3.2.63 \(\int (a+a \sin (e+f x))^{7/2} (A+B \sin (e+f x)) (c-c \sin (e+f x))^{3/2} \, dx\) [163]

Optimal. Leaf size=142 \[ \frac {(3 A+B) c^2 \cos (e+f x) (a+a \sin (e+f x))^{7/2}}{30 f \sqrt {c-c \sin (e+f x)}}+\frac {(3 A+B) c \cos (e+f x) (a+a \sin (e+f x))^{7/2} \sqrt {c-c \sin (e+f x)}}{15 f}-\frac {B \cos (e+f x) (a+a \sin (e+f x))^{7/2} (c-c \sin (e+f x))^{3/2}}{6 f} \]

[Out]

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

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Rubi [A]
time = 0.24, antiderivative size = 142, 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 = {3052, 2819, 2817} \begin {gather*} \frac {c^2 (3 A+B) \cos (e+f x) (a \sin (e+f x)+a)^{7/2}}{30 f \sqrt {c-c \sin (e+f x)}}+\frac {c (3 A+B) \cos (e+f x) (a \sin (e+f x)+a)^{7/2} \sqrt {c-c \sin (e+f x)}}{15 f}-\frac {B \cos (e+f x) (a \sin (e+f x)+a)^{7/2} (c-c \sin (e+f x))^{3/2}}{6 f} \end {gather*}

Antiderivative was successfully verified.

[In]

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

[Out]

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

Rule 2817

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 2819

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 - 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] &&
EqQ[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 3052

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

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Mathematica [A]
time = 1.26, size = 212, normalized size = 1.49 \begin {gather*} -\frac {a^3 c (-1+\sin (e+f x)) (1+\sin (e+f x))^3 \sqrt {a (1+\sin (e+f x))} \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)))}{960 f \left (\cos \left (\frac {1}{2} (e+f x)\right )-\sin \left (\frac {1}{2} (e+f x)\right )\right )^3 \left (\cos \left (\frac {1}{2} (e+f x)\right )+\sin \left (\frac {1}{2} (e+f x)\right )\right )^7} \end {gather*}

Antiderivative was successfully verified.

[In]

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

[Out]

-1/960*(a^3*c*(-1 + Sin[e + f*x])*(1 + Sin[e + f*x])^3*Sqrt[a*(1 + Sin[e + f*x])]*Sqrt[c - c*Sin[e + f*x]]*(-1
5*(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])^3*(Cos[(e + f*x)/2] + Sin[(e + f*x)/2])^7)

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Maple [A]
time = 0.44, size = 187, normalized size = 1.32

method	result	size
default	\(\frac {\left (-5 B \sin \left (f x +e \right ) \left (\cos ^{4}\left (f x +e \right )\right )-6 A \left (\cos ^{4}\left (f x +e \right )\right )-12 B \left (\cos ^{4}\left (f x +e \right )\right )+15 A \left (\cos ^{2}\left (f x +e \right )\right ) \sin \left (f x +e \right )+10 B \left (\cos ^{2}\left (f x +e \right )\right ) \sin \left (f x +e \right )+12 A \left (\cos ^{2}\left (f x +e \right )\right )+4 B \left (\cos ^{2}\left (f x +e \right )\right )+15 A \sin \left (f x +e \right )+10 B \sin \left (f x +e \right )+24 A +8 B \right ) \left (-c \left (\sin \left (f x +e \right )-1\right )\right )^{\frac {3}{2}} \sin \left (f x +e \right ) \left (a \left (1+\sin \left (f x +e \right )\right )\right )^{\frac {7}{2}}}{30 f \left (2 \sin \left (f x +e \right )-\left (\cos ^{2}\left (f x +e \right )\right )+2\right ) \cos \left (f x +e \right )^{3}}\)	\(187\)

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))^(3/2),x,method=_RETURNVERBOSE)

[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))^(3/2)*sin(f*x+e)*(a*(1+sin(f*x+e)))^(7/2)/(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 \begin {gather*} \text {Failed to integrate} \end {gather*}

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))^(3/2),x, algorithm="maxima")

[Out]

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

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Fricas [A]
time = 0.40, size = 150, normalized size = 1.06 \begin {gather*} \frac {{\left (5 \, B a^{3} c \cos \left (f x + e\right )^{6} - 15 \, {\left (A + B\right )} a^{3} c \cos \left (f x + e\right )^{4} + 5 \, {\left (3 \, A + 2 \, B\right )} a^{3} c - 2 \, {\left (3 \, {\left (A + 2 \, B\right )} a^{3} c \cos \left (f x + e\right )^{4} - 2 \, {\left (3 \, A + B\right )} a^{3} c \cos \left (f x + e\right )^{2} - 4 \, {\left (3 \, A + B\right )} a^{3} c\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 {gather*}

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))^(3/2),x, algorithm="fricas")

[Out]

1/30*(5*B*a^3*c*cos(f*x + e)^6 - 15*(A + B)*a^3*c*cos(f*x + e)^4 + 5*(3*A + 2*B)*a^3*c - 2*(3*(A + 2*B)*a^3*c*
cos(f*x + e)^4 - 2*(3*A + B)*a^3*c*cos(f*x + e)^2 - 4*(3*A + B)*a^3*c)*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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Sympy [F(-1)] Timed out
time = 0.00, size = 0, normalized size = 0.00 \begin {gather*} \text {Timed out} \end {gather*}

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))**(3/2),x)

[Out]

Timed out

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Giac [A]
time = 0.55, size = 262, normalized size = 1.85 \begin {gather*} \frac {8 \, {\left (20 \, B a^{3} c \cos \left (-\frac {1}{4} \, \pi + \frac {1}{2} \, f x + \frac {1}{2} \, e\right )^{12} \mathrm {sgn}\left (\cos \left (-\frac {1}{4} \, \pi + \frac {1}{2} \, f x + \frac {1}{2} \, e\right )\right ) \mathrm {sgn}\left (\sin \left (-\frac {1}{4} \, \pi + \frac {1}{2} \, f x + \frac {1}{2} \, e\right )\right ) + 12 \, A a^{3} c \cos \left (-\frac {1}{4} \, \pi + \frac {1}{2} \, f x + \frac {1}{2} \, e\right )^{10} \mathrm {sgn}\left (\cos \left (-\frac {1}{4} \, \pi + \frac {1}{2} \, f x + \frac {1}{2} \, e\right )\right ) \mathrm {sgn}\left (\sin \left (-\frac {1}{4} \, \pi + \frac {1}{2} \, f x + \frac {1}{2} \, e\right )\right ) - 36 \, B a^{3} c \cos \left (-\frac {1}{4} \, \pi + \frac {1}{2} \, f x + \frac {1}{2} \, e\right )^{10} \mathrm {sgn}\left (\cos \left (-\frac {1}{4} \, \pi + \frac {1}{2} \, f x + \frac {1}{2} \, e\right )\right ) \mathrm {sgn}\left (\sin \left (-\frac {1}{4} \, \pi + \frac {1}{2} \, f x + \frac {1}{2} \, e\right )\right ) - 15 \, A a^{3} c \cos \left (-\frac {1}{4} \, \pi + \frac {1}{2} \, f x + \frac {1}{2} \, e\right )^{8} \mathrm {sgn}\left (\cos \left (-\frac {1}{4} \, \pi + \frac {1}{2} \, f x + \frac {1}{2} \, e\right )\right ) \mathrm {sgn}\left (\sin \left (-\frac {1}{4} \, \pi + \frac {1}{2} \, f x + \frac {1}{2} \, e\right )\right ) + 15 \, B a^{3} c \cos \left (-\frac {1}{4} \, \pi + \frac {1}{2} \, f x + \frac {1}{2} \, e\right )^{8} \mathrm {sgn}\left (\cos \left (-\frac {1}{4} \, \pi + \frac {1}{2} \, f x + \frac {1}{2} \, e\right )\right ) \mathrm {sgn}\left (\sin \left (-\frac {1}{4} \, \pi + \frac {1}{2} \, f x + \frac {1}{2} \, e\right )\right )\right )} \sqrt {a} \sqrt {c}}{15 \, f} \end {gather*}

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))^(3/2),x, algorithm="giac")

[Out]

8/15*(20*B*a^3*c*cos(-1/4*pi + 1/2*f*x + 1/2*e)^12*sgn(cos(-1/4*pi + 1/2*f*x + 1/2*e))*sgn(sin(-1/4*pi + 1/2*f
*x + 1/2*e)) + 12*A*a^3*c*cos(-1/4*pi + 1/2*f*x + 1/2*e)^10*sgn(cos(-1/4*pi + 1/2*f*x + 1/2*e))*sgn(sin(-1/4*p
i + 1/2*f*x + 1/2*e)) - 36*B*a^3*c*cos(-1/4*pi + 1/2*f*x + 1/2*e)^10*sgn(cos(-1/4*pi + 1/2*f*x + 1/2*e))*sgn(s
in(-1/4*pi + 1/2*f*x + 1/2*e)) - 15*A*a^3*c*cos(-1/4*pi + 1/2*f*x + 1/2*e)^8*sgn(cos(-1/4*pi + 1/2*f*x + 1/2*e
))*sgn(sin(-1/4*pi + 1/2*f*x + 1/2*e)) + 15*B*a^3*c*cos(-1/4*pi + 1/2*f*x + 1/2*e)^8*sgn(cos(-1/4*pi + 1/2*f*x
+ 1/2*e))*sgn(sin(-1/4*pi + 1/2*f*x + 1/2*e)))*sqrt(a)*sqrt(c)/f

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Mupad [B]
time = 18.17, size = 321, normalized size = 2.26 \begin {gather*} -\frac {{\mathrm {e}}^{-e\,6{}\mathrm {i}-f\,x\,6{}\mathrm {i}}\,\sqrt {c-c\,\sin \left (e+f\,x\right )}\,\left (\frac {a^3\,c\,{\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 {B\,a^3\,c\,{\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^3\,c\,{\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^3\,c\,{\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^3\,c\,{\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^3\,c\,{\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 )} \end {gather*}

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))^(3/2),x)

[Out]

-(exp(- e*6i - f*x*6i)*(c - c*sin(e + f*x))^(1/2)*((a^3*c*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) - (B*a^3*c*exp(e*6i + f*x*6i)*cos(6*e + 6*f*x)*(a + a*sin(e + f*x))^(1/2))/(96*f)
+ (a^3*c*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^3*c*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^3*c*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^3*c*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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