3.139 \(\int (a+a \sin (e+f x))^{3/2} (A+B \sin (e+f x)) (c-c \sin (e+f x))^{7/2} \, dx\)

Optimal. Leaf size=146 \[ -\frac {a^2 (3 A-B) \cos (e+f x) (c-c \sin (e+f x))^{7/2}}{30 f \sqrt {a \sin (e+f x)+a}}-\frac {a (3 A-B) \cos (e+f x) \sqrt {a \sin (e+f x)+a} (c-c \sin (e+f x))^{7/2}}{15 f}-\frac {B \cos (e+f x) (a \sin (e+f x)+a)^{3/2} (c-c \sin (e+f x))^{7/2}}{6 f} \]

[Out]

-1/6*B*cos(f*x+e)*(a+a*sin(f*x+e))^(3/2)*(c-c*sin(f*x+e))^(7/2)/f-1/30*a^2*(3*A-B)*cos(f*x+e)*(c-c*sin(f*x+e))
^(7/2)/f/(a+a*sin(f*x+e))^(1/2)-1/15*a*(3*A-B)*cos(f*x+e)*(c-c*sin(f*x+e))^(7/2)*(a+a*sin(f*x+e))^(1/2)/f

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Rubi [A]  time = 0.36, antiderivative size = 146, normalized size of antiderivative = 1.00, number of steps used = 3, number of rules used = 3, integrand size = 40, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.075, Rules used = {2973, 2740, 2738} \[ -\frac {a^2 (3 A-B) \cos (e+f x) (c-c \sin (e+f x))^{7/2}}{30 f \sqrt {a \sin (e+f x)+a}}-\frac {a (3 A-B) \cos (e+f x) \sqrt {a \sin (e+f x)+a} (c-c \sin (e+f x))^{7/2}}{15 f}-\frac {B \cos (e+f x) (a \sin (e+f x)+a)^{3/2} (c-c \sin (e+f x))^{7/2}}{6 f} \]

Antiderivative was successfully verified.

[In]

Int[(a + a*Sin[e + f*x])^(3/2)*(A + B*Sin[e + f*x])*(c - c*Sin[e + f*x])^(7/2),x]

[Out]

-(a^2*(3*A - B)*Cos[e + f*x]*(c - c*Sin[e + f*x])^(7/2))/(30*f*Sqrt[a + a*Sin[e + f*x]]) - (a*(3*A - B)*Cos[e
+ f*x]*Sqrt[a + a*Sin[e + f*x]]*(c - c*Sin[e + f*x])^(7/2))/(15*f) - (B*Cos[e + f*x]*(a + a*Sin[e + f*x])^(3/2
)*(c - c*Sin[e + f*x])^(7/2))/(6*f)

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 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 2973

Int[((a_) + (b_.)*sin[(e_.) + (f_.)*(x_)])^(m_.)*((A_.) + (B_.)*sin[(e_.) + (f_.)*(x_)])*((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)/(f*(
m + n + 1)), x] - Dist[(B*c*(m - n) - A*d*(m + n + 1))/(d*(m + n + 1)), Int[(a + b*Sin[e + f*x])^m*(c + d*Sin[
e + f*x])^n, x], x] /; FreeQ[{a, b, c, d, e, f, A, B, m, n}, x] && EqQ[b*c + a*d, 0] && EqQ[a^2 - b^2, 0] &&
!LtQ[m, -2^(-1)] && NeQ[m + n + 1, 0]

Rubi steps

\begin {align*} \int (a+a \sin (e+f x))^{3/2} (A+B \sin (e+f x)) (c-c \sin (e+f x))^{7/2} \, dx &=-\frac {B \cos (e+f x) (a+a \sin (e+f x))^{3/2} (c-c \sin (e+f x))^{7/2}}{6 f}+\frac {1}{3} (3 A-B) \int (a+a \sin (e+f x))^{3/2} (c-c \sin (e+f x))^{7/2} \, dx\\ &=-\frac {a (3 A-B) \cos (e+f x) \sqrt {a+a \sin (e+f x)} (c-c \sin (e+f x))^{7/2}}{15 f}-\frac {B \cos (e+f x) (a+a \sin (e+f x))^{3/2} (c-c \sin (e+f x))^{7/2}}{6 f}+\frac {1}{15} (2 a (3 A-B)) \int \sqrt {a+a \sin (e+f x)} (c-c \sin (e+f x))^{7/2} \, dx\\ &=-\frac {a^2 (3 A-B) \cos (e+f x) (c-c \sin (e+f x))^{7/2}}{30 f \sqrt {a+a \sin (e+f x)}}-\frac {a (3 A-B) \cos (e+f x) \sqrt {a+a \sin (e+f x)} (c-c \sin (e+f x))^{7/2}}{15 f}-\frac {B \cos (e+f x) (a+a \sin (e+f x))^{3/2} (c-c \sin (e+f x))^{7/2}}{6 f}\\ \end {align*}

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Mathematica [A]  time = 1.49, size = 205, normalized size = 1.40 \[ -\frac {c^3 (\sin (e+f x)-1)^3 (a (\sin (e+f x)+1))^{3/2} \sqrt {c-c \sin (e+f x)} (15 (16 A-11 B) \cos (2 (e+f x))+30 (2 A-B) \cos (4 (e+f x))+840 A \sin (e+f x)+60 A \sin (3 (e+f x))-12 A \sin (5 (e+f x))-240 B \sin (e+f x)+40 B \sin (3 (e+f x))+24 B \sin (5 (e+f x))+5 B \cos (6 (e+f x)))}{960 f \left (\cos \left (\frac {1}{2} (e+f x)\right )-\sin \left (\frac {1}{2} (e+f x)\right )\right )^7 \left (\sin \left (\frac {1}{2} (e+f x)\right )+\cos \left (\frac {1}{2} (e+f x)\right )\right )^3} \]

Antiderivative was successfully verified.

[In]

Integrate[(a + a*Sin[e + f*x])^(3/2)*(A + B*Sin[e + f*x])*(c - c*Sin[e + f*x])^(7/2),x]

[Out]

-1/960*(c^3*(-1 + Sin[e + f*x])^3*(a*(1 + Sin[e + f*x]))^(3/2)*Sqrt[c - c*Sin[e + f*x]]*(15*(16*A - 11*B)*Cos[
2*(e + f*x)] + 30*(2*A - B)*Cos[4*(e + f*x)] + 5*B*Cos[6*(e + f*x)] + 840*A*Sin[e + f*x] - 240*B*Sin[e + f*x]
+ 60*A*Sin[3*(e + f*x)] + 40*B*Sin[3*(e + f*x)] - 12*A*Sin[5*(e + f*x)] + 24*B*Sin[5*(e + f*x)]))/(f*(Cos[(e +
f*x)/2] - Sin[(e + f*x)/2])^7*(Cos[(e + f*x)/2] + Sin[(e + f*x)/2])^3)

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fricas [A]  time = 0.45, size = 148, normalized size = 1.01 \[ \frac {{\left (5 \, B a c^{3} \cos \left (f x + e\right )^{6} + 15 \, {\left (A - B\right )} a c^{3} \cos \left (f x + e\right )^{4} - 5 \, {\left (3 \, A - 2 \, B\right )} a c^{3} - 2 \, {\left (3 \, {\left (A - 2 \, B\right )} a c^{3} \cos \left (f x + e\right )^{4} - 2 \, {\left (3 \, A - B\right )} a c^{3} \cos \left (f x + e\right )^{2} - 4 \, {\left (3 \, A - B\right )} a c^{3}\right )} \sin \left (f x + e\right )\right )} \sqrt {a \sin \left (f x + e\right ) + a} \sqrt {-c \sin \left (f x + e\right ) + c}}{30 \, f \cos \left (f x + e\right )} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate((a+a*sin(f*x+e))^(3/2)*(A+B*sin(f*x+e))*(c-c*sin(f*x+e))^(7/2),x, algorithm="fricas")

[Out]

1/30*(5*B*a*c^3*cos(f*x + e)^6 + 15*(A - B)*a*c^3*cos(f*x + e)^4 - 5*(3*A - 2*B)*a*c^3 - 2*(3*(A - 2*B)*a*c^3*
cos(f*x + e)^4 - 2*(3*A - B)*a*c^3*cos(f*x + e)^2 - 4*(3*A - B)*a*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))

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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))^(3/2)*(A+B*sin(f*x+e))*(c-c*sin(f*x+e))^(7/2),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)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)sqrt(2*a)*sqrt(2*c
)*(160*f*(A*a*c^3*sign(sin(1/2*(f*x+exp(1))-1/4*pi))*sign(cos(1/2*(f*x+exp(1))-1/4*pi))-2*B*a*c^3*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)^2-64*f*(4*A*a*c^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*c^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))/(64*f)^2-256*f*(8*A*a*c^3*sign(sin(1/2*(f*x+exp(1))-1/
4*pi))*sign(cos(1/2*(f*x+exp(1))-1/4*pi))-5*B*a*c^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))/(256*f)^2+96*f*(-3*A*a*c^3*sign(sin(1/2*(f*x+exp(1))-1/4*pi))*sign(cos(1/2*(f
*x+exp(1))-1/4*pi))-2*B*a*c^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+16*f*(-7*A*a*c^3*sign(sin(1/2*(f*x+exp(1))-1/4*pi))*sign(cos(1/2*(f*x+exp(1))-1/4*pi))+2*B
*a*c^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+128*f*(
-8*A*a*c^3*sign(sin(1/2*(f*x+exp(1))-1/4*pi))*sign(cos(1/2*(f*x+exp(1))-1/4*pi))+5*B*a*c^3*sign(sin(1/2*(f*x+e
xp(1))-1/4*pi))*sign(cos(1/2*(f*x+exp(1))-1/4*pi)))*cos(-2*f*x-2*exp(1))/(-128*f)^2-384*B*a*c^3*f*sign(sin(1/2
*(f*x+exp(1))-1/4*pi))*sign(cos(1/2*(f*x+exp(1))-1/4*pi))*cos(6*f*x+6*exp(1))/(384*f)^2-256*B*a*c^3*f*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))/(-256*f)^2)

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maple [A]  time = 0.84, size = 185, normalized size = 1.27 \[ \frac {\left (5 B \sin \left (f x +e \right ) \left (\cos ^{4}\left (f x +e \right )\right )+6 A \left (\cos ^{4}\left (f x +e \right )\right )-12 B \left (\cos ^{4}\left (f x +e \right )\right )+15 A \left (\cos ^{2}\left (f x +e \right )\right ) \sin \left (f x +e \right )-10 B \left (\cos ^{2}\left (f x +e \right )\right ) \sin \left (f x +e \right )-12 A \left (\cos ^{2}\left (f x +e \right )\right )+4 B \left (\cos ^{2}\left (f x +e \right )\right )+15 A \sin \left (f x +e \right )-10 B \sin \left (f x +e \right )-24 A +8 B \right ) \left (-c \left (\sin \left (f x +e \right )-1\right )\right )^{\frac {7}{2}} \sin \left (f x +e \right ) \left (a \left (1+\sin \left (f x +e \right )\right )\right )^{\frac {3}{2}}}{30 f \left (2 \sin \left (f x +e \right )+\cos ^{2}\left (f x +e \right )-2\right ) \cos \left (f x +e \right )^{3}} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

int((a+a*sin(f*x+e))^(3/2)*(A+B*sin(f*x+e))*(c-c*sin(f*x+e))^(7/2),x)

[Out]

1/30/f*(5*B*sin(f*x+e)*cos(f*x+e)^4+6*A*cos(f*x+e)^4-12*B*cos(f*x+e)^4+15*A*cos(f*x+e)^2*sin(f*x+e)-10*B*cos(f
*x+e)^2*sin(f*x+e)-12*A*cos(f*x+e)^2+4*B*cos(f*x+e)^2+15*A*sin(f*x+e)-10*B*sin(f*x+e)-24*A+8*B)*(-c*(sin(f*x+e
)-1))^(7/2)*sin(f*x+e)*(a*(1+sin(f*x+e)))^(3/2)/(2*sin(f*x+e)+cos(f*x+e)^2-2)/cos(f*x+e)^3

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maxima [F]  time = 0.00, size = 0, normalized size = 0.00 \[ \int {\left (B \sin \left (f x + e\right ) + A\right )} {\left (a \sin \left (f x + e\right ) + a\right )}^{\frac {3}{2}} {\left (-c \sin \left (f x + e\right ) + c\right )}^{\frac {7}{2}}\,{d x} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate((a+a*sin(f*x+e))^(3/2)*(A+B*sin(f*x+e))*(c-c*sin(f*x+e))^(7/2),x, algorithm="maxima")

[Out]

integrate((B*sin(f*x + e) + A)*(a*sin(f*x + e) + a)^(3/2)*(-c*sin(f*x + e) + c)^(7/2), x)

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mupad [B]  time = 17.53, size = 323, normalized size = 2.21 \[ \frac {{\mathrm {e}}^{-e\,6{}\mathrm {i}-f\,x\,6{}\mathrm {i}}\,\sqrt {c-c\,\sin \left (e+f\,x\right )}\,\left (\frac {B\,a\,c^3\,{\mathrm {e}}^{e\,6{}\mathrm {i}+f\,x\,6{}\mathrm {i}}\,\cos \left (6\,e+6\,f\,x\right )\,\sqrt {a+a\,\sin \left (e+f\,x\right )}}{96\,f}-\frac {a\,c^3\,{\mathrm {e}}^{e\,6{}\mathrm {i}+f\,x\,6{}\mathrm {i}}\,\sin \left (e+f\,x\right )\,\left (A\,7{}\mathrm {i}-B\,2{}\mathrm {i}\right )\,\sqrt {a+a\,\sin \left (e+f\,x\right )}\,1{}\mathrm {i}}{4\,f}+\frac {a\,c^3\,{\mathrm {e}}^{e\,6{}\mathrm {i}+f\,x\,6{}\mathrm {i}}\,\cos \left (4\,e+4\,f\,x\right )\,\left (2\,A-B\right )\,\sqrt {a+a\,\sin \left (e+f\,x\right )}}{16\,f}+\frac {a\,c^3\,{\mathrm {e}}^{e\,6{}\mathrm {i}+f\,x\,6{}\mathrm {i}}\,\cos \left (2\,e+2\,f\,x\right )\,\left (16\,A-11\,B\right )\,\sqrt {a+a\,\sin \left (e+f\,x\right )}}{32\,f}-\frac {a\,c^3\,{\mathrm {e}}^{e\,6{}\mathrm {i}+f\,x\,6{}\mathrm {i}}\,\sin \left (3\,e+3\,f\,x\right )\,\left (A\,3{}\mathrm {i}+B\,2{}\mathrm {i}\right )\,\sqrt {a+a\,\sin \left (e+f\,x\right )}\,1{}\mathrm {i}}{24\,f}+\frac {a\,c^3\,{\mathrm {e}}^{e\,6{}\mathrm {i}+f\,x\,6{}\mathrm {i}}\,\sin \left (5\,e+5\,f\,x\right )\,\left (A\,1{}\mathrm {i}-B\,2{}\mathrm {i}\right )\,\sqrt {a+a\,\sin \left (e+f\,x\right )}\,1{}\mathrm {i}}{40\,f}\right )}{2\,\cos \left (e+f\,x\right )} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

int((A + B*sin(e + f*x))*(a + a*sin(e + f*x))^(3/2)*(c - c*sin(e + f*x))^(7/2),x)

[Out]

(exp(- e*6i - f*x*6i)*(c - c*sin(e + f*x))^(1/2)*((B*a*c^3*exp(e*6i + f*x*6i)*cos(6*e + 6*f*x)*(a + a*sin(e +
f*x))^(1/2))/(96*f) - (a*c^3*exp(e*6i + f*x*6i)*sin(e + f*x)*(A*7i - B*2i)*(a + a*sin(e + f*x))^(1/2)*1i)/(4*f
) + (a*c^3*exp(e*6i + f*x*6i)*cos(4*e + 4*f*x)*(2*A - B)*(a + a*sin(e + f*x))^(1/2))/(16*f) + (a*c^3*exp(e*6i
+ f*x*6i)*cos(2*e + 2*f*x)*(16*A - 11*B)*(a + a*sin(e + f*x))^(1/2))/(32*f) - (a*c^3*exp(e*6i + f*x*6i)*sin(3*
e + 3*f*x)*(A*3i + B*2i)*(a + a*sin(e + f*x))^(1/2)*1i)/(24*f) + (a*c^3*exp(e*6i + f*x*6i)*sin(5*e + 5*f*x)*(A
*1i - B*2i)*(a + a*sin(e + f*x))^(1/2)*1i)/(40*f)))/(2*cos(e + f*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((a+a*sin(f*x+e))**(3/2)*(A+B*sin(f*x+e))*(c-c*sin(f*x+e))**(7/2),x)

[Out]

Timed out

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