3.213 \(\int (d \sin (e+f x))^m (a+b \sin (e+f x))^3 \, dx\)

Optimal. Leaf size=270 \[ \frac {b \left (3 a^2 (m+3)+b^2 (m+2)\right ) \cos (e+f x) (d \sin (e+f x))^{m+2} \, _2F_1\left (\frac {1}{2},\frac {m+2}{2};\frac {m+4}{2};\sin ^2(e+f x)\right )}{d^2 f (m+2) (m+3) \sqrt {\cos ^2(e+f x)}}+\frac {a \left (a^2 (m+2)+3 b^2 (m+1)\right ) \cos (e+f x) (d \sin (e+f x))^{m+1} \, _2F_1\left (\frac {1}{2},\frac {m+1}{2};\frac {m+3}{2};\sin ^2(e+f x)\right )}{d f (m+1) (m+2) \sqrt {\cos ^2(e+f x)}}-\frac {a b^2 (2 m+7) \cos (e+f x) (d \sin (e+f x))^{m+1}}{d f (m+2) (m+3)}-\frac {b^2 \cos (e+f x) (a+b \sin (e+f x)) (d \sin (e+f x))^{m+1}}{d f (m+3)} \]

[Out]

-a*b^2*(7+2*m)*cos(f*x+e)*(d*sin(f*x+e))^(1+m)/d/f/(2+m)/(3+m)-b^2*cos(f*x+e)*(d*sin(f*x+e))^(1+m)*(a+b*sin(f*
x+e))/d/f/(3+m)+a*(3*b^2*(1+m)+a^2*(2+m))*cos(f*x+e)*hypergeom([1/2, 1/2+1/2*m],[3/2+1/2*m],sin(f*x+e)^2)*(d*s
in(f*x+e))^(1+m)/d/f/(1+m)/(2+m)/(cos(f*x+e)^2)^(1/2)+b*(b^2*(2+m)+3*a^2*(3+m))*cos(f*x+e)*hypergeom([1/2, 1+1
/2*m],[2+1/2*m],sin(f*x+e)^2)*(d*sin(f*x+e))^(2+m)/d^2/f/(2+m)/(3+m)/(cos(f*x+e)^2)^(1/2)

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Rubi [A]  time = 0.38, antiderivative size = 270, normalized size of antiderivative = 1.00, number of steps used = 5, number of rules used = 4, integrand size = 23, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.174, Rules used = {2793, 3023, 2748, 2643} \[ \frac {b \left (3 a^2 (m+3)+b^2 (m+2)\right ) \cos (e+f x) (d \sin (e+f x))^{m+2} \, _2F_1\left (\frac {1}{2},\frac {m+2}{2};\frac {m+4}{2};\sin ^2(e+f x)\right )}{d^2 f (m+2) (m+3) \sqrt {\cos ^2(e+f x)}}+\frac {a \left (a^2 (m+2)+3 b^2 (m+1)\right ) \cos (e+f x) (d \sin (e+f x))^{m+1} \, _2F_1\left (\frac {1}{2},\frac {m+1}{2};\frac {m+3}{2};\sin ^2(e+f x)\right )}{d f (m+1) (m+2) \sqrt {\cos ^2(e+f x)}}-\frac {a b^2 (2 m+7) \cos (e+f x) (d \sin (e+f x))^{m+1}}{d f (m+2) (m+3)}-\frac {b^2 \cos (e+f x) (a+b \sin (e+f x)) (d \sin (e+f x))^{m+1}}{d f (m+3)} \]

Antiderivative was successfully verified.

[In]

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

[Out]

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

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

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Mathematica [A]  time = 0.80, size = 199, normalized size = 0.74 \[ \frac {\sin (e+f x) \cos (e+f x) (d \sin (e+f x))^m \left (\frac {b \left (3 a^2 (m+3)+b^2 (m+2)\right ) \sin (e+f x) \, _2F_1\left (\frac {1}{2},\frac {m+2}{2};\frac {m+4}{2};\sin ^2(e+f x)\right )}{(m+2) \sqrt {\cos ^2(e+f x)}}+\frac {a (m+3) \left (a^2 (m+2)+3 b^2 (m+1)\right ) \, _2F_1\left (\frac {1}{2},\frac {m+1}{2};\frac {m+3}{2};\sin ^2(e+f x)\right )}{(m+1) (m+2) \sqrt {\cos ^2(e+f x)}}-b^2 (a+b \sin (e+f x))-\frac {a b^2 (2 m+7)}{m+2}\right )}{f (m+3)} \]

Antiderivative was successfully verified.

[In]

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

[Out]

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

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fricas [F]  time = 2.59, size = 0, normalized size = 0.00 \[ {\rm integral}\left (-{\left (3 \, a b^{2} \cos \left (f x + e\right )^{2} - a^{3} - 3 \, a b^{2} + {\left (b^{3} \cos \left (f x + e\right )^{2} - 3 \, a^{2} b - b^{3}\right )} \sin \left (f x + e\right )\right )} \left (d \sin \left (f x + e\right )\right )^{m}, x\right ) \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate((d*sin(f*x+e))^m*(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))^m, x)

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

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

integrate((b*sin(f*x + e) + a)^3*(d*sin(f*x + e))^m, x)

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maple [F]  time = 5.11, size = 0, normalized size = 0.00 \[ \int \left (d \sin \left (f x +e \right )\right )^{m} \left (a +b \sin \left (f x +e \right )\right )^{3}\, dx \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

int((d*sin(f*x+e))^m*(a+b*sin(f*x+e))^3,x)

[Out]

int((d*sin(f*x+e))^m*(a+b*sin(f*x+e))^3,x)

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

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

integrate((b*sin(f*x + e) + a)^3*(d*sin(f*x + e))^m, x)

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

Verification of antiderivative is not currently implemented for this CAS.

[In]

int((d*sin(e + f*x))^m*(a + b*sin(e + f*x))^3,x)

[Out]

int((d*sin(e + f*x))^m*(a + b*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))**m*(a+b*sin(f*x+e))**3,x)

[Out]

Timed out

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