3.42 \(\int \frac{\cos ^2(e+f x) (a+a \sin (e+f x))^{7/2}}{(c-c \sin (e+f x))^{15/2}} \, dx\)

Optimal. Leaf size=97 \[ \frac{\cos (e+f x) (a \sin (e+f x)+a)^{9/2}}{120 a c^2 f (c-c \sin (e+f x))^{11/2}}+\frac{\cos (e+f x) (a \sin (e+f x)+a)^{9/2}}{12 a c f (c-c \sin (e+f x))^{13/2}} \]

[Out]

(Cos[e + f*x]*(a + a*Sin[e + f*x])^(9/2))/(12*a*c*f*(c - c*Sin[e + f*x])^(13/2)) + (Cos[e + f*x]*(a + a*Sin[e
+ f*x])^(9/2))/(120*a*c^2*f*(c - c*Sin[e + f*x])^(11/2))

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Rubi [A]  time = 0.444469, antiderivative size = 97, normalized size of antiderivative = 1., 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} \[ \frac{\cos (e+f x) (a \sin (e+f x)+a)^{9/2}}{120 a c^2 f (c-c \sin (e+f x))^{11/2}}+\frac{\cos (e+f x) (a \sin (e+f x)+a)^{9/2}}{12 a c f (c-c \sin (e+f x))^{13/2}} \]

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])^(15/2),x]

[Out]

(Cos[e + f*x]*(a + a*Sin[e + f*x])^(9/2))/(12*a*c*f*(c - c*Sin[e + f*x])^(13/2)) + (Cos[e + f*x]*(a + a*Sin[e
+ f*x])^(9/2))/(120*a*c^2*f*(c - c*Sin[e + f*x])^(11/2))

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 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 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)]

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))^{15/2}} \, dx &=\frac{\int \frac{(a+a \sin (e+f x))^{9/2}}{(c-c \sin (e+f x))^{13/2}} \, dx}{a c}\\ &=\frac{\cos (e+f x) (a+a \sin (e+f x))^{9/2}}{12 a c f (c-c \sin (e+f x))^{13/2}}+\frac{\int \frac{(a+a \sin (e+f x))^{9/2}}{(c-c \sin (e+f x))^{11/2}} \, dx}{12 a c^2}\\ &=\frac{\cos (e+f x) (a+a \sin (e+f x))^{9/2}}{12 a c f (c-c \sin (e+f x))^{13/2}}+\frac{\cos (e+f x) (a+a \sin (e+f x))^{9/2}}{120 a c^2 f (c-c \sin (e+f x))^{11/2}}\\ \end{align*}

Mathematica [B]  time = 6.75914, size = 419, normalized size = 4.32 \[ \frac{(a (\sin (e+f x)+1))^{7/2} \left (\cos \left (\frac{1}{2} (e+f x)\right )-\sin \left (\frac{1}{2} (e+f x)\right )\right )^{11}}{2 f (c-c \sin (e+f x))^{15/2} \left (\sin \left (\frac{1}{2} (e+f x)\right )+\cos \left (\frac{1}{2} (e+f x)\right )\right )^7}-\frac{8 (a (\sin (e+f x)+1))^{7/2} \left (\cos \left (\frac{1}{2} (e+f x)\right )-\sin \left (\frac{1}{2} (e+f x)\right )\right )^9}{3 f (c-c \sin (e+f x))^{15/2} \left (\sin \left (\frac{1}{2} (e+f x)\right )+\cos \left (\frac{1}{2} (e+f x)\right )\right )^7}+\frac{6 (a (\sin (e+f x)+1))^{7/2} \left (\cos \left (\frac{1}{2} (e+f x)\right )-\sin \left (\frac{1}{2} (e+f x)\right )\right )^7}{f (c-c \sin (e+f x))^{15/2} \left (\sin \left (\frac{1}{2} (e+f x)\right )+\cos \left (\frac{1}{2} (e+f x)\right )\right )^7}-\frac{32 (a (\sin (e+f x)+1))^{7/2} \left (\cos \left (\frac{1}{2} (e+f x)\right )-\sin \left (\frac{1}{2} (e+f x)\right )\right )^5}{5 f (c-c \sin (e+f x))^{15/2} \left (\sin \left (\frac{1}{2} (e+f x)\right )+\cos \left (\frac{1}{2} (e+f x)\right )\right )^7}+\frac{8 (a (\sin (e+f x)+1))^{7/2} \left (\cos \left (\frac{1}{2} (e+f x)\right )-\sin \left (\frac{1}{2} (e+f x)\right )\right )^3}{3 f (c-c \sin (e+f x))^{15/2} \left (\sin \left (\frac{1}{2} (e+f x)\right )+\cos \left (\frac{1}{2} (e+f x)\right )\right )^7} \]

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])^(15/2),x]

[Out]

(8*(Cos[(e + f*x)/2] - Sin[(e + f*x)/2])^3*(a*(1 + Sin[e + f*x]))^(7/2))/(3*f*(Cos[(e + f*x)/2] + Sin[(e + f*x
)/2])^7*(c - c*Sin[e + f*x])^(15/2)) - (32*(Cos[(e + f*x)/2] - Sin[(e + f*x)/2])^5*(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])^(15/2)) + (6*(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])^(15/
2)) - (8*(Cos[(e + f*x)/2] - Sin[(e + f*x)/2])^9*(a*(1 + Sin[e + f*x]))^(7/2))/(3*f*(Cos[(e + f*x)/2] + Sin[(e
+ f*x)/2])^7*(c - c*Sin[e + f*x])^(15/2)) + ((Cos[(e + f*x)/2] - Sin[(e + f*x)/2])^11*(a*(1 + Sin[e + f*x]))^
(7/2))/(2*f*(Cos[(e + f*x)/2] + Sin[(e + f*x)/2])^7*(c - c*Sin[e + f*x])^(15/2))

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Maple [B]  time = 0.225, size = 233, normalized size = 2.4 \begin{align*}{\frac{ \left ( 3\,\sin \left ( fx+e \right ) \left ( \cos \left ( fx+e \right ) \right ) ^{4}-18\, \left ( \cos \left ( fx+e \right ) \right ) ^{4}-36\, \left ( \cos \left ( fx+e \right ) \right ) ^{2}\sin \left ( fx+e \right ) +116\, \left ( \cos \left ( fx+e \right ) \right ) ^{2}+48\,\sin \left ( fx+e \right ) -128 \right ) \sin \left ( fx+e \right ) \left ( \sin \left ( fx+e \right ) \cos \left ( fx+e \right ) - \left ( \cos \left ( fx+e \right ) \right ) ^{2}-2\,\sin \left ( fx+e \right ) -\cos \left ( fx+e \right ) +2 \right ) }{30\,f \left ( \sin \left ( fx+e \right ) \left ( \cos \left ( fx+e \right ) \right ) ^{3}+ \left ( \cos \left ( fx+e \right ) \right ) ^{4}-4\, \left ( \cos \left ( fx+e \right ) \right ) ^{2}\sin \left ( fx+e \right ) +3\, \left ( \cos \left ( fx+e \right ) \right ) ^{3}-4\,\sin \left ( fx+e \right ) \cos \left ( fx+e \right ) -8\, \left ( \cos \left ( fx+e \right ) \right ) ^{2}+8\,\sin \left ( fx+e \right ) -4\,\cos \left ( fx+e \right ) +8 \right ) } \left ( a \left ( 1+\sin \left ( fx+e \right ) \right ) \right ) ^{{\frac{7}{2}}} \left ( -c \left ( -1+\sin \left ( fx+e \right ) \right ) \right ) ^{-{\frac{15}{2}}}} \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))^(7/2)/(c-c*sin(f*x+e))^(15/2),x)

[Out]

1/30/f*(3*sin(f*x+e)*cos(f*x+e)^4-18*cos(f*x+e)^4-36*cos(f*x+e)^2*sin(f*x+e)+116*cos(f*x+e)^2+48*sin(f*x+e)-12
8)*(a*(1+sin(f*x+e)))^(7/2)*sin(f*x+e)*(sin(f*x+e)*cos(f*x+e)-cos(f*x+e)^2-2*sin(f*x+e)-cos(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*(-1+sin(f*x+e)))^(15/2)

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Maxima [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))^(7/2)/(c-c*sin(f*x+e))^(15/2),x, algorithm="maxima")

[Out]

Timed out

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Fricas [B]  time = 1.91221, size = 478, normalized size = 4.93 \begin{align*} -\frac{{\left (15 \, a^{3} \cos \left (f x + e\right )^{4} - 60 \, a^{3} \cos \left (f x + e\right )^{2} + 48 \, a^{3} - 4 \,{\left (5 \, a^{3} \cos \left (f x + e\right )^{2} - 8 \, a^{3}\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}}{30 \,{\left (c^{8} f \cos \left (f x + e\right )^{7} - 18 \, c^{8} f \cos \left (f x + e\right )^{5} + 48 \, c^{8} f \cos \left (f x + e\right )^{3} - 32 \, c^{8} f \cos \left (f x + e\right ) + 2 \,{\left (3 \, c^{8} f \cos \left (f x + e\right )^{5} - 16 \, c^{8} f \cos \left (f x + e\right )^{3} + 16 \, c^{8} f \cos \left (f x + e\right )\right )} \sin \left (f x + e\right )\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))^(7/2)/(c-c*sin(f*x+e))^(15/2),x, algorithm="fricas")

[Out]

-1/30*(15*a^3*cos(f*x + e)^4 - 60*a^3*cos(f*x + e)^2 + 48*a^3 - 4*(5*a^3*cos(f*x + e)^2 - 8*a^3)*sin(f*x + e))
*sqrt(a*sin(f*x + e) + a)*sqrt(-c*sin(f*x + e) + c)/(c^8*f*cos(f*x + e)^7 - 18*c^8*f*cos(f*x + e)^5 + 48*c^8*f
*cos(f*x + e)^3 - 32*c^8*f*cos(f*x + e) + 2*(3*c^8*f*cos(f*x + e)^5 - 16*c^8*f*cos(f*x + e)^3 + 16*c^8*f*cos(f
*x + e))*sin(f*x + e))

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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))**(7/2)/(c-c*sin(f*x+e))**(15/2),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 )}^{\frac{7}{2}} \cos \left (f x + e\right )^{2}}{{\left (-c \sin \left (f x + e\right ) + c\right )}^{\frac{15}{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))^(7/2)/(c-c*sin(f*x+e))^(15/2),x, algorithm="giac")

[Out]

integrate((a*sin(f*x + e) + a)^(7/2)*cos(f*x + e)^2/(-c*sin(f*x + e) + c)^(15/2), x)