3.653 \(\int \frac{a+b \sin ^{-1}(c x)}{(d+e x^2)^{7/2}} \, dx\)

Optimal. Leaf size=226 \[ \frac{8 x \left (a+b \sin ^{-1}(c x)\right )}{15 d^3 \sqrt{d+e x^2}}+\frac{4 x \left (a+b \sin ^{-1}(c x)\right )}{15 d^2 \left (d+e x^2\right )^{3/2}}+\frac{x \left (a+b \sin ^{-1}(c x)\right )}{5 d \left (d+e x^2\right )^{5/2}}+\frac{2 b c \sqrt{1-c^2 x^2} \left (3 c^2 d+2 e\right )}{15 d^2 \left (c^2 d+e\right )^2 \sqrt{d+e x^2}}+\frac{8 b \tan ^{-1}\left (\frac{\sqrt{e} \sqrt{1-c^2 x^2}}{c \sqrt{d+e x^2}}\right )}{15 d^3 \sqrt{e}}+\frac{b c \sqrt{1-c^2 x^2}}{15 d \left (c^2 d+e\right ) \left (d+e x^2\right )^{3/2}} \]

[Out]

(b*c*Sqrt[1 - c^2*x^2])/(15*d*(c^2*d + e)*(d + e*x^2)^(3/2)) + (2*b*c*(3*c^2*d + 2*e)*Sqrt[1 - c^2*x^2])/(15*d
^2*(c^2*d + e)^2*Sqrt[d + e*x^2]) + (x*(a + b*ArcSin[c*x]))/(5*d*(d + e*x^2)^(5/2)) + (4*x*(a + b*ArcSin[c*x])
)/(15*d^2*(d + e*x^2)^(3/2)) + (8*x*(a + b*ArcSin[c*x]))/(15*d^3*Sqrt[d + e*x^2]) + (8*b*ArcTan[(Sqrt[e]*Sqrt[
1 - c^2*x^2])/(c*Sqrt[d + e*x^2])])/(15*d^3*Sqrt[e])

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Rubi [A]  time = 0.824567, antiderivative size = 226, normalized size of antiderivative = 1., number of steps used = 8, number of rules used = 10, integrand size = 20, \(\frac{\text{number of rules}}{\text{integrand size}}\) = 0.5, Rules used = {192, 191, 4665, 12, 6715, 949, 78, 63, 217, 203} \[ \frac{8 x \left (a+b \sin ^{-1}(c x)\right )}{15 d^3 \sqrt{d+e x^2}}+\frac{4 x \left (a+b \sin ^{-1}(c x)\right )}{15 d^2 \left (d+e x^2\right )^{3/2}}+\frac{x \left (a+b \sin ^{-1}(c x)\right )}{5 d \left (d+e x^2\right )^{5/2}}+\frac{2 b c \sqrt{1-c^2 x^2} \left (3 c^2 d+2 e\right )}{15 d^2 \left (c^2 d+e\right )^2 \sqrt{d+e x^2}}+\frac{8 b \tan ^{-1}\left (\frac{\sqrt{e} \sqrt{1-c^2 x^2}}{c \sqrt{d+e x^2}}\right )}{15 d^3 \sqrt{e}}+\frac{b c \sqrt{1-c^2 x^2}}{15 d \left (c^2 d+e\right ) \left (d+e x^2\right )^{3/2}} \]

Antiderivative was successfully verified.

[In]

Int[(a + b*ArcSin[c*x])/(d + e*x^2)^(7/2),x]

[Out]

(b*c*Sqrt[1 - c^2*x^2])/(15*d*(c^2*d + e)*(d + e*x^2)^(3/2)) + (2*b*c*(3*c^2*d + 2*e)*Sqrt[1 - c^2*x^2])/(15*d
^2*(c^2*d + e)^2*Sqrt[d + e*x^2]) + (x*(a + b*ArcSin[c*x]))/(5*d*(d + e*x^2)^(5/2)) + (4*x*(a + b*ArcSin[c*x])
)/(15*d^2*(d + e*x^2)^(3/2)) + (8*x*(a + b*ArcSin[c*x]))/(15*d^3*Sqrt[d + e*x^2]) + (8*b*ArcTan[(Sqrt[e]*Sqrt[
1 - c^2*x^2])/(c*Sqrt[d + e*x^2])])/(15*d^3*Sqrt[e])

Rule 192

Int[((a_) + (b_.)*(x_)^(n_))^(p_), x_Symbol] :> -Simp[(x*(a + b*x^n)^(p + 1))/(a*n*(p + 1)), x] + Dist[(n*(p +
1) + 1)/(a*n*(p + 1)), Int[(a + b*x^n)^(p + 1), x], x] /; FreeQ[{a, b, n, p}, x] && ILtQ[Simplify[1/n + p + 1
], 0] && NeQ[p, -1]

Rule 191

Int[((a_) + (b_.)*(x_)^(n_))^(p_), x_Symbol] :> Simp[(x*(a + b*x^n)^(p + 1))/a, x] /; FreeQ[{a, b, n, p}, x] &
& EqQ[1/n + p + 1, 0]

Rule 4665

Int[((a_.) + ArcSin[(c_.)*(x_)]*(b_.))*((d_) + (e_.)*(x_)^2)^(p_.), x_Symbol] :> With[{u = IntHide[(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]] /; F
reeQ[{a, b, c, d, e}, x] && NeQ[c^2*d + e, 0] && (IGtQ[p, 0] || ILtQ[p + 1/2, 0])

Rule 12

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

Rule 6715

Int[(u_)*(x_)^(m_.), x_Symbol] :> Dist[1/(m + 1), Subst[Int[SubstFor[x^(m + 1), u, x], x], x, x^(m + 1)], x] /
; FreeQ[m, x] && NeQ[m, -1] && FunctionOfQ[x^(m + 1), u, x]

Rule 949

Int[((d_.) + (e_.)*(x_))^(m_)*((f_.) + (g_.)*(x_))^(n_)*((a_.) + (b_.)*(x_) + (c_.)*(x_)^2)^(p_.), x_Symbol] :
> With[{Qx = PolynomialQuotient[(a + b*x + c*x^2)^p, d + e*x, x], R = PolynomialRemainder[(a + b*x + c*x^2)^p,
d + e*x, x]}, Simp[(R*(d + e*x)^(m + 1)*(f + g*x)^(n + 1))/((m + 1)*(e*f - d*g)), x] + Dist[1/((m + 1)*(e*f -
d*g)), Int[(d + e*x)^(m + 1)*(f + g*x)^n*ExpandToSum[(m + 1)*(e*f - d*g)*Qx - g*R*(m + n + 2), x], x], x]] /;
FreeQ[{a, b, c, d, e, f, g}, x] && NeQ[e*f - d*g, 0] && NeQ[b^2 - 4*a*c, 0] && NeQ[c*d^2 - b*d*e + a*e^2, 0]
&& IGtQ[p, 0] && LtQ[m, -1]

Rule 78

Int[((a_.) + (b_.)*(x_))*((c_.) + (d_.)*(x_))^(n_.)*((e_.) + (f_.)*(x_))^(p_.), x_Symbol] :> -Simp[((b*e - a*f
)*(c + d*x)^(n + 1)*(e + f*x)^(p + 1))/(f*(p + 1)*(c*f - d*e)), x] - Dist[(a*d*f*(n + p + 2) - b*(d*e*(n + 1)
+ c*f*(p + 1)))/(f*(p + 1)*(c*f - d*e)), Int[(c + d*x)^n*(e + f*x)^(p + 1), x], x] /; FreeQ[{a, b, c, d, e, f,
n}, x] && LtQ[p, -1] && ( !LtQ[n, -1] || IntegerQ[p] ||  !(IntegerQ[n] ||  !(EqQ[e, 0] ||  !(EqQ[c, 0] || LtQ
[p, n]))))

Rule 63

Int[((a_.) + (b_.)*(x_))^(m_)*((c_.) + (d_.)*(x_))^(n_), x_Symbol] :> With[{p = Denominator[m]}, Dist[p/b, Sub
st[Int[x^(p*(m + 1) - 1)*(c - (a*d)/b + (d*x^p)/b)^n, x], x, (a + b*x)^(1/p)], x]] /; FreeQ[{a, b, c, d}, x] &
& NeQ[b*c - a*d, 0] && LtQ[-1, m, 0] && LeQ[-1, n, 0] && LeQ[Denominator[n], Denominator[m]] && IntLinearQ[a,
b, c, d, m, n, x]

Rule 217

Int[1/Sqrt[(a_) + (b_.)*(x_)^2], x_Symbol] :> Subst[Int[1/(1 - b*x^2), x], x, x/Sqrt[a + b*x^2]] /; FreeQ[{a,
b}, x] &&  !GtQ[a, 0]

Rule 203

Int[((a_) + (b_.)*(x_)^2)^(-1), x_Symbol] :> Simp[(1*ArcTan[(Rt[b, 2]*x)/Rt[a, 2]])/(Rt[a, 2]*Rt[b, 2]), x] /;
FreeQ[{a, b}, x] && PosQ[a/b] && (GtQ[a, 0] || GtQ[b, 0])

Rubi steps

\begin{align*} \int \frac{a+b \sin ^{-1}(c x)}{\left (d+e x^2\right )^{7/2}} \, dx &=\frac{x \left (a+b \sin ^{-1}(c x)\right )}{5 d \left (d+e x^2\right )^{5/2}}+\frac{4 x \left (a+b \sin ^{-1}(c x)\right )}{15 d^2 \left (d+e x^2\right )^{3/2}}+\frac{8 x \left (a+b \sin ^{-1}(c x)\right )}{15 d^3 \sqrt{d+e x^2}}-(b c) \int \frac{x \left (15 d^2+20 d e x^2+8 e^2 x^4\right )}{15 d^3 \sqrt{1-c^2 x^2} \left (d+e x^2\right )^{5/2}} \, dx\\ &=\frac{x \left (a+b \sin ^{-1}(c x)\right )}{5 d \left (d+e x^2\right )^{5/2}}+\frac{4 x \left (a+b \sin ^{-1}(c x)\right )}{15 d^2 \left (d+e x^2\right )^{3/2}}+\frac{8 x \left (a+b \sin ^{-1}(c x)\right )}{15 d^3 \sqrt{d+e x^2}}-\frac{(b c) \int \frac{x \left (15 d^2+20 d e x^2+8 e^2 x^4\right )}{\sqrt{1-c^2 x^2} \left (d+e x^2\right )^{5/2}} \, dx}{15 d^3}\\ &=\frac{x \left (a+b \sin ^{-1}(c x)\right )}{5 d \left (d+e x^2\right )^{5/2}}+\frac{4 x \left (a+b \sin ^{-1}(c x)\right )}{15 d^2 \left (d+e x^2\right )^{3/2}}+\frac{8 x \left (a+b \sin ^{-1}(c x)\right )}{15 d^3 \sqrt{d+e x^2}}-\frac{(b c) \operatorname{Subst}\left (\int \frac{15 d^2+20 d e x+8 e^2 x^2}{\sqrt{1-c^2 x} (d+e x)^{5/2}} \, dx,x,x^2\right )}{30 d^3}\\ &=\frac{b c \sqrt{1-c^2 x^2}}{15 d \left (c^2 d+e\right ) \left (d+e x^2\right )^{3/2}}+\frac{x \left (a+b \sin ^{-1}(c x)\right )}{5 d \left (d+e x^2\right )^{5/2}}+\frac{4 x \left (a+b \sin ^{-1}(c x)\right )}{15 d^2 \left (d+e x^2\right )^{3/2}}+\frac{8 x \left (a+b \sin ^{-1}(c x)\right )}{15 d^3 \sqrt{d+e x^2}}+\frac{(b c) \operatorname{Subst}\left (\int \frac{-3 d \left (7 c^2 d+6 e\right )-12 e \left (c^2 d+e\right ) x}{\sqrt{1-c^2 x} (d+e x)^{3/2}} \, dx,x,x^2\right )}{45 d^3 \left (c^2 d+e\right )}\\ &=\frac{b c \sqrt{1-c^2 x^2}}{15 d \left (c^2 d+e\right ) \left (d+e x^2\right )^{3/2}}+\frac{2 b c \left (3 c^2 d+2 e\right ) \sqrt{1-c^2 x^2}}{15 d^2 \left (c^2 d+e\right )^2 \sqrt{d+e x^2}}+\frac{x \left (a+b \sin ^{-1}(c x)\right )}{5 d \left (d+e x^2\right )^{5/2}}+\frac{4 x \left (a+b \sin ^{-1}(c x)\right )}{15 d^2 \left (d+e x^2\right )^{3/2}}+\frac{8 x \left (a+b \sin ^{-1}(c x)\right )}{15 d^3 \sqrt{d+e x^2}}-\frac{(4 b c) \operatorname{Subst}\left (\int \frac{1}{\sqrt{1-c^2 x} \sqrt{d+e x}} \, dx,x,x^2\right )}{15 d^3}\\ &=\frac{b c \sqrt{1-c^2 x^2}}{15 d \left (c^2 d+e\right ) \left (d+e x^2\right )^{3/2}}+\frac{2 b c \left (3 c^2 d+2 e\right ) \sqrt{1-c^2 x^2}}{15 d^2 \left (c^2 d+e\right )^2 \sqrt{d+e x^2}}+\frac{x \left (a+b \sin ^{-1}(c x)\right )}{5 d \left (d+e x^2\right )^{5/2}}+\frac{4 x \left (a+b \sin ^{-1}(c x)\right )}{15 d^2 \left (d+e x^2\right )^{3/2}}+\frac{8 x \left (a+b \sin ^{-1}(c x)\right )}{15 d^3 \sqrt{d+e x^2}}+\frac{(8 b) \operatorname{Subst}\left (\int \frac{1}{\sqrt{d+\frac{e}{c^2}-\frac{e x^2}{c^2}}} \, dx,x,\sqrt{1-c^2 x^2}\right )}{15 c d^3}\\ &=\frac{b c \sqrt{1-c^2 x^2}}{15 d \left (c^2 d+e\right ) \left (d+e x^2\right )^{3/2}}+\frac{2 b c \left (3 c^2 d+2 e\right ) \sqrt{1-c^2 x^2}}{15 d^2 \left (c^2 d+e\right )^2 \sqrt{d+e x^2}}+\frac{x \left (a+b \sin ^{-1}(c x)\right )}{5 d \left (d+e x^2\right )^{5/2}}+\frac{4 x \left (a+b \sin ^{-1}(c x)\right )}{15 d^2 \left (d+e x^2\right )^{3/2}}+\frac{8 x \left (a+b \sin ^{-1}(c x)\right )}{15 d^3 \sqrt{d+e x^2}}+\frac{(8 b) \operatorname{Subst}\left (\int \frac{1}{1+\frac{e x^2}{c^2}} \, dx,x,\frac{\sqrt{1-c^2 x^2}}{\sqrt{d+e x^2}}\right )}{15 c d^3}\\ &=\frac{b c \sqrt{1-c^2 x^2}}{15 d \left (c^2 d+e\right ) \left (d+e x^2\right )^{3/2}}+\frac{2 b c \left (3 c^2 d+2 e\right ) \sqrt{1-c^2 x^2}}{15 d^2 \left (c^2 d+e\right )^2 \sqrt{d+e x^2}}+\frac{x \left (a+b \sin ^{-1}(c x)\right )}{5 d \left (d+e x^2\right )^{5/2}}+\frac{4 x \left (a+b \sin ^{-1}(c x)\right )}{15 d^2 \left (d+e x^2\right )^{3/2}}+\frac{8 x \left (a+b \sin ^{-1}(c x)\right )}{15 d^3 \sqrt{d+e x^2}}+\frac{8 b \tan ^{-1}\left (\frac{\sqrt{e} \sqrt{1-c^2 x^2}}{c \sqrt{d+e x^2}}\right )}{15 d^3 \sqrt{e}}\\ \end{align*}

Mathematica [C]  time = 0.433486, size = 188, normalized size = 0.83 \[ \frac{a x \left (15 d^2+20 d e x^2+8 e^2 x^4\right )-4 b c x^2 \sqrt{\frac{e x^2}{d}+1} \left (d+e x^2\right )^2 F_1\left (1;\frac{1}{2},\frac{1}{2};2;c^2 x^2,-\frac{e x^2}{d}\right )+\frac{b c d \sqrt{1-c^2 x^2} \left (d+e x^2\right ) \left (c^2 d \left (7 d+6 e x^2\right )+e \left (5 d+4 e x^2\right )\right )}{\left (c^2 d+e\right )^2}+b x \sin ^{-1}(c x) \left (15 d^2+20 d e x^2+8 e^2 x^4\right )}{15 d^3 \left (d+e x^2\right )^{5/2}} \]

Warning: Unable to verify antiderivative.

[In]

Integrate[(a + b*ArcSin[c*x])/(d + e*x^2)^(7/2),x]

[Out]

(a*x*(15*d^2 + 20*d*e*x^2 + 8*e^2*x^4) + (b*c*d*Sqrt[1 - c^2*x^2]*(d + e*x^2)*(e*(5*d + 4*e*x^2) + c^2*d*(7*d
+ 6*e*x^2)))/(c^2*d + e)^2 - 4*b*c*x^2*(d + e*x^2)^2*Sqrt[1 + (e*x^2)/d]*AppellF1[1, 1/2, 1/2, 2, c^2*x^2, -((
e*x^2)/d)] + b*x*(15*d^2 + 20*d*e*x^2 + 8*e^2*x^4)*ArcSin[c*x])/(15*d^3*(d + e*x^2)^(5/2))

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Maple [F]  time = 0.325, size = 0, normalized size = 0. \begin{align*} \int{(a+b\arcsin \left ( cx \right ) ) \left ( e{x}^{2}+d \right ) ^{-{\frac{7}{2}}}}\, dx \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

int((a+b*arcsin(c*x))/(e*x^2+d)^(7/2),x)

[Out]

int((a+b*arcsin(c*x))/(e*x^2+d)^(7/2),x)

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Maxima [F]  time = 0., size = 0, normalized size = 0. \begin{align*} \frac{1}{15} \, a{\left (\frac{8 \, x}{\sqrt{e x^{2} + d} d^{3}} + \frac{4 \, x}{{\left (e x^{2} + d\right )}^{\frac{3}{2}} d^{2}} + \frac{3 \, x}{{\left (e x^{2} + d\right )}^{\frac{5}{2}} d}\right )} + b \int \frac{\arctan \left (c x, \sqrt{c x + 1} \sqrt{-c x + 1}\right )}{{\left (e^{3} x^{6} + 3 \, d e^{2} x^{4} + 3 \, d^{2} e x^{2} + d^{3}\right )} \sqrt{e x^{2} + d}}\,{d x} \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate((a+b*arcsin(c*x))/(e*x^2+d)^(7/2),x, algorithm="maxima")

[Out]

1/15*a*(8*x/(sqrt(e*x^2 + d)*d^3) + 4*x/((e*x^2 + d)^(3/2)*d^2) + 3*x/((e*x^2 + d)^(5/2)*d)) + b*integrate(arc
tan2(c*x, sqrt(c*x + 1)*sqrt(-c*x + 1))/((e^3*x^6 + 3*d*e^2*x^4 + 3*d^2*e*x^2 + d^3)*sqrt(e*x^2 + d)), x)

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Fricas [B]  time = 3.49197, size = 2700, normalized size = 11.95 \begin{align*} \text{result too large to display} \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate((a+b*arcsin(c*x))/(e*x^2+d)^(7/2),x, algorithm="fricas")

[Out]

[-1/15*(2*(b*c^4*d^5 + 2*b*c^2*d^4*e + (b*c^4*d^2*e^3 + 2*b*c^2*d*e^4 + b*e^5)*x^6 + b*d^3*e^2 + 3*(b*c^4*d^3*
e^2 + 2*b*c^2*d^2*e^3 + b*d*e^4)*x^4 + 3*(b*c^4*d^4*e + 2*b*c^2*d^3*e^2 + b*d^2*e^3)*x^2)*sqrt(-e)*log(8*c^4*e
^2*x^4 + c^4*d^2 - 6*c^2*d*e + 8*(c^4*d*e - c^2*e^2)*x^2 + 4*(2*c^3*e*x^2 + c^3*d - c*e)*sqrt(-c^2*x^2 + 1)*sq
rt(e*x^2 + d)*sqrt(-e) + e^2) - (8*(a*c^4*d^2*e^3 + 2*a*c^2*d*e^4 + a*e^5)*x^5 + 20*(a*c^4*d^3*e^2 + 2*a*c^2*d
^2*e^3 + a*d*e^4)*x^3 + 15*(a*c^4*d^4*e + 2*a*c^2*d^3*e^2 + a*d^2*e^3)*x + (8*(b*c^4*d^2*e^3 + 2*b*c^2*d*e^4 +
b*e^5)*x^5 + 20*(b*c^4*d^3*e^2 + 2*b*c^2*d^2*e^3 + b*d*e^4)*x^3 + 15*(b*c^4*d^4*e + 2*b*c^2*d^3*e^2 + b*d^2*e
^3)*x)*arcsin(c*x) + (7*b*c^3*d^4*e + 5*b*c*d^3*e^2 + 2*(3*b*c^3*d^2*e^3 + 2*b*c*d*e^4)*x^4 + (13*b*c^3*d^3*e^
2 + 9*b*c*d^2*e^3)*x^2)*sqrt(-c^2*x^2 + 1))*sqrt(e*x^2 + d))/(c^4*d^8*e + 2*c^2*d^7*e^2 + d^6*e^3 + (c^4*d^5*e
^4 + 2*c^2*d^4*e^5 + d^3*e^6)*x^6 + 3*(c^4*d^6*e^3 + 2*c^2*d^5*e^4 + d^4*e^5)*x^4 + 3*(c^4*d^7*e^2 + 2*c^2*d^6
*e^3 + d^5*e^4)*x^2), 1/15*(4*(b*c^4*d^5 + 2*b*c^2*d^4*e + (b*c^4*d^2*e^3 + 2*b*c^2*d*e^4 + b*e^5)*x^6 + b*d^3
*e^2 + 3*(b*c^4*d^3*e^2 + 2*b*c^2*d^2*e^3 + b*d*e^4)*x^4 + 3*(b*c^4*d^4*e + 2*b*c^2*d^3*e^2 + b*d^2*e^3)*x^2)*
sqrt(e)*arctan(1/2*(2*c^2*e*x^2 + c^2*d - e)*sqrt(-c^2*x^2 + 1)*sqrt(e*x^2 + d)*sqrt(e)/(c^3*e^2*x^4 - c*d*e +
(c^3*d*e - c*e^2)*x^2)) + (8*(a*c^4*d^2*e^3 + 2*a*c^2*d*e^4 + a*e^5)*x^5 + 20*(a*c^4*d^3*e^2 + 2*a*c^2*d^2*e^
3 + a*d*e^4)*x^3 + 15*(a*c^4*d^4*e + 2*a*c^2*d^3*e^2 + a*d^2*e^3)*x + (8*(b*c^4*d^2*e^3 + 2*b*c^2*d*e^4 + b*e^
5)*x^5 + 20*(b*c^4*d^3*e^2 + 2*b*c^2*d^2*e^3 + b*d*e^4)*x^3 + 15*(b*c^4*d^4*e + 2*b*c^2*d^3*e^2 + b*d^2*e^3)*x
)*arcsin(c*x) + (7*b*c^3*d^4*e + 5*b*c*d^3*e^2 + 2*(3*b*c^3*d^2*e^3 + 2*b*c*d*e^4)*x^4 + (13*b*c^3*d^3*e^2 + 9
*b*c*d^2*e^3)*x^2)*sqrt(-c^2*x^2 + 1))*sqrt(e*x^2 + d))/(c^4*d^8*e + 2*c^2*d^7*e^2 + d^6*e^3 + (c^4*d^5*e^4 +
2*c^2*d^4*e^5 + d^3*e^6)*x^6 + 3*(c^4*d^6*e^3 + 2*c^2*d^5*e^4 + d^4*e^5)*x^4 + 3*(c^4*d^7*e^2 + 2*c^2*d^6*e^3
+ d^5*e^4)*x^2)]

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Sympy [F(-1)]  time = 0., size = 0, normalized size = 0. \begin{align*} \text{Timed out} \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate((a+b*asin(c*x))/(e*x**2+d)**(7/2),x)

[Out]

Timed out

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Giac [F]  time = 0., size = 0, normalized size = 0. \begin{align*} \int \frac{b \arcsin \left (c x\right ) + a}{{\left (e x^{2} + d\right )}^{\frac{7}{2}}}\,{d x} \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate((a+b*arcsin(c*x))/(e*x^2+d)^(7/2),x, algorithm="giac")

[Out]

integrate((b*arcsin(c*x) + a)/(e*x^2 + d)^(7/2), x)