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

Optimal. Leaf size=92 \[ -\frac{\cos (e+f x) \sqrt{a \sin (e+f x)+a} (c-c \sin (e+f x))^{9/2}}{6 c f}-\frac{a \cos (e+f x) (c-c \sin (e+f x))^{9/2}}{15 c f \sqrt{a \sin (e+f x)+a}} \]

[Out]

-(a*Cos[e + f*x]*(c - c*Sin[e + f*x])^(9/2))/(15*c*f*Sqrt[a + a*Sin[e + f*x]]) - (Cos[e + f*x]*Sqrt[a + a*Sin[
e + f*x]]*(c - c*Sin[e + f*x])^(9/2))/(6*c*f)

________________________________________________________________________________________

Rubi [A]  time = 0.397736, antiderivative size = 92, 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, 2740, 2738} \[ -\frac{\cos (e+f x) \sqrt{a \sin (e+f x)+a} (c-c \sin (e+f x))^{9/2}}{6 c f}-\frac{a \cos (e+f x) (c-c \sin (e+f x))^{9/2}}{15 c f \sqrt{a \sin (e+f x)+a}} \]

Antiderivative was successfully verified.

[In]

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

[Out]

-(a*Cos[e + f*x]*(c - c*Sin[e + f*x])^(9/2))/(15*c*f*Sqrt[a + a*Sin[e + f*x]]) - (Cos[e + f*x]*Sqrt[a + a*Sin[
e + f*x]]*(c - c*Sin[e + f*x])^(9/2))/(6*c*f)

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 2740

Int[((a_) + (b_.)*sin[(e_.) + (f_.)*(x_)])^(m_)*((c_) + (d_.)*sin[(e_.) + (f_.)*(x_)])^(n_), x_Symbol] :> -Sim
p[(b*Cos[e + f*x]*(a + b*Sin[e + f*x])^(m - 1)*(c + d*Sin[e + f*x])^n)/(f*(m + n)), x] + Dist[(a*(2*m - 1))/(m
+ n), 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, n}, x] && E
qQ[b*c + a*d, 0] && EqQ[a^2 - b^2, 0] && IGtQ[m - 1/2, 0] &&  !LtQ[n, -1] &&  !(IGtQ[n - 1/2, 0] && LtQ[n, m])
&&  !(ILtQ[m + n, 0] && GtQ[2*m + n + 1, 0])

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)]

Rubi steps

\begin{align*} \int \cos ^2(e+f x) \sqrt{a+a \sin (e+f x)} (c-c \sin (e+f x))^{7/2} \, dx &=\frac{\int (a+a \sin (e+f x))^{3/2} (c-c \sin (e+f x))^{9/2} \, dx}{a c}\\ &=-\frac{\cos (e+f x) \sqrt{a+a \sin (e+f x)} (c-c \sin (e+f x))^{9/2}}{6 c f}+\frac{\int \sqrt{a+a \sin (e+f x)} (c-c \sin (e+f x))^{9/2} \, dx}{3 c}\\ &=-\frac{a \cos (e+f x) (c-c \sin (e+f x))^{9/2}}{15 c f \sqrt{a+a \sin (e+f x)}}-\frac{\cos (e+f x) \sqrt{a+a \sin (e+f x)} (c-c \sin (e+f x))^{9/2}}{6 c f}\\ \end{align*}

Mathematica [A]  time = 0.571301, size = 104, normalized size = 1.13 \[ \frac{c^3 \sec (e+f x) \sqrt{a (\sin (e+f x)+1)} \sqrt{c-c \sin (e+f x)} (1080 \sin (e+f x)+20 \sin (3 (e+f x))-36 \sin (5 (e+f x))+405 \cos (2 (e+f x))+90 \cos (4 (e+f x))-5 \cos (6 (e+f x)))}{960 f} \]

Antiderivative was successfully verified.

[In]

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

[Out]

(c^3*Sec[e + f*x]*Sqrt[a*(1 + Sin[e + f*x])]*Sqrt[c - c*Sin[e + f*x]]*(405*Cos[2*(e + f*x)] + 90*Cos[4*(e + f*
x)] - 5*Cos[6*(e + f*x)] + 1080*Sin[e + f*x] + 20*Sin[3*(e + f*x)] - 36*Sin[5*(e + f*x)]))/(960*f)

________________________________________________________________________________________

Maple [A]  time = 0.274, size = 133, normalized size = 1.5 \begin{align*}{\frac{\sin \left ( fx+e \right ) \left ( 5\, \left ( \cos \left ( fx+e \right ) \right ) ^{8}+3\, \left ( \cos \left ( fx+e \right ) \right ) ^{6}\sin \left ( fx+e \right ) +4\, \left ( \cos \left ( fx+e \right ) \right ) ^{6}+7\,\sin \left ( fx+e \right ) \left ( \cos \left ( fx+e \right ) \right ) ^{4}+7\, \left ( \cos \left ( fx+e \right ) \right ) ^{2}\sin \left ( fx+e \right ) -7\, \left ( \cos \left ( fx+e \right ) \right ) ^{2}+28\,\sin \left ( fx+e \right ) +28 \right ) }{30\,f \left ( \cos \left ( fx+e \right ) \right ) ^{7}} \left ( -c \left ( -1+\sin \left ( fx+e \right ) \right ) \right ) ^{{\frac{7}{2}}}\sqrt{a \left ( 1+\sin \left ( fx+e \right ) \right ) }} \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

int(cos(f*x+e)^2*(c-c*sin(f*x+e))^(7/2)*(a+a*sin(f*x+e))^(1/2),x)

[Out]

1/30/f*(-c*(-1+sin(f*x+e)))^(7/2)*sin(f*x+e)*(a*(1+sin(f*x+e)))^(1/2)*(5*cos(f*x+e)^8+3*cos(f*x+e)^6*sin(f*x+e
)+4*cos(f*x+e)^6+7*sin(f*x+e)*cos(f*x+e)^4+7*cos(f*x+e)^2*sin(f*x+e)-7*cos(f*x+e)^2+28*sin(f*x+e)+28)/cos(f*x+
e)^7

________________________________________________________________________________________

Maxima [F]  time = 0., size = 0, normalized size = 0. \begin{align*} \int \sqrt{a \sin \left (f x + e\right ) + a}{\left (-c \sin \left (f x + e\right ) + c\right )}^{\frac{7}{2}} \cos \left (f x + e\right )^{2}\,{d x} \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate(cos(f*x+e)^2*(c-c*sin(f*x+e))^(7/2)*(a+a*sin(f*x+e))^(1/2),x, algorithm="maxima")

[Out]

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

________________________________________________________________________________________

Fricas [A]  time = 1.78385, size = 274, normalized size = 2.98 \begin{align*} -\frac{{\left (5 \, c^{3} \cos \left (f x + e\right )^{6} - 30 \, c^{3} \cos \left (f x + e\right )^{4} + 25 \, c^{3} + 2 \,{\left (9 \, c^{3} \cos \left (f x + e\right )^{4} - 8 \, c^{3} \cos \left (f x + e\right )^{2} - 16 \, c^{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 \, f \cos \left (f x + e\right )} \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate(cos(f*x+e)^2*(c-c*sin(f*x+e))^(7/2)*(a+a*sin(f*x+e))^(1/2),x, algorithm="fricas")

[Out]

-1/30*(5*c^3*cos(f*x + e)^6 - 30*c^3*cos(f*x + e)^4 + 25*c^3 + 2*(9*c^3*cos(f*x + e)^4 - 8*c^3*cos(f*x + e)^2
- 16*c^3)*sin(f*x + e))*sqrt(a*sin(f*x + e) + a)*sqrt(-c*sin(f*x + e) + c)/(f*cos(f*x + e))

________________________________________________________________________________________

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*(c-c*sin(f*x+e))**(7/2)*(a+a*sin(f*x+e))**(1/2),x)

[Out]

Timed out

________________________________________________________________________________________

Giac [F]  time = 0., size = 0, normalized size = 0. \begin{align*} \mathit{sage}_{2} \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate(cos(f*x+e)^2*(c-c*sin(f*x+e))^(7/2)*(a+a*sin(f*x+e))^(1/2),x, algorithm="giac")

[Out]

sage2