3.41 \(\int \frac {\cos ^2(e+f x) (a+a \sin (e+f x))^{7/2}}{(c-c \sin (e+f x))^{13/2}} \, dx\)

Optimal. Leaf size=48 \[ \frac {\cos (e+f x) (a \sin (e+f x)+a)^{9/2}}{10 a c f (c-c \sin (e+f x))^{11/2}} \]

[Out]

1/10*cos(f*x+e)*(a+a*sin(f*x+e))^(9/2)/a/c/f/(c-c*sin(f*x+e))^(11/2)

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Rubi [A]  time = 0.34, antiderivative size = 48, 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, 2742} \[ \frac {\cos (e+f x) (a \sin (e+f x)+a)^{9/2}}{10 a c f (c-c \sin (e+f x))^{11/2}} \]

Antiderivative was successfully verified.

[In]

Int[(Cos[e + f*x]^2*(a + a*Sin[e + f*x])^(7/2))/(c - c*Sin[e + f*x])^(13/2),x]

[Out]

(Cos[e + f*x]*(a + a*Sin[e + f*x])^(9/2))/(10*a*c*f*(c - c*Sin[e + f*x])^(11/2))

Rule 2742

Int[((a_) + (b_.)*sin[(e_.) + (f_.)*(x_)])^(m_)*((c_) + (d_.)*sin[(e_.) + (f_.)*(x_)])^(n_), x_Symbol] :> Simp
[(b*Cos[e + f*x]*(a + b*Sin[e + f*x])^m*(c + d*Sin[e + f*x])^n)/(a*f*(2*m + 1)), x] /; FreeQ[{a, b, c, d, e, f
, m, n}, x] && EqQ[b*c + a*d, 0] && EqQ[a^2 - b^2, 0] && EqQ[m + n + 1, 0] && NeQ[m, -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))^{7/2}}{(c-c \sin (e+f x))^{13/2}} \, dx &=\frac {\int \frac {(a+a \sin (e+f x))^{9/2}}{(c-c \sin (e+f x))^{11/2}} \, dx}{a c}\\ &=\frac {\cos (e+f x) (a+a \sin (e+f x))^{9/2}}{10 a c f (c-c \sin (e+f x))^{11/2}}\\ \end {align*}

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Mathematica [B]  time = 6.74, size = 412, normalized size = 8.58 \[ \frac {(a (\sin (e+f x)+1))^{7/2} \left (\cos \left (\frac {1}{2} (e+f x)\right )-\sin \left (\frac {1}{2} (e+f x)\right )\right )^{11}}{f (c-c \sin (e+f x))^{13/2} \left (\sin \left (\frac {1}{2} (e+f x)\right )+\cos \left (\frac {1}{2} (e+f x)\right )\right )^7}-\frac {4 (a (\sin (e+f x)+1))^{7/2} \left (\cos \left (\frac {1}{2} (e+f x)\right )-\sin \left (\frac {1}{2} (e+f x)\right )\right )^9}{f (c-c \sin (e+f x))^{13/2} \left (\sin \left (\frac {1}{2} (e+f x)\right )+\cos \left (\frac {1}{2} (e+f x)\right )\right )^7}+\frac {8 (a (\sin (e+f x)+1))^{7/2} \left (\cos \left (\frac {1}{2} (e+f x)\right )-\sin \left (\frac {1}{2} (e+f x)\right )\right )^7}{f (c-c \sin (e+f x))^{13/2} \left (\sin \left (\frac {1}{2} (e+f x)\right )+\cos \left (\frac {1}{2} (e+f x)\right )\right )^7}-\frac {8 (a (\sin (e+f x)+1))^{7/2} \left (\cos \left (\frac {1}{2} (e+f x)\right )-\sin \left (\frac {1}{2} (e+f x)\right )\right )^5}{f (c-c \sin (e+f x))^{13/2} \left (\sin \left (\frac {1}{2} (e+f x)\right )+\cos \left (\frac {1}{2} (e+f x)\right )\right )^7}+\frac {16 (a (\sin (e+f x)+1))^{7/2} \left (\cos \left (\frac {1}{2} (e+f x)\right )-\sin \left (\frac {1}{2} (e+f x)\right )\right )^3}{5 f (c-c \sin (e+f x))^{13/2} \left (\sin \left (\frac {1}{2} (e+f x)\right )+\cos \left (\frac {1}{2} (e+f x)\right )\right )^7} \]

Antiderivative was successfully verified.

[In]

Integrate[(Cos[e + f*x]^2*(a + a*Sin[e + f*x])^(7/2))/(c - c*Sin[e + f*x])^(13/2),x]

[Out]

(16*(Cos[(e + f*x)/2] - Sin[(e + f*x)/2])^3*(a*(1 + Sin[e + f*x]))^(7/2))/(5*f*(Cos[(e + f*x)/2] + Sin[(e + f*
x)/2])^7*(c - c*Sin[e + f*x])^(13/2)) - (8*(Cos[(e + f*x)/2] - Sin[(e + f*x)/2])^5*(a*(1 + Sin[e + f*x]))^(7/2
))/(f*(Cos[(e + f*x)/2] + Sin[(e + f*x)/2])^7*(c - c*Sin[e + f*x])^(13/2)) + (8*(Cos[(e + f*x)/2] - Sin[(e + f
*x)/2])^7*(a*(1 + Sin[e + f*x]))^(7/2))/(f*(Cos[(e + f*x)/2] + Sin[(e + f*x)/2])^7*(c - c*Sin[e + f*x])^(13/2)
) - (4*(Cos[(e + f*x)/2] - Sin[(e + f*x)/2])^9*(a*(1 + Sin[e + f*x]))^(7/2))/(f*(Cos[(e + f*x)/2] + Sin[(e + f
*x)/2])^7*(c - c*Sin[e + f*x])^(13/2)) + ((Cos[(e + f*x)/2] - Sin[(e + f*x)/2])^11*(a*(1 + Sin[e + f*x]))^(7/2
))/(f*(Cos[(e + f*x)/2] + Sin[(e + f*x)/2])^7*(c - c*Sin[e + f*x])^(13/2))

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fricas [B]  time = 0.47, size = 150, normalized size = 3.12 \[ \frac {{\left (5 \, a^{3} \cos \left (f x + e\right )^{4} - 20 \, a^{3} \cos \left (f x + e\right )^{2} + 16 \, a^{3}\right )} \sqrt {a \sin \left (f x + e\right ) + a} \sqrt {-c \sin \left (f x + e\right ) + c}}{5 \, {\left (5 \, c^{7} f \cos \left (f x + e\right )^{5} - 20 \, c^{7} f \cos \left (f x + e\right )^{3} + 16 \, c^{7} f \cos \left (f x + e\right ) - {\left (c^{7} f \cos \left (f x + e\right )^{5} - 12 \, c^{7} f \cos \left (f x + e\right )^{3} + 16 \, c^{7} f \cos \left (f x + e\right )\right )} \sin \left (f x + e\right )\right )}} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate(cos(f*x+e)^2*(a+a*sin(f*x+e))^(7/2)/(c-c*sin(f*x+e))^(13/2),x, algorithm="fricas")

[Out]

1/5*(5*a^3*cos(f*x + e)^4 - 20*a^3*cos(f*x + e)^2 + 16*a^3)*sqrt(a*sin(f*x + e) + a)*sqrt(-c*sin(f*x + e) + c)
/(5*c^7*f*cos(f*x + e)^5 - 20*c^7*f*cos(f*x + e)^3 + 16*c^7*f*cos(f*x + e) - (c^7*f*cos(f*x + e)^5 - 12*c^7*f*
cos(f*x + e)^3 + 16*c^7*f*cos(f*x + e))*sin(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))^(7/2)/(c-c*sin(f*x+e))^(13/2),x, algorithm="giac")

[Out]

Timed out

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maple [B]  time = 0.36, size = 188, normalized size = 3.92 \[ \frac {\left (\cos ^{4}\left (f x +e \right )-12 \left (\cos ^{2}\left (f x +e \right )\right )+16\right ) \left (a \left (1+\sin \left (f x +e \right )\right )\right )^{\frac {7}{2}} \sin \left (f x +e \right ) \left (\cos ^{2}\left (f x +e \right )-\sin \left (f x +e \right ) \cos \left (f x +e \right )+\cos \left (f x +e \right )+2 \sin \left (f x +e \right )-2\right )}{5 f \left (\sin \left (f x +e \right ) \left (\cos ^{3}\left (f x +e \right )\right )+\cos ^{4}\left (f x +e \right )-4 \left (\cos ^{2}\left (f x +e \right )\right ) \sin \left (f x +e \right )+3 \left (\cos ^{3}\left (f x +e \right )\right )-4 \sin \left (f x +e \right ) \cos \left (f x +e \right )-8 \left (\cos ^{2}\left (f x +e \right )\right )+8 \sin \left (f x +e \right )-4 \cos \left (f x +e \right )+8\right ) \left (-c \left (\sin \left (f x +e \right )-1\right )\right )^{\frac {13}{2}}} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

int(cos(f*x+e)^2*(a+a*sin(f*x+e))^(7/2)/(c-c*sin(f*x+e))^(13/2),x)

[Out]

1/5/f*(cos(f*x+e)^4-12*cos(f*x+e)^2+16)*(a*(1+sin(f*x+e)))^(7/2)*sin(f*x+e)*(cos(f*x+e)^2-sin(f*x+e)*cos(f*x+e
)+cos(f*x+e)+2*sin(f*x+e)-2)/(sin(f*x+e)*cos(f*x+e)^3+cos(f*x+e)^4-4*cos(f*x+e)^2*sin(f*x+e)+3*cos(f*x+e)^3-4*
sin(f*x+e)*cos(f*x+e)-8*cos(f*x+e)^2+8*sin(f*x+e)-4*cos(f*x+e)+8)/(-c*(sin(f*x+e)-1))^(13/2)

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maxima [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))^(7/2)/(c-c*sin(f*x+e))^(13/2),x, algorithm="maxima")

[Out]

Timed out

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mupad [F]  time = 0.00, size = -1, normalized size = -0.02 \[ \int \frac {{\cos \left (e+f\,x\right )}^2\,{\left (a+a\,\sin \left (e+f\,x\right )\right )}^{7/2}}{{\left (c-c\,\sin \left (e+f\,x\right )\right )}^{13/2}} \,d x \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

int((cos(e + f*x)^2*(a + a*sin(e + f*x))^(7/2))/(c - c*sin(e + f*x))^(13/2),x)

[Out]

int((cos(e + f*x)^2*(a + a*sin(e + f*x))^(7/2))/(c - c*sin(e + f*x))^(13/2), 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))**(7/2)/(c-c*sin(f*x+e))**(13/2),x)

[Out]

Timed out

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