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

Optimal. Leaf size=96 \[ \frac {c (A-B) \cos (e+f x) (a \sin (e+f x)+a)^{7/2}}{4 f \sqrt {c-c \sin (e+f x)}}+\frac {B c \cos (e+f x) (a \sin (e+f x)+a)^{9/2}}{5 a f \sqrt {c-c \sin (e+f x)}} \]

[Out]

1/4*(A-B)*c*cos(f*x+e)*(a+a*sin(f*x+e))^(7/2)/f/(c-c*sin(f*x+e))^(1/2)+1/5*B*c*cos(f*x+e)*(a+a*sin(f*x+e))^(9/
2)/a/f/(c-c*sin(f*x+e))^(1/2)

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Rubi [A]  time = 0.32, antiderivative size = 96, normalized size of antiderivative = 1.00, number of steps used = 3, number of rules used = 2, integrand size = 40, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.050, Rules used = {2971, 2738} \[ \frac {c (A-B) \cos (e+f x) (a \sin (e+f x)+a)^{7/2}}{4 f \sqrt {c-c \sin (e+f x)}}+\frac {B c \cos (e+f x) (a \sin (e+f x)+a)^{9/2}}{5 a f \sqrt {c-c \sin (e+f x)}} \]

Antiderivative was successfully verified.

[In]

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

[Out]

((A - B)*c*Cos[e + f*x]*(a + a*Sin[e + f*x])^(7/2))/(4*f*Sqrt[c - c*Sin[e + f*x]]) + (B*c*Cos[e + f*x]*(a + a*
Sin[e + f*x])^(9/2))/(5*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 2971

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

Rubi steps

\begin {align*} \int (a+a \sin (e+f x))^{7/2} (A+B \sin (e+f x)) \sqrt {c-c \sin (e+f x)} \, dx &=\frac {B \int (a+a \sin (e+f x))^{9/2} \sqrt {c-c \sin (e+f x)} \, dx}{a}-(-A+B) \int (a+a \sin (e+f x))^{7/2} \sqrt {c-c \sin (e+f x)} \, dx\\ &=\frac {(A-B) c \cos (e+f x) (a+a \sin (e+f x))^{7/2}}{4 f \sqrt {c-c \sin (e+f x)}}+\frac {B c \cos (e+f x) (a+a \sin (e+f x))^{9/2}}{5 a f \sqrt {c-c \sin (e+f x)}}\\ \end {align*}

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Mathematica [A]  time = 0.97, size = 121, normalized size = 1.26 \[ \frac {a^3 \sec (e+f x) \sqrt {a (\sin (e+f x)+1)} \sqrt {c-c \sin (e+f x)} (4 (60 A+23 B) \sin (e+f x)+\cos (4 (e+f x)) (5 A+4 B \sin (e+f x)+15 B)-4 \cos (2 (e+f x)) (4 (5 A+6 B) \sin (e+f x)+5 (7 A+5 B)))}{160 f} \]

Antiderivative was successfully verified.

[In]

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

[Out]

(a^3*Sec[e + f*x]*Sqrt[a*(1 + Sin[e + f*x])]*Sqrt[c - c*Sin[e + f*x]]*(4*(60*A + 23*B)*Sin[e + f*x] + Cos[4*(e
+ f*x)]*(5*A + 15*B + 4*B*Sin[e + f*x]) - 4*Cos[2*(e + f*x)]*(5*(7*A + 5*B) + 4*(5*A + 6*B)*Sin[e + f*x])))/(
160*f)

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fricas [A]  time = 0.45, size = 139, normalized size = 1.45 \[ \frac {{\left (5 \, {\left (A + 3 \, B\right )} a^{3} \cos \left (f x + e\right )^{4} - 40 \, {\left (A + B\right )} a^{3} \cos \left (f x + e\right )^{2} + 5 \, {\left (7 \, A + 5 \, B\right )} a^{3} + 4 \, {\left (B a^{3} \cos \left (f x + e\right )^{4} - {\left (5 \, A + 7 \, B\right )} a^{3} \cos \left (f x + e\right )^{2} + 2 \, {\left (5 \, A + 3 \, B\right )} a^{3}\right )} \sin \left (f x + e\right )\right )} \sqrt {a \sin \left (f x + e\right ) + a} \sqrt {-c \sin \left (f x + e\right ) + c}}{20 \, f \cos \left (f x + e\right )} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

1/20*(5*(A + 3*B)*a^3*cos(f*x + e)^4 - 40*(A + B)*a^3*cos(f*x + e)^2 + 5*(7*A + 5*B)*a^3 + 4*(B*a^3*cos(f*x +
e)^4 - (5*A + 7*B)*a^3*cos(f*x + e)^2 + 2*(5*A + 3*B)*a^3)*sin(f*x + e))*sqrt(a*sin(f*x + e) + a)*sqrt(-c*sin(
f*x + e) + c)/(f*cos(f*x + e))

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giac [F(-2)]  time = 0.00, size = 0, normalized size = 0.00 \[ \text {Exception raised: NotImplementedError} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

Exception raised: NotImplementedError >> Unable to parse Giac output: Unable to check sign: (4*pi/x/2)>(-4*pi/
x/2)Unable to check sign: (4*pi/x/2)>(-4*pi/x/2)Unable to check sign: (4*pi/x/2)>(-4*pi/x/2)Unable to check si
gn: (4*pi/x/2)>(-4*pi/x/2)Unable to check sign: (4*pi/x/2)>(-4*pi/x/2)Unable to check sign: (4*pi/x/2)>(-4*pi/
x/2)Unable to check sign: (4*pi/x/2)>(-4*pi/x/2)Unable to check sign: (4*pi/x/2)>(-4*pi/x/2)Unable to check si
gn: (4*pi/x/2)>(-4*pi/x/2)Unable to check sign: (4*pi/x/2)>(-4*pi/x/2)Unable to check sign: (4*pi/x/2)>(-4*pi/
x/2)Unable to check sign: (4*pi/x/2)>(-4*pi/x/2)Unable to check sign: (4*pi/x/2)>(-4*pi/x/2)Unable to check si
gn: (4*pi/x/2)>(-4*pi/x/2)Unable to check sign: (4*pi/x/2)>(-4*pi/x/2)Unable to check sign: (4*pi/x/2)>(-4*pi/
x/2)Unable to check sign: (4*pi/x/2)>(-4*pi/x/2)Unable to check sign: (4*pi/x/2)>(-4*pi/x/2)Unable to check si
gn: (4*pi/x/2)>(-4*pi/x/2)Unable to check sign: (4*pi/x/2)>(-4*pi/x/2)Unable to check sign: (4*pi/x/2)>(-4*pi/
x/2)Unable to check sign: (4*pi/x/2)>(-4*pi/x/2)sqrt(2*a)*sqrt(2*c)*(-64*f*(A*a^3*sign(sin(1/2*(f*x+exp(1))-1/
4*pi))*sign(cos(1/2*(f*x+exp(1))-1/4*pi))+3*B*a^3*sign(sin(1/2*(f*x+exp(1))-1/4*pi))*sign(cos(1/2*(f*x+exp(1))
-1/4*pi)))*cos(4*f*x+4*exp(1))/(64*f)^2+96*f*(12*A*a^3*sign(sin(1/2*(f*x+exp(1))-1/4*pi))*sign(cos(1/2*(f*x+ex
p(1))-1/4*pi))+15*B*a^3*sign(sin(1/2*(f*x+exp(1))-1/4*pi))*sign(cos(1/2*(f*x+exp(1))-1/4*pi)))*sin(3*f*x+3*exp
(1))/(96*f)^2+32*f*(-A*a^3*sign(sin(1/2*(f*x+exp(1))-1/4*pi))*sign(cos(1/2*(f*x+exp(1))-1/4*pi))-3*B*a^3*sign(
sin(1/2*(f*x+exp(1))-1/4*pi))*sign(cos(1/2*(f*x+exp(1))-1/4*pi)))*cos(-2*f*x-2*exp(1))/(-32*f)^2+16*f*(-14*A*a
^3*sign(sin(1/2*(f*x+exp(1))-1/4*pi))*sign(cos(1/2*(f*x+exp(1))-1/4*pi))-7*B*a^3*sign(sin(1/2*(f*x+exp(1))-1/4
*pi))*sign(cos(1/2*(f*x+exp(1))-1/4*pi)))*sin(f*x+exp(1))/(16*f)^2-32*f*(-15*A*a^3*sign(sin(1/2*(f*x+exp(1))-1
/4*pi))*sign(cos(1/2*(f*x+exp(1))-1/4*pi))-13*B*a^3*sign(sin(1/2*(f*x+exp(1))-1/4*pi))*sign(cos(1/2*(f*x+exp(1
))-1/4*pi)))*cos(2*f*x+2*exp(1))/(32*f)^2-160*B*a^3*f*sign(sin(1/2*(f*x+exp(1))-1/4*pi))*sign(cos(1/2*(f*x+exp
(1))-1/4*pi))*sin(5*f*x+5*exp(1))/(160*f)^2)

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maple [B]  time = 0.82, size = 174, normalized size = 1.81 \[ \frac {\left (-4 B \left (\cos ^{4}\left (f x +e \right )\right )+5 A \left (\cos ^{2}\left (f x +e \right )\right ) \sin \left (f x +e \right )+15 B \left (\cos ^{2}\left (f x +e \right )\right ) \sin \left (f x +e \right )+20 A \left (\cos ^{2}\left (f x +e \right )\right )+28 B \left (\cos ^{2}\left (f x +e \right )\right )-35 A \sin \left (f x +e \right )-25 B \sin \left (f x +e \right )-40 A -24 B \right ) \sqrt {-c \left (\sin \left (f x +e \right )-1\right )}\, \sin \left (f x +e \right ) \left (a \left (1+\sin \left (f x +e \right )\right )\right )^{\frac {7}{2}}}{20 f \left (\left (\cos ^{2}\left (f x +e \right )\right ) \sin \left (f x +e \right )+3 \left (\cos ^{2}\left (f x +e \right )\right )-4 \sin \left (f x +e \right )-4\right ) \cos \left (f x +e \right )} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

1/20/f*(-4*B*cos(f*x+e)^4+5*A*cos(f*x+e)^2*sin(f*x+e)+15*B*cos(f*x+e)^2*sin(f*x+e)+20*A*cos(f*x+e)^2+28*B*cos(
f*x+e)^2-35*A*sin(f*x+e)-25*B*sin(f*x+e)-40*A-24*B)*(-c*(sin(f*x+e)-1))^(1/2)*sin(f*x+e)*(a*(1+sin(f*x+e)))^(7
/2)/(cos(f*x+e)^2*sin(f*x+e)+3*cos(f*x+e)^2-4*sin(f*x+e)-4)/cos(f*x+e)

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maxima [F]  time = 0.00, size = 0, normalized size = 0.00 \[ \int {\left (B \sin \left (f x + e\right ) + A\right )} {\left (a \sin \left (f x + e\right ) + a\right )}^{\frac {7}{2}} \sqrt {-c \sin \left (f x + e\right ) + c}\,{d x} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

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

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mupad [B]  time = 16.68, size = 173, normalized size = 1.80 \[ -\frac {a^3\,\sqrt {a\,\left (\sin \left (e+f\,x\right )+1\right )}\,\sqrt {-c\,\left (\sin \left (e+f\,x\right )-1\right )}\,\left (140\,A\,\cos \left (e+f\,x\right )+100\,B\,\cos \left (e+f\,x\right )+135\,A\,\cos \left (3\,e+3\,f\,x\right )-5\,A\,\cos \left (5\,e+5\,f\,x\right )+85\,B\,\cos \left (3\,e+3\,f\,x\right )-15\,B\,\cos \left (5\,e+5\,f\,x\right )-240\,A\,\sin \left (2\,e+2\,f\,x\right )+40\,A\,\sin \left (4\,e+4\,f\,x\right )-90\,B\,\sin \left (2\,e+2\,f\,x\right )+48\,B\,\sin \left (4\,e+4\,f\,x\right )-2\,B\,\sin \left (6\,e+6\,f\,x\right )\right )}{160\,f\,\left (\cos \left (2\,e+2\,f\,x\right )+1\right )} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

-(a^3*(a*(sin(e + f*x) + 1))^(1/2)*(-c*(sin(e + f*x) - 1))^(1/2)*(140*A*cos(e + f*x) + 100*B*cos(e + f*x) + 13
5*A*cos(3*e + 3*f*x) - 5*A*cos(5*e + 5*f*x) + 85*B*cos(3*e + 3*f*x) - 15*B*cos(5*e + 5*f*x) - 240*A*sin(2*e +
2*f*x) + 40*A*sin(4*e + 4*f*x) - 90*B*sin(2*e + 2*f*x) + 48*B*sin(4*e + 4*f*x) - 2*B*sin(6*e + 6*f*x)))/(160*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((a+a*sin(f*x+e))**(7/2)*(A+B*sin(f*x+e))*(c-c*sin(f*x+e))**(1/2),x)

[Out]

Timed out

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