Integrand size = 31, antiderivative size = 410 \[ \int \frac {(f+g x)^3 (a+b \arcsin (c x))}{\left (d-c^2 d x^2\right )^{5/2}} \, dx=-\frac {b (f+g x) \left (c^2 f^2+g^2+2 c^2 f g x\right )}{6 c^3 d^2 \sqrt {1-c^2 x^2} \sqrt {d-c^2 d x^2}}+\frac {2 (c f-g) (c f+g) \left (g+c^2 f x\right ) (a+b \arcsin (c x))}{3 c^4 d^2 \sqrt {d-c^2 d x^2}}+\frac {\left (g+c^2 f x\right ) (f+g x)^2 (a+b \arcsin (c x))}{3 c^2 d^2 \left (1-c^2 x^2\right ) \sqrt {d-c^2 d x^2}}+\frac {b (c f-g) (c f+g)^2 \sqrt {1-c^2 x^2} \log (1-c x)}{3 c^4 d^2 \sqrt {d-c^2 d x^2}}-\frac {b g (c f+g)^2 \sqrt {1-c^2 x^2} \log (1-c x)}{12 c^4 d^2 \sqrt {d-c^2 d x^2}}+\frac {b (c f-g)^2 g \sqrt {1-c^2 x^2} \log (1+c x)}{12 c^4 d^2 \sqrt {d-c^2 d x^2}}+\frac {b (c f-g)^2 (c f+g) \sqrt {1-c^2 x^2} \log (1+c x)}{3 c^4 d^2 \sqrt {d-c^2 d x^2}} \]
2/3*(c*f-g)*(c*f+g)*(c^2*f*x+g)*(a+b*arcsin(c*x))/c^4/d^2/(-c^2*d*x^2+d)^( 1/2)+1/3*(c^2*f*x+g)*(g*x+f)^2*(a+b*arcsin(c*x))/c^2/d^2/(-c^2*x^2+1)/(-c^ 2*d*x^2+d)^(1/2)-1/6*b*(g*x+f)*(2*c^2*f*g*x+c^2*f^2+g^2)/c^3/d^2/(-c^2*x^2 +1)^(1/2)/(-c^2*d*x^2+d)^(1/2)+1/3*b*(c*f-g)*(c*f+g)^2*ln(-c*x+1)*(-c^2*x^ 2+1)^(1/2)/c^4/d^2/(-c^2*d*x^2+d)^(1/2)-1/12*b*g*(c*f+g)^2*ln(-c*x+1)*(-c^ 2*x^2+1)^(1/2)/c^4/d^2/(-c^2*d*x^2+d)^(1/2)+1/12*b*(c*f-g)^2*g*ln(c*x+1)*( -c^2*x^2+1)^(1/2)/c^4/d^2/(-c^2*d*x^2+d)^(1/2)+1/3*b*(c*f-g)^2*(c*f+g)*ln( c*x+1)*(-c^2*x^2+1)^(1/2)/c^4/d^2/(-c^2*d*x^2+d)^(1/2)
Result contains higher order function than in optimal. Order 4 vs. order 3 in optimal.
Time = 0.95 (sec) , antiderivative size = 366, normalized size of antiderivative = 0.89 \[ \int \frac {(f+g x)^3 (a+b \arcsin (c x))}{\left (d-c^2 d x^2\right )^{5/2}} \, dx=\frac {\sqrt {d-c^2 d x^2} \left (i b c g \left (3 c^2 f^2-5 g^2\right ) \left (1-c^2 x^2\right )^{3/2} \operatorname {EllipticF}\left (i \text {arcsinh}\left (\sqrt {-c^2} x\right ),1\right )-\sqrt {-c^2} \left (-6 a c^2 f^2 g+4 a g^3-6 a c^4 f^3 x-6 a c^2 g^3 x^2+4 a c^6 f^3 x^3-6 a c^4 f g^2 x^3+b c^3 f^3 \sqrt {1-c^2 x^2}+3 b c f g^2 \sqrt {1-c^2 x^2}+3 b c^3 f^2 g x \sqrt {1-c^2 x^2}+b c g^3 x \sqrt {1-c^2 x^2}+2 b \left (2 g^3+2 c^6 f^3 x^3-3 c^2 g \left (f^2+g^2 x^2\right )-3 c^4 f x \left (f^2+g^2 x^2\right )\right ) \arcsin (c x)-b c f \left (2 c^2 f^2-3 g^2\right ) \left (1-c^2 x^2\right )^{3/2} \log \left (-1+c^2 x^2\right )\right )\right )}{6 c^4 \sqrt {-c^2} d^3 \left (-1+c^2 x^2\right )^2} \]
(Sqrt[d - c^2*d*x^2]*(I*b*c*g*(3*c^2*f^2 - 5*g^2)*(1 - c^2*x^2)^(3/2)*Elli pticF[I*ArcSinh[Sqrt[-c^2]*x], 1] - Sqrt[-c^2]*(-6*a*c^2*f^2*g + 4*a*g^3 - 6*a*c^4*f^3*x - 6*a*c^2*g^3*x^2 + 4*a*c^6*f^3*x^3 - 6*a*c^4*f*g^2*x^3 + b *c^3*f^3*Sqrt[1 - c^2*x^2] + 3*b*c*f*g^2*Sqrt[1 - c^2*x^2] + 3*b*c^3*f^2*g *x*Sqrt[1 - c^2*x^2] + b*c*g^3*x*Sqrt[1 - c^2*x^2] + 2*b*(2*g^3 + 2*c^6*f^ 3*x^3 - 3*c^2*g*(f^2 + g^2*x^2) - 3*c^4*f*x*(f^2 + g^2*x^2))*ArcSin[c*x] - b*c*f*(2*c^2*f^2 - 3*g^2)*(1 - c^2*x^2)^(3/2)*Log[-1 + c^2*x^2])))/(6*c^4 *Sqrt[-c^2]*d^3*(-1 + c^2*x^2)^2)
Time = 0.66 (sec) , antiderivative size = 277, normalized size of antiderivative = 0.68, number of steps used = 3, number of rules used = 3, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.097, Rules used = {5276, 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.
\(\displaystyle \int \frac {(f+g x)^3 (a+b \arcsin (c x))}{\left (d-c^2 d x^2\right )^{5/2}} \, dx\) |
\(\Big \downarrow \) 5276 |
\(\displaystyle \frac {\sqrt {1-c^2 x^2} \int \frac {(f+g x)^3 (a+b \arcsin (c x))}{\left (1-c^2 x^2\right )^{5/2}}dx}{d^2 \sqrt {d-c^2 d x^2}}\) |
\(\Big \downarrow \) 5260 |
\(\displaystyle \frac {\sqrt {1-c^2 x^2} \left (-b c \int \left (\frac {\left (f x c^2+g\right ) (f+g x)^2}{3 c^2 \left (1-c^2 x^2\right )^2}+\frac {2 (c f-g) (c f+g) \left (f x c^2+g\right )}{3 c^4 \left (1-c^2 x^2\right )}\right )dx+\frac {(f+g x)^2 \left (c^2 f x+g\right ) (a+b \arcsin (c x))}{3 c^2 \left (1-c^2 x^2\right )^{3/2}}+\frac {2 (c f-g) (c f+g) \left (c^2 f x+g\right ) (a+b \arcsin (c x))}{3 c^4 \sqrt {1-c^2 x^2}}\right )}{d^2 \sqrt {d-c^2 d x^2}}\) |
\(\Big \downarrow \) 2009 |
\(\displaystyle \frac {\sqrt {1-c^2 x^2} \left (\frac {(f+g x)^2 \left (c^2 f x+g\right ) (a+b \arcsin (c x))}{3 c^2 \left (1-c^2 x^2\right )^{3/2}}+\frac {2 (c f-g) (c f+g) \left (c^2 f x+g\right ) (a+b \arcsin (c x))}{3 c^4 \sqrt {1-c^2 x^2}}-b c \left (\frac {2 g \text {arctanh}(c x) (c f+g) (c f-g)}{3 c^5}+\frac {(c f-g)^3}{12 c^5 (c x+1)}+\frac {(c f+g)^3}{12 c^5 (1-c x)}-\frac {g (c f-g)^2 \log (c x+1)}{12 c^5}+\frac {g (c f+g)^2 \log (1-c x)}{12 c^5}-\frac {f (c f+g) (c f-g) \log \left (1-c^2 x^2\right )}{3 c^4}\right )\right )}{d^2 \sqrt {d-c^2 d x^2}}\) |
(Sqrt[1 - c^2*x^2]*(((g + c^2*f*x)*(f + g*x)^2*(a + b*ArcSin[c*x]))/(3*c^2 *(1 - c^2*x^2)^(3/2)) + (2*(c*f - g)*(c*f + g)*(g + c^2*f*x)*(a + b*ArcSin [c*x]))/(3*c^4*Sqrt[1 - c^2*x^2]) - b*c*((c*f + g)^3/(12*c^5*(1 - c*x)) + (c*f - g)^3/(12*c^5*(1 + c*x)) + (2*(c*f - g)*g*(c*f + g)*ArcTanh[c*x])/(3 *c^5) + (g*(c*f + g)^2*Log[1 - c*x])/(12*c^5) - ((c*f - g)^2*g*Log[1 + c*x ])/(12*c^5) - (f*(c*f - g)*(c*f + g)*Log[1 - c^2*x^2])/(3*c^4))))/(d^2*Sqr t[d - c^2*d*x^2])
3.1.54.3.1 Defintions of rubi rules used
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] )
Int[((a_.) + ArcSin[(c_.)*(x_)]*(b_.))^(n_.)*((f_) + (g_.)*(x_))^(m_.)*((d_ ) + (e_.)*(x_)^2)^(p_), x_Symbol] :> Simp[Simp[(d + e*x^2)^p/(1 - c^2*x^2)^ p] Int[(f + g*x)^m*(1 - c^2*x^2)^p*(a + b*ArcSin[c*x])^n, x], x] /; FreeQ [{a, b, c, d, e, f, g, n}, x] && EqQ[c^2*d + e, 0] && IntegerQ[m] && Intege rQ[p - 1/2] && !GtQ[d, 0]
Result contains complex when optimal does not.
Time = 1.01 (sec) , antiderivative size = 5114, normalized size of antiderivative = 12.47
method | result | size |
default | \(\text {Expression too large to display}\) | \(5114\) |
parts | \(\text {Expression too large to display}\) | \(5114\) |
\[ \int \frac {(f+g x)^3 (a+b \arcsin (c x))}{\left (d-c^2 d x^2\right )^{5/2}} \, dx=\int { \frac {{\left (g x + f\right )}^{3} {\left (b \arcsin \left (c x\right ) + a\right )}}{{\left (-c^{2} d x^{2} + d\right )}^{\frac {5}{2}}} \,d x } \]
integral(-(a*g^3*x^3 + 3*a*f*g^2*x^2 + 3*a*f^2*g*x + a*f^3 + (b*g^3*x^3 + 3*b*f*g^2*x^2 + 3*b*f^2*g*x + b*f^3)*arcsin(c*x))*sqrt(-c^2*d*x^2 + d)/(c^ 6*d^3*x^6 - 3*c^4*d^3*x^4 + 3*c^2*d^3*x^2 - d^3), x)
\[ \int \frac {(f+g x)^3 (a+b \arcsin (c x))}{\left (d-c^2 d x^2\right )^{5/2}} \, dx=\int \frac {\left (a + b \operatorname {asin}{\left (c x \right )}\right ) \left (f + g x\right )^{3}}{\left (- d \left (c x - 1\right ) \left (c x + 1\right )\right )^{\frac {5}{2}}}\, dx \]
\[ \int \frac {(f+g x)^3 (a+b \arcsin (c x))}{\left (d-c^2 d x^2\right )^{5/2}} \, dx=\int { \frac {{\left (g x + f\right )}^{3} {\left (b \arcsin \left (c x\right ) + a\right )}}{{\left (-c^{2} d x^{2} + d\right )}^{\frac {5}{2}}} \,d x } \]
1/6*b*c*f^3*(1/(c^4*d^(5/2)*x^2 - c^2*d^(5/2)) + 2*log(c*x + 1)/(c^2*d^(5/ 2)) + 2*log(c*x - 1)/(c^2*d^(5/2))) + 1/3*b*f^3*(2*x/(sqrt(-c^2*d*x^2 + d) *d^2) + x/((-c^2*d*x^2 + d)^(3/2)*d))*arcsin(c*x) + 1/3*a*f^3*(2*x/(sqrt(- c^2*d*x^2 + d)*d^2) + x/((-c^2*d*x^2 + d)^(3/2)*d)) + 1/3*a*g^3*(3*x^2/((- c^2*d*x^2 + d)^(3/2)*c^2*d) - 2/((-c^2*d*x^2 + d)^(3/2)*c^4*d)) - a*f*g^2* (x/(sqrt(-c^2*d*x^2 + d)*c^2*d^2) - x/((-c^2*d*x^2 + d)^(3/2)*c^2*d)) + in tegrate((b*g^3*x^3 + 3*b*f*g^2*x^2 + 3*b*f^2*g*x)*arctan2(c*x, sqrt(c*x + 1)*sqrt(-c*x + 1))/((c^4*d^2*x^4 - 2*c^2*d^2*x^2 + d^2)*sqrt(c*x + 1)*sqrt (-c*x + 1)), x)/sqrt(d) + a*f^2*g/((-c^2*d*x^2 + d)^(3/2)*c^2*d)
Exception generated. \[ \int \frac {(f+g x)^3 (a+b \arcsin (c x))}{\left (d-c^2 d x^2\right )^{5/2}} \, dx=\text {Exception raised: TypeError} \]
Exception raised: TypeError >> an error occurred running a Giac command:IN PUT:sage2:=int(sage0,sageVARx):;OUTPUT:index.cc index_m i_lex_is_greater E rror: Bad Argument Value
Timed out. \[ \int \frac {(f+g x)^3 (a+b \arcsin (c x))}{\left (d-c^2 d x^2\right )^{5/2}} \, dx=\int \frac {{\left (f+g\,x\right )}^3\,\left (a+b\,\mathrm {asin}\left (c\,x\right )\right )}{{\left (d-c^2\,d\,x^2\right )}^{5/2}} \,d x \]