Integrand size = 31, antiderivative size = 528 \[ \int \frac {(f+g x)^4 (a+b \arcsin (c x))}{\left (d-c^2 d x^2\right )^{5/2}} \, dx=-\frac {b (f+g x)^2 \left (c^2 f^2+g^2+2 c^2 f g x\right )}{6 c^3 d^2 \sqrt {1-c^2 x^2} \sqrt {d-c^2 d x^2}}-\frac {b f g^3 x \sqrt {1-c^2 x^2}}{3 c^3 d^2 \sqrt {d-c^2 d x^2}}-\frac {b g^4 \sqrt {1-c^2 x^2} \arcsin (c x)^2}{2 c^5 d^2 \sqrt {d-c^2 d x^2}}+\frac {(f+g x) \left (g \left (c^2 f^2-3 g^2\right )+2 c^2 f \left (c^2 f^2-2 g^2\right ) x\right ) (a+b \arcsin (c x))}{3 c^4 d^2 \sqrt {d-c^2 d x^2}}+\frac {\left (g+c^2 f x\right ) (f+g x)^3 (a+b \arcsin (c x))}{3 c^2 d^2 \left (1-c^2 x^2\right ) \sqrt {d-c^2 d x^2}}+\frac {f g \left (2 c^2 f^2-5 g^2\right ) \left (1-c^2 x^2\right ) (a+b \arcsin (c x))}{3 c^4 d^2 \sqrt {d-c^2 d x^2}}+\frac {g^4 \sqrt {1-c^2 x^2} \arcsin (c x) (a+b \arcsin (c x))}{c^5 d^2 \sqrt {d-c^2 d x^2}}+\frac {b (c f-2 g) (c f+g)^3 \sqrt {1-c^2 x^2} \log (1-c x)}{3 c^5 d^2 \sqrt {d-c^2 d x^2}}+\frac {b (c f-g)^3 (c f+2 g) \sqrt {1-c^2 x^2} \log (1+c x)}{3 c^5 d^2 \sqrt {d-c^2 d x^2}} \]
1/3*(g*x+f)*(g*(c^2*f^2-3*g^2)+2*c^2*f*(c^2*f^2-2*g^2)*x)*(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)^3*(a+b*arcsin(c*x)) /c^2/d^2/(-c^2*x^2+1)/(-c^2*d*x^2+d)^(1/2)+1/3*f*g*(2*c^2*f^2-5*g^2)*(-c^2 *x^2+1)*(a+b*arcsin(c*x))/c^4/d^2/(-c^2*d*x^2+d)^(1/2)-1/6*b*(g*x+f)^2*(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*f*g^3*x*(-c^2*x^2+1)^(1/2)/c^3/d^2/(-c^2*d*x^2+d)^(1/2)-1/2*b*g^4*arcsi n(c*x)^2*(-c^2*x^2+1)^(1/2)/c^5/d^2/(-c^2*d*x^2+d)^(1/2)+g^4*arcsin(c*x)*( a+b*arcsin(c*x))*(-c^2*x^2+1)^(1/2)/c^5/d^2/(-c^2*d*x^2+d)^(1/2)+1/3*b*(c* f-2*g)*(c*f+g)^3*ln(-c*x+1)*(-c^2*x^2+1)^(1/2)/c^5/d^2/(-c^2*d*x^2+d)^(1/2 )+1/3*b*(c*f-g)^3*(c*f+2*g)*ln(c*x+1)*(-c^2*x^2+1)^(1/2)/c^5/d^2/(-c^2*d*x ^2+d)^(1/2)
Time = 2.55 (sec) , antiderivative size = 868, normalized size of antiderivative = 1.64 \[ \int \frac {(f+g x)^4 (a+b \arcsin (c x))}{\left (d-c^2 d x^2\right )^{5/2}} \, dx=\sqrt {-d \left (-1+c^2 x^2\right )} \left (\frac {4 a c^2 f^3 g+4 a f g^3+a c^4 f^4 x+6 a c^2 f^2 g^2 x+a g^4 x}{3 c^4 d^3 \left (-1+c^2 x^2\right )^2}-\frac {2 a \left (-6 f g^3+c^4 f^4 x-3 c^2 f^2 g^2 x-2 g^4 x\right )}{3 c^4 d^3 \left (-1+c^2 x^2\right )}\right )-\frac {a g^4 \arctan \left (\frac {c x \sqrt {-d \left (-1+c^2 x^2\right )}}{\sqrt {d} \left (-1+c^2 x^2\right )}\right )}{c^5 d^{5/2}}+\frac {b f^2 g^2 \left (-2 c x \arcsin (c x)+\frac {-1+\frac {2 c x \arcsin (c x)}{\sqrt {1-c^2 x^2}}}{\sqrt {1-c^2 x^2}}-2 \sqrt {1-c^2 x^2} \log \left (\sqrt {1-c^2 x^2}\right )\right )}{c^3 d^2 \sqrt {d \left (1-c^2 x^2\right )}}+\frac {b f^4 \left (4 c x \arcsin (c x)+\frac {-1+\frac {2 c x \arcsin (c x)}{\sqrt {1-c^2 x^2}}}{\sqrt {1-c^2 x^2}}+4 \sqrt {1-c^2 x^2} \log \left (\sqrt {1-c^2 x^2}\right )\right )}{6 c d^2 \sqrt {d \left (1-c^2 x^2\right )}}+\frac {b f^3 g \left (8 \arcsin (c x)+3 \sqrt {1-c^2 x^2} \left (\log \left (\cos \left (\frac {1}{2} \arcsin (c x)\right )-\sin \left (\frac {1}{2} \arcsin (c x)\right )\right )-\log \left (\cos \left (\frac {1}{2} \arcsin (c x)\right )+\sin \left (\frac {1}{2} \arcsin (c x)\right )\right )\right )+\cos (3 \arcsin (c x)) \left (\log \left (\cos \left (\frac {1}{2} \arcsin (c x)\right )-\sin \left (\frac {1}{2} \arcsin (c x)\right )\right )-\log \left (\cos \left (\frac {1}{2} \arcsin (c x)\right )+\sin \left (\frac {1}{2} \arcsin (c x)\right )\right )\right )-2 \sin (2 \arcsin (c x))\right )}{6 c^2 d \left (d \left (1-c^2 x^2\right )\right )^{3/2}}-\frac {b f g^3 \left (4 \arcsin (c x)+12 \arcsin (c x) \cos (2 \arcsin (c x))+5 \cos (3 \arcsin (c x)) \log \left (\cos \left (\frac {1}{2} \arcsin (c x)\right )-\sin \left (\frac {1}{2} \arcsin (c x)\right )\right )+15 \sqrt {1-c^2 x^2} \left (\log \left (\cos \left (\frac {1}{2} \arcsin (c x)\right )-\sin \left (\frac {1}{2} \arcsin (c x)\right )\right )-\log \left (\cos \left (\frac {1}{2} \arcsin (c x)\right )+\sin \left (\frac {1}{2} \arcsin (c x)\right )\right )\right )-5 \cos (3 \arcsin (c x)) \log \left (\cos \left (\frac {1}{2} \arcsin (c x)\right )+\sin \left (\frac {1}{2} \arcsin (c x)\right )\right )+2 \sin (2 \arcsin (c x))\right )}{6 c^4 d \left (d \left (1-c^2 x^2\right )\right )^{3/2}}+\frac {b g^4 \left (\sqrt {1-c^2 x^2} \left (3 \arcsin (c x)^2-8 \log \left (\sqrt {1-c^2 x^2}\right )\right )-\frac {1+\frac {2 \arcsin (c x) \sin (3 \arcsin (c x))}{\sqrt {1-c^2 x^2}}}{\sqrt {1-c^2 x^2}}\right )}{6 c^5 d^2 \sqrt {d \left (1-c^2 x^2\right )}} \]
Sqrt[-(d*(-1 + c^2*x^2))]*((4*a*c^2*f^3*g + 4*a*f*g^3 + a*c^4*f^4*x + 6*a* c^2*f^2*g^2*x + a*g^4*x)/(3*c^4*d^3*(-1 + c^2*x^2)^2) - (2*a*(-6*f*g^3 + c ^4*f^4*x - 3*c^2*f^2*g^2*x - 2*g^4*x))/(3*c^4*d^3*(-1 + c^2*x^2))) - (a*g^ 4*ArcTan[(c*x*Sqrt[-(d*(-1 + c^2*x^2))])/(Sqrt[d]*(-1 + c^2*x^2))])/(c^5*d ^(5/2)) + (b*f^2*g^2*(-2*c*x*ArcSin[c*x] + (-1 + (2*c*x*ArcSin[c*x])/Sqrt[ 1 - c^2*x^2])/Sqrt[1 - c^2*x^2] - 2*Sqrt[1 - c^2*x^2]*Log[Sqrt[1 - c^2*x^2 ]]))/(c^3*d^2*Sqrt[d*(1 - c^2*x^2)]) + (b*f^4*(4*c*x*ArcSin[c*x] + (-1 + ( 2*c*x*ArcSin[c*x])/Sqrt[1 - c^2*x^2])/Sqrt[1 - c^2*x^2] + 4*Sqrt[1 - c^2*x ^2]*Log[Sqrt[1 - c^2*x^2]]))/(6*c*d^2*Sqrt[d*(1 - c^2*x^2)]) + (b*f^3*g*(8 *ArcSin[c*x] + 3*Sqrt[1 - c^2*x^2]*(Log[Cos[ArcSin[c*x]/2] - Sin[ArcSin[c* x]/2]] - Log[Cos[ArcSin[c*x]/2] + Sin[ArcSin[c*x]/2]]) + Cos[3*ArcSin[c*x] ]*(Log[Cos[ArcSin[c*x]/2] - Sin[ArcSin[c*x]/2]] - Log[Cos[ArcSin[c*x]/2] + Sin[ArcSin[c*x]/2]]) - 2*Sin[2*ArcSin[c*x]]))/(6*c^2*d*(d*(1 - c^2*x^2))^ (3/2)) - (b*f*g^3*(4*ArcSin[c*x] + 12*ArcSin[c*x]*Cos[2*ArcSin[c*x]] + 5*C os[3*ArcSin[c*x]]*Log[Cos[ArcSin[c*x]/2] - Sin[ArcSin[c*x]/2]] + 15*Sqrt[1 - c^2*x^2]*(Log[Cos[ArcSin[c*x]/2] - Sin[ArcSin[c*x]/2]] - Log[Cos[ArcSin [c*x]/2] + Sin[ArcSin[c*x]/2]]) - 5*Cos[3*ArcSin[c*x]]*Log[Cos[ArcSin[c*x] /2] + Sin[ArcSin[c*x]/2]] + 2*Sin[2*ArcSin[c*x]]))/(6*c^4*d*(d*(1 - c^2*x^ 2))^(3/2)) + (b*g^4*(Sqrt[1 - c^2*x^2]*(3*ArcSin[c*x]^2 - 8*Log[Sqrt[1 - c ^2*x^2]]) - (1 + (2*ArcSin[c*x]*Sin[3*ArcSin[c*x]])/Sqrt[1 - c^2*x^2])/...
Time = 0.90 (sec) , antiderivative size = 452, normalized size of antiderivative = 0.86, 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.
| \(\displaystyle \int \frac {(f+g x)^4 (a+b \arcsin (c x))}{\left (d-c^2 d x^2\right )^{5/2}} \, dx\) |
| \(\Big \downarrow \) 5276 |
| \(\displaystyle \frac {\sqrt {1-c^2 x^2} \int \frac {(f+g x)^4 (a+b \arcsin (c x))}{\left (1-c^2 x^2\right )^{5/2}}dx}{d^2 \sqrt {d-c^2 d x^2}}\) |
| \(\Big \downarrow \) 5260 |
| \(\displaystyle \frac {\sqrt {1-c^2 x^2} \left (-b c \int \left (\frac {\arcsin (c x) g^4}{c^5 \sqrt {1-c^2 x^2}}+\frac {f \left (2 c^2 f^2-5 g^2\right ) g}{3 c^4}+\frac {(f+g x) \left (2 f \left (c^2 f^2-2 g^2\right ) x c^2+g \left (c^2 f^2-3 g^2\right )\right )}{3 c^4 \left (1-c^2 x^2\right )}+\frac {\left (f x c^2+g\right ) (f+g x)^3}{3 c^2 \left (1-c^2 x^2\right )^2}\right )dx+\frac {g^4 \arcsin (c x) (a+b \arcsin (c x))}{c^5}+\frac {(f+g x)^3 \left (c^2 f x+g\right ) (a+b \arcsin (c x))}{3 c^2 \left (1-c^2 x^2\right )^{3/2}}+\frac {f g \sqrt {1-c^2 x^2} \left (2 c^2 f^2-5 g^2\right ) (a+b \arcsin (c x))}{3 c^4}+\frac {(f+g x) \left (2 c^2 f x \left (c^2 f^2-2 g^2\right )+g \left (c^2 f^2-3 g^2\right )\right ) (a+b \arcsin (c x))}{3 c^4 \sqrt {1-c^2 x^2}}\right )}{d^2 \sqrt {d-c^2 d x^2}}\) |
| \(\Big \downarrow \) 2009 |
| \(\displaystyle \frac {\sqrt {1-c^2 x^2} \left (\frac {g^4 \arcsin (c x) (a+b \arcsin (c x))}{c^5}+\frac {(f+g x)^3 \left (c^2 f x+g\right ) (a+b \arcsin (c x))}{3 c^2 \left (1-c^2 x^2\right )^{3/2}}+\frac {f g \sqrt {1-c^2 x^2} \left (2 c^2 f^2-5 g^2\right ) (a+b \arcsin (c x))}{3 c^4}+\frac {(f+g x) \left (2 c^2 f x \left (c^2 f^2-2 g^2\right )+g \left (c^2 f^2-3 g^2\right )\right ) (a+b \arcsin (c x))}{3 c^4 \sqrt {1-c^2 x^2}}-b c \left (\frac {g^4 \arcsin (c x)^2}{2 c^6}+\frac {f g \text {arctanh}(c x) \left (3 c^2 f^2-7 g^2\right )}{3 c^5}+\frac {(c f-g)^4}{12 c^6 (c x+1)}+\frac {(c f+g)^4}{12 c^6 (1-c x)}-\frac {g (c f-g)^3 \log (c x+1)}{6 c^6}+\frac {g (c f+g)^3 \log (1-c x)}{6 c^6}+\frac {f g^3 x}{3 c^4}+\frac {f g x \left (2 c^2 f^2-5 g^2\right )}{3 c^4}-\frac {2 f g x \left (c^2 f^2-2 g^2\right )}{3 c^4}-\frac {\left (2 c^4 f^4-3 c^2 f^2 g^2-3 g^4\right ) \log \left (1-c^2 x^2\right )}{6 c^6}\right )\right )}{d^2 \sqrt {d-c^2 d x^2}}\) |
(Sqrt[1 - c^2*x^2]*(((g + c^2*f*x)*(f + g*x)^3*(a + b*ArcSin[c*x]))/(3*c^2 *(1 - c^2*x^2)^(3/2)) + ((f + g*x)*(g*(c^2*f^2 - 3*g^2) + 2*c^2*f*(c^2*f^2 - 2*g^2)*x)*(a + b*ArcSin[c*x]))/(3*c^4*Sqrt[1 - c^2*x^2]) + (f*g*(2*c^2* f^2 - 5*g^2)*Sqrt[1 - c^2*x^2]*(a + b*ArcSin[c*x]))/(3*c^4) + (g^4*ArcSin[ c*x]*(a + b*ArcSin[c*x]))/c^5 - b*c*((f*g^3*x)/(3*c^4) + (f*g*(2*c^2*f^2 - 5*g^2)*x)/(3*c^4) - (2*f*g*(c^2*f^2 - 2*g^2)*x)/(3*c^4) + (c*f + g)^4/(12 *c^6*(1 - c*x)) + (c*f - g)^4/(12*c^6*(1 + c*x)) + (g^4*ArcSin[c*x]^2)/(2* c^6) + (f*g*(3*c^2*f^2 - 7*g^2)*ArcTanh[c*x])/(3*c^5) + (g*(c*f + g)^3*Log [1 - c*x])/(6*c^6) - ((c*f - g)^3*g*Log[1 + c*x])/(6*c^6) - ((2*c^4*f^4 - 3*c^2*f^2*g^2 - 3*g^4)*Log[1 - c^2*x^2])/(6*c^6))))/(d^2*Sqrt[d - c^2*d*x^ 2])
3.1.53.3.1 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 = 1.02 (sec) , antiderivative size = 6743, normalized size of antiderivative = 12.77
| method | result | size |
| default | \(\text {Expression too large to display}\) | \(6743\) |
| parts | \(\text {Expression too large to display}\) | \(6743\) |
\[ \int \frac {(f+g x)^4 (a+b \arcsin (c x))}{\left (d-c^2 d x^2\right )^{5/2}} \, dx=\int { \frac {{\left (g x + f\right )}^{4} {\left (b \arcsin \left (c x\right ) + a\right )}}{{\left (-c^{2} d x^{2} + d\right )}^{\frac {5}{2}}} \,d x } \]
integral(-(a*g^4*x^4 + 4*a*f*g^3*x^3 + 6*a*f^2*g^2*x^2 + 4*a*f^3*g*x + a*f ^4 + (b*g^4*x^4 + 4*b*f*g^3*x^3 + 6*b*f^2*g^2*x^2 + 4*b*f^3*g*x + b*f^4)*a rcsin(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)
\[ \int \frac {(f+g x)^4 (a+b \arcsin (c x))}{\left (d-c^2 d x^2\right )^{5/2}} \, dx=\int \frac {\left (a + b \operatorname {asin}{\left (c x \right )}\right ) \left (f + g x\right )^{4}}{\left (- d \left (c x - 1\right ) \left (c x + 1\right )\right )^{\frac {5}{2}}}\, dx \]
\[ \int \frac {(f+g x)^4 (a+b \arcsin (c x))}{\left (d-c^2 d x^2\right )^{5/2}} \, dx=\int { \frac {{\left (g x + f\right )}^{4} {\left (b \arcsin \left (c x\right ) + a\right )}}{{\left (-c^{2} d x^{2} + d\right )}^{\frac {5}{2}}} \,d x } \]
1/6*b*c*f^4*(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^4*(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*(x*(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)) - x/(sqrt(-c^2* d*x^2 + d)*c^4*d^2) + 3*arcsin(c*x)/(c^5*d^(5/2)))*a*g^4 + 1/3*a*f^4*(2*x/ (sqrt(-c^2*d*x^2 + d)*d^2) + x/((-c^2*d*x^2 + d)^(3/2)*d)) + 4/3*a*f*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)) - 2*a*f^2*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)) - sqrt(d)*integrate((b*g^4*x^4 + 4*b*f*g^3*x^3 + 6*b*f^2*g^2*x^2 + 4*b*f^3*g*x)*sqrt(c*x + 1)*sqrt(-c*x + 1)*arctan2(c*x, sqrt(c*x + 1)*sqr t(-c*x + 1))/(c^6*d^3*x^6 - 3*c^4*d^3*x^4 + 3*c^2*d^3*x^2 - d^3), x) + 4/3 *a*f^3*g/((-c^2*d*x^2 + d)^(3/2)*c^2*d)
Exception generated. \[ \int \frac {(f+g x)^4 (a+b \arcsin (c x))}{\left (d-c^2 d x^2\right )^{5/2}} \, dx=\text {Exception raised: TypeError} \]
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. \[ \int \frac {(f+g x)^4 (a+b \arcsin (c x))}{\left (d-c^2 d x^2\right )^{5/2}} \, dx=\int \frac {{\left (f+g\,x\right )}^4\,\left (a+b\,\mathrm {asin}\left (c\,x\right )\right )}{{\left (d-c^2\,d\,x^2\right )}^{5/2}} \,d x \]