\(\int (f x)^m (d+e x^2) (a+b \arcsin (c x)) \, dx\) [656]

Optimal result

Integrand size = 21, antiderivative size = 161 \[ \int (f x)^m \left (d+e x^2\right ) (a+b \arcsin (c x)) \, dx=\frac {b e (f x)^{2+m} \sqrt {1-c^2 x^2}}{c f^2 (3+m)^2}+\frac {d (f x)^{1+m} (a+b \arcsin (c x))}{f (1+m)}+\frac {e (f x)^{3+m} (a+b \arcsin (c x))}{f^3 (3+m)}-\frac {b \left (e (1+m) (2+m)+c^2 d (3+m)^2\right ) (f x)^{2+m} \operatorname {Hypergeometric2F1}\left (\frac {1}{2},\frac {2+m}{2},\frac {4+m}{2},c^2 x^2\right )}{c f^2 (1+m) (2+m) (3+m)^2} \]

[Out]

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*(1+m)*(2+m)+c^2*d*(3+m)
^2)*(f*x)^(2+m)*hypergeom([1/2, 1+1/2*m],[2+1/2*m],c^2*x^2)/c/f^2/(1+m)/(2+m)/(3+m)^2+b*e*(f*x)^(2+m)*(-c^2*x^
2+1)^(1/2)/c/f^2/(3+m)^2

Rubi [A] (verified)

Time = 0.12 (sec) , antiderivative size = 148, normalized size of antiderivative = 0.92, number of steps used = 4, number of rules used = 5, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.238, Rules used = {14, 4815, 12, 470, 371} \[ \int (f x)^m \left (d+e x^2\right ) (a+b \arcsin (c x)) \, dx=\frac {d (f x)^{m+1} (a+b \arcsin (c x))}{f (m+1)}+\frac {e (f x)^{m+3} (a+b \arcsin (c x))}{f^3 (m+3)}-\frac {b c (f x)^{m+2} \left (\frac {e}{c^2 (m+3)^2}+\frac {d}{m^2+3 m+2}\right ) \operatorname {Hypergeometric2F1}\left (\frac {1}{2},\frac {m+2}{2},\frac {m+4}{2},c^2 x^2\right )}{f^2}+\frac {b e \sqrt {1-c^2 x^2} (f x)^{m+2}}{c f^2 (m+3)^2} \]

[In]

Int[(f*x)^m*(d + e*x^2)*(a + b*ArcSin[c*x]),x]

[Out]

(b*e*(f*x)^(2 + m)*Sqrt[1 - c^2*x^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*c*(e/(c^2*(3 + m)^2) + d/(2 + 3*m + m^2))*(f*x)^(2 +
m)*Hypergeometric2F1[1/2, (2 + m)/2, (4 + m)/2, c^2*x^2])/f^2

Rule 12

Int[(a_)*(u_), x_Symbol] :> Dist[a, Int[u, x], x] /; FreeQ[a, x] &&  !MatchQ[u, (b_)*(v_) /; FreeQ[b, x]]

Rule 14

Int[(u_)*((c_.)*(x_))^(m_.), x_Symbol] :> Int[ExpandIntegrand[(c*x)^m*u, x], x] /; FreeQ[{c, m}, x] && SumQ[u]
&&  !LinearQ[u, x] &&  !MatchQ[u, (a_) + (b_.)*(v_) /; FreeQ[{a, b}, x] && InverseFunctionQ[v]]

Rule 371

Int[((c_.)*(x_))^(m_.)*((a_) + (b_.)*(x_)^(n_))^(p_), x_Symbol] :> Simp[a^p*((c*x)^(m + 1)/(c*(m + 1)))*Hyperg
eometric2F1[-p, (m + 1)/n, (m + 1)/n + 1, (-b)*(x^n/a)], x] /; FreeQ[{a, b, c, m, n, p}, x] &&  !IGtQ[p, 0] &&
(ILtQ[p, 0] || GtQ[a, 0])

Rule 470

Int[((e_.)*(x_))^(m_.)*((a_) + (b_.)*(x_)^(n_))^(p_.)*((c_) + (d_.)*(x_)^(n_)), x_Symbol] :> Simp[d*(e*x)^(m +
1)*((a + b*x^n)^(p + 1)/(b*e*(m + n*(p + 1) + 1))), x] - Dist[(a*d*(m + 1) - b*c*(m + n*(p + 1) + 1))/(b*(m +
n*(p + 1) + 1)), Int[(e*x)^m*(a + b*x^n)^p, x], x] /; FreeQ[{a, b, c, d, e, m, n, p}, x] && NeQ[b*c - a*d, 0]
&& NeQ[m + n*(p + 1) + 1, 0]

Rule 4815

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]}, Dist[a + b*ArcSin[c*x], u, x] - Dist[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]))

Rubi steps \begin{align*} \text {integral}& = \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)}-(b c) \int \frac {(f x)^{1+m} \left (d (3+m)+e (1+m) x^2\right )}{f (1+m) (3+m) \sqrt {1-c^2 x^2}} \, dx \\ & = \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 c) \int \frac {(f x)^{1+m} \left (d (3+m)+e (1+m) x^2\right )}{\sqrt {1-c^2 x^2}} \, dx}{f \left (3+4 m+m^2\right )} \\ & = \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 {\left (b c \left (\frac {e (1+m) (2+m)}{c^2 (3+m)}+d (3+m)\right )\right ) \int \frac {(f x)^{1+m}}{\sqrt {1-c^2 x^2}} \, dx}{f \left (3+4 m+m^2\right )} \\ & = \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 c \left (\frac {e (1+m) (2+m)}{c^2 (3+m)}+d (3+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 )}{f^2 (2+m) \left (3+4 m+m^2\right )} \\ \end{align*}

Mathematica [A] (verified)

Time = 0.13 (sec) , antiderivative size = 122, normalized size of antiderivative = 0.76 \[ \int (f x)^m \left (d+e x^2\right ) (a+b \arcsin (c x)) \, dx=x (f x)^m \left (-\frac {b c d x \operatorname {Hypergeometric2F1}\left (\frac {1}{2},1+\frac {m}{2},2+\frac {m}{2},c^2 x^2\right )}{2+3 m+m^2}+\frac {\frac {\left (d (3+m)+e (1+m) x^2\right ) (a+b \arcsin (c x))}{1+m}-\frac {b c e x^3 \operatorname {Hypergeometric2F1}\left (\frac {1}{2},2+\frac {m}{2},3+\frac {m}{2},c^2 x^2\right )}{4+m}}{3+m}\right ) \]

[In]

Integrate[(f*x)^m*(d + e*x^2)*(a + b*ArcSin[c*x]),x]

[Out]

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))

Maple [F]

\[\int \left (f x \right )^{m} \left (e \,x^{2}+d \right ) \left (a +b \arcsin \left (c x \right )\right )d x\]

[In]

int((f*x)^m*(e*x^2+d)*(a+b*arcsin(c*x)),x)

[Out]

int((f*x)^m*(e*x^2+d)*(a+b*arcsin(c*x)),x)

Fricas [F]

\[ \int (f x)^m \left (d+e x^2\right ) (a+b \arcsin (c x)) \, dx=\int { {\left (e x^{2} + d\right )} {\left (b \arcsin \left (c x\right ) + a\right )} \left (f x\right )^{m} \,d x } \]

[In]

integrate((f*x)^m*(e*x^2+d)*(a+b*arcsin(c*x)),x, algorithm="fricas")

[Out]

integral((a*e*x^2 + a*d + (b*e*x^2 + b*d)*arcsin(c*x))*(f*x)^m, x)

Sympy [F]

\[ \int (f x)^m \left (d+e x^2\right ) (a+b \arcsin (c x)) \, dx=\int \left (f x\right )^{m} \left (a + b \operatorname {asin}{\left (c x \right )}\right ) \left (d + e x^{2}\right )\, dx \]

[In]

integrate((f*x)**m*(e*x**2+d)*(a+b*asin(c*x)),x)

[Out]

Integral((f*x)**m*(a + b*asin(c*x))*(d + e*x**2), x)

Maxima [F]

\[ \int (f x)^m \left (d+e x^2\right ) (a+b \arcsin (c x)) \, dx=\int { {\left (e x^{2} + d\right )} {\left (b \arcsin \left (c x\right ) + a\right )} \left (f x\right )^{m} \,d x } \]

[In]

integrate((f*x)^m*(e*x^2+d)*(a+b*arcsin(c*x)),x, algorithm="maxima")

[Out]

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)*sqrt(-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)

Giac [F]

\[ \int (f x)^m \left (d+e x^2\right ) (a+b \arcsin (c x)) \, dx=\int { {\left (e x^{2} + d\right )} {\left (b \arcsin \left (c x\right ) + a\right )} \left (f x\right )^{m} \,d x } \]

[In]

integrate((f*x)^m*(e*x^2+d)*(a+b*arcsin(c*x)),x, algorithm="giac")

[Out]

integrate((e*x^2 + d)*(b*arcsin(c*x) + a)*(f*x)^m, x)

Mupad [F(-1)]

Timed out. \[ \int (f x)^m \left (d+e x^2\right ) (a+b \arcsin (c x)) \, dx=\int \left (a+b\,\mathrm {asin}\left (c\,x\right )\right )\,{\left (f\,x\right )}^m\,\left (e\,x^2+d\right ) \,d x \]

[In]

int((a + b*asin(c*x))*(f*x)^m*(d + e*x^2),x)

[Out]

int((a + b*asin(c*x))*(f*x)^m*(d + e*x^2), x)