3.84 \(\int \cos ^2(e+f x) (a+a \sin (e+f x))^m (c-c \sin (e+f x))^{-2-m} \, dx\)

Optimal. Leaf size=113 \[ \frac {2^{-m-\frac {1}{2}} \cos ^3(e+f x) (1-\sin (e+f x))^{m+\frac {1}{2}} (a \sin (e+f x)+a)^m (c-c \sin (e+f x))^{-m-2} \, _2F_1\left (\frac {1}{2} (2 m+3),\frac {1}{2} (2 m+3);\frac {1}{2} (2 m+5);\frac {1}{2} (\sin (e+f x)+1)\right )}{f (2 m+3)} \]

[Out]

2^(-1/2-m)*cos(f*x+e)^3*hypergeom([3/2+m, 3/2+m],[5/2+m],1/2+1/2*sin(f*x+e))*(1-sin(f*x+e))^(1/2+m)*(a+a*sin(f
*x+e))^m*(c-c*sin(f*x+e))^(-2-m)/f/(3+2*m)

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Rubi [A]  time = 0.38, antiderivative size = 113, normalized size of antiderivative = 1.00, number of steps used = 5, number of rules used = 5, integrand size = 38, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.132, Rules used = {2841, 2745, 2689, 70, 69} \[ \frac {2^{-m-\frac {1}{2}} \cos ^3(e+f x) (1-\sin (e+f x))^{m+\frac {1}{2}} (a \sin (e+f x)+a)^m (c-c \sin (e+f x))^{-m-2} \, _2F_1\left (\frac {1}{2} (2 m+3),\frac {1}{2} (2 m+3);\frac {1}{2} (2 m+5);\frac {1}{2} (\sin (e+f x)+1)\right )}{f (2 m+3)} \]

Antiderivative was successfully verified.

[In]

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

[Out]

(2^(-1/2 - m)*Cos[e + f*x]^3*Hypergeometric2F1[(3 + 2*m)/2, (3 + 2*m)/2, (5 + 2*m)/2, (1 + Sin[e + f*x])/2]*(1
- Sin[e + f*x])^(1/2 + m)*(a + a*Sin[e + f*x])^m*(c - c*Sin[e + f*x])^(-2 - m))/(f*(3 + 2*m))

Rule 69

Int[((a_) + (b_.)*(x_))^(m_)*((c_) + (d_.)*(x_))^(n_), x_Symbol] :> Simp[((a + b*x)^(m + 1)*Hypergeometric2F1[
-n, m + 1, m + 2, -((d*(a + b*x))/(b*c - a*d))])/(b*(m + 1)*(b/(b*c - a*d))^n), x] /; FreeQ[{a, b, c, d, m, n}
, x] && NeQ[b*c - a*d, 0] &&  !IntegerQ[m] &&  !IntegerQ[n] && GtQ[b/(b*c - a*d), 0] && (RationalQ[m] ||  !(Ra
tionalQ[n] && GtQ[-(d/(b*c - a*d)), 0]))

Rule 70

Int[((a_) + (b_.)*(x_))^(m_)*((c_) + (d_.)*(x_))^(n_), 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*Simp[(b*c)/(b*c - a*d) + (b*d*x)/(b*c -
a*d), x]^n, x], x] /; FreeQ[{a, b, c, d, m, n}, x] && NeQ[b*c - a*d, 0] &&  !IntegerQ[m] &&  !IntegerQ[n] &&
(RationalQ[m] ||  !SimplerQ[n + 1, m + 1])

Rule 2689

Int[(cos[(e_.) + (f_.)*(x_)]*(g_.))^(p_)*((a_) + (b_.)*sin[(e_.) + (f_.)*(x_)])^(m_.), x_Symbol] :> Dist[(a^2*
(g*Cos[e + f*x])^(p + 1))/(f*g*(a + b*Sin[e + f*x])^((p + 1)/2)*(a - b*Sin[e + f*x])^((p + 1)/2)), Subst[Int[(
a + b*x)^(m + (p - 1)/2)*(a - b*x)^((p - 1)/2), x], x, Sin[e + f*x]], x] /; FreeQ[{a, b, e, f, g, m, p}, x] &&
EqQ[a^2 - b^2, 0] &&  !IntegerQ[m]

Rule 2745

Int[((a_) + (b_.)*sin[(e_.) + (f_.)*(x_)])^(m_)*((c_) + (d_.)*sin[(e_.) + (f_.)*(x_)])^(n_), x_Symbol] :> Dist
[(a^IntPart[m]*c^IntPart[m]*(a + b*Sin[e + f*x])^FracPart[m]*(c + d*Sin[e + f*x])^FracPart[m])/Cos[e + f*x]^(2
*FracPart[m]), Int[Cos[e + f*x]^(2*m)*(c + d*Sin[e + f*x])^(n - m), x], x] /; FreeQ[{a, b, c, d, e, f, m, n},
x] && EqQ[b*c + a*d, 0] && EqQ[a^2 - b^2, 0] && (FractionQ[m] ||  !FractionQ[n])

Rule 2841

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

Rubi steps

\begin {align*} \int \cos ^2(e+f x) (a+a \sin (e+f x))^m (c-c \sin (e+f x))^{-2-m} \, dx &=\frac {\int (a+a \sin (e+f x))^{1+m} (c-c \sin (e+f x))^{-1-m} \, dx}{a c}\\ &=\left (\cos ^{-2 m}(e+f x) (a+a \sin (e+f x))^m (c-c \sin (e+f x))^m\right ) \int \cos ^{2 (1+m)}(e+f x) (c-c \sin (e+f x))^{-2-2 m} \, dx\\ &=\frac {\left (c^2 \cos ^{1-2 m+2 (1+m)}(e+f x) (a+a \sin (e+f x))^m (c-c \sin (e+f x))^{m+\frac {1}{2} (-1-2 (1+m))} (c+c \sin (e+f x))^{\frac {1}{2} (-1-2 (1+m))}\right ) \operatorname {Subst}\left (\int (c-c x)^{-2-2 m+\frac {1}{2} (-1+2 (1+m))} (c+c x)^{\frac {1}{2} (-1+2 (1+m))} \, dx,x,\sin (e+f x)\right )}{f}\\ &=\frac {\left (2^{-\frac {3}{2}-m} c \cos ^{1-2 m+2 (1+m)}(e+f x) (a+a \sin (e+f x))^m (c-c \sin (e+f x))^{-\frac {1}{2}+\frac {1}{2} (-1-2 (1+m))} \left (\frac {c-c \sin (e+f x)}{c}\right )^{\frac {1}{2}+m} (c+c \sin (e+f x))^{\frac {1}{2} (-1-2 (1+m))}\right ) \operatorname {Subst}\left (\int \left (\frac {1}{2}-\frac {x}{2}\right )^{-2-2 m+\frac {1}{2} (-1+2 (1+m))} (c+c x)^{\frac {1}{2} (-1+2 (1+m))} \, dx,x,\sin (e+f x)\right )}{f}\\ &=\frac {2^{-\frac {1}{2}-m} \cos ^3(e+f x) \, _2F_1\left (\frac {1}{2} (3+2 m),\frac {1}{2} (3+2 m);\frac {1}{2} (5+2 m);\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 (c-c \sin (e+f x))^{-2-m}}{f (3+2 m)}\\ \end {align*}

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Mathematica [C]  time = 21.56, size = 589, normalized size = 5.21 \[ -\frac {2^{1-m} (2 m-3) \cos ^2\left (\frac {1}{2} \left (-e-f x+\frac {\pi }{2}\right )\right ) \cot \left (\frac {1}{4} \left (-e-f x+\frac {\pi }{2}\right )\right ) \csc ^2\left (\frac {1}{4} \left (-e-f x+\frac {\pi }{2}\right )\right ) \sin ^{-2 m}\left (\frac {1}{2} \left (-e-f x+\frac {\pi }{2}\right )\right ) (a \sin (e+f x)+a)^m (c-c \sin (e+f x))^{-m-2} \left (\cos \left (\frac {1}{2} (e+f x)\right )-\sin \left (\frac {1}{2} (e+f x)\right )\right )^{-2 (-m-2)} \left ((2 m-1) \cot ^2\left (\frac {1}{4} \left (-e-f x+\frac {\pi }{2}\right )\right ) \, _2F_1\left (-m-\frac {1}{2},-2 (m+1);\frac {1}{2}-m;\tan ^2\left (\frac {1}{4} \left (-e-f x+\frac {\pi }{2}\right )\right )\right )-(2 m+1) F_1\left (\frac {1}{2}-m;-2 (m+1),1;\frac {3}{2}-m;\tan ^2\left (\frac {1}{4} \left (-e-f x+\frac {\pi }{2}\right )\right ),-\tan ^2\left (\frac {1}{4} \left (-e-f x+\frac {\pi }{2}\right )\right )\right )\right )}{f \left (4 m^2-1\right ) \left (8 (m+1) F_1\left (\frac {3}{2}-m;-2 m-1,1;\frac {5}{2}-m;\tan ^2\left (\frac {1}{4} \left (-e-f x+\frac {\pi }{2}\right )\right ),-\tan ^2\left (\frac {1}{4} \left (-e-f x+\frac {\pi }{2}\right )\right )\right )+4 F_1\left (\frac {3}{2}-m;-2 (m+1),2;\frac {5}{2}-m;\tan ^2\left (\frac {1}{4} \left (-e-f x+\frac {\pi }{2}\right )\right ),-\tan ^2\left (\frac {1}{4} \left (-e-f x+\frac {\pi }{2}\right )\right )\right )+(2 m-3) \left (2 \cot ^2\left (\frac {1}{4} \left (-e-f x+\frac {\pi }{2}\right )\right ) F_1\left (\frac {1}{2}-m;-2 (m+1),1;\frac {3}{2}-m;\tan ^2\left (\frac {1}{4} \left (-e-f x+\frac {\pi }{2}\right )\right ),-\tan ^2\left (\frac {1}{4} \left (-e-f x+\frac {\pi }{2}\right )\right )\right )-(\sin (e+f x)+1) \csc ^4\left (\frac {1}{4} \left (-e-f x+\frac {\pi }{2}\right )\right ) \left (1-\tan ^2\left (\frac {1}{4} \left (-e-f x+\frac {\pi }{2}\right )\right )\right )^{2 m}\right )\right )} \]

Warning: Unable to verify antiderivative.

[In]

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

[Out]

-((2^(1 - m)*(-3 + 2*m)*Cos[(-e + Pi/2 - f*x)/2]^2*Cot[(-e + Pi/2 - f*x)/4]*Csc[(-e + Pi/2 - f*x)/4]^2*(-((1 +
2*m)*AppellF1[1/2 - m, -2*(1 + m), 1, 3/2 - m, Tan[(-e + Pi/2 - f*x)/4]^2, -Tan[(-e + Pi/2 - f*x)/4]^2]) + (-
1 + 2*m)*Cot[(-e + Pi/2 - f*x)/4]^2*Hypergeometric2F1[-1/2 - m, -2*(1 + m), 1/2 - m, Tan[(-e + Pi/2 - f*x)/4]^
2])*(a + a*Sin[e + f*x])^m*(c - c*Sin[e + f*x])^(-2 - m))/(f*(-1 + 4*m^2)*Sin[(-e + Pi/2 - f*x)/2]^(2*m)*(Cos[
(e + f*x)/2] - Sin[(e + f*x)/2])^(2*(-2 - m))*(8*(1 + m)*AppellF1[3/2 - m, -1 - 2*m, 1, 5/2 - m, Tan[(-e + Pi/
2 - f*x)/4]^2, -Tan[(-e + Pi/2 - f*x)/4]^2] + 4*AppellF1[3/2 - m, -2*(1 + m), 2, 5/2 - m, Tan[(-e + Pi/2 - f*x
)/4]^2, -Tan[(-e + Pi/2 - f*x)/4]^2] + (-3 + 2*m)*(2*AppellF1[1/2 - m, -2*(1 + m), 1, 3/2 - m, Tan[(-e + Pi/2
- f*x)/4]^2, -Tan[(-e + Pi/2 - f*x)/4]^2]*Cot[(-e + Pi/2 - f*x)/4]^2 - Csc[(-e + Pi/2 - f*x)/4]^4*(1 + Sin[e +
f*x])*(1 - Tan[(-e + Pi/2 - f*x)/4]^2)^(2*m)))))

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

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate(cos(f*x+e)^2*(a+a*sin(f*x+e))^m*(c-c*sin(f*x+e))^(-2-m),x, algorithm="fricas")

[Out]

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

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

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate(cos(f*x+e)^2*(a+a*sin(f*x+e))^m*(c-c*sin(f*x+e))^(-2-m),x, algorithm="giac")

[Out]

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

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maple [F]  time = 3.58, size = 0, normalized size = 0.00 \[ \int \left (\cos ^{2}\left (f x +e \right )\right ) \left (a +a \sin \left (f x +e \right )\right )^{m} \left (c -c \sin \left (f x +e \right )\right )^{-2-m}\, dx \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

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

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

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate(cos(f*x+e)^2*(a+a*sin(f*x+e))^m*(c-c*sin(f*x+e))^(-2-m),x, algorithm="maxima")

[Out]

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

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

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

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

[Out]

Timed out

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