3.802 \(\int (a+b \sin (e+f x))^m (c+d \sin (e+f x))^2 \, dx\)

Optimal. Leaf size=311 \[ -\frac {\sqrt {2} \cos (e+f x) \left (a d (a d-2 b c (m+2))+b^2 \left (c^2 (m+2)+d^2 (m+1)\right )\right ) (a+b \sin (e+f x))^m \left (\frac {a+b \sin (e+f x)}{a+b}\right )^{-m} F_1\left (\frac {1}{2};\frac {1}{2},-m;\frac {3}{2};\frac {1}{2} (1-\sin (e+f x)),\frac {b (1-\sin (e+f x))}{a+b}\right )}{b^2 f (m+2) \sqrt {\sin (e+f x)+1}}+\frac {\sqrt {2} d (a+b) \cos (e+f x) (a d-2 b c (m+2)) (a+b \sin (e+f x))^m \left (\frac {a+b \sin (e+f x)}{a+b}\right )^{-m} F_1\left (\frac {1}{2};\frac {1}{2},-m-1;\frac {3}{2};\frac {1}{2} (1-\sin (e+f x)),\frac {b (1-\sin (e+f x))}{a+b}\right )}{b^2 f (m+2) \sqrt {\sin (e+f x)+1}}-\frac {d^2 \cos (e+f x) (a+b \sin (e+f x))^{m+1}}{b f (m+2)} \]

[Out]

-d^2*cos(f*x+e)*(a+b*sin(f*x+e))^(1+m)/b/f/(2+m)+(a+b)*d*(a*d-2*b*c*(2+m))*AppellF1(1/2,-1-m,1/2,3/2,b*(1-sin(
f*x+e))/(a+b),1/2-1/2*sin(f*x+e))*cos(f*x+e)*(a+b*sin(f*x+e))^m*2^(1/2)/b^2/f/(2+m)/(((a+b*sin(f*x+e))/(a+b))^
m)/(1+sin(f*x+e))^(1/2)-(a*d*(a*d-2*b*c*(2+m))+b^2*(d^2*(1+m)+c^2*(2+m)))*AppellF1(1/2,-m,1/2,3/2,b*(1-sin(f*x
+e))/(a+b),1/2-1/2*sin(f*x+e))*cos(f*x+e)*(a+b*sin(f*x+e))^m*2^(1/2)/b^2/f/(2+m)/(((a+b*sin(f*x+e))/(a+b))^m)/
(1+sin(f*x+e))^(1/2)

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Rubi [A]  time = 0.44, antiderivative size = 311, normalized size of antiderivative = 1.00, number of steps used = 8, number of rules used = 5, integrand size = 25, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.200, Rules used = {2791, 2756, 2665, 139, 138} \[ -\frac {\sqrt {2} \cos (e+f x) \left (a d (a d-2 b c (m+2))+b^2 \left (c^2 (m+2)+d^2 (m+1)\right )\right ) (a+b \sin (e+f x))^m \left (\frac {a+b \sin (e+f x)}{a+b}\right )^{-m} F_1\left (\frac {1}{2};\frac {1}{2},-m;\frac {3}{2};\frac {1}{2} (1-\sin (e+f x)),\frac {b (1-\sin (e+f x))}{a+b}\right )}{b^2 f (m+2) \sqrt {\sin (e+f x)+1}}+\frac {\sqrt {2} d (a+b) \cos (e+f x) (a d-2 b c (m+2)) (a+b \sin (e+f x))^m \left (\frac {a+b \sin (e+f x)}{a+b}\right )^{-m} F_1\left (\frac {1}{2};\frac {1}{2},-m-1;\frac {3}{2};\frac {1}{2} (1-\sin (e+f x)),\frac {b (1-\sin (e+f x))}{a+b}\right )}{b^2 f (m+2) \sqrt {\sin (e+f x)+1}}-\frac {d^2 \cos (e+f x) (a+b \sin (e+f x))^{m+1}}{b f (m+2)} \]

Antiderivative was successfully verified.

[In]

Int[(a + b*Sin[e + f*x])^m*(c + d*Sin[e + f*x])^2,x]

[Out]

-((d^2*Cos[e + f*x]*(a + b*Sin[e + f*x])^(1 + m))/(b*f*(2 + m))) + (Sqrt[2]*(a + b)*d*(a*d - 2*b*c*(2 + m))*Ap
pellF1[1/2, 1/2, -1 - m, 3/2, (1 - Sin[e + f*x])/2, (b*(1 - Sin[e + f*x]))/(a + b)]*Cos[e + f*x]*(a + b*Sin[e
+ f*x])^m)/(b^2*f*(2 + m)*Sqrt[1 + Sin[e + f*x]]*((a + b*Sin[e + f*x])/(a + b))^m) - (Sqrt[2]*(a*d*(a*d - 2*b*
c*(2 + m)) + b^2*(d^2*(1 + m) + c^2*(2 + m)))*AppellF1[1/2, 1/2, -m, 3/2, (1 - Sin[e + f*x])/2, (b*(1 - Sin[e
+ f*x]))/(a + b)]*Cos[e + f*x]*(a + b*Sin[e + f*x])^m)/(b^2*f*(2 + m)*Sqrt[1 + Sin[e + f*x]]*((a + b*Sin[e + f
*x])/(a + b))^m)

Rule 138

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

Rule 139

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

Rule 2665

Int[((a_) + (b_.)*sin[(c_.) + (d_.)*(x_)])^(n_), x_Symbol] :> Dist[Cos[c + d*x]/(d*Sqrt[1 + Sin[c + d*x]]*Sqrt
[1 - Sin[c + d*x]]), Subst[Int[(a + b*x)^n/(Sqrt[1 + x]*Sqrt[1 - x]), x], x, Sin[c + d*x]], x] /; FreeQ[{a, b,
c, d, n}, x] && NeQ[a^2 - b^2, 0] &&  !IntegerQ[2*n]

Rule 2756

Int[((a_) + (b_.)*sin[(e_.) + (f_.)*(x_)])^(m_)*((c_.) + (d_.)*sin[(e_.) + (f_.)*(x_)]), x_Symbol] :> Dist[(b*
c - a*d)/b, Int[(a + b*Sin[e + f*x])^m, x], x] + Dist[d/b, Int[(a + b*Sin[e + f*x])^(m + 1), x], x] /; FreeQ[{
a, b, c, d, e, f, m}, x] && NeQ[b*c - a*d, 0] && NeQ[a^2 - b^2, 0]

Rule 2791

Int[((a_) + (b_.)*sin[(e_.) + (f_.)*(x_)])^(m_)*((c_.) + (d_.)*sin[(e_.) + (f_.)*(x_)])^2, x_Symbol] :> -Simp[
(d^2*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[b*(d^2*(m + 1) + c^2*(m + 2)) - d*(a*d - 2*b*c*(m + 2))*Sin[e + f*x], x], x], x] /; FreeQ[{a, b, c,
d, e, f, m}, x] && NeQ[b*c - a*d, 0] && NeQ[a^2 - b^2, 0] &&  !LtQ[m, -1]

Rubi steps

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

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Mathematica [F]  time = 18.73, size = 0, normalized size = 0.00 \[ \int (a+b \sin (e+f x))^m (c+d \sin (e+f x))^2 \, dx \]

Verification is Not applicable to the result.

[In]

Integrate[(a + b*Sin[e + f*x])^m*(c + d*Sin[e + f*x])^2,x]

[Out]

Integrate[(a + b*Sin[e + f*x])^m*(c + d*Sin[e + f*x])^2, x]

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fricas [F]  time = 0.48, size = 0, normalized size = 0.00 \[ {\rm integral}\left (-{\left (d^{2} \cos \left (f x + e\right )^{2} - 2 \, c d \sin \left (f x + e\right ) - c^{2} - d^{2}\right )} {\left (b \sin \left (f x + e\right ) + a\right )}^{m}, x\right ) \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate((a+b*sin(f*x+e))^m*(c+d*sin(f*x+e))^2,x, algorithm="fricas")

[Out]

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

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

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate((a+b*sin(f*x+e))^m*(c+d*sin(f*x+e))^2,x, algorithm="giac")

[Out]

integrate((d*sin(f*x + e) + c)^2*(b*sin(f*x + e) + a)^m, x)

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maple [F]  time = 1.31, size = 0, normalized size = 0.00 \[ \int \left (a +b \sin \left (f x +e \right )\right )^{m} \left (c +d \sin \left (f x +e \right )\right )^{2}\, dx \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

int((a+b*sin(f*x+e))^m*(c+d*sin(f*x+e))^2,x)

[Out]

int((a+b*sin(f*x+e))^m*(c+d*sin(f*x+e))^2,x)

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

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate((a+b*sin(f*x+e))^m*(c+d*sin(f*x+e))^2,x, algorithm="maxima")

[Out]

integrate((d*sin(f*x + e) + c)^2*(b*sin(f*x + e) + a)^m, x)

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

Verification of antiderivative is not currently implemented for this CAS.

[In]

int((a + b*sin(e + f*x))^m*(c + d*sin(e + f*x))^2,x)

[Out]

int((a + b*sin(e + f*x))^m*(c + d*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((a+b*sin(f*x+e))**m*(c+d*sin(f*x+e))**2,x)

[Out]

Timed out

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