Integrand size = 33, antiderivative size = 393 \[ \int x^3 \sqrt {a-a \sin (e+f x)} (c+c \sin (e+f x))^{3/2} \, dx=-\frac {6 c \sqrt {a-a \sin (e+f x)} \sqrt {c+c \sin (e+f x)}}{f^4}+\frac {3 c x^2 \sqrt {a-a \sin (e+f x)} \sqrt {c+c \sin (e+f x)}}{f^2}+\frac {3 c x \sec (e+f x) \sqrt {a-a \sin (e+f x)} \sqrt {c+c \sin (e+f x)}}{8 f^3}-\frac {3 c x^3 \sec (e+f x) \sqrt {a-a \sin (e+f x)} \sqrt {c+c \sin (e+f x)}}{4 f}-\frac {3 c \sin (e+f x) \sqrt {a-a \sin (e+f x)} \sqrt {c+c \sin (e+f x)}}{8 f^4}+\frac {3 c x^2 \sin (e+f x) \sqrt {a-a \sin (e+f x)} \sqrt {c+c \sin (e+f x)}}{4 f^2}+\frac {x^3 \sec (e+f x) \sqrt {a-a \sin (e+f x)} (c+c \sin (e+f x))^{5/2}}{2 c f}-\frac {6 c x \sqrt {a-a \sin (e+f x)} \sqrt {c+c \sin (e+f x)} \tan (e+f x)}{f^3}-\frac {3 c x \sin (e+f x) \sqrt {a-a \sin (e+f x)} \sqrt {c+c \sin (e+f x)} \tan (e+f x)}{4 f^3} \] Output:
-6*c*(a-a*sin(f*x+e))^(1/2)*(c+c*sin(f*x+e))^(1/2)/f^4+3*c*x^2*(a-a*sin(f* x+e))^(1/2)*(c+c*sin(f*x+e))^(1/2)/f^2+3/8*c*x*sec(f*x+e)*(a-a*sin(f*x+e)) ^(1/2)*(c+c*sin(f*x+e))^(1/2)/f^3-3/4*c*x^3*sec(f*x+e)*(a-a*sin(f*x+e))^(1 /2)*(c+c*sin(f*x+e))^(1/2)/f-3/8*c*sin(f*x+e)*(a-a*sin(f*x+e))^(1/2)*(c+c* sin(f*x+e))^(1/2)/f^4+3/4*c*x^2*sin(f*x+e)*(a-a*sin(f*x+e))^(1/2)*(c+c*sin (f*x+e))^(1/2)/f^2+1/2*x^3*sec(f*x+e)*(a-a*sin(f*x+e))^(1/2)*(c+c*sin(f*x+ e))^(5/2)/c/f-6*c*x*(a-a*sin(f*x+e))^(1/2)*(c+c*sin(f*x+e))^(1/2)*tan(f*x+ e)/f^3-3/4*c*x*sin(f*x+e)*(a-a*sin(f*x+e))^(1/2)*(c+c*sin(f*x+e))^(1/2)*ta n(f*x+e)/f^3
Time = 1.27 (sec) , antiderivative size = 113, normalized size of antiderivative = 0.29 \[ \int x^3 \sqrt {a-a \sin (e+f x)} (c+c \sin (e+f x))^{3/2} \, dx=\frac {c \sqrt {c (1+\sin (e+f x))} \sqrt {a-a \sin (e+f x)} \left (-f x \left (-3+2 f^2 x^2\right ) \cos (2 (e+f x)) \sec (e+f x)+\left (-3+6 f^2 x^2\right ) \sin (e+f x)+8 \left (-6+3 f^2 x^2+f x \left (-6+f^2 x^2\right ) \tan (e+f x)\right )\right )}{8 f^4} \] Input:
Integrate[x^3*Sqrt[a - a*Sin[e + f*x]]*(c + c*Sin[e + f*x])^(3/2),x]
Output:
(c*Sqrt[c*(1 + Sin[e + f*x])]*Sqrt[a - a*Sin[e + f*x]]*(-(f*x*(-3 + 2*f^2* x^2)*Cos[2*(e + f*x)]*Sec[e + f*x]) + (-3 + 6*f^2*x^2)*Sin[e + f*x] + 8*(- 6 + 3*f^2*x^2 + f*x*(-6 + f^2*x^2)*Tan[e + f*x])))/(8*f^4)
Time = 0.66 (sec) , antiderivative size = 206, normalized size of antiderivative = 0.52, number of steps used = 5, number of rules used = 5, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.152, Rules used = {5115, 4922, 3042, 3798, 2009}
Below are the steps used by Rubi to obtain the solution. The rule number used for the transformation is given above next to the arrow. The rules definitions used are listed below.
\(\displaystyle \int x^3 \sqrt {a-a \sin (e+f x)} (c \sin (e+f x)+c)^{3/2} \, dx\) |
\(\Big \downarrow \) 5115 |
\(\displaystyle \sec (e+f x) \sqrt {a-a \sin (e+f x)} \sqrt {c \sin (e+f x)+c} \int x^3 \cos (e+f x) (\sin (e+f x) c+c)dx\) |
\(\Big \downarrow \) 4922 |
\(\displaystyle \sec (e+f x) \sqrt {a-a \sin (e+f x)} \sqrt {c \sin (e+f x)+c} \left (\frac {x^3 (c \sin (e+f x)+c)^2}{2 c f}-\frac {3 \int x^2 (\sin (e+f x) c+c)^2dx}{2 c f}\right )\) |
\(\Big \downarrow \) 3042 |
\(\displaystyle \sec (e+f x) \sqrt {a-a \sin (e+f x)} \sqrt {c \sin (e+f x)+c} \left (\frac {x^3 (c \sin (e+f x)+c)^2}{2 c f}-\frac {3 \int x^2 (\sin (e+f x) c+c)^2dx}{2 c f}\right )\) |
\(\Big \downarrow \) 3798 |
\(\displaystyle \sec (e+f x) \sqrt {a-a \sin (e+f x)} \sqrt {c \sin (e+f x)+c} \left (\frac {x^3 (c \sin (e+f x)+c)^2}{2 c f}-\frac {3 \int \left (c^2 x^2+c^2 \sin ^2(e+f x) x^2+2 c^2 \sin (e+f x) x^2\right )dx}{2 c f}\right )\) |
\(\Big \downarrow \) 2009 |
\(\displaystyle \sec (e+f x) \sqrt {a-a \sin (e+f x)} \sqrt {c \sin (e+f x)+c} \left (\frac {x^3 (c \sin (e+f x)+c)^2}{2 c f}-\frac {3 \left (\frac {4 c^2 \cos (e+f x)}{f^3}+\frac {c^2 \sin (e+f x) \cos (e+f x)}{4 f^3}+\frac {c^2 x \sin ^2(e+f x)}{2 f^2}+\frac {4 c^2 x \sin (e+f x)}{f^2}-\frac {2 c^2 x^2 \cos (e+f x)}{f}-\frac {c^2 x^2 \sin (e+f x) \cos (e+f x)}{2 f}-\frac {c^2 x}{4 f^2}+\frac {c^2 x^3}{2}\right )}{2 c f}\right )\) |
Input:
Int[x^3*Sqrt[a - a*Sin[e + f*x]]*(c + c*Sin[e + f*x])^(3/2),x]
Output:
Sec[e + f*x]*Sqrt[a - a*Sin[e + f*x]]*Sqrt[c + c*Sin[e + f*x]]*((x^3*(c + c*Sin[e + f*x])^2)/(2*c*f) - (3*(-1/4*(c^2*x)/f^2 + (c^2*x^3)/2 + (4*c^2*C os[e + f*x])/f^3 - (2*c^2*x^2*Cos[e + f*x])/f + (4*c^2*x*Sin[e + f*x])/f^2 + (c^2*Cos[e + f*x]*Sin[e + f*x])/(4*f^3) - (c^2*x^2*Cos[e + f*x]*Sin[e + f*x])/(2*f) + (c^2*x*Sin[e + f*x]^2)/(2*f^2)))/(2*c*f))
Int[((c_.) + (d_.)*(x_))^(m_.)*((a_) + (b_.)*sin[(e_.) + (f_.)*(x_)])^(n_.) , x_Symbol] :> Int[ExpandIntegrand[(c + d*x)^m, (a + b*Sin[e + f*x])^n, x], x] /; FreeQ[{a, b, c, d, e, f, m}, x] && IGtQ[n, 0] && (EqQ[n, 1] || IGtQ[ m, 0] || NeQ[a^2 - b^2, 0])
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] - Simp[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]
Int[((g_.) + (h_.)*(x_))^(p_.)*((a_) + (b_.)*Sin[(e_.) + (f_.)*(x_)])^(m_)* ((c_) + (d_.)*Sin[(e_.) + (f_.)*(x_)])^(n_), x_Symbol] :> Simp[a^IntPart[m] *c^IntPart[m]*(a + b*Sin[e + f*x])^FracPart[m]*((c + d*Sin[e + f*x])^FracPa rt[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]
\[\int x^{3} \sqrt {a -a \sin \left (f x +e \right )}\, \left (c +c \sin \left (f x +e \right )\right )^{\frac {3}{2}}d x\]
Input:
int(x^3*(a-a*sin(f*x+e))^(1/2)*(c+c*sin(f*x+e))^(3/2),x)
Output:
int(x^3*(a-a*sin(f*x+e))^(1/2)*(c+c*sin(f*x+e))^(3/2),x)
Exception generated. \[ \int x^3 \sqrt {a-a \sin (e+f x)} (c+c \sin (e+f x))^{3/2} \, dx=\text {Exception raised: TypeError} \] Input:
integrate(x^3*(a-a*sin(f*x+e))^(1/2)*(c+c*sin(f*x+e))^(3/2),x, algorithm=" fricas")
Output:
Exception raised: TypeError >> Error detected within library code: inte grate: implementation incomplete (has polynomial part)
Timed out. \[ \int x^3 \sqrt {a-a \sin (e+f x)} (c+c \sin (e+f x))^{3/2} \, dx=\text {Timed out} \] Input:
integrate(x**3*(a-a*sin(f*x+e))**(1/2)*(c+c*sin(f*x+e))**(3/2),x)
Output:
Timed out
\[ \int x^3 \sqrt {a-a \sin (e+f x)} (c+c \sin (e+f x))^{3/2} \, dx=\int { \sqrt {-a \sin \left (f x + e\right ) + a} {\left (c \sin \left (f x + e\right ) + c\right )}^{\frac {3}{2}} x^{3} \,d x } \] Input:
integrate(x^3*(a-a*sin(f*x+e))^(1/2)*(c+c*sin(f*x+e))^(3/2),x, algorithm=" maxima")
Output:
integrate(sqrt(-a*sin(f*x + e) + a)*(c*sin(f*x + e) + c)^(3/2)*x^3, x)
Leaf count of result is larger than twice the leaf count of optimal. 1487 vs. \(2 (345) = 690\).
Time = 0.45 (sec) , antiderivative size = 1487, normalized size of antiderivative = 3.78 \[ \int x^3 \sqrt {a-a \sin (e+f x)} (c+c \sin (e+f x))^{3/2} \, dx=\text {Too large to display} \] Input:
integrate(x^3*(a-a*sin(f*x+e))^(1/2)*(c+c*sin(f*x+e))^(3/2),x, algorithm=" giac")
Output:
1/32*sqrt(a)*sqrt(c)*((pi^3*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)) - 3*pi^2*(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)) + 3*pi*(pi - 2*f*x - 2*e)^2*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)^3*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)) - 6*pi^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)) + 12*pi*(pi - 2*f*x - 2*e)*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)) - 6*(pi - 2*f*x - 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)) + 12*pi*c*e^2*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*(pi - 2*f*x - 2*e)*c*e^2*sgn(co s(-1/4*pi + 1/2*f*x + 1/2*e))*sgn(sin(-1/4*pi + 1/2*f*x + 1/2*e)) - 8*c*e^ 3*sgn(cos(-1/4*pi + 1/2*f*x + 1/2*e))*sgn(sin(-1/4*pi + 1/2*f*x + 1/2*e)) - 6*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)) + 6*(pi - 2*f*x - 2*e)*c*sgn(cos(-1/4*pi + 1/2*f*x + 1/2*e))*sgn(si n(-1/4*pi + 1/2*f*x + 1/2*e)) + 12*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^3 - 24*(pi^2*c*sg n(cos(-1/4*pi + 1/2*f*x + 1/2*e))*sgn(sin(-1/4*pi + 1/2*f*x + 1/2*e)) - 2* pi*(pi - 2*f*x - 2*e)*c*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)) + (pi - 2*f*x - 2*e)^2*c*sgn(cos(-1/4*pi + 1/2*f*...
Time = 17.94 (sec) , antiderivative size = 216, normalized size of antiderivative = 0.55 \[ \int x^3 \sqrt {a-a \sin (e+f x)} (c+c \sin (e+f x))^{3/2} \, dx=-\frac {c\,\sqrt {-a\,\left (\sin \left (e+f\,x\right )-1\right )}\,\sqrt {c\,\left (\sin \left (e+f\,x\right )+1\right )}\,\left (3\,\sin \left (e+f\,x\right )+96\,\cos \left (2\,e+2\,f\,x\right )+3\,\sin \left (3\,e+3\,f\,x\right )-48\,f^2\,x^2-6\,f\,x\,\cos \left (3\,e+3\,f\,x\right )+96\,f\,x\,\sin \left (2\,e+2\,f\,x\right )+4\,f^3\,x^3\,\cos \left (e+f\,x\right )-6\,f^2\,x^2\,\sin \left (e+f\,x\right )-6\,f\,x\,\cos \left (e+f\,x\right )-48\,f^2\,x^2\,\cos \left (2\,e+2\,f\,x\right )+4\,f^3\,x^3\,\cos \left (3\,e+3\,f\,x\right )-6\,f^2\,x^2\,\sin \left (3\,e+3\,f\,x\right )-16\,f^3\,x^3\,\sin \left (2\,e+2\,f\,x\right )+96\right )}{16\,f^4\,\left (\cos \left (2\,e+2\,f\,x\right )+1\right )} \] Input:
int(x^3*(a - a*sin(e + f*x))^(1/2)*(c + c*sin(e + f*x))^(3/2),x)
Output:
-(c*(-a*(sin(e + f*x) - 1))^(1/2)*(c*(sin(e + f*x) + 1))^(1/2)*(3*sin(e + f*x) + 96*cos(2*e + 2*f*x) + 3*sin(3*e + 3*f*x) - 48*f^2*x^2 - 6*f*x*cos(3 *e + 3*f*x) + 96*f*x*sin(2*e + 2*f*x) + 4*f^3*x^3*cos(e + f*x) - 6*f^2*x^2 *sin(e + f*x) - 6*f*x*cos(e + f*x) - 48*f^2*x^2*cos(2*e + 2*f*x) + 4*f^3*x ^3*cos(3*e + 3*f*x) - 6*f^2*x^2*sin(3*e + 3*f*x) - 16*f^3*x^3*sin(2*e + 2* f*x) + 96))/(16*f^4*(cos(2*e + 2*f*x) + 1))
\[ \int x^3 \sqrt {a-a \sin (e+f x)} (c+c \sin (e+f x))^{3/2} \, dx=\sqrt {c}\, \sqrt {a}\, c \left (\int \sqrt {\sin \left (f x +e \right )+1}\, \sqrt {-\sin \left (f x +e \right )+1}\, x^{3}d x +\int \sqrt {\sin \left (f x +e \right )+1}\, \sqrt {-\sin \left (f x +e \right )+1}\, \sin \left (f x +e \right ) x^{3}d x \right ) \] Input:
int(x^3*(a-a*sin(f*x+e))^(1/2)*(c+c*sin(f*x+e))^(3/2),x)
Output:
sqrt(c)*sqrt(a)*c*(int(sqrt(sin(e + f*x) + 1)*sqrt( - sin(e + f*x) + 1)*x* *3,x) + int(sqrt(sin(e + f*x) + 1)*sqrt( - sin(e + f*x) + 1)*sin(e + f*x)* x**3,x))