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

Optimal. Leaf size=45 \[ \frac {\cos (e+f x) (a \sin (e+f x)+a)^{5/2}}{3 a f \sqrt {c-c \sin (e+f x)}} \]

[Out]

1/3*cos(f*x+e)*(a+a*sin(f*x+e))^(5/2)/a/f/(c-c*sin(f*x+e))^(1/2)

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Rubi [A]  time = 0.31, antiderivative size = 45, normalized size of antiderivative = 1.00, number of steps used = 2, number of rules used = 2, integrand size = 38, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.053, Rules used = {2841, 2738} \[ \frac {\cos (e+f x) (a \sin (e+f x)+a)^{5/2}}{3 a f \sqrt {c-c \sin (e+f x)}} \]

Antiderivative was successfully verified.

[In]

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

[Out]

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

Rule 2738

Int[Sqrt[(a_) + (b_.)*sin[(e_.) + (f_.)*(x_)]]*((c_) + (d_.)*sin[(e_.) + (f_.)*(x_)])^(n_), x_Symbol] :> Simp[
(-2*b*Cos[e + f*x]*(c + d*Sin[e + f*x])^n)/(f*(2*n + 1)*Sqrt[a + b*Sin[e + f*x]]), x] /; FreeQ[{a, b, c, d, e,
f, n}, x] && EqQ[b*c + a*d, 0] && EqQ[a^2 - b^2, 0] && NeQ[n, -2^(-1)]

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))^{3/2}}{\sqrt {c-c \sin (e+f x)}} \, dx &=\frac {\int (a+a \sin (e+f x))^{5/2} \sqrt {c-c \sin (e+f x)} \, dx}{a c}\\ &=\frac {\cos (e+f x) (a+a \sin (e+f x))^{5/2}}{3 a f \sqrt {c-c \sin (e+f x)}}\\ \end {align*}

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Mathematica [B]  time = 0.55, size = 111, normalized size = 2.47 \[ \frac {(a (\sin (e+f x)+1))^{3/2} \left (\cos \left (\frac {1}{2} (e+f x)\right )-\sin \left (\frac {1}{2} (e+f x)\right )\right ) (15 \sin (e+f x)-\sin (3 (e+f x))-6 \cos (2 (e+f x)))}{12 f \sqrt {c-c \sin (e+f x)} \left (\sin \left (\frac {1}{2} (e+f x)\right )+\cos \left (\frac {1}{2} (e+f x)\right )\right )^3} \]

Antiderivative was successfully verified.

[In]

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

[Out]

((Cos[(e + f*x)/2] - Sin[(e + f*x)/2])*(a*(1 + Sin[e + f*x]))^(3/2)*(-6*Cos[2*(e + f*x)] + 15*Sin[e + f*x] - S
in[3*(e + f*x)]))/(12*f*(Cos[(e + f*x)/2] + Sin[(e + f*x)/2])^3*Sqrt[c - c*Sin[e + f*x]])

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fricas [A]  time = 0.44, size = 77, normalized size = 1.71 \[ -\frac {{\left (3 \, a \cos \left (f x + e\right )^{2} + {\left (a \cos \left (f x + e\right )^{2} - 4 \, a\right )} \sin \left (f x + e\right ) - 3 \, a\right )} \sqrt {a \sin \left (f x + e\right ) + a} \sqrt {-c \sin \left (f x + e\right ) + c}}{3 \, c f \cos \left (f x + e\right )} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate(cos(f*x+e)^2*(a+a*sin(f*x+e))^(3/2)/(c-c*sin(f*x+e))^(1/2),x, algorithm="fricas")

[Out]

-1/3*(3*a*cos(f*x + e)^2 + (a*cos(f*x + e)^2 - 4*a)*sin(f*x + e) - 3*a)*sqrt(a*sin(f*x + e) + a)*sqrt(-c*sin(f
*x + e) + c)/(c*f*cos(f*x + e))

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giac [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))^(3/2)/(c-c*sin(f*x+e))^(1/2),x, algorithm="giac")

[Out]

Timed out

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maple [B]  time = 0.36, size = 141, normalized size = 3.13 \[ \frac {\sin \left (f x +e \right ) \left (a \left (1+\sin \left (f x +e \right )\right )\right )^{\frac {3}{2}} \left (\cos ^{3}\left (f x +e \right )+\left (\cos ^{2}\left (f x +e \right )\right ) \sin \left (f x +e \right )+2 \left (\cos ^{2}\left (f x +e \right )\right )-3 \sin \left (f x +e \right ) \cos \left (f x +e \right )-4 \cos \left (f x +e \right )-\sin \left (f x +e \right )+1\right )}{3 f \sqrt {-c \left (\sin \left (f x +e \right )-1\right )}\, \left (\cos ^{2}\left (f x +e \right )+\sin \left (f x +e \right ) \cos \left (f x +e \right )+\cos \left (f x +e \right )-2 \sin \left (f x +e \right )-2\right )} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

1/3/f*sin(f*x+e)*(a*(1+sin(f*x+e)))^(3/2)*(cos(f*x+e)^3+cos(f*x+e)^2*sin(f*x+e)+2*cos(f*x+e)^2-3*sin(f*x+e)*co
s(f*x+e)-4*cos(f*x+e)-sin(f*x+e)+1)/(-c*(sin(f*x+e)-1))^(1/2)/(cos(f*x+e)^2+sin(f*x+e)*cos(f*x+e)+cos(f*x+e)-2
*sin(f*x+e)-2)

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

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate(cos(f*x+e)^2*(a+a*sin(f*x+e))^(3/2)/(c-c*sin(f*x+e))^(1/2),x, algorithm="maxima")

[Out]

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

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mupad [B]  time = 9.40, size = 87, normalized size = 1.93 \[ -\frac {a\,\sqrt {a\,\left (\sin \left (e+f\,x\right )+1\right )}\,\sqrt {-c\,\left (\sin \left (e+f\,x\right )-1\right )}\,\left (6\,\cos \left (e+f\,x\right )+6\,\cos \left (3\,e+3\,f\,x\right )-14\,\sin \left (2\,e+2\,f\,x\right )+\sin \left (4\,e+4\,f\,x\right )\right )}{12\,c\,f\,\left (\cos \left (2\,e+2\,f\,x\right )+1\right )} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

-(a*(a*(sin(e + f*x) + 1))^(1/2)*(-c*(sin(e + f*x) - 1))^(1/2)*(6*cos(e + f*x) + 6*cos(3*e + 3*f*x) - 14*sin(2
*e + 2*f*x) + sin(4*e + 4*f*x)))/(12*c*f*(cos(2*e + 2*f*x) + 1))

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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))**(3/2)/(c-c*sin(f*x+e))**(1/2),x)

[Out]

Timed out

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