3.78 \(\int \frac{\cos ^2(e+f x) (a+a \sin (e+f x))^m}{(c-c \sin (e+f x))^{5/2}} \, dx\)
Optimal. Leaf size=79 \[ \frac{\cos (e+f x) (a \sin (e+f x)+a)^{m+1} \, _2F_1\left (2,m+\frac{3}{2};m+\frac{5}{2};\frac{1}{2} (\sin (e+f x)+1)\right )}{2 a c^2 f (2 m+3) \sqrt{c-c \sin (e+f x)}} \]
[Out]
(Cos[e + f*x]*Hypergeometric2F1[2, 3/2 + m, 5/2 + m, (1 + Sin[e + f*x])/2]*(a + a*Sin[e + f*x])^(1 + m))/(2*a*
c^2*f*(3 + 2*m)*Sqrt[c - c*Sin[e + f*x]])
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Rubi [A] time = 0.372359, antiderivative size = 79, normalized size of antiderivative = 1.,
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} \[ \frac{\cos (e+f x) (a \sin (e+f x)+a)^{m+1} \, _2F_1\left (2,m+\frac{3}{2};m+\frac{5}{2};\frac{1}{2} (\sin (e+f x)+1)\right )}{2 a c^2 f (2 m+3) \sqrt{c-c \sin (e+f x)}} \]
Antiderivative was successfully verified.
[In]
Int[(Cos[e + f*x]^2*(a + a*Sin[e + f*x])^m)/(c - c*Sin[e + f*x])^(5/2),x]
[Out]
(Cos[e + f*x]*Hypergeometric2F1[2, 3/2 + m, 5/2 + m, (1 + Sin[e + f*x])/2]*(a + a*Sin[e + f*x])^(1 + m))/(2*a*
c^2*f*(3 + 2*m)*Sqrt[c - c*Sin[e + f*x]])
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]
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 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 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]
Rubi steps
\begin{align*} \int \frac{\cos ^2(e+f x) (a+a \sin (e+f x))^m}{(c-c \sin (e+f x))^{5/2}} \, dx &=\frac{\int \frac{(a+a \sin (e+f x))^{1+m}}{(c-c \sin (e+f x))^{3/2}} \, dx}{a c}\\ &=\frac{\cos (e+f x) \int \sec ^3(e+f x) (a+a \sin (e+f x))^{\frac{5}{2}+m} \, dx}{a^2 c^2 \sqrt{a+a \sin (e+f x)} \sqrt{c-c \sin (e+f x)}}\\ &=\frac{(a \cos (e+f x)) \operatorname{Subst}\left (\int \frac{(a+x)^{\frac{1}{2}+m}}{(a-x)^2} \, dx,x,a \sin (e+f x)\right )}{c^2 f \sqrt{a+a \sin (e+f x)} \sqrt{c-c \sin (e+f x)}}\\ &=\frac{\cos (e+f x) \, _2F_1\left (2,\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}}{2 a c^2 f (3+2 m) \sqrt{c-c \sin (e+f x)}}\\ \end{align*}
Mathematica [C] time = 41.4602, size = 3157, normalized size = 39.96 \[ \text{Result too large to show} \]
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])^(5/2),x]
[Out]
-((Cos[(e + f*x)/2] - Sin[(e + f*x)/2])^5*(a + a*Sin[e + f*x])^m*((Cos[(-e + Pi/2 - f*x)/2]^(2*m)*Cos[Pi/4 + (
e - Pi/2 + f*x)/2]^2)/(Cos[Pi/4 + (e - Pi/2 + f*x)/2] - Sin[Pi/4 + (e - Pi/2 + f*x)/2])^3 + (2*Cos[(-e + Pi/2
- f*x)/2]^(2*m)*Cos[Pi/4 + (e - Pi/2 + f*x)/2]*Sin[Pi/4 + (e - Pi/2 + f*x)/2])/(Cos[Pi/4 + (e - Pi/2 + f*x)/2]
- Sin[Pi/4 + (e - Pi/2 + f*x)/2])^3 + (Cos[(-e + Pi/2 - f*x)/2]^(2*m)*Sin[Pi/4 + (e - Pi/2 + f*x)/2]^2)/(Cos[
Pi/4 + (e - Pi/2 + f*x)/2] - Sin[Pi/4 + (e - Pi/2 + f*x)/2])^3)*(-((AppellF1[1, -2*m, 2*m, 2, Cot[(-e + Pi/2 -
f*x)/4]^2, -Cot[(-e + Pi/2 - f*x)/4]^2]*Cot[(-e + Pi/2 - f*x)/4]^2*(Csc[(-e + Pi/2 - f*x)/4]^2)^(2*m))/(1 - C
ot[(-e + Pi/2 - f*x)/4]^2)^(2*m)) + ((Sec[(-e + Pi/2 - f*x)/4]^2)^(2*m)*(4^m*(1 + 2*m)*AppellF1[1, -2*m, 2*m,
2, Tan[(-e + Pi/2 - f*x)/4]^2, -Tan[(-e + Pi/2 - f*x)/4]^2]*Tan[(-e + Pi/2 - f*x)/4]^2 + 2*AppellF1[1 + 2*m, 2
*m, 1, 2 + 2*m, (1 - Tan[(-e + Pi/2 - f*x)/4]^2)/2, 1 - Tan[(-e + Pi/2 - f*x)/4]^2]*(1 - Tan[(-e + Pi/2 - f*x)
/4]^2)^(1 + 2*m)))/((1 + 2*m)*(2 - 2*Tan[(-e + Pi/2 - f*x)/4]^2)^(2*m))))/(4*Sqrt[2]*f*(c - c*Sin[e + f*x])^(5
/2)*(-(m*Cos[(-e + Pi/2 - f*x)/2]^(-1 + 2*m)*Sin[(-e + Pi/2 - f*x)/2]*(-((AppellF1[1, -2*m, 2*m, 2, Cot[(-e +
Pi/2 - f*x)/4]^2, -Cot[(-e + Pi/2 - f*x)/4]^2]*Cot[(-e + Pi/2 - f*x)/4]^2*(Csc[(-e + Pi/2 - f*x)/4]^2)^(2*m))/
(1 - Cot[(-e + Pi/2 - f*x)/4]^2)^(2*m)) + ((Sec[(-e + Pi/2 - f*x)/4]^2)^(2*m)*(4^m*(1 + 2*m)*AppellF1[1, -2*m,
2*m, 2, Tan[(-e + Pi/2 - f*x)/4]^2, -Tan[(-e + Pi/2 - f*x)/4]^2]*Tan[(-e + Pi/2 - f*x)/4]^2 + 2*AppellF1[1 +
2*m, 2*m, 1, 2 + 2*m, (1 - Tan[(-e + Pi/2 - f*x)/4]^2)/2, 1 - Tan[(-e + Pi/2 - f*x)/4]^2]*(1 - Tan[(-e + Pi/2
- f*x)/4]^2)^(1 + 2*m)))/((1 + 2*m)*(2 - 2*Tan[(-e + Pi/2 - f*x)/4]^2)^(2*m))))/(4*Sqrt[2]) + (Cos[(-e + Pi/2
- f*x)/2]^(2*m)*((m*AppellF1[1, -2*m, 2*m, 2, Cot[(-e + Pi/2 - f*x)/4]^2, -Cot[(-e + Pi/2 - f*x)/4]^2]*Cot[(-e
+ Pi/2 - f*x)/4]^3*(Csc[(-e + Pi/2 - f*x)/4]^2)^(2*m))/(1 - Cot[(-e + Pi/2 - f*x)/4]^2)^(2*m) + m*AppellF1[1,
-2*m, 2*m, 2, Cot[(-e + Pi/2 - f*x)/4]^2, -Cot[(-e + Pi/2 - f*x)/4]^2]*Cot[(-e + Pi/2 - f*x)/4]^3*(1 - Cot[(-
e + Pi/2 - f*x)/4]^2)^(-1 - 2*m)*(Csc[(-e + Pi/2 - f*x)/4]^2)^(1 + 2*m) + (AppellF1[1, -2*m, 2*m, 2, Cot[(-e +
Pi/2 - f*x)/4]^2, -Cot[(-e + Pi/2 - f*x)/4]^2]*Cot[(-e + Pi/2 - f*x)/4]*(Csc[(-e + Pi/2 - f*x)/4]^2)^(1 + 2*m
))/(2*(1 - Cot[(-e + Pi/2 - f*x)/4]^2)^(2*m)) - (Cot[(-e + Pi/2 - f*x)/4]^2*(Csc[(-e + Pi/2 - f*x)/4]^2)^(2*m)
*((m*AppellF1[2, 1 - 2*m, 2*m, 3, Cot[(-e + Pi/2 - f*x)/4]^2, -Cot[(-e + Pi/2 - f*x)/4]^2]*Cot[(-e + Pi/2 - f*
x)/4]*Csc[(-e + Pi/2 - f*x)/4]^2)/2 + (m*AppellF1[2, -2*m, 1 + 2*m, 3, Cot[(-e + Pi/2 - f*x)/4]^2, -Cot[(-e +
Pi/2 - f*x)/4]^2]*Cot[(-e + Pi/2 - f*x)/4]*Csc[(-e + Pi/2 - f*x)/4]^2)/2))/(1 - Cot[(-e + Pi/2 - f*x)/4]^2)^(2
*m) + (2*m*(Sec[(-e + Pi/2 - f*x)/4]^2)^(1 + 2*m)*Tan[(-e + Pi/2 - f*x)/4]*(2 - 2*Tan[(-e + Pi/2 - f*x)/4]^2)^
(-1 - 2*m)*(4^m*(1 + 2*m)*AppellF1[1, -2*m, 2*m, 2, Tan[(-e + Pi/2 - f*x)/4]^2, -Tan[(-e + Pi/2 - f*x)/4]^2]*T
an[(-e + Pi/2 - f*x)/4]^2 + 2*AppellF1[1 + 2*m, 2*m, 1, 2 + 2*m, (1 - Tan[(-e + Pi/2 - f*x)/4]^2)/2, 1 - Tan[(
-e + Pi/2 - f*x)/4]^2]*(1 - Tan[(-e + Pi/2 - f*x)/4]^2)^(1 + 2*m)))/(1 + 2*m) + (m*(Sec[(-e + Pi/2 - f*x)/4]^2
)^(2*m)*Tan[(-e + Pi/2 - f*x)/4]*(4^m*(1 + 2*m)*AppellF1[1, -2*m, 2*m, 2, Tan[(-e + Pi/2 - f*x)/4]^2, -Tan[(-e
+ Pi/2 - f*x)/4]^2]*Tan[(-e + Pi/2 - f*x)/4]^2 + 2*AppellF1[1 + 2*m, 2*m, 1, 2 + 2*m, (1 - Tan[(-e + Pi/2 - f
*x)/4]^2)/2, 1 - Tan[(-e + Pi/2 - f*x)/4]^2]*(1 - Tan[(-e + Pi/2 - f*x)/4]^2)^(1 + 2*m)))/((1 + 2*m)*(2 - 2*Ta
n[(-e + Pi/2 - f*x)/4]^2)^(2*m)) + ((Sec[(-e + Pi/2 - f*x)/4]^2)^(2*m)*(2^(-1 + 2*m)*(1 + 2*m)*AppellF1[1, -2*
m, 2*m, 2, Tan[(-e + Pi/2 - f*x)/4]^2, -Tan[(-e + Pi/2 - f*x)/4]^2]*Sec[(-e + Pi/2 - f*x)/4]^2*Tan[(-e + Pi/2
- f*x)/4] + 4^m*(1 + 2*m)*Tan[(-e + Pi/2 - f*x)/4]^2*(-(m*AppellF1[2, 1 - 2*m, 2*m, 3, Tan[(-e + Pi/2 - f*x)/4
]^2, -Tan[(-e + Pi/2 - f*x)/4]^2]*Sec[(-e + Pi/2 - f*x)/4]^2*Tan[(-e + Pi/2 - f*x)/4])/2 - (m*AppellF1[2, -2*m
, 1 + 2*m, 3, Tan[(-e + Pi/2 - f*x)/4]^2, -Tan[(-e + Pi/2 - f*x)/4]^2]*Sec[(-e + Pi/2 - f*x)/4]^2*Tan[(-e + Pi
/2 - f*x)/4])/2) - (1 + 2*m)*AppellF1[1 + 2*m, 2*m, 1, 2 + 2*m, (1 - Tan[(-e + Pi/2 - f*x)/4]^2)/2, 1 - Tan[(-
e + Pi/2 - f*x)/4]^2]*Sec[(-e + Pi/2 - f*x)/4]^2*Tan[(-e + Pi/2 - f*x)/4]*(1 - Tan[(-e + Pi/2 - f*x)/4]^2)^(2*
m) + 2*(-((1 + 2*m)*AppellF1[2 + 2*m, 2*m, 2, 3 + 2*m, (1 - Tan[(-e + Pi/2 - f*x)/4]^2)/2, 1 - Tan[(-e + Pi/2
- f*x)/4]^2]*Sec[(-e + Pi/2 - f*x)/4]^2*Tan[(-e + Pi/2 - f*x)/4])/(2*(2 + 2*m)) - (m*(1 + 2*m)*AppellF1[2 + 2*
m, 1 + 2*m, 1, 3 + 2*m, (1 - Tan[(-e + Pi/2 - f*x)/4]^2)/2, 1 - Tan[(-e + Pi/2 - f*x)/4]^2]*Sec[(-e + Pi/2 - f
*x)/4]^2*Tan[(-e + Pi/2 - f*x)/4])/(2*(2 + 2*m)))*(1 - Tan[(-e + Pi/2 - f*x)/4]^2)^(1 + 2*m)))/((1 + 2*m)*(2 -
2*Tan[(-e + Pi/2 - f*x)/4]^2)^(2*m))))/(4*Sqrt[2])))
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Maple [F] time = 0.375, size = 0, normalized size = 0. \begin{align*} \int{ \left ( \cos \left ( fx+e \right ) \right ) ^{2} \left ( a+a\sin \left ( fx+e \right ) \right ) ^{m} \left ( c-c\sin \left ( fx+e \right ) \right ) ^{-{\frac{5}{2}}}}\, dx \end{align*}
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))^(5/2),x)
[Out]
int(cos(f*x+e)^2*(a+a*sin(f*x+e))^m/(c-c*sin(f*x+e))^(5/2),x)
________________________________________________________________________________________
Maxima [F] time = 0., size = 0, normalized size = 0. \begin{align*} \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{5}{2}}}\,{d x} \end{align*}
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))^(5/2),x, algorithm="maxima")
[Out]
integrate((a*sin(f*x + e) + a)^m*cos(f*x + e)^2/(-c*sin(f*x + e) + c)^(5/2), x)
________________________________________________________________________________________
Fricas [F] time = 0., size = 0, normalized size = 0. \begin{align*}{\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}}{3 \, c^{3} \cos \left (f x + e\right )^{2} - 4 \, c^{3} -{\left (c^{3} \cos \left (f x + e\right )^{2} - 4 \, c^{3}\right )} \sin \left (f x + e\right )}, x\right ) \end{align*}
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))^(5/2),x, algorithm="fricas")
[Out]
integral(-sqrt(-c*sin(f*x + e) + c)*(a*sin(f*x + e) + a)^m*cos(f*x + e)^2/(3*c^3*cos(f*x + e)^2 - 4*c^3 - (c^3
*cos(f*x + e)^2 - 4*c^3)*sin(f*x + e)), x)
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Sympy [F(-1)] time = 0., size = 0, normalized size = 0. \begin{align*} \text{Timed out} \end{align*}
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))**(5/2),x)
[Out]
Timed out
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Giac [F] time = 0., size = 0, normalized size = 0. \begin{align*} \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{5}{2}}}\,{d x} \end{align*}
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))^(5/2),x, algorithm="giac")
[Out]
integrate((a*sin(f*x + e) + a)^m*cos(f*x + e)^2/(-c*sin(f*x + e) + c)^(5/2), x)