A - B)*Sin[2*(e + f*x)])/(1 + Sin[e + f*x])^3 + ((A*(5 - 2*n) + 2*B*n)*Sin[2*(e + f*x)])/(1 + Sin[e + f*x])^2
+ ((-1 + n)*(A*(-7 + 4*n) - B*(3 + 4*n))*Sin[2*(e + f*x)])/(1 + Sin[e + f*x])))/(30*a^3*f)

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fricas [F]  time = 0.45, size = 0, normalized size = 0.00 \[ {\rm integral}\left (-\frac {{\left (B \sin \left (f x + e\right ) + A\right )} \left (d \sin \left (f x + e\right )\right )^{n}}{3 \, a^{3} \cos \left (f x + e\right )^{2} - 4 \, a^{3} + {\left (a^{3} \cos \left (f x + e\right )^{2} - 4 \, a^{3}\right )} \sin \left (f x + e\right )}, x\right ) \]

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))^3,x, algorithm="fricas")

[Out]

integral(-(B*sin(f*x + e) + A)*(d*sin(f*x + e))^n/(3*a^3*cos(f*x + e)^2 - 4*a^3 + (a^3*cos(f*x + e)^2 - 4*a^3)
*sin(f*x + e)), x)

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giac [F]  time = 0.00, size = 0, normalized size = 0.00 \[ \int \frac {{\left (B \sin \left (f x + e\right ) + A\right )} \left (d \sin \left (f x + e\right )\right )^{n}}{{\left (a \sin \left (f x + e\right ) + a\right )}^{3}}\,{d x} \]

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))^3,x, algorithm="giac")

[Out]

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

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maple [F]  time = 11.62, size = 0, normalized size = 0.00 \[ \int \frac {\left (d \sin \left (f x +e \right )\right )^{n} \left (A +B \sin \left (f x +e \right )\right )}{\left (a +a \sin \left (f x +e \right )\right )^{3}}\, dx \]

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))^3,x)

[Out]

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

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maxima [F]  time = 0.00, size = 0, normalized size = 0.00 \[ \int \frac {{\left (B \sin \left (f x + e\right ) + A\right )} \left (d \sin \left (f x + e\right )\right )^{n}}{{\left (a \sin \left (f x + e\right ) + a\right )}^{3}}\,{d x} \]

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))^3,x, algorithm="maxima")

[Out]

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

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mupad [F]  time = 0.00, size = -1, normalized size = -0.00 \[ \int \frac {{\left (d\,\sin \left (e+f\,x\right )\right )}^n\,\left (A+B\,\sin \left (e+f\,x\right )\right )}{{\left (a+a\,\sin \left (e+f\,x\right )\right )}^3} \,d x \]

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))^3,x)

[Out]

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

[Out]

Timed out

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