3.7.53 \(\int \frac {a+b \text {ArcSin}(c x)}{(d+e x^2)^{7/2}} \, dx\) [653]

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

[Out]

1/5*x*(a+b*arcsin(c*x))/d/(e*x^2+d)^(5/2)+4/15*x*(a+b*arcsin(c*x))/d^2/(e*x^2+d)^(3/2)+8/15*b*arctan(e^(1/2)*(
-c^2*x^2+1)^(1/2)/c/(e*x^2+d)^(1/2))/d^3/e^(1/2)+1/15*b*c*(-c^2*x^2+1)^(1/2)/d/(c^2*d+e)/(e*x^2+d)^(3/2)+8/15*
x*(a+b*arcsin(c*x))/d^3/(e*x^2+d)^(1/2)+2/15*b*c*(3*c^2*d+2*e)*(-c^2*x^2+1)^(1/2)/d^2/(c^2*d+e)^2/(e*x^2+d)^(1
/2)

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Rubi [A]
time = 0.58, antiderivative size = 226, normalized size of antiderivative = 1.00, number of steps used = 8, number of rules used = 10, integrand size = 20, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.500, Rules used = {198, 197, 4755, 12, 6847, 963, 79, 65, 223, 209} \begin {gather*} \frac {8 x (a+b \text {ArcSin}(c x))}{15 d^3 \sqrt {d+e x^2}}+\frac {4 x (a+b \text {ArcSin}(c x))}{15 d^2 \left (d+e x^2\right )^{3/2}}+\frac {x (a+b \text {ArcSin}(c x))}{5 d \left (d+e x^2\right )^{5/2}}+\frac {8 b \text {ArcTan}\left (\frac {\sqrt {e} \sqrt {1-c^2 x^2}}{c \sqrt {d+e x^2}}\right )}{15 d^3 \sqrt {e}}+\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 {b c \sqrt {1-c^2 x^2}}{15 d \left (c^2 d+e\right ) \left (d+e x^2\right )^{3/2}} \end {gather*}

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 12

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

Rule 65

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 79

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] || L
tQ[p, n]))))

Rule 197

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 198

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 209

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

Rule 223

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 963

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 4755

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 6847

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]

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) \text {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) \text {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) \text {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) \text {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) \text {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*}

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Mathematica [C] Result contains higher order function than in optimal. Order 6 vs. order 3 in optimal.
time = 0.27, size = 188, normalized size = 0.83 \begin {gather*} \frac {a x \left (15 d^2+20 d e x^2+8 e^2 x^4\right )+\frac {b c d \sqrt {1-c^2 x^2} \left (d+e x^2\right ) \left (e \left (5 d+4 e x^2\right )+c^2 d \left (7 d+6 e x^2\right )\right )}{\left (c^2 d+e\right )^2}-4 b c x^2 \left (d+e x^2\right )^2 \sqrt {1+\frac {e x^2}{d}} F_1\left (1;\frac {1}{2},\frac {1}{2};2;c^2 x^2,-\frac {e x^2}{d}\right )+b x \left (15 d^2+20 d e x^2+8 e^2 x^4\right ) \text {ArcSin}(c x)}{15 d^3 \left (d+e x^2\right )^{5/2}} \end {gather*}

Antiderivative was successfully verified.

[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 = 0.82, size = 0, normalized size = 0.00 \[\int \frac {a +b \arcsin \left (c x \right )}{\left (e \,x^{2}+d \right )^{\frac {7}{2}}}\, dx\]

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 = 0.00, size = 0, normalized size = 0.00 \begin {gather*} \text {Failed to integrate} \end {gather*}

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

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Fricas [B] Leaf count of result is larger than twice the leaf count of optimal. 660 vs. \(2 (199) = 398\).
time = 3.09, size = 660, normalized size = 2.92 \begin {gather*} \frac {4 \, {\left (b c^{4} d^{5} + b x^{6} e^{5} + {\left (2 \, b c^{2} d x^{6} + 3 \, b d x^{4}\right )} e^{4} + {\left (b c^{4} d^{2} x^{6} + 6 \, b c^{2} d^{2} x^{4} + 3 \, b d^{2} x^{2}\right )} e^{3} + {\left (3 \, b c^{4} d^{3} x^{4} + 6 \, b c^{2} d^{3} x^{2} + b d^{3}\right )} e^{2} + {\left (3 \, b c^{4} d^{4} x^{2} + 2 \, b c^{2} d^{4}\right )} e\right )} \arctan \left (\frac {\sqrt {-c^{2} x^{2} + 1} {\left (c^{2} d + {\left (2 \, c^{2} x^{2} - 1\right )} e\right )} \sqrt {x^{2} e + d} e^{\frac {1}{2}}}{2 \, {\left ({\left (c^{3} x^{4} - c x^{2}\right )} e^{2} + {\left (c^{3} d x^{2} - c d\right )} e\right )}}\right ) e^{\frac {1}{2}} + {\left (15 \, a c^{4} d^{4} x e + 8 \, a x^{5} e^{5} + {\left (15 \, b c^{4} d^{4} x e + 8 \, b x^{5} e^{5} + 4 \, {\left (4 \, b c^{2} d x^{5} + 5 \, b d x^{3}\right )} e^{4} + {\left (8 \, b c^{4} d^{2} x^{5} + 40 \, b c^{2} d^{2} x^{3} + 15 \, b d^{2} x\right )} e^{3} + 10 \, {\left (2 \, b c^{4} d^{3} x^{3} + 3 \, b c^{2} d^{3} x\right )} e^{2}\right )} \arcsin \left (c x\right ) + 4 \, {\left (4 \, a c^{2} d x^{5} + 5 \, a d x^{3}\right )} e^{4} + {\left (8 \, a c^{4} d^{2} x^{5} + 40 \, a c^{2} d^{2} x^{3} + 15 \, a d^{2} x\right )} e^{3} + 10 \, {\left (2 \, a c^{4} d^{3} x^{3} + 3 \, a c^{2} d^{3} x\right )} e^{2} + {\left (7 \, b c^{3} d^{4} e + 4 \, b c d x^{4} e^{4} + 3 \, {\left (2 \, b c^{3} d^{2} x^{4} + 3 \, b c d^{2} x^{2}\right )} e^{3} + {\left (13 \, b c^{3} d^{3} x^{2} + 5 \, b c d^{3}\right )} e^{2}\right )} \sqrt {-c^{2} x^{2} + 1}\right )} \sqrt {x^{2} e + d}}{15 \, {\left (c^{4} d^{8} e + d^{3} x^{6} e^{6} + {\left (2 \, c^{2} d^{4} x^{6} + 3 \, d^{4} x^{4}\right )} e^{5} + {\left (c^{4} d^{5} x^{6} + 6 \, c^{2} d^{5} x^{4} + 3 \, d^{5} x^{2}\right )} e^{4} + {\left (3 \, c^{4} d^{6} x^{4} + 6 \, c^{2} d^{6} x^{2} + d^{6}\right )} e^{3} + {\left (3 \, c^{4} d^{7} x^{2} + 2 \, c^{2} d^{7}\right )} e^{2}\right )}} \end {gather*}

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*(4*(b*c^4*d^5 + b*x^6*e^5 + (2*b*c^2*d*x^6 + 3*b*d*x^4)*e^4 + (b*c^4*d^2*x^6 + 6*b*c^2*d^2*x^4 + 3*b*d^2*
x^2)*e^3 + (3*b*c^4*d^3*x^4 + 6*b*c^2*d^3*x^2 + b*d^3)*e^2 + (3*b*c^4*d^4*x^2 + 2*b*c^2*d^4)*e)*arctan(1/2*sqr
t(-c^2*x^2 + 1)*(c^2*d + (2*c^2*x^2 - 1)*e)*sqrt(x^2*e + d)*e^(1/2)/((c^3*x^4 - c*x^2)*e^2 + (c^3*d*x^2 - c*d)
*e))*e^(1/2) + (15*a*c^4*d^4*x*e + 8*a*x^5*e^5 + (15*b*c^4*d^4*x*e + 8*b*x^5*e^5 + 4*(4*b*c^2*d*x^5 + 5*b*d*x^
3)*e^4 + (8*b*c^4*d^2*x^5 + 40*b*c^2*d^2*x^3 + 15*b*d^2*x)*e^3 + 10*(2*b*c^4*d^3*x^3 + 3*b*c^2*d^3*x)*e^2)*arc
sin(c*x) + 4*(4*a*c^2*d*x^5 + 5*a*d*x^3)*e^4 + (8*a*c^4*d^2*x^5 + 40*a*c^2*d^2*x^3 + 15*a*d^2*x)*e^3 + 10*(2*a
*c^4*d^3*x^3 + 3*a*c^2*d^3*x)*e^2 + (7*b*c^3*d^4*e + 4*b*c*d*x^4*e^4 + 3*(2*b*c^3*d^2*x^4 + 3*b*c*d^2*x^2)*e^3
+ (13*b*c^3*d^3*x^2 + 5*b*c*d^3)*e^2)*sqrt(-c^2*x^2 + 1))*sqrt(x^2*e + d))/(c^4*d^8*e + d^3*x^6*e^6 + (2*c^2*
d^4*x^6 + 3*d^4*x^4)*e^5 + (c^4*d^5*x^6 + 6*c^2*d^5*x^4 + 3*d^5*x^2)*e^4 + (3*c^4*d^6*x^4 + 6*c^2*d^6*x^2 + d^
6)*e^3 + (3*c^4*d^7*x^2 + 2*c^2*d^7)*e^2)

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Sympy [F(-1)] Timed out
time = 0.00, size = 0, normalized size = 0.00 \begin {gather*} \text {Timed out} \end {gather*}

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 = 0.00, size = 0, normalized size = 0.00 \begin {gather*} \text {could not integrate} \end {gather*}

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)

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Mupad [F]
time = 0.00, size = -1, normalized size = -0.00 \begin {gather*} \int \frac {a+b\,\mathrm {asin}\left (c\,x\right )}{{\left (e\,x^2+d\right )}^{7/2}} \,d x \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

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

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