3.19 \(\int (a+b \sin (e+f x))^m (A+B \sin (e+f x)+C \sin ^2(e+f x)) \, dx\)

Optimal. Leaf size=304 \[ -\frac {\sqrt {2} \cos (e+f x) \left (a^2 C-a b B (m+2)+A b^2 (m+2)+b^2 C (m+1)\right ) (a+b \sin (e+f x))^m \left (\frac {a+b \sin (e+f x)}{a+b}\right )^{-m} F_1\left (\frac {1}{2};\frac {1}{2},-m;\frac {3}{2};\frac {1}{2} (1-\sin (e+f x)),\frac {b (1-\sin (e+f x))}{a+b}\right )}{b^2 f (m+2) \sqrt {\sin (e+f x)+1}}+\frac {\sqrt {2} (a+b) \cos (e+f x) (a C-b B (m+2)) (a+b \sin (e+f x))^m \left (\frac {a+b \sin (e+f x)}{a+b}\right )^{-m} F_1\left (\frac {1}{2};\frac {1}{2},-m-1;\frac {3}{2};\frac {1}{2} (1-\sin (e+f x)),\frac {b (1-\sin (e+f x))}{a+b}\right )}{b^2 f (m+2) \sqrt {\sin (e+f x)+1}}-\frac {C \cos (e+f x) (a+b \sin (e+f x))^{m+1}}{b f (m+2)} \]

[Out]

-C*cos(f*x+e)*(a+b*sin(f*x+e))^(1+m)/b/f/(2+m)+(a+b)*(a*C-b*B*(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^2*C+b^2*C*(1+m)+A*b^2*(2+m)-a*b*B*(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 = 0.36, antiderivative size = 304, normalized size of antiderivative = 1.00, number of steps used = 8, number of rules used = 5, integrand size = 33, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.152, Rules used = {3023, 2756, 2665, 139, 138} \[ -\frac {\sqrt {2} \cos (e+f x) \left (a^2 C-a b B (m+2)+A b^2 (m+2)+b^2 C (m+1)\right ) (a+b \sin (e+f x))^m \left (\frac {a+b \sin (e+f x)}{a+b}\right )^{-m} F_1\left (\frac {1}{2};\frac {1}{2},-m;\frac {3}{2};\frac {1}{2} (1-\sin (e+f x)),\frac {b (1-\sin (e+f x))}{a+b}\right )}{b^2 f (m+2) \sqrt {\sin (e+f x)+1}}+\frac {\sqrt {2} (a+b) \cos (e+f x) (a C-b B (m+2)) (a+b \sin (e+f x))^m \left (\frac {a+b \sin (e+f x)}{a+b}\right )^{-m} F_1\left (\frac {1}{2};\frac {1}{2},-m-1;\frac {3}{2};\frac {1}{2} (1-\sin (e+f x)),\frac {b (1-\sin (e+f x))}{a+b}\right )}{b^2 f (m+2) \sqrt {\sin (e+f x)+1}}-\frac {C \cos (e+f x) (a+b \sin (e+f x))^{m+1}}{b f (m+2)} \]

Antiderivative was successfully verified.

[In]

Int[(a + b*Sin[e + f*x])^m*(A + B*Sin[e + f*x] + C*Sin[e + f*x]^2),x]

[Out]

-((C*Cos[e + f*x]*(a + b*Sin[e + f*x])^(1 + m))/(b*f*(2 + m))) + (Sqrt[2]*(a + b)*(a*C - b*B*(2 + m))*AppellF1
[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^2*C + b^2*C*(1 + m)
+ A*b^2*(2 + m) - a*b*B*(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 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 (a+b \sin (e+f x))^m \left (A+B \sin (e+f x)+C \sin ^2(e+f x)\right ) \, dx &=-\frac {C \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 (b (C (1+m)+A (2+m))-(a C-b B (2+m)) \sin (e+f x)) \, dx}{b (2+m)}\\ &=-\frac {C \cos (e+f x) (a+b \sin (e+f x))^{1+m}}{b f (2+m)}+\frac {(-a C+b B (2+m)) \int (a+b \sin (e+f x))^{1+m} \, dx}{b^2 (2+m)}+\frac {\left (b^2 (C (1+m)+A (2+m))-a (-a C+b B (2+m))\right ) \int (a+b \sin (e+f x))^m \, dx}{b^2 (2+m)}\\ &=-\frac {C \cos (e+f x) (a+b \sin (e+f x))^{1+m}}{b f (2+m)}+\frac {((-a C+b B (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 (b^2 (C (1+m)+A (2+m))-a (-a C+b B (2+m))\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 {C \cos (e+f x) (a+b \sin (e+f x))^{1+m}}{b f (2+m)}-\frac {\left ((-a-b) (-a C+b B (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 (b^2 (C (1+m)+A (2+m))-a (-a C+b B (2+m))\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 {C \cos (e+f x) (a+b \sin (e+f x))^{1+m}}{b f (2+m)}+\frac {\sqrt {2} (a+b) (a C-b B (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^2 C+b^2 C (1+m)+A b^2 (2+m)-a b B (2+m)\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 = 13.08, size = 0, normalized size = 0.00 \[ \int (a+b \sin (e+f x))^m \left (A+B \sin (e+f x)+C \sin ^2(e+f x)\right ) \, dx \]

Verification is Not applicable to the result.

[In]

Integrate[(a + b*Sin[e + f*x])^m*(A + B*Sin[e + f*x] + C*Sin[e + f*x]^2),x]

[Out]

Integrate[(a + b*Sin[e + f*x])^m*(A + B*Sin[e + f*x] + C*Sin[e + f*x]^2), x]

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fricas [F]  time = 0.47, size = 0, normalized size = 0.00 \[ {\rm integral}\left (-{\left (C \cos \left (f x + e\right )^{2} - B \sin \left (f x + e\right ) - A - C\right )} {\left (b \sin \left (f x + e\right ) + a\right )}^{m}, x\right ) \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate((a+b*sin(f*x+e))^m*(A+B*sin(f*x+e)+C*sin(f*x+e)^2),x, algorithm="fricas")

[Out]

integral(-(C*cos(f*x + e)^2 - B*sin(f*x + e) - A - C)*(b*sin(f*x + e) + a)^m, x)

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

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate((a+b*sin(f*x+e))^m*(A+B*sin(f*x+e)+C*sin(f*x+e)^2),x, algorithm="giac")

[Out]

integrate((C*sin(f*x + e)^2 + B*sin(f*x + e) + A)*(b*sin(f*x + e) + a)^m, x)

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maple [F]  time = 2.94, size = 0, normalized size = 0.00 \[ \int \left (a +b \sin \left (f x +e \right )\right )^{m} \left (A +B \sin \left (f x +e \right )+C \left (\sin ^{2}\left (f x +e \right )\right )\right )\, dx \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

int((a+b*sin(f*x+e))^m*(A+B*sin(f*x+e)+C*sin(f*x+e)^2),x)

[Out]

int((a+b*sin(f*x+e))^m*(A+B*sin(f*x+e)+C*sin(f*x+e)^2),x)

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

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate((a+b*sin(f*x+e))^m*(A+B*sin(f*x+e)+C*sin(f*x+e)^2),x, algorithm="maxima")

[Out]

integrate((C*sin(f*x + e)^2 + B*sin(f*x + e) + A)*(b*sin(f*x + e) + a)^m, x)

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

Verification of antiderivative is not currently implemented for this CAS.

[In]

int((a + b*sin(e + f*x))^m*(A + B*sin(e + f*x) + C*sin(e + f*x)^2),x)

[Out]

int((a + b*sin(e + f*x))^m*(A + B*sin(e + f*x) + C*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*(A+B*sin(f*x+e)+C*sin(f*x+e)**2),x)

[Out]

Timed out

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