3.832 \(\int (c (d \sin (e+f x))^p)^n (a+b \sin (e+f x))^3 \, dx\)

Optimal. Leaf size=323 \[ \frac {b \left (3 a^2 (n p+3)+b^2 (n p+2)\right ) \sin ^2(e+f x) \cos (e+f x) \, _2F_1\left (\frac {1}{2},\frac {1}{2} (n p+2);\frac {1}{2} (n p+4);\sin ^2(e+f x)\right ) \left (c (d \sin (e+f x))^p\right )^n}{f (n p+2) (n p+3) \sqrt {\cos ^2(e+f x)}}+\frac {a \left (a^2 (n p+2)+3 b^2 (n p+1)\right ) \sin (e+f x) \cos (e+f x) \, _2F_1\left (\frac {1}{2},\frac {1}{2} (n p+1);\frac {1}{2} (n p+3);\sin ^2(e+f x)\right ) \left (c (d \sin (e+f x))^p\right )^n}{f (n p+1) (n p+2) \sqrt {\cos ^2(e+f x)}}-\frac {a b^2 (2 n p+7) \sin (e+f x) \cos (e+f x) \left (c (d \sin (e+f x))^p\right )^n}{f (n p+2) (n p+3)}-\frac {b^2 \sin (e+f x) \cos (e+f x) (a+b \sin (e+f x)) \left (c (d \sin (e+f x))^p\right )^n}{f (n p+3)} \]

[Out]

-a*b^2*(2*n*p+7)*cos(f*x+e)*sin(f*x+e)*(c*(d*sin(f*x+e))^p)^n/f/(n*p+2)/(n*p+3)-b^2*cos(f*x+e)*sin(f*x+e)*(c*(
d*sin(f*x+e))^p)^n*(a+b*sin(f*x+e))/f/(n*p+3)+a*(3*b^2*(n*p+1)+a^2*(n*p+2))*cos(f*x+e)*hypergeom([1/2, 1/2*n*p
+1/2],[1/2*n*p+3/2],sin(f*x+e)^2)*sin(f*x+e)*(c*(d*sin(f*x+e))^p)^n/f/(n*p+1)/(n*p+2)/(cos(f*x+e)^2)^(1/2)+b*(
b^2*(n*p+2)+3*a^2*(n*p+3))*cos(f*x+e)*hypergeom([1/2, 1/2*n*p+1],[1/2*n*p+2],sin(f*x+e)^2)*sin(f*x+e)^2*(c*(d*
sin(f*x+e))^p)^n/f/(n*p+2)/(n*p+3)/(cos(f*x+e)^2)^(1/2)

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Rubi [A]  time = 0.55, antiderivative size = 303, normalized size of antiderivative = 0.94, number of steps used = 6, number of rules used = 5, integrand size = 27, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.185, Rules used = {2826, 2793, 3023, 2748, 2643} \[ \frac {b \left (\frac {3 a^2}{n p+2}+\frac {b^2}{n p+3}\right ) \sin ^2(e+f x) \cos (e+f x) \, _2F_1\left (\frac {1}{2},\frac {1}{2} (n p+2);\frac {1}{2} (n p+4);\sin ^2(e+f x)\right ) \left (c (d \sin (e+f x))^p\right )^n}{f \sqrt {\cos ^2(e+f x)}}+\frac {a \left (\frac {a^2}{n p+1}+\frac {3 b^2}{n p+2}\right ) \sin (e+f x) \cos (e+f x) \, _2F_1\left (\frac {1}{2},\frac {1}{2} (n p+1);\frac {1}{2} (n p+3);\sin ^2(e+f x)\right ) \left (c (d \sin (e+f x))^p\right )^n}{f \sqrt {\cos ^2(e+f x)}}-\frac {a b^2 (2 n p+7) \sin (e+f x) \cos (e+f x) \left (c (d \sin (e+f x))^p\right )^n}{f (n p+2) (n p+3)}-\frac {b^2 \sin (e+f x) \cos (e+f x) (a+b \sin (e+f x)) \left (c (d \sin (e+f x))^p\right )^n}{f (n p+3)} \]

Antiderivative was successfully verified.

[In]

Int[(c*(d*Sin[e + f*x])^p)^n*(a + b*Sin[e + f*x])^3,x]

[Out]

-((a*b^2*(7 + 2*n*p)*Cos[e + f*x]*Sin[e + f*x]*(c*(d*Sin[e + f*x])^p)^n)/(f*(2 + n*p)*(3 + n*p))) + (a*(a^2/(1
+ n*p) + (3*b^2)/(2 + n*p))*Cos[e + f*x]*Hypergeometric2F1[1/2, (1 + n*p)/2, (3 + n*p)/2, Sin[e + f*x]^2]*Sin
[e + f*x]*(c*(d*Sin[e + f*x])^p)^n)/(f*Sqrt[Cos[e + f*x]^2]) + (b*((3*a^2)/(2 + n*p) + b^2/(3 + n*p))*Cos[e +
f*x]*Hypergeometric2F1[1/2, (2 + n*p)/2, (4 + n*p)/2, Sin[e + f*x]^2]*Sin[e + f*x]^2*(c*(d*Sin[e + f*x])^p)^n)
/(f*Sqrt[Cos[e + f*x]^2]) - (b^2*Cos[e + f*x]*Sin[e + f*x]*(c*(d*Sin[e + f*x])^p)^n*(a + b*Sin[e + f*x]))/(f*(
3 + n*p))

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 2793

Int[((a_.) + (b_.)*sin[(e_.) + (f_.)*(x_)])^(m_)*((c_.) + (d_.)*sin[(e_.) + (f_.)*(x_)])^(n_), x_Symbol] :> -S
imp[(b^2*Cos[e + f*x]*(a + b*Sin[e + f*x])^(m - 2)*(c + d*Sin[e + f*x])^(n + 1))/(d*f*(m + n)), x] + Dist[1/(d
*(m + n)), Int[(a + b*Sin[e + f*x])^(m - 3)*(c + d*Sin[e + f*x])^n*Simp[a^3*d*(m + n) + b^2*(b*c*(m - 2) + a*d
*(n + 1)) - b*(a*b*c - b^2*d*(m + n - 1) - 3*a^2*d*(m + n))*Sin[e + f*x] - b^2*(b*c*(m - 1) - a*d*(3*m + 2*n -
2))*Sin[e + f*x]^2, x], x], x] /; FreeQ[{a, b, c, d, e, f, n}, x] && NeQ[b*c - a*d, 0] && NeQ[a^2 - b^2, 0] &
& NeQ[c^2 - d^2, 0] && GtQ[m, 2] && (IntegerQ[m] || IntegersQ[2*m, 2*n]) &&  !(IGtQ[n, 2] && ( !IntegerQ[m] ||
(EqQ[a, 0] && NeQ[c, 0])))

Rule 2826

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

Rule 3023

Int[((a_.) + (b_.)*sin[(e_.) + (f_.)*(x_)])^(m_.)*((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 + 1))/(b*f*(m + 2)), x] + Dist[1/(b*
(m + 2)), Int[(a + b*Sin[e + f*x])^m*Simp[A*b*(m + 2) + b*C*(m + 1) + (b*B*(m + 2) - a*C)*Sin[e + f*x], x], x]
, x] /; FreeQ[{a, b, e, f, A, B, C, m}, x] &&  !LtQ[m, -1]

Rubi steps

\begin {align*} \int \left (c (d \sin (e+f x))^p\right )^n (a+b \sin (e+f x))^3 \, dx &=\left ((d \sin (e+f x))^{-n p} \left (c (d \sin (e+f x))^p\right )^n\right ) \int (d \sin (e+f x))^{n p} (a+b \sin (e+f x))^3 \, dx\\ &=-\frac {b^2 \cos (e+f x) \sin (e+f x) \left (c (d \sin (e+f x))^p\right )^n (a+b \sin (e+f x))}{f (3+n p)}+\frac {\left ((d \sin (e+f x))^{-n p} \left (c (d \sin (e+f x))^p\right )^n\right ) \int (d \sin (e+f x))^{n p} \left (a d \left (b^2 (1+n p)+a^2 (3+n p)\right )+b d \left (b^2 (2+n p)+3 a^2 (3+n p)\right ) \sin (e+f x)+a b^2 d (7+2 n p) \sin ^2(e+f x)\right ) \, dx}{d (3+n p)}\\ &=-\frac {a b^2 (7+2 n p) \cos (e+f x) \sin (e+f x) \left (c (d \sin (e+f x))^p\right )^n}{f (2+n p) (3+n p)}-\frac {b^2 \cos (e+f x) \sin (e+f x) \left (c (d \sin (e+f x))^p\right )^n (a+b \sin (e+f x))}{f (3+n p)}+\frac {\left ((d \sin (e+f x))^{-n p} \left (c (d \sin (e+f x))^p\right )^n\right ) \int (d \sin (e+f x))^{n p} \left (a d^2 (3+n p) \left (3 b^2 (1+n p)+a^2 (2+n p)\right )+b d^2 (2+n p) \left (b^2 (2+n p)+3 a^2 (3+n p)\right ) \sin (e+f x)\right ) \, dx}{d^2 (2+n p) (3+n p)}\\ &=-\frac {a b^2 (7+2 n p) \cos (e+f x) \sin (e+f x) \left (c (d \sin (e+f x))^p\right )^n}{f (2+n p) (3+n p)}-\frac {b^2 \cos (e+f x) \sin (e+f x) \left (c (d \sin (e+f x))^p\right )^n (a+b \sin (e+f x))}{f (3+n p)}+\frac {\left (a \left (3 b^2 (1+n p)+a^2 (2+n p)\right ) (d \sin (e+f x))^{-n p} \left (c (d \sin (e+f x))^p\right )^n\right ) \int (d \sin (e+f x))^{n p} \, dx}{2+n p}+\frac {\left (b \left (b^2 (2+n p)+3 a^2 (3+n p)\right ) (d \sin (e+f x))^{-n p} \left (c (d \sin (e+f x))^p\right )^n\right ) \int (d \sin (e+f x))^{1+n p} \, dx}{d (3+n p)}\\ &=-\frac {a b^2 (7+2 n p) \cos (e+f x) \sin (e+f x) \left (c (d \sin (e+f x))^p\right )^n}{f (2+n p) (3+n p)}+\frac {a \left (\frac {a^2}{1+n p}+\frac {3 b^2}{2+n p}\right ) \cos (e+f x) \, _2F_1\left (\frac {1}{2},\frac {1}{2} (1+n p);\frac {1}{2} (3+n p);\sin ^2(e+f x)\right ) \sin (e+f x) \left (c (d \sin (e+f x))^p\right )^n}{f \sqrt {\cos ^2(e+f x)}}+\frac {b \left (\frac {3 a^2}{2+n p}+\frac {b^2}{3+n p}\right ) \cos (e+f x) \, _2F_1\left (\frac {1}{2},\frac {1}{2} (2+n p);\frac {1}{2} (4+n p);\sin ^2(e+f x)\right ) \sin ^2(e+f x) \left (c (d \sin (e+f x))^p\right )^n}{f \sqrt {\cos ^2(e+f x)}}-\frac {b^2 \cos (e+f x) \sin (e+f x) \left (c (d \sin (e+f x))^p\right )^n (a+b \sin (e+f x))}{f (3+n p)}\\ \end {align*}

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Mathematica [A]  time = 1.04, size = 230, normalized size = 0.71 \[ \frac {\sin (e+f x) \cos (e+f x) \left (c (d \sin (e+f x))^p\right )^n \left (\frac {b \left (3 a^2 (n p+3)+b^2 (n p+2)\right ) \sin (e+f x) \, _2F_1\left (\frac {1}{2},\frac {n p}{2}+1;\frac {n p}{2}+2;\sin ^2(e+f x)\right )}{(n p+2) \sqrt {\cos ^2(e+f x)}}+\frac {a (n p+3) \left (a^2 (n p+2)+3 b^2 (n p+1)\right ) \, _2F_1\left (\frac {1}{2},\frac {1}{2} (n p+1);\frac {1}{2} (n p+3);\sin ^2(e+f x)\right )}{(n p+1) (n p+2) \sqrt {\cos ^2(e+f x)}}-b^2 (a+b \sin (e+f x))-\frac {a b^2 (2 n p+7)}{n p+2}\right )}{f (n p+3)} \]

Antiderivative was successfully verified.

[In]

Integrate[(c*(d*Sin[e + f*x])^p)^n*(a + b*Sin[e + f*x])^3,x]

[Out]

(Cos[e + f*x]*Sin[e + f*x]*(c*(d*Sin[e + f*x])^p)^n*(-((a*b^2*(7 + 2*n*p))/(2 + n*p)) + (a*(3 + n*p)*(3*b^2*(1
+ n*p) + a^2*(2 + n*p))*Hypergeometric2F1[1/2, (1 + n*p)/2, (3 + n*p)/2, Sin[e + f*x]^2])/((1 + n*p)*(2 + n*p
)*Sqrt[Cos[e + f*x]^2]) + (b*(b^2*(2 + n*p) + 3*a^2*(3 + n*p))*Hypergeometric2F1[1/2, 1 + (n*p)/2, 2 + (n*p)/2
, Sin[e + f*x]^2]*Sin[e + f*x])/((2 + n*p)*Sqrt[Cos[e + f*x]^2]) - b^2*(a + b*Sin[e + f*x])))/(f*(3 + n*p))

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fricas [F]  time = 0.46, size = 0, normalized size = 0.00 \[ {\rm integral}\left (-{\left (3 \, a b^{2} \cos \left (f x + e\right )^{2} - a^{3} - 3 \, a b^{2} + {\left (b^{3} \cos \left (f x + e\right )^{2} - 3 \, a^{2} b - b^{3}\right )} \sin \left (f x + e\right )\right )} \left (\left (d \sin \left (f x + e\right )\right )^{p} c\right )^{n}, x\right ) \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate((c*(d*sin(f*x+e))^p)^n*(a+b*sin(f*x+e))^3,x, algorithm="fricas")

[Out]

integral(-(3*a*b^2*cos(f*x + e)^2 - a^3 - 3*a*b^2 + (b^3*cos(f*x + e)^2 - 3*a^2*b - b^3)*sin(f*x + e))*((d*sin
(f*x + e))^p*c)^n, x)

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

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate((c*(d*sin(f*x+e))^p)^n*(a+b*sin(f*x+e))^3,x, algorithm="giac")

[Out]

integrate((b*sin(f*x + e) + a)^3*((d*sin(f*x + e))^p*c)^n, x)

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maple [F]  time = 1.27, size = 0, normalized size = 0.00 \[ \int \left (c \left (d \sin \left (f x +e \right )\right )^{p}\right )^{n} \left (a +b \sin \left (f x +e \right )\right )^{3}\, dx \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

int((c*(d*sin(f*x+e))^p)^n*(a+b*sin(f*x+e))^3,x)

[Out]

int((c*(d*sin(f*x+e))^p)^n*(a+b*sin(f*x+e))^3,x)

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

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate((c*(d*sin(f*x+e))^p)^n*(a+b*sin(f*x+e))^3,x, algorithm="maxima")

[Out]

integrate((b*sin(f*x + e) + a)^3*((d*sin(f*x + e))^p*c)^n, x)

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

Verification of antiderivative is not currently implemented for this CAS.

[In]

int((c*(d*sin(e + f*x))^p)^n*(a + b*sin(e + f*x))^3,x)

[Out]

int((c*(d*sin(e + f*x))^p)^n*(a + b*sin(e + f*x))^3, x)

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sympy [F]  time = 0.00, size = 0, normalized size = 0.00 \[ \int \left (c \left (d \sin {\left (e + f x \right )}\right )^{p}\right )^{n} \left (a + b \sin {\left (e + f x \right )}\right )^{3}\, dx \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate((c*(d*sin(f*x+e))**p)**n*(a+b*sin(f*x+e))**3,x)

[Out]

Integral((c*(d*sin(e + f*x))**p)**n*(a + b*sin(e + f*x))**3, x)

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