Integrand size = 30, antiderivative size = 332 \[ \int \frac {(f-c f x)^{3/2} (a+b \arcsin (c x))}{(d+c d x)^{5/2}} \, dx=-\frac {4 b f^2 \sqrt {1-c^2 x^2}}{3 c d^2 (1+c x) \sqrt {d+c d x} \sqrt {f-c f x}}-\frac {b f^2 \sqrt {1-c^2 x^2} \arcsin (c x)^2}{2 c d^2 \sqrt {d+c d x} \sqrt {f-c f x}}+\frac {2 f^2 (1-c x) (a+b \arcsin (c x))}{c d^2 \sqrt {d+c d x} \sqrt {f-c f x}}-\frac {2 f^2 (1-c x)^3 (a+b \arcsin (c x))}{3 c d^2 \sqrt {d+c d x} \sqrt {f-c f x} \left (1-c^2 x^2\right )}+\frac {f^2 \sqrt {1-c^2 x^2} \arcsin (c x) (a+b \arcsin (c x))}{c d^2 \sqrt {d+c d x} \sqrt {f-c f x}}-\frac {8 b f^2 \sqrt {1-c^2 x^2} \log (1+c x)}{3 c d^2 \sqrt {d+c d x} \sqrt {f-c f x}} \] Output:
-4/3*b*f^2*(-c^2*x^2+1)^(1/2)/c/d^2/(c*x+1)/(c*d*x+d)^(1/2)/(-c*f*x+f)^(1/ 2)-1/2*b*f^2*(-c^2*x^2+1)^(1/2)*arcsin(c*x)^2/c/d^2/(c*d*x+d)^(1/2)/(-c*f* x+f)^(1/2)+2*f^2*(-c*x+1)*(a+b*arcsin(c*x))/c/d^2/(c*d*x+d)^(1/2)/(-c*f*x+ f)^(1/2)-2/3*f^2*(-c*x+1)^3*(a+b*arcsin(c*x))/c/d^2/(c*d*x+d)^(1/2)/(-c*f* x+f)^(1/2)/(-c^2*x^2+1)+f^2*(-c^2*x^2+1)^(1/2)*arcsin(c*x)*(a+b*arcsin(c*x ))/c/d^2/(c*d*x+d)^(1/2)/(-c*f*x+f)^(1/2)-8/3*b*f^2*(-c^2*x^2+1)^(1/2)*ln( c*x+1)/c/d^2/(c*d*x+d)^(1/2)/(-c*f*x+f)^(1/2)
Time = 7.40 (sec) , antiderivative size = 599, normalized size of antiderivative = 1.80 \[ \int \frac {(f-c f x)^{3/2} (a+b \arcsin (c x))}{(d+c d x)^{5/2}} \, dx =\text {Too large to display} \] Input:
Integrate[((f - c*f*x)^(3/2)*(a + b*ArcSin[c*x]))/(d + c*d*x)^(5/2),x]
Output:
(f*((16*a*(1 + 2*c*x)*Sqrt[d + c*d*x]*Sqrt[f - c*f*x])/(1 + c*x)^2 - 12*a* Sqrt[d]*Sqrt[f]*ArcTan[(c*x*Sqrt[d + c*d*x]*Sqrt[f - c*f*x])/(Sqrt[d]*Sqrt [f]*(-1 + c^2*x^2))] - (b*Sqrt[d + c*d*x]*Sqrt[f - c*f*x]*(Cos[ArcSin[c*x] /2] - Sin[ArcSin[c*x]/2])*(Cos[ArcSin[c*x]/2]*(-8 + 6*ArcSin[c*x] + 9*ArcS in[c*x]^2 - 84*Log[Cos[ArcSin[c*x]/2] + Sin[ArcSin[c*x]/2]]) + Cos[(3*ArcS in[c*x])/2]*((14 - 3*ArcSin[c*x])*ArcSin[c*x] + 28*Log[Cos[ArcSin[c*x]/2] + Sin[ArcSin[c*x]/2]]) + 2*(-4 + 2*(2 + 7*Sqrt[1 - c^2*x^2])*ArcSin[c*x] + 3*(2 + Sqrt[1 - c^2*x^2])*ArcSin[c*x]^2 - 28*(2 + Sqrt[1 - c^2*x^2])*Log[ Cos[ArcSin[c*x]/2] + Sin[ArcSin[c*x]/2]])*Sin[ArcSin[c*x]/2]))/((-1 + c*x) *(Cos[ArcSin[c*x]/2] + Sin[ArcSin[c*x]/2])^4) - (2*b*Sqrt[d + c*d*x]*Sqrt[ f - c*f*x]*(Cos[ArcSin[c*x]/2] - Sin[ArcSin[c*x]/2])*(Cos[(3*ArcSin[c*x])/ 2]*(ArcSin[c*x] + 2*Log[Cos[ArcSin[c*x]/2] + Sin[ArcSin[c*x]/2]]) - Cos[Ar cSin[c*x]/2]*(4 + 3*ArcSin[c*x] + 6*Log[Cos[ArcSin[c*x]/2] + Sin[ArcSin[c* x]/2]]) + 2*(-2 + (2 + Sqrt[1 - c^2*x^2])*ArcSin[c*x] - 2*(2 + Sqrt[1 - c^ 2*x^2])*Log[Cos[ArcSin[c*x]/2] + Sin[ArcSin[c*x]/2]])*Sin[ArcSin[c*x]/2])) /((-1 + c*x)*(Cos[ArcSin[c*x]/2] + Sin[ArcSin[c*x]/2])^4)))/(12*c*d^3)
Time = 0.62 (sec) , antiderivative size = 171, normalized size of antiderivative = 0.52, number of steps used = 4, number of rules used = 4, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.133, Rules used = {5178, 27, 5260, 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.
\(\displaystyle \int \frac {(f-c f x)^{3/2} (a+b \arcsin (c x))}{(c d x+d)^{5/2}} \, dx\) |
\(\Big \downarrow \) 5178 |
\(\displaystyle \frac {\left (1-c^2 x^2\right )^{5/2} \int \frac {f^4 (1-c x)^4 (a+b \arcsin (c x))}{\left (1-c^2 x^2\right )^{5/2}}dx}{(c d x+d)^{5/2} (f-c f x)^{5/2}}\) |
\(\Big \downarrow \) 27 |
\(\displaystyle \frac {f^4 \left (1-c^2 x^2\right )^{5/2} \int \frac {(1-c x)^4 (a+b \arcsin (c x))}{\left (1-c^2 x^2\right )^{5/2}}dx}{(c d x+d)^{5/2} (f-c f x)^{5/2}}\) |
\(\Big \downarrow \) 5260 |
\(\displaystyle \frac {f^4 \left (1-c^2 x^2\right )^{5/2} \left (-b c \int \left (-\frac {2 (1-c x)^3}{3 c \left (1-c^2 x^2\right )^2}+\frac {2 (1-c x)}{c \left (1-c^2 x^2\right )}+\frac {\arcsin (c x)}{c \sqrt {1-c^2 x^2}}\right )dx-\frac {2 (1-c x)^3 (a+b \arcsin (c x))}{3 c \left (1-c^2 x^2\right )^{3/2}}+\frac {2 (1-c x) (a+b \arcsin (c x))}{c \sqrt {1-c^2 x^2}}+\frac {\arcsin (c x) (a+b \arcsin (c x))}{c}\right )}{(c d x+d)^{5/2} (f-c f x)^{5/2}}\) |
\(\Big \downarrow \) 2009 |
\(\displaystyle \frac {f^4 \left (1-c^2 x^2\right )^{5/2} \left (-\frac {2 (1-c x)^3 (a+b \arcsin (c x))}{3 c \left (1-c^2 x^2\right )^{3/2}}+\frac {2 (1-c x) (a+b \arcsin (c x))}{c \sqrt {1-c^2 x^2}}+\frac {\arcsin (c x) (a+b \arcsin (c x))}{c}-b c \left (\frac {\arcsin (c x)^2}{2 c^2}+\frac {4}{3 c^2 (c x+1)}+\frac {8 \log (c x+1)}{3 c^2}\right )\right )}{(c d x+d)^{5/2} (f-c f x)^{5/2}}\) |
Input:
Int[((f - c*f*x)^(3/2)*(a + b*ArcSin[c*x]))/(d + c*d*x)^(5/2),x]
Output:
(f^4*(1 - c^2*x^2)^(5/2)*((-2*(1 - c*x)^3*(a + b*ArcSin[c*x]))/(3*c*(1 - c ^2*x^2)^(3/2)) + (2*(1 - c*x)*(a + b*ArcSin[c*x]))/(c*Sqrt[1 - c^2*x^2]) + (ArcSin[c*x]*(a + b*ArcSin[c*x]))/c - b*c*(4/(3*c^2*(1 + c*x)) + ArcSin[c *x]^2/(2*c^2) + (8*Log[1 + c*x])/(3*c^2))))/((d + c*d*x)^(5/2)*(f - c*f*x) ^(5/2))
Int[(a_)*(Fx_), x_Symbol] :> Simp[a Int[Fx, x], x] /; FreeQ[a, x] && !Ma tchQ[Fx, (b_)*(Gx_) /; FreeQ[b, x]]
Int[((a_.) + ArcSin[(c_.)*(x_)]*(b_.))^(n_.)*((d_) + (e_.)*(x_))^(p_)*((f_) + (g_.)*(x_))^(q_), x_Symbol] :> Simp[(d + e*x)^q*((f + g*x)^q/(1 - c^2*x^ 2)^q) Int[(d + e*x)^(p - q)*(1 - c^2*x^2)^q*(a + b*ArcSin[c*x])^n, x], x] /; FreeQ[{a, b, c, d, e, f, g, n}, x] && EqQ[e*f + d*g, 0] && EqQ[c^2*d^2 - e^2, 0] && HalfIntegerQ[p, q] && GeQ[p - q, 0]
Int[((a_.) + ArcSin[(c_.)*(x_)]*(b_.))*((f_) + (g_.)*(x_))^(m_.)*((d_) + (e _.)*(x_)^2)^(p_), x_Symbol] :> With[{u = IntHide[(f + g*x)^m*(d + e*x^2)^p, x]}, Simp[(a + b*ArcSin[c*x]) u, x] - Simp[b*c Int[1/Sqrt[1 - c^2*x^2] u, x], x]] /; FreeQ[{a, b, c, d, e, f, g}, x] && EqQ[c^2*d + e, 0] && IG tQ[m, 0] && ILtQ[p + 1/2, 0] && GtQ[d, 0] && (LtQ[m, -2*p - 1] || GtQ[m, 3] )
Result contains complex when optimal does not.
Time = 4.15 (sec) , antiderivative size = 348, normalized size of antiderivative = 1.05
method | result | size |
default | \(-\frac {\sqrt {-f \left (c x -1\right )}\, \sqrt {d \left (c x +1\right )}\, \sqrt {-c^{2} x^{2}+1}\, \left (32 i \arcsin \left (c x \right ) b c x +3 \arcsin \left (c x \right )^{2} b \,c^{2} x^{2}-32 i a c x +6 \arcsin \left (c x \right ) a \,c^{2} x^{2}-32 \ln \left (i c x +\sqrt {-c^{2} x^{2}+1}+i\right ) b \,c^{2} x^{2}+16 i \arcsin \left (c x \right ) b \,c^{2} x^{2}+6 b c x \arcsin \left (c x \right )^{2}+16 \arcsin \left (c x \right ) \sqrt {-c^{2} x^{2}+1}\, b c x -16 i a \,c^{2} x^{2}+12 \arcsin \left (c x \right ) a c x +16 \sqrt {-c^{2} x^{2}+1}\, a c x -64 \ln \left (i c x +\sqrt {-c^{2} x^{2}+1}+i\right ) b c x -16 i a +3 \arcsin \left (c x \right )^{2} b +8 \arcsin \left (c x \right ) \sqrt {-c^{2} x^{2}+1}\, b -8 x b c +16 i b \arcsin \left (c x \right )+6 \arcsin \left (c x \right ) a +8 \sqrt {-c^{2} x^{2}+1}\, a -32 b \ln \left (i c x +\sqrt {-c^{2} x^{2}+1}+i\right )-8 b \right ) f}{6 \left (c^{3} x^{3}+3 c^{2} x^{2}+3 c x +1\right ) d^{3} \left (c x -1\right ) c}\) | \(348\) |
Input:
int((-c*f*x+f)^(3/2)*(a+b*arcsin(c*x))/(c*d*x+d)^(5/2),x,method=_RETURNVER BOSE)
Output:
-1/6*(-f*(c*x-1))^(1/2)*(d*(c*x+1))^(1/2)*(-c^2*x^2+1)^(1/2)/(c^3*x^3+3*c^ 2*x^2+3*c*x+1)/d^3/(c*x-1)/c*(32*I*arcsin(c*x)*b*c*x+3*arcsin(c*x)^2*b*c^2 *x^2-32*I*a*c*x+6*arcsin(c*x)*a*c^2*x^2-32*ln(I*c*x+(-c^2*x^2+1)^(1/2)+I)* b*c^2*x^2+16*I*arcsin(c*x)*b*c^2*x^2+6*b*c*x*arcsin(c*x)^2+16*arcsin(c*x)* (-c^2*x^2+1)^(1/2)*b*c*x-16*I*a*c^2*x^2+12*arcsin(c*x)*a*c*x+16*(-c^2*x^2+ 1)^(1/2)*a*c*x-64*ln(I*c*x+(-c^2*x^2+1)^(1/2)+I)*b*c*x-16*I*a+3*arcsin(c*x )^2*b+8*arcsin(c*x)*(-c^2*x^2+1)^(1/2)*b-8*x*b*c+16*I*b*arcsin(c*x)+6*arcs in(c*x)*a+8*(-c^2*x^2+1)^(1/2)*a-32*b*ln(I*c*x+(-c^2*x^2+1)^(1/2)+I)-8*b)* f
\[ \int \frac {(f-c f x)^{3/2} (a+b \arcsin (c x))}{(d+c d x)^{5/2}} \, dx=\int { \frac {{\left (-c f x + f\right )}^{\frac {3}{2}} {\left (b \arcsin \left (c x\right ) + a\right )}}{{\left (c d x + d\right )}^{\frac {5}{2}}} \,d x } \] Input:
integrate((-c*f*x+f)^(3/2)*(a+b*arcsin(c*x))/(c*d*x+d)^(5/2),x, algorithm= "fricas")
Output:
integral(-(a*c*f*x - a*f + (b*c*f*x - b*f)*arcsin(c*x))*sqrt(c*d*x + d)*sq rt(-c*f*x + f)/(c^3*d^3*x^3 + 3*c^2*d^3*x^2 + 3*c*d^3*x + d^3), x)
\[ \int \frac {(f-c f x)^{3/2} (a+b \arcsin (c x))}{(d+c d x)^{5/2}} \, dx=\int \frac {\left (- f \left (c x - 1\right )\right )^{\frac {3}{2}} \left (a + b \operatorname {asin}{\left (c x \right )}\right )}{\left (d \left (c x + 1\right )\right )^{\frac {5}{2}}}\, dx \] Input:
integrate((-c*f*x+f)**(3/2)*(a+b*asin(c*x))/(c*d*x+d)**(5/2),x)
Output:
Integral((-f*(c*x - 1))**(3/2)*(a + b*asin(c*x))/(d*(c*x + 1))**(5/2), x)
\[ \int \frac {(f-c f x)^{3/2} (a+b \arcsin (c x))}{(d+c d x)^{5/2}} \, dx=\int { \frac {{\left (-c f x + f\right )}^{\frac {3}{2}} {\left (b \arcsin \left (c x\right ) + a\right )}}{{\left (c d x + d\right )}^{\frac {5}{2}}} \,d x } \] Input:
integrate((-c*f*x+f)^(3/2)*(a+b*arcsin(c*x))/(c*d*x+d)^(5/2),x, algorithm= "maxima")
Output:
-b*sqrt(d)*sqrt(f)*integrate((c*f*x - f)*sqrt(c*x + 1)*sqrt(-c*x + 1)*arct an2(c*x, sqrt(c*x + 1)*sqrt(-c*x + 1))/(c^3*d^3*x^3 + 3*c^2*d^3*x^2 + 3*c* d^3*x + d^3), x) - 1/3*a*((-c^2*d*f*x^2 + d*f)^(3/2)/(c^4*d^4*x^3 + 3*c^3* d^4*x^2 + 3*c^2*d^4*x + c*d^4) + 2*sqrt(-c^2*d*f*x^2 + d*f)*f/(c^3*d^3*x^2 + 2*c^2*d^3*x + c*d^3) - 7*sqrt(-c^2*d*f*x^2 + d*f)*f/(c^2*d^3*x + c*d^3) - 3*f^2*arcsin(c*x)/(c*d^3*sqrt(f/d)))
\[ \int \frac {(f-c f x)^{3/2} (a+b \arcsin (c x))}{(d+c d x)^{5/2}} \, dx=\int { \frac {{\left (-c f x + f\right )}^{\frac {3}{2}} {\left (b \arcsin \left (c x\right ) + a\right )}}{{\left (c d x + d\right )}^{\frac {5}{2}}} \,d x } \] Input:
integrate((-c*f*x+f)^(3/2)*(a+b*arcsin(c*x))/(c*d*x+d)^(5/2),x, algorithm= "giac")
Output:
integrate((-c*f*x + f)^(3/2)*(b*arcsin(c*x) + a)/(c*d*x + d)^(5/2), x)
Timed out. \[ \int \frac {(f-c f x)^{3/2} (a+b \arcsin (c x))}{(d+c d x)^{5/2}} \, dx=\int \frac {\left (a+b\,\mathrm {asin}\left (c\,x\right )\right )\,{\left (f-c\,f\,x\right )}^{3/2}}{{\left (d+c\,d\,x\right )}^{5/2}} \,d x \] Input:
int(((a + b*asin(c*x))*(f - c*f*x)^(3/2))/(d + c*d*x)^(5/2),x)
Output:
int(((a + b*asin(c*x))*(f - c*f*x)^(3/2))/(d + c*d*x)^(5/2), x)
\[ \int \frac {(f-c f x)^{3/2} (a+b \arcsin (c x))}{(d+c d x)^{5/2}} \, dx=\frac {\sqrt {f}\, f \left (-6 \sqrt {c x +1}\, \mathit {asin} \left (\frac {\sqrt {-c x +1}}{\sqrt {2}}\right ) a c x -6 \sqrt {c x +1}\, \mathit {asin} \left (\frac {\sqrt {-c x +1}}{\sqrt {2}}\right ) a +8 \sqrt {-c x +1}\, a c x +4 \sqrt {-c x +1}\, a -3 \sqrt {c x +1}\, \left (\int \frac {\sqrt {-c x +1}\, \mathit {asin} \left (c x \right ) x}{\sqrt {c x +1}\, c^{2} x^{2}+2 \sqrt {c x +1}\, c x +\sqrt {c x +1}}d x \right ) b \,c^{3} x -3 \sqrt {c x +1}\, \left (\int \frac {\sqrt {-c x +1}\, \mathit {asin} \left (c x \right ) x}{\sqrt {c x +1}\, c^{2} x^{2}+2 \sqrt {c x +1}\, c x +\sqrt {c x +1}}d x \right ) b \,c^{2}+3 \sqrt {c x +1}\, \left (\int \frac {\sqrt {-c x +1}\, \mathit {asin} \left (c x \right )}{\sqrt {c x +1}\, c^{2} x^{2}+2 \sqrt {c x +1}\, c x +\sqrt {c x +1}}d x \right ) b \,c^{2} x +3 \sqrt {c x +1}\, \left (\int \frac {\sqrt {-c x +1}\, \mathit {asin} \left (c x \right )}{\sqrt {c x +1}\, c^{2} x^{2}+2 \sqrt {c x +1}\, c x +\sqrt {c x +1}}d x \right ) b c \right )}{3 \sqrt {d}\, \sqrt {c x +1}\, c \,d^{2} \left (c x +1\right )} \] Input:
int((-c*f*x+f)^(3/2)*(a+b*asin(c*x))/(c*d*x+d)^(5/2),x)
Output:
(sqrt(f)*f*( - 6*sqrt(c*x + 1)*asin(sqrt( - c*x + 1)/sqrt(2))*a*c*x - 6*sq rt(c*x + 1)*asin(sqrt( - c*x + 1)/sqrt(2))*a + 8*sqrt( - c*x + 1)*a*c*x + 4*sqrt( - c*x + 1)*a - 3*sqrt(c*x + 1)*int((sqrt( - c*x + 1)*asin(c*x)*x)/ (sqrt(c*x + 1)*c**2*x**2 + 2*sqrt(c*x + 1)*c*x + sqrt(c*x + 1)),x)*b*c**3* x - 3*sqrt(c*x + 1)*int((sqrt( - c*x + 1)*asin(c*x)*x)/(sqrt(c*x + 1)*c**2 *x**2 + 2*sqrt(c*x + 1)*c*x + sqrt(c*x + 1)),x)*b*c**2 + 3*sqrt(c*x + 1)*i nt((sqrt( - c*x + 1)*asin(c*x))/(sqrt(c*x + 1)*c**2*x**2 + 2*sqrt(c*x + 1) *c*x + sqrt(c*x + 1)),x)*b*c**2*x + 3*sqrt(c*x + 1)*int((sqrt( - c*x + 1)* asin(c*x))/(sqrt(c*x + 1)*c**2*x**2 + 2*sqrt(c*x + 1)*c*x + sqrt(c*x + 1)) ,x)*b*c))/(3*sqrt(d)*sqrt(c*x + 1)*c*d**2*(c*x + 1))