Integrand size = 21, antiderivative size = 161 \[ \int (f x)^m \left (d+e x^2\right ) (a+b \arcsin (c x)) \, dx=\frac {b e (f x)^{2+m} \sqrt {1-c^2 x^2}}{c f^2 (3+m)^2}+\frac {d (f x)^{1+m} (a+b \arcsin (c x))}{f (1+m)}+\frac {e (f x)^{3+m} (a+b \arcsin (c x))}{f^3 (3+m)}-\frac {b \left (\frac {e (2+m)}{3+m}+\frac {c^2 d (3+m)}{1+m}\right ) (f x)^{2+m} \operatorname {Hypergeometric2F1}\left (\frac {1}{2},\frac {2+m}{2},\frac {4+m}{2},c^2 x^2\right )}{c f^2 (2+m) (3+m)} \] Output:
b*e*(f*x)^(2+m)*(-c^2*x^2+1)^(1/2)/c/f^2/(3+m)^2+d*(f*x)^(1+m)*(a+b*arcsin (c*x))/f/(1+m)+e*(f*x)^(3+m)*(a+b*arcsin(c*x))/f^3/(3+m)-b*(e*(2+m)/(3+m)+ c^2*d*(3+m)/(1+m))*(f*x)^(2+m)*hypergeom([1/2, 1+1/2*m],[2+1/2*m],c^2*x^2) /c/f^2/(2+m)/(3+m)
Time = 0.11 (sec) , antiderivative size = 122, normalized size of antiderivative = 0.76 \[ \int (f x)^m \left (d+e x^2\right ) (a+b \arcsin (c x)) \, dx=x (f x)^m \left (-\frac {b c d x \operatorname {Hypergeometric2F1}\left (\frac {1}{2},1+\frac {m}{2},2+\frac {m}{2},c^2 x^2\right )}{2+3 m+m^2}+\frac {\frac {\left (d (3+m)+e (1+m) x^2\right ) (a+b \arcsin (c x))}{1+m}-\frac {b c e x^3 \operatorname {Hypergeometric2F1}\left (\frac {1}{2},2+\frac {m}{2},3+\frac {m}{2},c^2 x^2\right )}{4+m}}{3+m}\right ) \] Input:
Integrate[(f*x)^m*(d + e*x^2)*(a + b*ArcSin[c*x]),x]
Output:
x*(f*x)^m*(-((b*c*d*x*Hypergeometric2F1[1/2, 1 + m/2, 2 + m/2, c^2*x^2])/( 2 + 3*m + m^2)) + (((d*(3 + m) + e*(1 + m)*x^2)*(a + b*ArcSin[c*x]))/(1 + m) - (b*c*e*x^3*Hypergeometric2F1[1/2, 2 + m/2, 3 + m/2, c^2*x^2])/(4 + m) )/(3 + m))
Time = 0.39 (sec) , antiderivative size = 170, normalized size of antiderivative = 1.06, number of steps used = 4, number of rules used = 4, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.190, Rules used = {5230, 27, 363, 278}
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.
\(\displaystyle \int \left (d+e x^2\right ) (f x)^m (a+b \arcsin (c x)) \, dx\) |
\(\Big \downarrow \) 5230 |
\(\displaystyle -b c \int \frac {(f x)^{m+1} \left (e (m+1) x^2+d (m+3)\right )}{f \left (m^2+4 m+3\right ) \sqrt {1-c^2 x^2}}dx+\frac {d (f x)^{m+1} (a+b \arcsin (c x))}{f (m+1)}+\frac {e (f x)^{m+3} (a+b \arcsin (c x))}{f^3 (m+3)}\) |
\(\Big \downarrow \) 27 |
\(\displaystyle -\frac {b c \int \frac {(f x)^{m+1} \left (e (m+1) x^2+d (m+3)\right )}{\sqrt {1-c^2 x^2}}dx}{f \left (m^2+4 m+3\right )}+\frac {d (f x)^{m+1} (a+b \arcsin (c x))}{f (m+1)}+\frac {e (f x)^{m+3} (a+b \arcsin (c x))}{f^3 (m+3)}\) |
\(\Big \downarrow \) 363 |
\(\displaystyle -\frac {b c \left (\left (\frac {e (m+1) (m+2)}{c^2 (m+3)}+d (m+3)\right ) \int \frac {(f x)^{m+1}}{\sqrt {1-c^2 x^2}}dx-\frac {e (m+1) \sqrt {1-c^2 x^2} (f x)^{m+2}}{c^2 f (m+3)}\right )}{f \left (m^2+4 m+3\right )}+\frac {d (f x)^{m+1} (a+b \arcsin (c x))}{f (m+1)}+\frac {e (f x)^{m+3} (a+b \arcsin (c x))}{f^3 (m+3)}\) |
\(\Big \downarrow \) 278 |
\(\displaystyle \frac {d (f x)^{m+1} (a+b \arcsin (c x))}{f (m+1)}+\frac {e (f x)^{m+3} (a+b \arcsin (c x))}{f^3 (m+3)}-\frac {b c \left (\frac {(f x)^{m+2} \left (\frac {e (m+1) (m+2)}{c^2 (m+3)}+d (m+3)\right ) \operatorname {Hypergeometric2F1}\left (\frac {1}{2},\frac {m+2}{2},\frac {m+4}{2},c^2 x^2\right )}{f (m+2)}-\frac {e (m+1) \sqrt {1-c^2 x^2} (f x)^{m+2}}{c^2 f (m+3)}\right )}{f \left (m^2+4 m+3\right )}\) |
Input:
Int[(f*x)^m*(d + e*x^2)*(a + b*ArcSin[c*x]),x]
Output:
(d*(f*x)^(1 + m)*(a + b*ArcSin[c*x]))/(f*(1 + m)) + (e*(f*x)^(3 + m)*(a + b*ArcSin[c*x]))/(f^3*(3 + m)) - (b*c*(-((e*(1 + m)*(f*x)^(2 + m)*Sqrt[1 - c^2*x^2])/(c^2*f*(3 + m))) + (((e*(1 + m)*(2 + m))/(c^2*(3 + m)) + d*(3 + m))*(f*x)^(2 + m)*Hypergeometric2F1[1/2, (2 + m)/2, (4 + m)/2, c^2*x^2])/( f*(2 + m))))/(f*(3 + 4*m + m^2))
Int[(a_)*(Fx_), x_Symbol] :> Simp[a Int[Fx, x], x] /; FreeQ[a, x] && !Ma tchQ[Fx, (b_)*(Gx_) /; FreeQ[b, x]]
Int[((c_.)*(x_))^(m_.)*((a_) + (b_.)*(x_)^2)^(p_), x_Symbol] :> Simp[a^p*(( c*x)^(m + 1)/(c*(m + 1)))*Hypergeometric2F1[-p, (m + 1)/2, (m + 1)/2 + 1, ( -b)*(x^2/a)], x] /; FreeQ[{a, b, c, m, p}, x] && !IGtQ[p, 0] && (ILtQ[p, 0 ] || GtQ[a, 0])
Int[((e_.)*(x_))^(m_.)*((a_) + (b_.)*(x_)^2)^(p_.)*((c_) + (d_.)*(x_)^2), x _Symbol] :> Simp[d*(e*x)^(m + 1)*((a + b*x^2)^(p + 1)/(b*e*(m + 2*p + 3))), x] - Simp[(a*d*(m + 1) - b*c*(m + 2*p + 3))/(b*(m + 2*p + 3)) Int[(e*x)^ m*(a + b*x^2)^p, x], x] /; FreeQ[{a, b, c, d, e, m, p}, x] && NeQ[b*c - a*d , 0] && NeQ[m + 2*p + 3, 0]
Int[((a_.) + ArcSin[(c_.)*(x_)]*(b_.))*((f_.)*(x_))^(m_.)*((d_) + (e_.)*(x_ )^2)^(p_.), x_Symbol] :> With[{u = IntHide[(f*x)^m*(d + e*x^2)^p, x]}, Simp [(a + b*ArcSin[c*x]) u, x] - Simp[b*c Int[SimplifyIntegrand[u/Sqrt[1 - c^2*x^2], x], x], x]] /; FreeQ[{a, b, c, d, e, f, m}, x] && NeQ[c^2*d + e, 0] && IntegerQ[p] && (GtQ[p, 0] || (IGtQ[(m - 1)/2, 0] && LeQ[m + p, 0]))
\[\int \left (f x \right )^{m} \left (e \,x^{2}+d \right ) \left (a +b \arcsin \left (c x \right )\right )d x\]
Input:
int((f*x)^m*(e*x^2+d)*(a+b*arcsin(c*x)),x)
Output:
int((f*x)^m*(e*x^2+d)*(a+b*arcsin(c*x)),x)
\[ \int (f x)^m \left (d+e x^2\right ) (a+b \arcsin (c x)) \, dx=\int { {\left (e x^{2} + d\right )} {\left (b \arcsin \left (c x\right ) + a\right )} \left (f x\right )^{m} \,d x } \] Input:
integrate((f*x)^m*(e*x^2+d)*(a+b*arcsin(c*x)),x, algorithm="fricas")
Output:
integral((a*e*x^2 + a*d + (b*e*x^2 + b*d)*arcsin(c*x))*(f*x)^m, x)
\[ \int (f x)^m \left (d+e x^2\right ) (a+b \arcsin (c x)) \, dx=\int \left (f x\right )^{m} \left (a + b \operatorname {asin}{\left (c x \right )}\right ) \left (d + e x^{2}\right )\, dx \] Input:
integrate((f*x)**m*(e*x**2+d)*(a+b*asin(c*x)),x)
Output:
Integral((f*x)**m*(a + b*asin(c*x))*(d + e*x**2), x)
\[ \int (f x)^m \left (d+e x^2\right ) (a+b \arcsin (c x)) \, dx=\int { {\left (e x^{2} + d\right )} {\left (b \arcsin \left (c x\right ) + a\right )} \left (f x\right )^{m} \,d x } \] Input:
integrate((f*x)^m*(e*x^2+d)*(a+b*arcsin(c*x)),x, algorithm="maxima")
Output:
a*e*f^m*x^3*x^m/(m + 3) + (f*x)^(m + 1)*a*d/(f*(m + 1)) + (((b*e*f^m*m + b *e*f^m)*x^3 + (b*d*f^m*m + 3*b*d*f^m)*x)*x^m*arctan2(c*x, sqrt(c*x + 1)*sq rt(-c*x + 1)) + (m^2 + 4*m + 3)*integrate(((b*c*e*f^m*m + b*c*e*f^m)*x^3 + (b*c*d*f^m*m + 3*b*c*d*f^m)*x)*sqrt(c*x + 1)*sqrt(-c*x + 1)*x^m/((c^2*m^2 + 4*c^2*m + 3*c^2)*x^2 - m^2 - 4*m - 3), x))/(m^2 + 4*m + 3)
\[ \int (f x)^m \left (d+e x^2\right ) (a+b \arcsin (c x)) \, dx=\int { {\left (e x^{2} + d\right )} {\left (b \arcsin \left (c x\right ) + a\right )} \left (f x\right )^{m} \,d x } \] Input:
integrate((f*x)^m*(e*x^2+d)*(a+b*arcsin(c*x)),x, algorithm="giac")
Output:
integrate((e*x^2 + d)*(b*arcsin(c*x) + a)*(f*x)^m, x)
Timed out. \[ \int (f x)^m \left (d+e x^2\right ) (a+b \arcsin (c x)) \, dx=\int \left (a+b\,\mathrm {asin}\left (c\,x\right )\right )\,{\left (f\,x\right )}^m\,\left (e\,x^2+d\right ) \,d x \] Input:
int((a + b*asin(c*x))*(f*x)^m*(d + e*x^2),x)
Output:
int((a + b*asin(c*x))*(f*x)^m*(d + e*x^2), x)
\[ \int (f x)^m \left (d+e x^2\right ) (a+b \arcsin (c x)) \, dx=\frac {f^{m} \left (x^{m} a d m x +3 x^{m} a d x +x^{m} a e m \,x^{3}+x^{m} a e \,x^{3}+\left (\int x^{m} \mathit {asin} \left (c x \right ) x^{2}d x \right ) b e \,m^{2}+4 \left (\int x^{m} \mathit {asin} \left (c x \right ) x^{2}d x \right ) b e m +3 \left (\int x^{m} \mathit {asin} \left (c x \right ) x^{2}d x \right ) b e +\left (\int x^{m} \mathit {asin} \left (c x \right )d x \right ) b d \,m^{2}+4 \left (\int x^{m} \mathit {asin} \left (c x \right )d x \right ) b d m +3 \left (\int x^{m} \mathit {asin} \left (c x \right )d x \right ) b d \right )}{m^{2}+4 m +3} \] Input:
int((f*x)^m*(e*x^2+d)*(a+b*asin(c*x)),x)
Output:
(f**m*(x**m*a*d*m*x + 3*x**m*a*d*x + x**m*a*e*m*x**3 + x**m*a*e*x**3 + int (x**m*asin(c*x)*x**2,x)*b*e*m**2 + 4*int(x**m*asin(c*x)*x**2,x)*b*e*m + 3* int(x**m*asin(c*x)*x**2,x)*b*e + int(x**m*asin(c*x),x)*b*d*m**2 + 4*int(x* *m*asin(c*x),x)*b*d*m + 3*int(x**m*asin(c*x),x)*b*d))/(m**2 + 4*m + 3)