3.531 \(\int (a+a \sin (e+f x))^{3/2} (c+d \sin (e+f x)) \, dx\)

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

[Out]

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

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Rubi [A]  time = 0.08, 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 = {2751, 2647, 2646} \[ -\frac {8 a^2 (5 c+3 d) \cos (e+f x)}{15 f \sqrt {a \sin (e+f x)+a}}-\frac {2 a (5 c+3 d) \cos (e+f x) \sqrt {a \sin (e+f x)+a}}{15 f}-\frac {2 d \cos (e+f x) (a \sin (e+f x)+a)^{3/2}}{5 f} \]

Antiderivative was successfully verified.

[In]

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

[Out]

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

Rule 2646

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 2647

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] && Eq
Q[a^2 - b^2, 0] && IGtQ[n - 1/2, 0]

Rule 2751

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

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Mathematica [A]  time = 0.39, size = 101, normalized size = 1.00 \[ -\frac {a \sqrt {a (\sin (e+f x)+1)} \left (\cos \left (\frac {1}{2} (e+f x)\right )-\sin \left (\frac {1}{2} (e+f x)\right )\right ) (2 (5 c+9 d) \sin (e+f x)+50 c-3 d \cos (2 (e+f x))+39 d)}{15 f \left (\sin \left (\frac {1}{2} (e+f x)\right )+\cos \left (\frac {1}{2} (e+f x)\right )\right )} \]

Antiderivative was successfully verified.

[In]

Integrate[(a + a*Sin[e + f*x])^(3/2)*(c + d*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*c + 39*d - 3*d*Cos[2*(e + f*x)]
+ 2*(5*c + 9*d)*Sin[e + f*x]))/(f*(Cos[(e + f*x)/2] + Sin[(e + f*x)/2]))

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fricas [A]  time = 0.45, size = 136, normalized size = 1.35 \[ \frac {2 \, {\left (3 \, a d \cos \left (f x + e\right )^{3} - {\left (5 \, a c + 6 \, a d\right )} \cos \left (f x + e\right )^{2} - 20 \, a c - 12 \, a d - {\left (25 \, a c + 21 \, a d\right )} \cos \left (f x + e\right ) - {\left (3 \, a d \cos \left (f x + e\right )^{2} - 20 \, a c - 12 \, a d + {\left (5 \, a c + 9 \, a d\right )} \cos \left (f x + e\right )\right )} \sin \left (f x + e\right )\right )} \sqrt {a \sin \left (f x + e\right ) + a}}{15 \, {\left (f \cos \left (f x + e\right ) + f \sin \left (f x + e\right ) + f\right )}} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

2/15*(3*a*d*cos(f*x + e)^3 - (5*a*c + 6*a*d)*cos(f*x + e)^2 - 20*a*c - 12*a*d - (25*a*c + 21*a*d)*cos(f*x + e)
- (3*a*d*cos(f*x + e)^2 - 20*a*c - 12*a*d + (5*a*c + 9*a*d)*cos(f*x + e))*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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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)*(c+d*sin(f*x+e)),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)sqrt(2*a)*(-2*f*(2
*a*c*sign(cos(1/2*(f*x+exp(1))-1/4*pi))+2*a*d*sign(cos(1/2*(f*x+exp(1))-1/4*pi)))*cos(1/4*(2*f*x+2*exp(1)+pi))
/(2*f)^2-6*f*(2*a*c*sign(cos(1/2*(f*x+exp(1))-1/4*pi))+2*a*d*sign(cos(1/2*(f*x+exp(1))-1/4*pi)))*cos(1/4*(6*f*
x+6*exp(1)-pi))/(6*f)^2+2*f*(4*a*c*sign(cos(1/2*(f*x+exp(1))-1/4*pi))+2*a*d*sign(cos(1/2*(f*x+exp(1))-1/4*pi))
)*sin(1/4*(2*f*x-pi)+1/2*exp(1))/(2*f)^2-24*a*d*f*sign(cos(1/2*(f*x+exp(1))-1/4*pi))*sin(1/4*(6*f*x+6*exp(1)+p
i))/(12*f)^2-40*a*d*f*sign(cos(1/2*(f*x+exp(1))-1/4*pi))*sin(1/4*(10*f*x+10*exp(1)-pi))/(20*f)^2)

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maple [A]  time = 0.71, size = 77, normalized size = 0.76 \[ \frac {2 \left (1+\sin \left (f x +e \right )\right ) a^{2} \left (\sin \left (f x +e \right )-1\right ) \left (\sin \left (f x +e \right ) \left (5 c +9 d \right )-3 \left (\cos ^{2}\left (f x +e \right )\right ) d +25 c +21 d \right )}{15 \cos \left (f x +e \right ) \sqrt {a +a \sin \left (f x +e \right )}\, f} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

int((a+a*sin(f*x+e))^(3/2)*(c+d*sin(f*x+e)),x)

[Out]

2/15*(1+sin(f*x+e))*a^2*(sin(f*x+e)-1)*(sin(f*x+e)*(5*c+9*d)-3*cos(f*x+e)^2*d+25*c+21*d)/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 \[ \int {\left (a \sin \left (f x + e\right ) + a\right )}^{\frac {3}{2}} {\left (d \sin \left (f x + e\right ) + c\right )}\,{d x} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

integrate((a*sin(f*x + e) + a)^(3/2)*(d*sin(f*x + e) + c), x)

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mupad [F]  time = 0.00, size = -1, normalized size = -0.01 \[ \int {\left (a+a\,\sin \left (e+f\,x\right )\right )}^{3/2}\,\left (c+d\,\sin \left (e+f\,x\right )\right ) \,d x \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

int((a + a*sin(e + f*x))^(3/2)*(c + d*sin(e + f*x)),x)

[Out]

int((a + a*sin(e + f*x))^(3/2)*(c + d*sin(e + f*x)), x)

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sympy [F]  time = 0.00, size = 0, normalized size = 0.00 \[ \int \left (a \left (\sin {\left (e + f x \right )} + 1\right )\right )^{\frac {3}{2}} \left (c + d \sin {\left (e + f x \right )}\right )\, dx \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate((a+a*sin(f*x+e))**(3/2)*(c+d*sin(f*x+e)),x)

[Out]

Integral((a*(sin(e + f*x) + 1))**(3/2)*(c + d*sin(e + f*x)), x)

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