3.3.96 \(\int (a+a \sin (e+f x))^{3/2} (A+B \sin (e+f x)) \, dx\) [296]

Optimal. Leaf size=101 \[ -\frac {8 a^2 (5 A+3 B) \cos (e+f x)}{15 f \sqrt {a+a \sin (e+f x)}}-\frac {2 a (5 A+3 B) \cos (e+f x) \sqrt {a+a \sin (e+f x)}}{15 f}-\frac {2 B \cos (e+f x) (a+a \sin (e+f x))^{3/2}}{5 f} \]

[Out]

-2/5*B*cos(f*x+e)*(a+a*sin(f*x+e))^(3/2)/f-8/15*a^2*(5*A+3*B)*cos(f*x+e)/f/(a+a*sin(f*x+e))^(1/2)-2/15*a*(5*A+
3*B)*cos(f*x+e)*(a+a*sin(f*x+e))^(1/2)/f

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Rubi [A]
time = 0.06, antiderivative size = 101, normalized size of antiderivative = 1.00, number of steps used = 3, number of rules used = 3, integrand size = 25, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.120, Rules used = {2830, 2726, 2725} \begin {gather*} -\frac {8 a^2 (5 A+3 B) \cos (e+f x)}{15 f \sqrt {a \sin (e+f x)+a}}-\frac {2 a (5 A+3 B) \cos (e+f x) \sqrt {a \sin (e+f x)+a}}{15 f}-\frac {2 B \cos (e+f x) (a \sin (e+f x)+a)^{3/2}}{5 f} \end {gather*}

Antiderivative was successfully verified.

[In]

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

[Out]

(-8*a^2*(5*A + 3*B)*Cos[e + f*x])/(15*f*Sqrt[a + a*Sin[e + f*x]]) - (2*a*(5*A + 3*B)*Cos[e + f*x]*Sqrt[a + a*S
in[e + f*x]])/(15*f) - (2*B*Cos[e + f*x]*(a + a*Sin[e + f*x])^(3/2))/(5*f)

Rule 2725

Int[Sqrt[(a_) + (b_.)*sin[(c_.) + (d_.)*(x_)]], x_Symbol] :> Simp[-2*b*(Cos[c + d*x]/(d*Sqrt[a + b*Sin[c + d*x
]])), x] /; FreeQ[{a, b, c, d}, x] && EqQ[a^2 - b^2, 0]

Rule 2726

Int[((a_) + (b_.)*sin[(c_.) + (d_.)*(x_)])^(n_), x_Symbol] :> Simp[(-b)*Cos[c + d*x]*((a + b*Sin[c + d*x])^(n
- 1)/(d*n)), x] + Dist[a*((2*n - 1)/n), Int[(a + b*Sin[c + d*x])^(n - 1), x], x] /; FreeQ[{a, b, c, d}, x] &&
EqQ[a^2 - b^2, 0] && IGtQ[n - 1/2, 0]

Rule 2830

Int[((a_) + (b_.)*sin[(e_.) + (f_.)*(x_)])^(m_)*((c_.) + (d_.)*sin[(e_.) + (f_.)*(x_)]), x_Symbol] :> Simp[(-d
)*Cos[e + f*x]*((a + b*Sin[e + f*x])^m/(f*(m + 1))), x] + Dist[(a*d*m + b*c*(m + 1))/(b*(m + 1)), Int[(a + b*S
in[e + f*x])^m, x], x] /; FreeQ[{a, b, c, d, e, f, m}, x] && NeQ[b*c - a*d, 0] && EqQ[a^2 - b^2, 0] &&  !LtQ[m
, -2^(-1)]

Rubi steps

\begin {align*} \int (a+a \sin (e+f x))^{3/2} (A+B \sin (e+f x)) \, dx &=-\frac {2 B \cos (e+f x) (a+a \sin (e+f x))^{3/2}}{5 f}+\frac {1}{5} (5 A+3 B) \int (a+a \sin (e+f x))^{3/2} \, dx\\ &=-\frac {2 a (5 A+3 B) \cos (e+f x) \sqrt {a+a \sin (e+f x)}}{15 f}-\frac {2 B \cos (e+f x) (a+a \sin (e+f x))^{3/2}}{5 f}+\frac {1}{15} (4 a (5 A+3 B)) \int \sqrt {a+a \sin (e+f x)} \, dx\\ &=-\frac {8 a^2 (5 A+3 B) \cos (e+f x)}{15 f \sqrt {a+a \sin (e+f x)}}-\frac {2 a (5 A+3 B) \cos (e+f x) \sqrt {a+a \sin (e+f x)}}{15 f}-\frac {2 B \cos (e+f x) (a+a \sin (e+f x))^{3/2}}{5 f}\\ \end {align*}

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Mathematica [A]
time = 0.28, size = 101, normalized size = 1.00 \begin {gather*} -\frac {a \left (\cos \left (\frac {1}{2} (e+f x)\right )-\sin \left (\frac {1}{2} (e+f x)\right )\right ) \sqrt {a (1+\sin (e+f x))} (50 A+39 B-3 B \cos (2 (e+f x))+2 (5 A+9 B) \sin (e+f x))}{15 f \left (\cos \left (\frac {1}{2} (e+f x)\right )+\sin \left (\frac {1}{2} (e+f x)\right )\right )} \end {gather*}

Antiderivative was successfully verified.

[In]

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

[Out]

-1/15*(a*(Cos[(e + f*x)/2] - Sin[(e + f*x)/2])*Sqrt[a*(1 + Sin[e + f*x])]*(50*A + 39*B - 3*B*Cos[2*(e + f*x)]
+ 2*(5*A + 9*B)*Sin[e + f*x]))/(f*(Cos[(e + f*x)/2] + Sin[(e + f*x)/2]))

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Maple [A]
time = 5.49, size = 77, normalized size = 0.76

method	result	size
default	\(\frac {2 \left (1+\sin \left (f x +e \right )\right ) a^{2} \left (\sin \left (f x +e \right )-1\right ) \left (\sin \left (f x +e \right ) \left (5 A +9 B \right )-3 B \left (\cos ^{2}\left (f x +e \right )\right )+25 A +21 B \right )}{15 \cos \left (f x +e \right ) \sqrt {a +a \sin \left (f x +e \right )}\, f}\)	\(77\)

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)),x,method=_RETURNVERBOSE)

[Out]

2/15*(1+sin(f*x+e))*a^2*(sin(f*x+e)-1)*(sin(f*x+e)*(5*A+9*B)-3*B*cos(f*x+e)^2+25*A+21*B)/cos(f*x+e)/(a+a*sin(f
*x+e))^(1/2)/f

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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))^(3/2)*(A+B*sin(f*x+e)),x, algorithm="maxima")

[Out]

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

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Fricas [A]
time = 0.36, size = 146, normalized size = 1.45 \begin {gather*} \frac {2 \, {\left (3 \, B a \cos \left (f x + e\right )^{3} - {\left (5 \, A + 6 \, B\right )} a \cos \left (f x + e\right )^{2} - {\left (25 \, A + 21 \, B\right )} a \cos \left (f x + e\right ) - 4 \, {\left (5 \, A + 3 \, B\right )} a - {\left (3 \, B a \cos \left (f x + e\right )^{2} + {\left (5 \, A + 9 \, B\right )} a \cos \left (f x + e\right ) - 4 \, {\left (5 \, A + 3 \, B\right )} a\right )} \sin \left (f x + e\right )\right )} \sqrt {a \sin \left (f x + e\right ) + a}}{15 \, {\left (f \cos \left (f x + e\right ) + f \sin \left (f x + e\right ) + f\right )}} \end {gather*}

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)),x, algorithm="fricas")

[Out]

2/15*(3*B*a*cos(f*x + e)^3 - (5*A + 6*B)*a*cos(f*x + e)^2 - (25*A + 21*B)*a*cos(f*x + e) - 4*(5*A + 3*B)*a - (
3*B*a*cos(f*x + e)^2 + (5*A + 9*B)*a*cos(f*x + e) - 4*(5*A + 3*B)*a)*sin(f*x + e))*sqrt(a*sin(f*x + e) + a)/(f
*cos(f*x + e) + f*sin(f*x + e) + f)

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Sympy [F]
time = 0.00, size = 0, normalized size = 0.00 \begin {gather*} \int \left (a \left (\sin {\left (e + f x \right )} + 1\right )\right )^{\frac {3}{2}} \left (A + B \sin {\left (e + f x \right )}\right )\, dx \end {gather*}

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)),x)

[Out]

Integral((a*(sin(e + f*x) + 1))**(3/2)*(A + B*sin(e + f*x)), x)

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Giac [A]
time = 0.52, size = 147, normalized size = 1.46 \begin {gather*} \frac {\sqrt {2} {\left (3 \, B a \mathrm {sgn}\left (\cos \left (-\frac {1}{4} \, \pi + \frac {1}{2} \, f x + \frac {1}{2} \, e\right )\right ) \sin \left (-\frac {5}{4} \, \pi + \frac {5}{2} \, f x + \frac {5}{2} \, e\right ) + 30 \, {\left (3 \, A a \mathrm {sgn}\left (\cos \left (-\frac {1}{4} \, \pi + \frac {1}{2} \, f x + \frac {1}{2} \, e\right )\right ) + 2 \, B a \mathrm {sgn}\left (\cos \left (-\frac {1}{4} \, \pi + \frac {1}{2} \, f x + \frac {1}{2} \, e\right )\right )\right )} \sin \left (-\frac {1}{4} \, \pi + \frac {1}{2} \, f x + \frac {1}{2} \, e\right ) + 5 \, {\left (2 \, A a \mathrm {sgn}\left (\cos \left (-\frac {1}{4} \, \pi + \frac {1}{2} \, f x + \frac {1}{2} \, e\right )\right ) + 3 \, B a \mathrm {sgn}\left (\cos \left (-\frac {1}{4} \, \pi + \frac {1}{2} \, f x + \frac {1}{2} \, e\right )\right )\right )} \sin \left (-\frac {3}{4} \, \pi + \frac {3}{2} \, f x + \frac {3}{2} \, e\right )\right )} \sqrt {a}}{30 \, f} \end {gather*}

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)),x, algorithm="giac")

[Out]

1/30*sqrt(2)*(3*B*a*sgn(cos(-1/4*pi + 1/2*f*x + 1/2*e))*sin(-5/4*pi + 5/2*f*x + 5/2*e) + 30*(3*A*a*sgn(cos(-1/
4*pi + 1/2*f*x + 1/2*e)) + 2*B*a*sgn(cos(-1/4*pi + 1/2*f*x + 1/2*e)))*sin(-1/4*pi + 1/2*f*x + 1/2*e) + 5*(2*A*
a*sgn(cos(-1/4*pi + 1/2*f*x + 1/2*e)) + 3*B*a*sgn(cos(-1/4*pi + 1/2*f*x + 1/2*e)))*sin(-3/4*pi + 3/2*f*x + 3/2
*e))*sqrt(a)/f

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Mupad [F]
time = 0.00, size = -1, normalized size = -0.01 \begin {gather*} \int \left (A+B\,\sin \left (e+f\,x\right )\right )\,{\left (a+a\,\sin \left (e+f\,x\right )\right )}^{3/2} \,d x \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))^(3/2),x)

[Out]

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

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