3.7.11 \(\int \frac {(a+a \sin (e+f x))^m}{c+d \sin (e+f x)} \, dx\) [611]

Optimal. Leaf size=100 \[ \frac {\sqrt {2} F_1\left (\frac {1}{2}+m;\frac {1}{2},1;\frac {3}{2}+m;\frac {1}{2} (1+\sin (e+f x)),-\frac {d (1+\sin (e+f x))}{c-d}\right ) \cos (e+f x) (a+a \sin (e+f x))^m}{(c-d) f (1+2 m) \sqrt {1-\sin (e+f x)}} \]

[Out]

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

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Rubi [A]
time = 0.09, antiderivative size = 100, normalized size of antiderivative = 1.00, number of steps used = 3, number of rules used = 3, integrand size = 25, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.120, Rules used = {2867, 142, 141} \begin {gather*} \frac {\sqrt {2} \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 )}{f (2 m+1) (c-d) \sqrt {1-\sin (e+f x)}} \end {gather*}

Antiderivative was successfully verified.

[In]

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

[Out]

(Sqrt[2]*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)*f*(1 + 2*m)*Sqrt[1 - Sin[e + f*x]])

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 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]

Rubi steps

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

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Mathematica [B] Leaf count is larger than twice the leaf count of optimal. \(363\) vs. \(2(100)=200\).
time = 1.22, size = 363, normalized size = 3.63 \begin {gather*} -\frac {6 (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 ) \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 ) (a (1+\sin (e+f x)))^m \sin ^2\left (\frac {1}{4} (2 e+\pi +2 f x)\right )^{\frac {1}{2}-m}}{f (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 )} \end {gather*}

Warning: Unable to verify antiderivative.

[In]

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

[Out]

(-6*(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)]*(Cos[(2*e - Pi + 2*f*x)/4]^2)^(-1/2 + m)*Cot[(2*e + Pi + 2*f*x)/4]*(a*(1 + Sin[e + f*x]))^m*(Sin[(2*e + P
i + 2*f*x)/4]^2)^(1/2 - m))/(f*(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, Co
s[(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))

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

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

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

[Out]

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

[Out]

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

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Sympy [F]
time = 0.00, size = 0, normalized size = 0.00 \begin {gather*} \int \frac {\left (a \left (\sin {\left (e + f x \right )} + 1\right )\right )^{m}}{c + d \sin {\left (e + f x \right )}}\, dx \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

Integral((a*(sin(e + f*x) + 1))**m/(c + d*sin(e + f*x)), x)

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

[Out]

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

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

[Out]

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

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