3.77 \(\int \frac {\cos ^2(e+f x) (a+a \sin (e+f x))^m}{(c-c \sin (e+f x))^{3/2}} \, dx\)

Optimal. Leaf size=76 \[ \frac {\cos (e+f x) (a \sin (e+f x)+a)^{m+1} \, _2F_1\left (1,m+\frac {3}{2};m+\frac {5}{2};\frac {1}{2} (\sin (e+f x)+1)\right )}{a c f (2 m+3) \sqrt {c-c \sin (e+f x)}} \]

[Out]

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

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Rubi [A]  time = 0.36, antiderivative size = 76, normalized size of antiderivative = 1.00, number of steps used = 4, number of rules used = 4, integrand size = 36, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.111, Rules used = {2841, 2745, 2667, 68} \[ \frac {\cos (e+f x) (a \sin (e+f x)+a)^{m+1} \, _2F_1\left (1,m+\frac {3}{2};m+\frac {5}{2};\frac {1}{2} (\sin (e+f x)+1)\right )}{a c f (2 m+3) \sqrt {c-c \sin (e+f x)}} \]

Antiderivative was successfully verified.

[In]

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

[Out]

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

Rule 68

Int[((a_) + (b_.)*(x_))^(m_)*((c_) + (d_.)*(x_))^(n_), x_Symbol] :> Simp[((b*c - a*d)^n*(a + b*x)^(m + 1)*Hype
rgeometric2F1[-n, m + 1, m + 2, -((d*(a + b*x))/(b*c - a*d))])/(b^(n + 1)*(m + 1)), x] /; FreeQ[{a, b, c, d, m
}, x] && NeQ[b*c - a*d, 0] &&  !IntegerQ[m] && IntegerQ[n]

Rule 2667

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

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

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Mathematica [B]  time = 6.66, size = 218, normalized size = 2.87 \[ -\frac {2^{-2 m-\frac {5}{2}} \cos ^2\left (\frac {1}{2} \left (-e-f x+\frac {\pi }{2}\right )\right ) \left (\cos \left (\frac {1}{2} (e+f x)\right )-\sin \left (\frac {1}{2} (e+f x)\right )\right )^3 (a \sin (e+f x)+a)^m \left (\sec ^4\left (\frac {1}{4} \left (-e-f x+\frac {\pi }{2}\right )\right ) \sec ^2\left (\frac {1}{4} \left (-e-f x+\frac {\pi }{2}\right )\right )^{2 m} \, _2F_1\left (2 m+2,2 m+2;2 m+3;\frac {1}{2} \left (1-\tan ^2\left (\frac {1}{4} \left (-e-f x+\frac {\pi }{2}\right )\right )\right )\right )-4^{m+1} \, _2F_1\left (1,2 m+2;2 m+3;\cos \left (\frac {1}{2} \left (-e-f x+\frac {\pi }{2}\right )\right )\right )\right )}{f (m+1) (c-c \sin (e+f x))^{3/2}} \]

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])^(3/2),x]

[Out]

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

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fricas [F]  time = 0.46, size = 0, normalized size = 0.00 \[ {\rm integral}\left (-\frac {\sqrt {-c \sin \left (f x + e\right ) + c} {\left (a \sin \left (f x + e\right ) + a\right )}^{m} \cos \left (f x + e\right )^{2}}{c^{2} \cos \left (f x + e\right )^{2} + 2 \, c^{2} \sin \left (f x + e\right ) - 2 \, c^{2}}, x\right ) \]

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))^(3/2),x, algorithm="fricas")

[Out]

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

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giac [F]  time = 0.00, size = 0, normalized size = 0.00 \[ \int \frac {{\left (a \sin \left (f x + e\right ) + a\right )}^{m} \cos \left (f x + e\right )^{2}}{{\left (-c \sin \left (f x + e\right ) + c\right )}^{\frac {3}{2}}}\,{d x} \]

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))^(3/2),x, algorithm="giac")

[Out]

integrate((a*sin(f*x + e) + a)^m*cos(f*x + e)^2/(-c*sin(f*x + e) + c)^(3/2), x)

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maple [F]  time = 0.65, size = 0, normalized size = 0.00 \[ \int \frac {\left (\cos ^{2}\left (f x +e \right )\right ) \left (a +a \sin \left (f x +e \right )\right )^{m}}{\left (c -c \sin \left (f x +e \right )\right )^{\frac {3}{2}}}\, dx \]

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))^(3/2),x)

[Out]

int(cos(f*x+e)^2*(a+a*sin(f*x+e))^m/(c-c*sin(f*x+e))^(3/2),x)

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maxima [F]  time = 0.00, size = 0, normalized size = 0.00 \[ \int \frac {{\left (a \sin \left (f x + e\right ) + a\right )}^{m} \cos \left (f x + e\right )^{2}}{{\left (-c \sin \left (f x + e\right ) + c\right )}^{\frac {3}{2}}}\,{d x} \]

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))^(3/2),x, algorithm="maxima")

[Out]

integrate((a*sin(f*x + e) + a)^m*cos(f*x + e)^2/(-c*sin(f*x + e) + c)^(3/2), x)

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mupad [F]  time = 0.00, size = -1, normalized size = -0.01 \[ \int \frac {{\cos \left (e+f\,x\right )}^2\,{\left (a+a\,\sin \left (e+f\,x\right )\right )}^m}{{\left (c-c\,\sin \left (e+f\,x\right )\right )}^{3/2}} \,d x \]

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))^(3/2),x)

[Out]

int((cos(e + f*x)^2*(a + a*sin(e + f*x))^m)/(c - c*sin(e + f*x))^(3/2), x)

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sympy [F]  time = 0.00, size = 0, normalized size = 0.00 \[ \int \frac {\left (a \left (\sin {\left (e + f x \right )} + 1\right )\right )^{m} \cos ^{2}{\left (e + f x \right )}}{\left (- c \left (\sin {\left (e + f x \right )} - 1\right )\right )^{\frac {3}{2}}}\, dx \]

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))**(3/2),x)

[Out]

Integral((a*(sin(e + f*x) + 1))**m*cos(e + f*x)**2/(-c*(sin(e + f*x) - 1))**(3/2), x)

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