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

Optimal. Leaf size=229 \[ -\frac{\sqrt{2} (b c-a d) \cos (e+f x) (a+b \sin (e+f x))^m \left (\frac{a+b \sin (e+f x)}{a+b}\right )^{-m} F_1\left (\frac{1}{2};\frac{1}{2},-m;\frac{3}{2};\frac{1}{2} (1-\sin (e+f x)),\frac{b (1-\sin (e+f x))}{a+b}\right )}{b f \sqrt{\sin (e+f x)+1}}-\frac{\sqrt{2} d (a+b) \cos (e+f x) (a+b \sin (e+f x))^m \left (\frac{a+b \sin (e+f x)}{a+b}\right )^{-m} F_1\left (\frac{1}{2};\frac{1}{2},-m-1;\frac{3}{2};\frac{1}{2} (1-\sin (e+f x)),\frac{b (1-\sin (e+f x))}{a+b}\right )}{b f \sqrt{\sin (e+f x)+1}} \]

[Out]

-((Sqrt[2]*(a + b)*d*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*f*Sqrt[1 + Sin[e + f*x]]*((a + b*Sin[e + f*x])/(a + b))^m)) - (Sqrt[2]*(b
*c - a*d)*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*f*Sqrt[1 + Sin[e + f*x]]*((a + b*Sin[e + f*x])/(a + b))^m)

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Rubi [A]  time = 0.203456, antiderivative size = 229, normalized size of antiderivative = 1., number of steps used = 7, number of rules used = 4, integrand size = 23, \(\frac{\text{number of rules}}{\text{integrand size}}\) = 0.174, Rules used = {2756, 2665, 139, 138} \[ -\frac{\sqrt{2} (b c-a d) \cos (e+f x) (a+b \sin (e+f x))^m \left (\frac{a+b \sin (e+f x)}{a+b}\right )^{-m} F_1\left (\frac{1}{2};\frac{1}{2},-m;\frac{3}{2};\frac{1}{2} (1-\sin (e+f x)),\frac{b (1-\sin (e+f x))}{a+b}\right )}{b f \sqrt{\sin (e+f x)+1}}-\frac{\sqrt{2} d (a+b) \cos (e+f x) (a+b \sin (e+f x))^m \left (\frac{a+b \sin (e+f x)}{a+b}\right )^{-m} F_1\left (\frac{1}{2};\frac{1}{2},-m-1;\frac{3}{2};\frac{1}{2} (1-\sin (e+f x)),\frac{b (1-\sin (e+f x))}{a+b}\right )}{b f \sqrt{\sin (e+f x)+1}} \]

Antiderivative was successfully verified.

[In]

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

[Out]

-((Sqrt[2]*(a + b)*d*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*f*Sqrt[1 + Sin[e + f*x]]*((a + b*Sin[e + f*x])/(a + b))^m)) - (Sqrt[2]*(b
*c - a*d)*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*f*Sqrt[1 + Sin[e + f*x]]*((a + b*Sin[e + f*x])/(a + b))^m)

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 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 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 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])

Rubi steps

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

Mathematica [A]  time = 0.57007, size = 200, normalized size = 0.87 \[ \frac{\sec (e+f x) \sqrt{-\frac{b (\sin (e+f x)-1)}{a+b}} \sqrt{\frac{b (\sin (e+f x)+1)}{b-a}} (a+b \sin (e+f x))^{m+1} \left ((m+2) (b c-a d) F_1\left (m+1;\frac{1}{2},\frac{1}{2};m+2;\frac{a+b \sin (e+f x)}{a-b},\frac{a+b \sin (e+f x)}{a+b}\right )+d (m+1) (a+b \sin (e+f x)) F_1\left (m+2;\frac{1}{2},\frac{1}{2};m+3;\frac{a+b \sin (e+f x)}{a-b},\frac{a+b \sin (e+f x)}{a+b}\right )\right )}{b^2 f (m+1) (m+2)} \]

Warning: Unable to verify antiderivative.

[In]

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

[Out]

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

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Maple [F]  time = 0.202, size = 0, normalized size = 0. \begin{align*} \int \left ( a+b\sin \left ( fx+e \right ) \right ) ^{m} \left ( c+d\sin \left ( fx+e \right ) \right ) \, dx \end{align*}

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)),x)

[Out]

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

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Maxima [F]  time = 0., size = 0, normalized size = 0. \begin{align*} \int{\left (d \sin \left (f x + e\right ) + c\right )}{\left (b \sin \left (f x + e\right ) + a\right )}^{m}\,{d x} \end{align*}

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)),x, algorithm="maxima")

[Out]

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

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Fricas [F]  time = 0., size = 0, normalized size = 0. \begin{align*}{\rm integral}\left ({\left (d \sin \left (f x + e\right ) + c\right )}{\left (b \sin \left (f x + e\right ) + a\right )}^{m}, x\right ) \end{align*}

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)),x, algorithm="fricas")

[Out]

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

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Sympy [F(-1)]  time = 0., size = 0, normalized size = 0. \begin{align*} \text{Timed out} \end{align*}

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)),x)

[Out]

Timed out

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Giac [F]  time = 0., size = 0, normalized size = 0. \begin{align*} \int{\left (d \sin \left (f x + e\right ) + c\right )}{\left (b \sin \left (f x + e\right ) + a\right )}^{m}\,{d x} \end{align*}

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)),x, algorithm="giac")

[Out]

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