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

Optimal. Leaf size=215 \[ -\frac {2 \sqrt {2} (A-C) \cos (e+f x) (a+b \sin (e+f x))^m \left (\frac {a+b \sin (e+f x)}{a+b}\right )^{-m} F_1\left (\frac {1}{2};-\frac {1}{2},-m;\frac {3}{2};\frac {1}{2} (1-\sin (e+f x)),\frac {b (1-\sin (e+f x))}{a+b}\right )}{f \sqrt {\sin (e+f x)+1}}-\frac {4 \sqrt {2} C \cos (e+f x) (a+b \sin (e+f x))^m \left (\frac {a+b \sin (e+f x)}{a+b}\right )^{-m} F_1\left (\frac {1}{2};-\frac {3}{2},-m;\frac {3}{2};\frac {1}{2} (1-\sin (e+f x)),\frac {b (1-\sin (e+f x))}{a+b}\right )}{f \sqrt {\sin (e+f x)+1}} \]

[Out]

-4*C*AppellF1(1/2,-m,-3/2,3/2,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)
/f/(((a+b*sin(f*x+e))/(a+b))^m)/(1+sin(f*x+e))^(1/2)-2*(A-C)*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)/f/(((a+b*sin(f*x+e))/(a+b))^m)/(1+sin(f*x+e))^(1/2)

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Rubi [A]  time = 0.23, antiderivative size = 215, normalized size of antiderivative = 1.00, number of steps used = 7, number of rules used = 5, integrand size = 35, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.143, Rules used = {3017, 2755, 139, 138, 2784} \[ -\frac {2 \sqrt {2} (A-C) \cos (e+f x) (a+b \sin (e+f x))^m \left (\frac {a+b \sin (e+f x)}{a+b}\right )^{-m} F_1\left (\frac {1}{2};-\frac {1}{2},-m;\frac {3}{2};\frac {1}{2} (1-\sin (e+f x)),\frac {b (1-\sin (e+f x))}{a+b}\right )}{f \sqrt {\sin (e+f x)+1}}-\frac {4 \sqrt {2} C \cos (e+f x) (a+b \sin (e+f x))^m \left (\frac {a+b \sin (e+f x)}{a+b}\right )^{-m} F_1\left (\frac {1}{2};-\frac {3}{2},-m;\frac {3}{2};\frac {1}{2} (1-\sin (e+f x)),\frac {b (1-\sin (e+f x))}{a+b}\right )}{f \sqrt {\sin (e+f x)+1}} \]

Antiderivative was successfully verified.

[In]

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

[Out]

(-4*Sqrt[2]*C*AppellF1[1/2, -3/2, -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)/(f*Sqrt[1 + Sin[e + f*x]]*((a + b*Sin[e + f*x])/(a + b))^m) - (2*Sqrt[2]*(A - C)*Appel
lF1[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)/(f*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 2755

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

Rule 2784

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

Rule 3017

Int[((a_) + (b_.)*sin[(e_.) + (f_.)*(x_)])^(m_.)*((A_.) + (B_.)*sin[(e_.) + (f_.)*(x_)] + (C_.)*sin[(e_.) + (f
_.)*(x_)]^2), x_Symbol] :> Dist[A - C, Int[(a + b*Sin[e + f*x])^m*(1 + Sin[e + f*x]), x], x] + Dist[C, Int[(a
+ b*Sin[e + f*x])^m*(1 + Sin[e + f*x])^2, x], x] /; FreeQ[{a, b, e, f, A, B, C, m}, x] && EqQ[A - B + C, 0] &&
!IntegerQ[2*m]

Rubi steps

\begin {align*} \int (a+b \sin (e+f x))^m \left (A+(A+C) \sin (e+f x)+C \sin ^2(e+f x)\right ) \, dx &=(A-C) \int (1+\sin (e+f x)) (a+b \sin (e+f x))^m \, dx+C \int (1+\sin (e+f x))^2 (a+b \sin (e+f x))^m \, dx\\ &=\frac {((A-C) \cos (e+f x)) \operatorname {Subst}\left (\int \frac {\sqrt {1+x} (a+b x)^m}{\sqrt {1-x}} \, dx,x,\sin (e+f x)\right )}{f \sqrt {1-\sin (e+f x)} \sqrt {1+\sin (e+f x)}}+\frac {(C \cos (e+f x)) \operatorname {Subst}\left (\int \frac {(1+x)^{3/2} (a+b x)^m}{\sqrt {1-x}} \, dx,x,\sin (e+f x)\right )}{f \sqrt {1-\sin (e+f x)} \sqrt {1+\sin (e+f x)}}\\ &=\frac {\left ((A-C) \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 {\sqrt {1+x} \left (-\frac {a}{-a-b}-\frac {b x}{-a-b}\right )^m}{\sqrt {1-x}} \, dx,x,\sin (e+f x)\right )}{f \sqrt {1-\sin (e+f x)} \sqrt {1+\sin (e+f x)}}+\frac {\left (C \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 {(1+x)^{3/2} \left (-\frac {a}{-a-b}-\frac {b x}{-a-b}\right )^m}{\sqrt {1-x}} \, dx,x,\sin (e+f x)\right )}{f \sqrt {1-\sin (e+f x)} \sqrt {1+\sin (e+f x)}}\\ &=-\frac {4 \sqrt {2} C F_1\left (\frac {1}{2};-\frac {3}{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}}{f \sqrt {1+\sin (e+f x)}}-\frac {2 \sqrt {2} (A-C) 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}}{f \sqrt {1+\sin (e+f x)}}\\ \end {align*}

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Mathematica [F]  time = 12.51, size = 0, normalized size = 0.00 \[ \int (a+b \sin (e+f x))^m \left (A+(A+C) \sin (e+f x)+C \sin ^2(e+f x)\right ) \, dx \]

Verification is Not applicable to the result.

[In]

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

[Out]

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

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

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

integral(-(C*cos(f*x + e)^2 - (A + C)*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 \[ \int {\left (C \sin \left (f x + e\right )^{2} + {\left (A + C\right )} \sin \left (f x + e\right ) + A\right )} {\left (b \sin \left (f x + e\right ) + a\right )}^{m}\,{d x} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

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

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maple [F]  time = 3.59, size = 0, normalized size = 0.00 \[ \int \left (a +b \sin \left (f x +e \right )\right )^{m} \left (A +\left (A +C \right ) \sin \left (f x +e \right )+C \left (\sin ^{2}\left (f x +e \right )\right )\right )\, dx \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

int((a+b*sin(f*x+e))^m*(A+(A+C)*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 \[ \int {\left (C \sin \left (f x + e\right )^{2} + {\left (A + C\right )} \sin \left (f x + e\right ) + A\right )} {\left (b \sin \left (f x + e\right ) + a\right )}^{m}\,{d x} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

integrate((C*sin(f*x + e)^2 + (A + C)*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 \[ \int {\left (a+b\,\sin \left (e+f\,x\right )\right )}^m\,\left (C\,{\sin \left (e+f\,x\right )}^2+\left (A+C\right )\,\sin \left (e+f\,x\right )+A\right ) \,d x \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

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

[Out]

Timed out

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