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

Optimal. Leaf size=157 \[ -\frac {8 a^2 \left (35 c^2+42 c d+19 d^2\right ) \cos (e+f x)}{105 f \sqrt {a \sin (e+f x)+a}}-\frac {2 a \left (35 c^2+42 c d+19 d^2\right ) \cos (e+f x) \sqrt {a \sin (e+f x)+a}}{105 f}-\frac {4 d (7 c-d) \cos (e+f x) (a \sin (e+f x)+a)^{3/2}}{35 f}-\frac {2 d^2 \cos (e+f x) (a \sin (e+f x)+a)^{5/2}}{7 a f} \]

[Out]

-4/35*(7*c-d)*d*cos(f*x+e)*(a+a*sin(f*x+e))^(3/2)/f-2/7*d^2*cos(f*x+e)*(a+a*sin(f*x+e))^(5/2)/a/f-8/105*a^2*(3
5*c^2+42*c*d+19*d^2)*cos(f*x+e)/f/(a+a*sin(f*x+e))^(1/2)-2/105*a*(35*c^2+42*c*d+19*d^2)*cos(f*x+e)*(a+a*sin(f*
x+e))^(1/2)/f

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Rubi [A]  time = 0.23, antiderivative size = 157, normalized size of antiderivative = 1.00, number of steps used = 4, number of rules used = 4, integrand size = 27, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.148, Rules used = {2761, 2751, 2647, 2646} \[ -\frac {8 a^2 \left (35 c^2+42 c d+19 d^2\right ) \cos (e+f x)}{105 f \sqrt {a \sin (e+f x)+a}}-\frac {2 a \left (35 c^2+42 c d+19 d^2\right ) \cos (e+f x) \sqrt {a \sin (e+f x)+a}}{105 f}-\frac {4 d (7 c-d) \cos (e+f x) (a \sin (e+f x)+a)^{3/2}}{35 f}-\frac {2 d^2 \cos (e+f x) (a \sin (e+f x)+a)^{5/2}}{7 a f} \]

Antiderivative was successfully verified.

[In]

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

[Out]

(-8*a^2*(35*c^2 + 42*c*d + 19*d^2)*Cos[e + f*x])/(105*f*Sqrt[a + a*Sin[e + f*x]]) - (2*a*(35*c^2 + 42*c*d + 19
*d^2)*Cos[e + f*x]*Sqrt[a + a*Sin[e + f*x]])/(105*f) - (4*(7*c - d)*d*Cos[e + f*x]*(a + a*Sin[e + f*x])^(3/2))
/(35*f) - (2*d^2*Cos[e + f*x]*(a + a*Sin[e + f*x])^(5/2))/(7*a*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)]

Rule 2761

Int[((a_) + (b_.)*sin[(e_.) + (f_.)*(x_)])^(m_)*((c_) + (d_.)*sin[(e_.) + (f_.)*(x_)])^2, x_Symbol] :> -Simp[(
d^2*Cos[e + f*x]*(a + b*Sin[e + f*x])^(m + 1))/(b*f*(m + 2)), x] + Dist[1/(b*(m + 2)), Int[(a + b*Sin[e + f*x]
)^m*Simp[b*(d^2*(m + 1) + c^2*(m + 2)) - d*(a*d - 2*b*c*(m + 2))*Sin[e + f*x], x], 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, -1]

Rubi steps

\begin {align*} \int (a+a \sin (e+f x))^{3/2} (c+d \sin (e+f x))^2 \, dx &=-\frac {2 d^2 \cos (e+f x) (a+a \sin (e+f x))^{5/2}}{7 a f}+\frac {2 \int (a+a \sin (e+f x))^{3/2} \left (\frac {1}{2} a \left (7 c^2+5 d^2\right )+a (7 c-d) d \sin (e+f x)\right ) \, dx}{7 a}\\ &=-\frac {4 (7 c-d) d \cos (e+f x) (a+a \sin (e+f x))^{3/2}}{35 f}-\frac {2 d^2 \cos (e+f x) (a+a \sin (e+f x))^{5/2}}{7 a f}+\frac {1}{35} \left (35 c^2+42 c d+19 d^2\right ) \int (a+a \sin (e+f x))^{3/2} \, dx\\ &=-\frac {2 a \left (35 c^2+42 c d+19 d^2\right ) \cos (e+f x) \sqrt {a+a \sin (e+f x)}}{105 f}-\frac {4 (7 c-d) d \cos (e+f x) (a+a \sin (e+f x))^{3/2}}{35 f}-\frac {2 d^2 \cos (e+f x) (a+a \sin (e+f x))^{5/2}}{7 a f}+\frac {1}{105} \left (4 a \left (35 c^2+42 c d+19 d^2\right )\right ) \int \sqrt {a+a \sin (e+f x)} \, dx\\ &=-\frac {8 a^2 \left (35 c^2+42 c d+19 d^2\right ) \cos (e+f x)}{105 f \sqrt {a+a \sin (e+f x)}}-\frac {2 a \left (35 c^2+42 c d+19 d^2\right ) \cos (e+f x) \sqrt {a+a \sin (e+f x)}}{105 f}-\frac {4 (7 c-d) d \cos (e+f x) (a+a \sin (e+f x))^{3/2}}{35 f}-\frac {2 d^2 \cos (e+f x) (a+a \sin (e+f x))^{5/2}}{7 a f}\\ \end {align*}

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Mathematica [A]  time = 0.88, size = 136, normalized size = 0.87 \[ -\frac {a \sqrt {a (\sin (e+f x)+1)} \left (\cos \left (\frac {1}{2} (e+f x)\right )-\sin \left (\frac {1}{2} (e+f x)\right )\right ) \left (\left (140 c^2+504 c d+253 d^2\right ) \sin (e+f x)+700 c^2-6 d (14 c+13 d) \cos (2 (e+f x))+1092 c d-15 d^2 \sin (3 (e+f x))+494 d^2\right )}{210 f \left (\sin \left (\frac {1}{2} (e+f x)\right )+\cos \left (\frac {1}{2} (e+f x)\right )\right )} \]

Antiderivative was successfully verified.

[In]

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

[Out]

-1/210*(a*(Cos[(e + f*x)/2] - Sin[(e + f*x)/2])*Sqrt[a*(1 + Sin[e + f*x])]*(700*c^2 + 1092*c*d + 494*d^2 - 6*d
*(14*c + 13*d)*Cos[2*(e + f*x)] + (140*c^2 + 504*c*d + 253*d^2)*Sin[e + f*x] - 15*d^2*Sin[3*(e + f*x)]))/(f*(C
os[(e + f*x)/2] + Sin[(e + f*x)/2]))

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fricas [A]  time = 0.45, size = 229, normalized size = 1.46 \[ \frac {2 \, {\left (15 \, a d^{2} \cos \left (f x + e\right )^{4} + 3 \, {\left (14 \, a c d + 13 \, a d^{2}\right )} \cos \left (f x + e\right )^{3} - 140 \, a c^{2} - 168 \, a c d - 76 \, a d^{2} - {\left (35 \, a c^{2} + 84 \, a c d + 43 \, a d^{2}\right )} \cos \left (f x + e\right )^{2} - {\left (175 \, a c^{2} + 294 \, a c d + 143 \, a d^{2}\right )} \cos \left (f x + e\right ) + {\left (15 \, a d^{2} \cos \left (f x + e\right )^{3} + 140 \, a c^{2} + 168 \, a c d + 76 \, a d^{2} - 6 \, {\left (7 \, a c d + 4 \, a d^{2}\right )} \cos \left (f x + e\right )^{2} - {\left (35 \, a c^{2} + 126 \, a c d + 67 \, a d^{2}\right )} \cos \left (f x + e\right )\right )} \sin \left (f x + e\right )\right )} \sqrt {a \sin \left (f x + e\right ) + a}}{105 \, {\left (f \cos \left (f x + e\right ) + f \sin \left (f x + e\right ) + f\right )}} \]

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))^2,x, algorithm="fricas")

[Out]

2/105*(15*a*d^2*cos(f*x + e)^4 + 3*(14*a*c*d + 13*a*d^2)*cos(f*x + e)^3 - 140*a*c^2 - 168*a*c*d - 76*a*d^2 - (
35*a*c^2 + 84*a*c*d + 43*a*d^2)*cos(f*x + e)^2 - (175*a*c^2 + 294*a*c*d + 143*a*d^2)*cos(f*x + e) + (15*a*d^2*
cos(f*x + e)^3 + 140*a*c^2 + 168*a*c*d + 76*a*d^2 - 6*(7*a*c*d + 4*a*d^2)*cos(f*x + e)^2 - (35*a*c^2 + 126*a*c
*d + 67*a*d^2)*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))^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)sqrt(2*a)*(12*f*(-2*a*d^2*sign(cos(1/2*(
f*x+exp(1))-1/4*pi))-4*a*c*d*sign(cos(1/2*(f*x+exp(1))-1/4*pi)))*sin(1/4*(6*f*x+6*exp(1)+pi))/(12*f)^2+20*f*(-
2*a*d^2*sign(cos(1/2*(f*x+exp(1))-1/4*pi))-4*a*c*d*sign(cos(1/2*(f*x+exp(1))-1/4*pi)))*sin(1/4*(10*f*x+10*exp(
1)-pi))/(20*f)^2+2*f*(4*a*c^2*sign(cos(1/2*(f*x+exp(1))-1/4*pi))+2*a*d^2*sign(cos(1/2*(f*x+exp(1))-1/4*pi))+4*
a*c*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)^2-8*f*(8*a*c^2*sign(cos(1/2*(f*
x+exp(1))-1/4*pi))+6*a*d^2*sign(cos(1/2*(f*x+exp(1))-1/4*pi))+16*a*c*d*sign(cos(1/2*(f*x+exp(1))-1/4*pi)))*cos
(1/4*(2*f*x+2*exp(1)+pi))/(8*f)^2-24*f*(8*a*c^2*sign(cos(1/2*(f*x+exp(1))-1/4*pi))+6*a*d^2*sign(cos(1/2*(f*x+e
xp(1))-1/4*pi))+16*a*c*d*sign(cos(1/2*(f*x+exp(1))-1/4*pi)))*cos(1/4*(6*f*x+6*exp(1)-pi))/(24*f)^2+80*a*d^2*f*
sign(cos(1/2*(f*x+exp(1))-1/4*pi))*cos(1/4*(10*f*x+10*exp(1)+pi))/(40*f)^2+112*a*d^2*f*sign(cos(1/2*(f*x+exp(1
))-1/4*pi))*cos(1/4*(14*f*x+14*exp(1)-pi))/(56*f)^2)

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maple [A]  time = 0.89, size = 130, normalized size = 0.83 \[ \frac {2 \left (1+\sin \left (f x +e \right )\right ) a^{2} \left (\sin \left (f x +e \right )-1\right ) \left (15 d^{2} \left (\sin ^{3}\left (f x +e \right )\right )+42 c d \left (\sin ^{2}\left (f x +e \right )\right )+39 d^{2} \left (\sin ^{2}\left (f x +e \right )\right )+35 c^{2} \sin \left (f x +e \right )+126 c d \sin \left (f x +e \right )+52 d^{2} \sin \left (f x +e \right )+175 c^{2}+252 c d +104 d^{2}\right )}{105 \cos \left (f x +e \right ) \sqrt {a +a \sin \left (f x +e \right )}\, f} \]

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))^2,x)

[Out]

2/105*(1+sin(f*x+e))*a^2*(sin(f*x+e)-1)*(15*d^2*sin(f*x+e)^3+42*c*d*sin(f*x+e)^2+39*d^2*sin(f*x+e)^2+35*c^2*si
n(f*x+e)+126*c*d*sin(f*x+e)+52*d^2*sin(f*x+e)+175*c^2+252*c*d+104*d^2)/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 \[ \int {\left (a \sin \left (f x + e\right ) + a\right )}^{\frac {3}{2}} {\left (d \sin \left (f x + e\right ) + c\right )}^{2}\,{d x} \]

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))^2,x, algorithm="maxima")

[Out]

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

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

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))^2,x)

[Out]

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

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

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))**2,x)

[Out]

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

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