\(\int x \sqrt {a-a \sin (e+f x)} (c+c \sin (e+f x))^{3/2} \, dx\) [173]

Optimal result

Integrand size = 31, antiderivative size = 168 \[ \int x \sqrt {a-a \sin (e+f x)} (c+c \sin (e+f x))^{3/2} \, dx=\frac {c \sqrt {a-a \sin (e+f x)} \sqrt {c+c \sin (e+f x)}}{f^2}-\frac {3 c x \sec (e+f x) \sqrt {a-a \sin (e+f x)} \sqrt {c+c \sin (e+f x)}}{4 f}+\frac {c \sin (e+f x) \sqrt {a-a \sin (e+f x)} \sqrt {c+c \sin (e+f x)}}{4 f^2}+\frac {x \sec (e+f x) \sqrt {a-a \sin (e+f x)} (c+c \sin (e+f x))^{5/2}}{2 c f} \]

[Out]

1/2*x*sec(f*x+e)*(c+c*sin(f*x+e))^(5/2)*(a-a*sin(f*x+e))^(1/2)/c/f+c*(a-a*sin(f*x+e))^(1/2)*(c+c*sin(f*x+e))^(
1/2)/f^2-3/4*c*x*sec(f*x+e)*(a-a*sin(f*x+e))^(1/2)*(c+c*sin(f*x+e))^(1/2)/f+1/4*c*sin(f*x+e)*(a-a*sin(f*x+e))^
(1/2)*(c+c*sin(f*x+e))^(1/2)/f^2

Rubi [A] (verified)

Time = 0.18 (sec) , antiderivative size = 168, normalized size of antiderivative = 1.00, number of steps used = 3, number of rules used = 3, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.097, Rules used = {4700, 4507, 2723} \[ \int x \sqrt {a-a \sin (e+f x)} (c+c \sin (e+f x))^{3/2} \, dx=\frac {c \sin (e+f x) \sqrt {a-a \sin (e+f x)} \sqrt {c \sin (e+f x)+c}}{4 f^2}+\frac {c \sqrt {a-a \sin (e+f x)} \sqrt {c \sin (e+f x)+c}}{f^2}+\frac {x \sec (e+f x) \sqrt {a-a \sin (e+f x)} (c \sin (e+f x)+c)^{5/2}}{2 c f}-\frac {3 c x \sec (e+f x) \sqrt {a-a \sin (e+f x)} \sqrt {c \sin (e+f x)+c}}{4 f} \]

[In]

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

[Out]

(c*Sqrt[a - a*Sin[e + f*x]]*Sqrt[c + c*Sin[e + f*x]])/f^2 - (3*c*x*Sec[e + f*x]*Sqrt[a - a*Sin[e + f*x]]*Sqrt[
c + c*Sin[e + f*x]])/(4*f) + (c*Sin[e + f*x]*Sqrt[a - a*Sin[e + f*x]]*Sqrt[c + c*Sin[e + f*x]])/(4*f^2) + (x*S
ec[e + f*x]*Sqrt[a - a*Sin[e + f*x]]*(c + c*Sin[e + f*x])^(5/2))/(2*c*f)

Rule 2723

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

Rule 4507

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

Rule 4700

Int[((g_.) + (h_.)*(x_))^(p_.)*((a_) + (b_.)*Sin[(e_.) + (f_.)*(x_)])^(m_)*((c_) + (d_.)*Sin[(e_.) + (f_.)*(x_
)])^(n_), x_Symbol] :> Dist[a^IntPart[m]*c^IntPart[m]*(a + b*Sin[e + f*x])^FracPart[m]*((c + d*Sin[e + f*x])^F
racPart[m]/Cos[e + f*x]^(2*FracPart[m])), Int[(g + h*x)^p*Cos[e + f*x]^(2*m)*(c + d*Sin[e + f*x])^(n - m), x],
x] /; FreeQ[{a, b, c, d, e, f, g, h}, x] && EqQ[b*c + a*d, 0] && EqQ[a^2 - b^2, 0] && IntegerQ[p] && IntegerQ
[2*m] && IGeQ[n - m, 0]

Rubi steps \begin{align*} \text {integral}& = \left (\sec (e+f x) \sqrt {a-a \sin (e+f x)} \sqrt {c+c \sin (e+f x)}\right ) \int x \cos (e+f x) (c+c \sin (e+f x)) \, dx \\ & = \frac {x \sec (e+f x) \sqrt {a-a \sin (e+f x)} (c+c \sin (e+f x))^{5/2}}{2 c f}-\frac {\left (\sec (e+f x) \sqrt {a-a \sin (e+f x)} \sqrt {c+c \sin (e+f x)}\right ) \int (c+c \sin (e+f x))^2 \, dx}{2 c f} \\ & = \frac {c \sqrt {a-a \sin (e+f x)} \sqrt {c+c \sin (e+f x)}}{f^2}-\frac {3 c x \sec (e+f x) \sqrt {a-a \sin (e+f x)} \sqrt {c+c \sin (e+f x)}}{4 f}+\frac {c \sin (e+f x) \sqrt {a-a \sin (e+f x)} \sqrt {c+c \sin (e+f x)}}{4 f^2}+\frac {x \sec (e+f x) \sqrt {a-a \sin (e+f x)} (c+c \sin (e+f x))^{5/2}}{2 c f} \\ \end{align*}

Mathematica [A] (verified)

Time = 0.89 (sec) , antiderivative size = 73, normalized size of antiderivative = 0.43 \[ \int x \sqrt {a-a \sin (e+f x)} (c+c \sin (e+f x))^{3/2} \, dx=\frac {c \sqrt {c (1+\sin (e+f x))} \sqrt {a-a \sin (e+f x)} (4-f x \cos (2 (e+f x)) \sec (e+f x)+\sin (e+f x)+4 f x \tan (e+f x))}{4 f^2} \]

[In]

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

[Out]

(c*Sqrt[c*(1 + Sin[e + f*x])]*Sqrt[a - a*Sin[e + f*x]]*(4 - f*x*Cos[2*(e + f*x)]*Sec[e + f*x] + Sin[e + f*x] +
4*f*x*Tan[e + f*x]))/(4*f^2)

Maple [F]

\[\int x \left (c +c \sin \left (f x +e \right )\right )^{\frac {3}{2}} \sqrt {a -\sin \left (f x +e \right ) a}d x\]

[In]

int(x*(c+c*sin(f*x+e))^(3/2)*(a-sin(f*x+e)*a)^(1/2),x)

[Out]

int(x*(c+c*sin(f*x+e))^(3/2)*(a-sin(f*x+e)*a)^(1/2),x)

Fricas [F(-2)]

Exception generated. \[ \int x \sqrt {a-a \sin (e+f x)} (c+c \sin (e+f x))^{3/2} \, dx=\text {Exception raised: TypeError} \]

[In]

integrate(x*(c+c*sin(f*x+e))^(3/2)*(a-a*sin(f*x+e))^(1/2),x, algorithm="fricas")

[Out]

Exception raised: TypeError >>  Error detected within library code:   integrate: implementation incomplete (ha
s polynomial part)

Sympy [F]

\[ \int x \sqrt {a-a \sin (e+f x)} (c+c \sin (e+f x))^{3/2} \, dx=\int x \left (c \left (\sin {\left (e + f x \right )} + 1\right )\right )^{\frac {3}{2}} \sqrt {- a \left (\sin {\left (e + f x \right )} - 1\right )}\, dx \]

[In]

integrate(x*(c+c*sin(f*x+e))**(3/2)*(a-a*sin(f*x+e))**(1/2),x)

[Out]

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

Maxima [F]

\[ \int x \sqrt {a-a \sin (e+f x)} (c+c \sin (e+f x))^{3/2} \, dx=\int { \sqrt {-a \sin \left (f x + e\right ) + a} {\left (c \sin \left (f x + e\right ) + c\right )}^{\frac {3}{2}} x \,d x } \]

[In]

integrate(x*(c+c*sin(f*x+e))^(3/2)*(a-a*sin(f*x+e))^(1/2),x, algorithm="maxima")

[Out]

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

Giac [B] (verification not implemented)

Leaf count of result is larger than twice the leaf count of optimal. 311 vs. \(2 (146) = 292\).

Time = 0.34 (sec) , antiderivative size = 311, normalized size of antiderivative = 1.85 \[ \int x \sqrt {a-a \sin (e+f x)} (c+c \sin (e+f x))^{3/2} \, dx=-\frac {{\left (\frac {8 \, c \cos \left (f x + e\right ) \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 )}{f} + \frac {c \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 ) \sin \left (2 \, f x + 2 \, e\right )}{f} - \frac {{\left (\pi c \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 ) - {\left (\pi - 2 \, f x - 2 \, e\right )} c \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 ) - 2 \, c e \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 )} \cos \left (2 \, f x + 2 \, e\right )}{f} + \frac {4 \, {\left (\pi c \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 ) - {\left (\pi - 2 \, f x - 2 \, e\right )} c \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 ) - 2 \, c e \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 )} \sin \left (f x + e\right )}{f}\right )} \sqrt {a} \sqrt {c}}{8 \, f} \]

[In]

integrate(x*(c+c*sin(f*x+e))^(3/2)*(a-a*sin(f*x+e))^(1/2),x, algorithm="giac")

[Out]

-1/8*(8*c*cos(f*x + e)*sgn(cos(-1/4*pi + 1/2*f*x + 1/2*e))*sgn(sin(-1/4*pi + 1/2*f*x + 1/2*e))/f + c*sgn(cos(-
1/4*pi + 1/2*f*x + 1/2*e))*sgn(sin(-1/4*pi + 1/2*f*x + 1/2*e))*sin(2*f*x + 2*e)/f - (pi*c*sgn(cos(-1/4*pi + 1/
2*f*x + 1/2*e))*sgn(sin(-1/4*pi + 1/2*f*x + 1/2*e)) - (pi - 2*f*x - 2*e)*c*sgn(cos(-1/4*pi + 1/2*f*x + 1/2*e))
*sgn(sin(-1/4*pi + 1/2*f*x + 1/2*e)) - 2*c*e*sgn(cos(-1/4*pi + 1/2*f*x + 1/2*e))*sgn(sin(-1/4*pi + 1/2*f*x + 1
/2*e)))*cos(2*f*x + 2*e)/f + 4*(pi*c*sgn(cos(-1/4*pi + 1/2*f*x + 1/2*e))*sgn(sin(-1/4*pi + 1/2*f*x + 1/2*e)) -
(pi - 2*f*x - 2*e)*c*sgn(cos(-1/4*pi + 1/2*f*x + 1/2*e))*sgn(sin(-1/4*pi + 1/2*f*x + 1/2*e)) - 2*c*e*sgn(cos(
-1/4*pi + 1/2*f*x + 1/2*e))*sgn(sin(-1/4*pi + 1/2*f*x + 1/2*e)))*sin(f*x + e)/f)*sqrt(a)*sqrt(c)/f

Mupad [B] (verification not implemented)

Time = 29.20 (sec) , antiderivative size = 123, normalized size of antiderivative = 0.73 \[ \int x \sqrt {a-a \sin (e+f x)} (c+c \sin (e+f x))^{3/2} \, dx=-\frac {c\,\sqrt {-a\,\left (\sin \left (e+f\,x\right )-1\right )}\,\sqrt {c\,\left (\sin \left (e+f\,x\right )+1\right )}\,\left (-16\,{\sin \left (e+f\,x\right )}^2+\sin \left (e+f\,x\right )+\sin \left (3\,e+3\,f\,x\right )+8\,f\,x\,\sin \left (2\,e+2\,f\,x\right )+2\,f\,x\,\left (2\,{\sin \left (\frac {e}{2}+\frac {f\,x}{2}\right )}^2-1\right )+2\,f\,x\,\left (2\,{\sin \left (\frac {3\,e}{2}+\frac {3\,f\,x}{2}\right )}^2-1\right )+16\right )}{8\,f^2\,\left (2\,{\sin \left (e+f\,x\right )}^2-2\right )} \]

[In]

int(x*(a - a*sin(e + f*x))^(1/2)*(c + c*sin(e + f*x))^(3/2),x)

[Out]

-(c*(-a*(sin(e + f*x) - 1))^(1/2)*(c*(sin(e + f*x) + 1))^(1/2)*(sin(e + f*x) + sin(3*e + 3*f*x) - 16*sin(e + f
*x)^2 + 8*f*x*sin(2*e + 2*f*x) + 2*f*x*(2*sin(e/2 + (f*x)/2)^2 - 1) + 2*f*x*(2*sin((3*e)/2 + (3*f*x)/2)^2 - 1)
+ 16))/(8*f^2*(2*sin(e + f*x)^2 - 2))