3.29 \(\int \frac {\cos ^2(e+f x) (a+a \sin (e+f x))^{5/2}}{(c-c \sin (e+f x))^{13/2}} \, dx\)

Optimal. Leaf size=97 \[ \frac {\cos (e+f x) (a \sin (e+f x)+a)^{7/2}}{80 a c^2 f (c-c \sin (e+f x))^{9/2}}+\frac {\cos (e+f x) (a \sin (e+f x)+a)^{7/2}}{10 a c f (c-c \sin (e+f x))^{11/2}} \]

[Out]

1/10*cos(f*x+e)*(a+a*sin(f*x+e))^(7/2)/a/c/f/(c-c*sin(f*x+e))^(11/2)+1/80*cos(f*x+e)*(a+a*sin(f*x+e))^(7/2)/a/
c^2/f/(c-c*sin(f*x+e))^(9/2)

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Rubi [A]  time = 0.44, antiderivative size = 97, normalized size of antiderivative = 1.00, number of steps used = 3, number of rules used = 3, integrand size = 38, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.079, Rules used = {2841, 2743, 2742} \[ \frac {\cos (e+f x) (a \sin (e+f x)+a)^{7/2}}{80 a c^2 f (c-c \sin (e+f x))^{9/2}}+\frac {\cos (e+f x) (a \sin (e+f x)+a)^{7/2}}{10 a c f (c-c \sin (e+f x))^{11/2}} \]

Antiderivative was successfully verified.

[In]

Int[(Cos[e + f*x]^2*(a + a*Sin[e + f*x])^(5/2))/(c - c*Sin[e + f*x])^(13/2),x]

[Out]

(Cos[e + f*x]*(a + a*Sin[e + f*x])^(7/2))/(10*a*c*f*(c - c*Sin[e + f*x])^(11/2)) + (Cos[e + f*x]*(a + a*Sin[e
+ f*x])^(7/2))/(80*a*c^2*f*(c - c*Sin[e + f*x])^(9/2))

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 2743

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] + Dist[(m + n + 1)/(a*(2*m
+ 1)), Int[(a + b*Sin[e + f*x])^(m + 1)*(c + d*Sin[e + f*x])^n, x], x] /; FreeQ[{a, b, c, d, e, f, m, n}, x]
&& EqQ[b*c + a*d, 0] && EqQ[a^2 - b^2, 0] && ILtQ[Simplify[m + n + 1], 0] && NeQ[m, -2^(-1)] && (SumSimplerQ[m
, 1] ||  !SumSimplerQ[n, 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))^{5/2}}{(c-c \sin (e+f x))^{13/2}} \, dx &=\frac {\int \frac {(a+a \sin (e+f x))^{7/2}}{(c-c \sin (e+f x))^{11/2}} \, dx}{a c}\\ &=\frac {\cos (e+f x) (a+a \sin (e+f x))^{7/2}}{10 a c f (c-c \sin (e+f x))^{11/2}}+\frac {\int \frac {(a+a \sin (e+f x))^{7/2}}{(c-c \sin (e+f x))^{9/2}} \, dx}{10 a c^2}\\ &=\frac {\cos (e+f x) (a+a \sin (e+f x))^{7/2}}{10 a c f (c-c \sin (e+f x))^{11/2}}+\frac {\cos (e+f x) (a+a \sin (e+f x))^{7/2}}{80 a c^2 f (c-c \sin (e+f x))^{9/2}}\\ \end {align*}

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Mathematica [A]  time = 6.27, size = 130, normalized size = 1.34 \[ \frac {a^2 \sqrt {a (\sin (e+f x)+1)} \left (\cos \left (\frac {1}{2} (e+f x)\right )-\sin \left (\frac {1}{2} (e+f x)\right )\right )^3 (35 \sin (e+f x)-5 \sin (3 (e+f x))-10 \cos (2 (e+f x))+14)}{40 c^6 f (\sin (e+f x)-1)^6 \sqrt {c-c \sin (e+f x)} \left (\sin \left (\frac {1}{2} (e+f x)\right )+\cos \left (\frac {1}{2} (e+f x)\right )\right )} \]

Antiderivative was successfully verified.

[In]

Integrate[(Cos[e + f*x]^2*(a + a*Sin[e + f*x])^(5/2))/(c - c*Sin[e + f*x])^(13/2),x]

[Out]

(a^2*(Cos[(e + f*x)/2] - Sin[(e + f*x)/2])^3*Sqrt[a*(1 + Sin[e + f*x])]*(14 - 10*Cos[2*(e + f*x)] + 35*Sin[e +
f*x] - 5*Sin[3*(e + f*x)]))/(40*c^6*f*(Cos[(e + f*x)/2] + Sin[(e + f*x)/2])*(-1 + Sin[e + f*x])^6*Sqrt[c - c*
Sin[e + f*x]])

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fricas [A]  time = 0.53, size = 163, normalized size = 1.68 \[ -\frac {{\left (5 \, a^{2} \cos \left (f x + e\right )^{2} - 6 \, a^{2} + 5 \, {\left (a^{2} \cos \left (f x + e\right )^{2} - 2 \, a^{2}\right )} \sin \left (f x + e\right )\right )} \sqrt {a \sin \left (f x + e\right ) + a} \sqrt {-c \sin \left (f x + e\right ) + c}}{10 \, {\left (5 \, c^{7} f \cos \left (f x + e\right )^{5} - 20 \, c^{7} f \cos \left (f x + e\right )^{3} + 16 \, c^{7} f \cos \left (f x + e\right ) - {\left (c^{7} f \cos \left (f x + e\right )^{5} - 12 \, c^{7} f \cos \left (f x + e\right )^{3} + 16 \, c^{7} f \cos \left (f x + e\right )\right )} \sin \left (f x + e\right )\right )}} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate(cos(f*x+e)^2*(a+a*sin(f*x+e))^(5/2)/(c-c*sin(f*x+e))^(13/2),x, algorithm="fricas")

[Out]

-1/10*(5*a^2*cos(f*x + e)^2 - 6*a^2 + 5*(a^2*cos(f*x + e)^2 - 2*a^2)*sin(f*x + e))*sqrt(a*sin(f*x + e) + a)*sq
rt(-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))^(5/2)/(c-c*sin(f*x+e))^(13/2),x, algorithm="giac")

[Out]

Timed out

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maple [B]  time = 0.37, size = 187, normalized size = 1.93 \[ \frac {\left (\cos ^{4}\left (f x +e \right )+5 \left (\cos ^{2}\left (f x +e \right )\right ) \sin \left (f x +e \right )-17 \left (\cos ^{2}\left (f x +e \right )\right )-10 \sin \left (f x +e \right )+26\right ) \left (a \left (1+\sin \left (f x +e \right )\right )\right )^{\frac {5}{2}} \sin \left (f x +e \right ) \left (\cos ^{2}\left (f x +e \right )-\sin \left (f x +e \right ) \cos \left (f x +e \right )+\cos \left (f x +e \right )+2 \sin \left (f x +e \right )-2\right )}{10 f \left (\cos ^{3}\left (f x +e \right )-\left (\cos ^{2}\left (f x +e \right )\right ) \sin \left (f x +e \right )-3 \left (\cos ^{2}\left (f x +e \right )\right )-2 \sin \left (f x +e \right ) \cos \left (f x +e \right )-2 \cos \left (f x +e \right )+4 \sin \left (f x +e \right )+4\right ) \left (-c \left (\sin \left (f x +e \right )-1\right )\right )^{\frac {13}{2}}} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

int(cos(f*x+e)^2*(a+a*sin(f*x+e))^(5/2)/(c-c*sin(f*x+e))^(13/2),x)

[Out]

1/10/f*(cos(f*x+e)^4+5*cos(f*x+e)^2*sin(f*x+e)-17*cos(f*x+e)^2-10*sin(f*x+e)+26)*(a*(1+sin(f*x+e)))^(5/2)*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)/(cos(f*x+e)^3-cos(f*x+e)^2*sin(f*x+e)-3*
cos(f*x+e)^2-2*sin(f*x+e)*cos(f*x+e)-2*cos(f*x+e)+4*sin(f*x+e)+4)/(-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))^(5/2)/(c-c*sin(f*x+e))^(13/2),x, algorithm="maxima")

[Out]

Timed out

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mupad [B]  time = 14.86, size = 317, normalized size = 3.27 \[ \frac {\sqrt {c-c\,\sin \left (e+f\,x\right )}\,\left (\frac {a^2\,{\mathrm {e}}^{e\,6{}\mathrm {i}+f\,x\,6{}\mathrm {i}}\,\sqrt {a+a\,\sin \left (e+f\,x\right )}\,112{}\mathrm {i}}{5\,c^7\,f}+\frac {a^2\,{\mathrm {e}}^{e\,6{}\mathrm {i}+f\,x\,6{}\mathrm {i}}\,\sin \left (e+f\,x\right )\,\sqrt {a+a\,\sin \left (e+f\,x\right )}\,56{}\mathrm {i}}{c^7\,f}-\frac {a^2\,{\mathrm {e}}^{e\,6{}\mathrm {i}+f\,x\,6{}\mathrm {i}}\,\cos \left (2\,e+2\,f\,x\right )\,\sqrt {a+a\,\sin \left (e+f\,x\right )}\,16{}\mathrm {i}}{c^7\,f}-\frac {a^2\,{\mathrm {e}}^{e\,6{}\mathrm {i}+f\,x\,6{}\mathrm {i}}\,\sin \left (3\,e+3\,f\,x\right )\,\sqrt {a+a\,\sin \left (e+f\,x\right )}\,8{}\mathrm {i}}{c^7\,f}\right )}{\cos \left (e+f\,x\right )\,{\mathrm {e}}^{e\,6{}\mathrm {i}+f\,x\,6{}\mathrm {i}}\,264{}\mathrm {i}-{\mathrm {e}}^{e\,6{}\mathrm {i}+f\,x\,6{}\mathrm {i}}\,\cos \left (3\,e+3\,f\,x\right )\,220{}\mathrm {i}+{\mathrm {e}}^{e\,6{}\mathrm {i}+f\,x\,6{}\mathrm {i}}\,\cos \left (5\,e+5\,f\,x\right )\,20{}\mathrm {i}-{\mathrm {e}}^{e\,6{}\mathrm {i}+f\,x\,6{}\mathrm {i}}\,\sin \left (2\,e+2\,f\,x\right )\,330{}\mathrm {i}+{\mathrm {e}}^{e\,6{}\mathrm {i}+f\,x\,6{}\mathrm {i}}\,\sin \left (4\,e+4\,f\,x\right )\,88{}\mathrm {i}-{\mathrm {e}}^{e\,6{}\mathrm {i}+f\,x\,6{}\mathrm {i}}\,\sin \left (6\,e+6\,f\,x\right )\,2{}\mathrm {i}} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

int((cos(e + f*x)^2*(a + a*sin(e + f*x))^(5/2))/(c - c*sin(e + f*x))^(13/2),x)

[Out]

((c - c*sin(e + f*x))^(1/2)*((a^2*exp(e*6i + f*x*6i)*(a + a*sin(e + f*x))^(1/2)*112i)/(5*c^7*f) + (a^2*exp(e*6
i + f*x*6i)*sin(e + f*x)*(a + a*sin(e + f*x))^(1/2)*56i)/(c^7*f) - (a^2*exp(e*6i + f*x*6i)*cos(2*e + 2*f*x)*(a
+ a*sin(e + f*x))^(1/2)*16i)/(c^7*f) - (a^2*exp(e*6i + f*x*6i)*sin(3*e + 3*f*x)*(a + a*sin(e + f*x))^(1/2)*8i
)/(c^7*f)))/(cos(e + f*x)*exp(e*6i + f*x*6i)*264i - exp(e*6i + f*x*6i)*cos(3*e + 3*f*x)*220i + exp(e*6i + f*x*
6i)*cos(5*e + 5*f*x)*20i - exp(e*6i + f*x*6i)*sin(2*e + 2*f*x)*330i + exp(e*6i + f*x*6i)*sin(4*e + 4*f*x)*88i
- exp(e*6i + f*x*6i)*sin(6*e + 6*f*x)*2i)

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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))**(5/2)/(c-c*sin(f*x+e))**(13/2),x)

[Out]

Timed out

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