3.4.39 \(\int \frac {(a+a \sin (e+f x))^m (A+B \sin (e+f x))}{c+d \sin (e+f x)} \, dx\) [339]

Optimal. Leaf size=191 \[ -\frac {\sqrt {2} (B c-A d) F_1\left (\frac {1}{2}+m;\frac {1}{2},1;\frac {3}{2}+m;\frac {1}{2} (1+\sin (e+f x)),-\frac {d (1+\sin (e+f x))}{c-d}\right ) \cos (e+f x) (a+a \sin (e+f x))^m}{(c-d) d f (1+2 m) \sqrt {1-\sin (e+f x)}}-\frac {2^{\frac {1}{2}+m} B \cos (e+f x) \, _2F_1\left (\frac {1}{2},\frac {1}{2}-m;\frac {3}{2};\frac {1}{2} (1-\sin (e+f x))\right ) (1+\sin (e+f x))^{-\frac {1}{2}-m} (a+a \sin (e+f x))^m}{d f} \]

[Out]

-2^(1/2+m)*B*cos(f*x+e)*hypergeom([1/2, 1/2-m],[3/2],1/2-1/2*sin(f*x+e))*(1+sin(f*x+e))^(-1/2-m)*(a+a*sin(f*x+
e))^m/d/f-(-A*d+B*c)*AppellF1(1/2+m,1,1/2,3/2+m,-d*(1+sin(f*x+e))/(c-d),1/2+1/2*sin(f*x+e))*cos(f*x+e)*(a+a*si
n(f*x+e))^m*2^(1/2)/(c-d)/d/f/(1+2*m)/(1-sin(f*x+e))^(1/2)

________________________________________________________________________________________

Rubi [A]
time = 0.20, antiderivative size = 191, normalized size of antiderivative = 1.00, number of steps used = 6, number of rules used = 6, integrand size = 35, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.171, Rules used = {3065, 2731, 2730, 2867, 142, 141} \begin {gather*} -\frac {\sqrt {2} (B c-A d) \cos (e+f x) (a \sin (e+f x)+a)^m F_1\left (m+\frac {1}{2};\frac {1}{2},1;m+\frac {3}{2};\frac {1}{2} (\sin (e+f x)+1),-\frac {d (\sin (e+f x)+1)}{c-d}\right )}{d f (2 m+1) (c-d) \sqrt {1-\sin (e+f x)}}-\frac {B 2^{m+\frac {1}{2}} \cos (e+f x) (\sin (e+f x)+1)^{-m-\frac {1}{2}} (a \sin (e+f x)+a)^m \, _2F_1\left (\frac {1}{2},\frac {1}{2}-m;\frac {3}{2};\frac {1}{2} (1-\sin (e+f x))\right )}{d f} \end {gather*}

Antiderivative was successfully verified.

[In]

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

[Out]

-((Sqrt[2]*(B*c - A*d)*AppellF1[1/2 + m, 1/2, 1, 3/2 + m, (1 + Sin[e + f*x])/2, -((d*(1 + Sin[e + f*x]))/(c -
d))]*Cos[e + f*x]*(a + a*Sin[e + f*x])^m)/((c - d)*d*f*(1 + 2*m)*Sqrt[1 - Sin[e + f*x]])) - (2^(1/2 + m)*B*Cos
[e + f*x]*Hypergeometric2F1[1/2, 1/2 - m, 3/2, (1 - Sin[e + f*x])/2]*(1 + Sin[e + f*x])^(-1/2 - m)*(a + a*Sin[
e + f*x])^m)/(d*f)

Rule 141

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

Rule 142

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

Rule 2730

Int[((a_) + (b_.)*sin[(c_.) + (d_.)*(x_)])^(n_), x_Symbol] :> Simp[(-2^(n + 1/2))*a^(n - 1/2)*b*(Cos[c + d*x]/
(d*Sqrt[a + b*Sin[c + d*x]]))*Hypergeometric2F1[1/2, 1/2 - n, 3/2, (1/2)*(1 - b*(Sin[c + d*x]/a))], x] /; Free
Q[{a, b, c, d, n}, x] && EqQ[a^2 - b^2, 0] &&  !IntegerQ[2*n] && GtQ[a, 0]

Rule 2731

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

Rule 2867

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

Rule 3065

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

Rubi steps

\begin {align*} \int \frac {(a+a \sin (e+f x))^m (A+B \sin (e+f x))}{c+d \sin (e+f x)} \, dx &=\frac {B \int (a+a \sin (e+f x))^m \, dx}{d}-\frac {(B c-A d) \int \frac {(a+a \sin (e+f x))^m}{c+d \sin (e+f x)} \, dx}{d}\\ &=-\frac {\left (a^2 (B c-A d) \cos (e+f x)\right ) \text {Subst}\left (\int \frac {(a+a x)^{-\frac {1}{2}+m}}{\sqrt {a-a x} (c+d x)} \, dx,x,\sin (e+f x)\right )}{d f \sqrt {a-a \sin (e+f x)} \sqrt {a+a \sin (e+f x)}}+\frac {\left (B (1+\sin (e+f x))^{-m} (a+a \sin (e+f x))^m\right ) \int (1+\sin (e+f x))^m \, dx}{d}\\ &=-\frac {2^{\frac {1}{2}+m} B \cos (e+f x) \, _2F_1\left (\frac {1}{2},\frac {1}{2}-m;\frac {3}{2};\frac {1}{2} (1-\sin (e+f x))\right ) (1+\sin (e+f x))^{-\frac {1}{2}-m} (a+a \sin (e+f x))^m}{d f}-\frac {\left (a^2 (B c-A d) \cos (e+f x) \sqrt {\frac {a-a \sin (e+f x)}{a}}\right ) \text {Subst}\left (\int \frac {(a+a x)^{-\frac {1}{2}+m}}{\sqrt {\frac {1}{2}-\frac {x}{2}} (c+d x)} \, dx,x,\sin (e+f x)\right )}{\sqrt {2} d f (a-a \sin (e+f x)) \sqrt {a+a \sin (e+f x)}}\\ &=-\frac {\sqrt {2} (B c-A d) F_1\left (\frac {1}{2}+m;\frac {1}{2},1;\frac {3}{2}+m;\frac {1}{2} (1+\sin (e+f x)),-\frac {d (1+\sin (e+f x))}{c-d}\right ) \cos (e+f x) (a+a \sin (e+f x))^m}{(c-d) d f (1+2 m) \sqrt {1-\sin (e+f x)}}-\frac {2^{\frac {1}{2}+m} B \cos (e+f x) \, _2F_1\left (\frac {1}{2},\frac {1}{2}-m;\frac {3}{2};\frac {1}{2} (1-\sin (e+f x))\right ) (1+\sin (e+f x))^{-\frac {1}{2}-m} (a+a \sin (e+f x))^m}{d f}\\ \end {align*}

________________________________________________________________________________________

Mathematica [B] Leaf count is larger than twice the leaf count of optimal. \(473\) vs. \(2(191)=382\).
time = 5.01, size = 473, normalized size = 2.48 \begin {gather*} \frac {(a (1+\sin (e+f x)))^m \left (\frac {\sqrt {2} B \cos (e+f x) \, _2F_1\left (\frac {1}{2},\frac {1}{2}+m;\frac {3}{2}+m;\frac {1}{4} \cos ^2(e+f x) \csc ^2\left (\frac {1}{4} (2 e-\pi +2 f x)\right )\right )}{(d+2 d m) \sqrt {1-\sin (e+f x)}}+\frac {6 (c+d) (B c-A d) F_1\left (\frac {1}{2};\frac {1}{2}-m,1;\frac {3}{2};\cos ^2\left (\frac {1}{4} (2 e+\pi +2 f x)\right ),\frac {2 d \sin ^2\left (\frac {1}{4} (2 e-\pi +2 f x)\right )}{c+d}\right ) \cos ^2\left (\frac {1}{4} (2 e-\pi +2 f x)\right )^{\frac {1}{2}+m} \cot \left (\frac {1}{4} (2 e+\pi +2 f x)\right ) \sec ^2\left (\frac {1}{4} (2 e-\pi +2 f x)\right ) \sin ^2\left (\frac {1}{4} (2 e+\pi +2 f x)\right )^{\frac {1}{2}-m}}{d (c+d \sin (e+f x)) \left (3 (c+d) F_1\left (\frac {1}{2};\frac {1}{2}-m,1;\frac {3}{2};\cos ^2\left (\frac {1}{4} (2 e+\pi +2 f x)\right ),\frac {2 d \sin ^2\left (\frac {1}{4} (2 e-\pi +2 f x)\right )}{c+d}\right )+\left (4 d F_1\left (\frac {3}{2};\frac {1}{2}-m,2;\frac {5}{2};\cos ^2\left (\frac {1}{4} (2 e+\pi +2 f x)\right ),\frac {2 d \sin ^2\left (\frac {1}{4} (2 e-\pi +2 f x)\right )}{c+d}\right )-(c+d) (-1+2 m) F_1\left (\frac {3}{2};\frac {3}{2}-m,1;\frac {5}{2};\cos ^2\left (\frac {1}{4} (2 e+\pi +2 f x)\right ),\frac {2 d \sin ^2\left (\frac {1}{4} (2 e-\pi +2 f x)\right )}{c+d}\right )\right ) \sin ^2\left (\frac {1}{4} (2 e-\pi +2 f x)\right )\right )}\right )}{f} \end {gather*}

Warning: Unable to verify antiderivative.

[In]

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

[Out]

((a*(1 + Sin[e + f*x]))^m*((Sqrt[2]*B*Cos[e + f*x]*Hypergeometric2F1[1/2, 1/2 + m, 3/2 + m, (Cos[e + f*x]^2*Cs
c[(2*e - Pi + 2*f*x)/4]^2)/4])/((d + 2*d*m)*Sqrt[1 - Sin[e + f*x]]) + (6*(c + d)*(B*c - A*d)*AppellF1[1/2, 1/2
- m, 1, 3/2, Cos[(2*e + Pi + 2*f*x)/4]^2, (2*d*Sin[(2*e - Pi + 2*f*x)/4]^2)/(c + d)]*(Cos[(2*e - Pi + 2*f*x)/
4]^2)^(1/2 + m)*Cot[(2*e + Pi + 2*f*x)/4]*Sec[(2*e - Pi + 2*f*x)/4]^2*(Sin[(2*e + Pi + 2*f*x)/4]^2)^(1/2 - m))
/(d*(c + d*Sin[e + f*x])*(3*(c + d)*AppellF1[1/2, 1/2 - m, 1, 3/2, Cos[(2*e + Pi + 2*f*x)/4]^2, (2*d*Sin[(2*e
- Pi + 2*f*x)/4]^2)/(c + d)] + (4*d*AppellF1[3/2, 1/2 - m, 2, 5/2, Cos[(2*e + Pi + 2*f*x)/4]^2, (2*d*Sin[(2*e
- Pi + 2*f*x)/4]^2)/(c + d)] - (c + d)*(-1 + 2*m)*AppellF1[3/2, 3/2 - m, 1, 5/2, Cos[(2*e + Pi + 2*f*x)/4]^2,
(2*d*Sin[(2*e - Pi + 2*f*x)/4]^2)/(c + d)])*Sin[(2*e - Pi + 2*f*x)/4]^2))))/f

________________________________________________________________________________________

Maple [F]
time = 0.45, size = 0, normalized size = 0.00 \[\int \frac {\left (a +a \sin \left (f x +e \right )\right )^{m} \left (A +B \sin \left (f x +e \right )\right )}{c +d \sin \left (f x +e \right )}\, dx\]

Verification of antiderivative is not currently implemented for this CAS.

[In]

int((a+a*sin(f*x+e))^m*(A+B*sin(f*x+e))/(c+d*sin(f*x+e)),x)

[Out]

int((a+a*sin(f*x+e))^m*(A+B*sin(f*x+e))/(c+d*sin(f*x+e)),x)

________________________________________________________________________________________

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+a*sin(f*x+e))^m*(A+B*sin(f*x+e))/(c+d*sin(f*x+e)),x, algorithm="maxima")

[Out]

integrate((B*sin(f*x + e) + A)*(a*sin(f*x + e) + a)^m/(d*sin(f*x + e) + c), x)

________________________________________________________________________________________

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+a*sin(f*x+e))^m*(A+B*sin(f*x+e))/(c+d*sin(f*x+e)),x, algorithm="fricas")

[Out]

integral((B*sin(f*x + e) + A)*(a*sin(f*x + e) + a)^m/(d*sin(f*x + e) + c), x)

________________________________________________________________________________________

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+a*sin(f*x+e))**m*(A+B*sin(f*x+e))/(c+d*sin(f*x+e)),x)

[Out]

Timed out

________________________________________________________________________________________

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+a*sin(f*x+e))^m*(A+B*sin(f*x+e))/(c+d*sin(f*x+e)),x, algorithm="giac")

[Out]

integrate((B*sin(f*x + e) + A)*(a*sin(f*x + e) + a)^m/(d*sin(f*x + e) + c), x)

________________________________________________________________________________________

Mupad [F]
time = 0.00, size = -1, normalized size = -0.01 \begin {gather*} \int \frac {\left (A+B\,\sin \left (e+f\,x\right )\right )\,{\left (a+a\,\sin \left (e+f\,x\right )\right )}^m}{c+d\,\sin \left (e+f\,x\right )} \,d x \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

int(((A + B*sin(e + f*x))*(a + a*sin(e + f*x))^m)/(c + d*sin(e + f*x)),x)

[Out]

int(((A + B*sin(e + f*x))*(a + a*sin(e + f*x))^m)/(c + d*sin(e + f*x)), x)

________________________________________________________________________________________