3.9.3 \(\int (a+b \sin (e+f x))^m (c+d \sin (e+f x)) \, dx\) [803]

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

[Out]

-(a+b)*d*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/f/(((a+b*sin(f*x+e))/(a+b))^m)/(1+sin(f*x+e))^(1/2)-(-a*d+b*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)/b/f/(((a+b*sin(f*x+e))/(a+b))^m)/(1+sin(f*x+
e))^(1/2)

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Rubi [A]
time = 0.13, antiderivative size = 229, normalized size of antiderivative = 1.00, 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 = {2835, 2744, 144, 143} \begin {gather*} -\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}} \end {gather*}

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 143

Int[((a_) + (b_.)*(x_))^(m_)*((c_.) + (d_.)*(x_))^(n_)*((e_.) + (f_.)*(x_))^(p_), x_Symbol] :> Simp[((a + b*x)
^(m + 1)/(b*(m + 1)*(b/(b*c - a*d))^n*(b/(b*e - a*f))^p))*AppellF1[m + 1, -n, -p, m + 2, (-d)*((a + b*x)/(b*c
- a*d)), (-f)*((a + b*x)/(b*e - a*f))], 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 144

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 2744

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 2835

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]

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)) \text {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)) \text {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 ) \text {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 ) \text {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*}

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Mathematica [A]
time = 0.38, size = 200, normalized size = 0.87 \begin {gather*} \frac {\sec (e+f x) \sqrt {-\frac {b (-1+\sin (e+f x))}{a+b}} \sqrt {\frac {b (1+\sin (e+f x))}{-a+b}} (a+b \sin (e+f x))^{1+m} \left ((b c-a d) (2+m) F_1\left (1+m;\frac {1}{2},\frac {1}{2};2+m;\frac {a+b \sin (e+f x)}{a-b},\frac {a+b \sin (e+f x)}{a+b}\right )+d (1+m) F_1\left (2+m;\frac {1}{2},\frac {1}{2};3+m;\frac {a+b \sin (e+f x)}{a-b},\frac {a+b \sin (e+f x)}{a+b}\right ) (a+b \sin (e+f x))\right )}{b^2 f (1+m) (2+m)} \end {gather*}

Antiderivative was successfully verified.

[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 = 0.12, size = 0, normalized size = 0.00 \[\int \left (a +b \sin \left (f x +e \right )\right )^{m} \left (c +d \sin \left (f x +e \right )\right )\, dx\]

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)),x)

[Out]

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

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Maxima [F]
time = 0.00, size = 0, normalized size = 0.00 \begin {gather*} \text {Failed to integrate} \end {gather*}

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 = 0.00, size = 0, normalized size = 0.00 \begin {gather*} \text {could not integrate} \end {gather*}

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)] Timed out
time = 0.00, size = 0, normalized size = 0.00 \begin {gather*} \text {Timed out} \end {gather*}

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 = 0.00, size = 0, normalized size = 0.00 \begin {gather*} \text {could not integrate} \end {gather*}

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)

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

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)),x)

[Out]

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

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