3.612 \(\int \frac {(a+a \sin (e+f x))^m}{(c+d \sin (e+f x))^2} \, dx\)

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

[Out]

AppellF1(1/2+m,2,1/2,3/2+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)^2/f/(1+2*m)/(1-sin(f*x+e))^(1/2)

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Rubi [A]  time = 0.13, 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 = {2788, 137, 136} \[ \frac {\sqrt {2} \cos (e+f x) (a \sin (e+f x)+a)^m F_1\left (m+\frac {1}{2};\frac {1}{2},2;m+\frac {3}{2};\frac {1}{2} (\sin (e+f x)+1),-\frac {d (\sin (e+f x)+1)}{c-d}\right )}{f (2 m+1) (c-d)^2 \sqrt {1-\sin (e+f x)}} \]

Antiderivative was successfully verified.

[In]

Int[(a + a*Sin[e + f*x])^m/(c + d*Sin[e + f*x])^2,x]

[Out]

(Sqrt[2]*AppellF1[1/2 + m, 1/2, 2, 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)^2*f*(1 + 2*m)*Sqrt[1 - Sin[e + f*x]])

Rule 136

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)*AppellF1[m + 1, -n, -p, m + 2, -((d*(a + b*x))/(b*c - a*d)), -((f*(a + b*x))/(b*e - a*f
))])/(b^(p + 1)*(m + 1)*(b/(b*c - a*d))^n), 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 137

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 2788

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

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Mathematica [B]  time = 1.27, size = 363, normalized size = 3.63 \[ -\frac {6 (c+d) \cot \left (\frac {1}{4} (2 e+2 f x+\pi )\right ) \sin ^2\left (\frac {1}{4} (2 e+2 f x+\pi )\right )^{\frac {1}{2}-m} \cos ^2\left (\frac {1}{4} (2 e+2 f x-\pi )\right )^{m-\frac {1}{2}} (a (\sin (e+f x)+1))^m F_1\left (\frac {1}{2};\frac {1}{2}-m,2;\frac {3}{2};\cos ^2\left (\frac {1}{4} (2 e+2 f x+\pi )\right ),\frac {2 d \sin ^2\left (\frac {1}{4} (2 e+2 f x-\pi )\right )}{c+d}\right )}{f (c+d \sin (e+f x))^2 \left (\sin ^2\left (\frac {1}{4} (2 e+2 f x-\pi )\right ) \left (8 d F_1\left (\frac {3}{2};\frac {1}{2}-m,3;\frac {5}{2};\cos ^2\left (\frac {1}{4} (2 e+2 f x+\pi )\right ),\frac {2 d \sin ^2\left (\frac {1}{4} (2 e+2 f x-\pi )\right )}{c+d}\right )-(2 m-1) (c+d) F_1\left (\frac {3}{2};\frac {3}{2}-m,2;\frac {5}{2};\cos ^2\left (\frac {1}{4} (2 e+2 f x+\pi )\right ),\frac {2 d \sin ^2\left (\frac {1}{4} (2 e+2 f x-\pi )\right )}{c+d}\right )\right )+3 (c+d) F_1\left (\frac {1}{2};\frac {1}{2}-m,2;\frac {3}{2};\cos ^2\left (\frac {1}{4} (2 e+2 f x+\pi )\right ),\frac {2 d \sin ^2\left (\frac {1}{4} (2 e+2 f x-\pi )\right )}{c+d}\right )\right )} \]

Warning: Unable to verify antiderivative.

[In]

Integrate[(a + a*Sin[e + f*x])^m/(c + d*Sin[e + f*x])^2,x]

[Out]

(-6*(c + d)*AppellF1[1/2, 1/2 - m, 2, 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])^2*(3*(c + d)*AppellF1[1/2, 1/2 - m, 2, 3/2, Cos[(2*e + Pi
+ 2*f*x)/4]^2, (2*d*Sin[(2*e - Pi + 2*f*x)/4]^2)/(c + d)] + (8*d*AppellF1[3/2, 1/2 - m, 3, 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, 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)])*Sin[(2*e - Pi + 2*f*x)/4]^2))

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fricas [F]  time = 0.49, size = 0, normalized size = 0.00 \[ {\rm integral}\left (-\frac {{\left (a \sin \left (f x + e\right ) + a\right )}^{m}}{d^{2} \cos \left (f x + e\right )^{2} - 2 \, c d \sin \left (f x + e\right ) - c^{2} - d^{2}}, x\right ) \]

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))^2,x, algorithm="fricas")

[Out]

integral(-(a*sin(f*x + e) + a)^m/(d^2*cos(f*x + e)^2 - 2*c*d*sin(f*x + e) - c^2 - d^2), x)

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giac [F]  time = 0.00, size = 0, normalized size = 0.00 \[ \int \frac {{\left (a \sin \left (f x + e\right ) + a\right )}^{m}}{{\left (d \sin \left (f x + e\right ) + c\right )}^{2}}\,{d x} \]

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))^2,x, algorithm="giac")

[Out]

integrate((a*sin(f*x + e) + a)^m/(d*sin(f*x + e) + c)^2, x)

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maple [F]  time = 1.84, size = 0, normalized size = 0.00 \[ \int \frac {\left (a +a \sin \left (f x +e \right )\right )^{m}}{\left (c +d \sin \left (f x +e \right )\right )^{2}}\, dx \]

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))^2,x)

[Out]

int((a+a*sin(f*x+e))^m/(c+d*sin(f*x+e))^2,x)

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maxima [F]  time = 0.00, size = 0, normalized size = 0.00 \[ \int \frac {{\left (a \sin \left (f x + e\right ) + a\right )}^{m}}{{\left (d \sin \left (f x + e\right ) + c\right )}^{2}}\,{d x} \]

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))^2,x, algorithm="maxima")

[Out]

integrate((a*sin(f*x + e) + a)^m/(d*sin(f*x + e) + c)^2, x)

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mupad [F]  time = 0.00, size = -1, normalized size = -0.01 \[ \int \frac {{\left (a+a\,\sin \left (e+f\,x\right )\right )}^m}{{\left (c+d\,\sin \left (e+f\,x\right )\right )}^2} \,d x \]

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))^2,x)

[Out]

int((a + a*sin(e + f*x))^m/(c + d*sin(e + f*x))^2, 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+a*sin(f*x+e))**m/(c+d*sin(f*x+e))**2,x)

[Out]

Timed out

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