3.7.11 \(\int \frac {(d+e x^2)^2 (a+b \text {ArcSin}(c x))}{x^2} \, dx\) [611]

Optimal. Leaf size=126 \[ \frac {b e \left (6 c^2 d+e\right ) \sqrt {1-c^2 x^2}}{3 c^3}-\frac {b e^2 \left (1-c^2 x^2\right )^{3/2}}{9 c^3}-\frac {d^2 (a+b \text {ArcSin}(c x))}{x}+2 d e x (a+b \text {ArcSin}(c x))+\frac {1}{3} e^2 x^3 (a+b \text {ArcSin}(c x))-b c d^2 \tanh ^{-1}\left (\sqrt {1-c^2 x^2}\right ) \]

[Out]

-1/9*b*e^2*(-c^2*x^2+1)^(3/2)/c^3-d^2*(a+b*arcsin(c*x))/x+2*d*e*x*(a+b*arcsin(c*x))+1/3*e^2*x^3*(a+b*arcsin(c*
x))-b*c*d^2*arctanh((-c^2*x^2+1)^(1/2))+1/3*b*e*(6*c^2*d+e)*(-c^2*x^2+1)^(1/2)/c^3

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Rubi [A]
time = 0.13, antiderivative size = 126, normalized size of antiderivative = 1.00, number of steps used = 6, number of rules used = 6, integrand size = 21, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.286, Rules used = {276, 4815, 1265, 911, 1167, 214} \begin {gather*} -\frac {d^2 (a+b \text {ArcSin}(c x))}{x}+2 d e x (a+b \text {ArcSin}(c x))+\frac {1}{3} e^2 x^3 (a+b \text {ArcSin}(c x))-b c d^2 \tanh ^{-1}\left (\sqrt {1-c^2 x^2}\right )+\frac {b e \sqrt {1-c^2 x^2} \left (6 c^2 d+e\right )}{3 c^3}-\frac {b e^2 \left (1-c^2 x^2\right )^{3/2}}{9 c^3} \end {gather*}

Antiderivative was successfully verified.

[In]

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

[Out]

(b*e*(6*c^2*d + e)*Sqrt[1 - c^2*x^2])/(3*c^3) - (b*e^2*(1 - c^2*x^2)^(3/2))/(9*c^3) - (d^2*(a + b*ArcSin[c*x])
)/x + 2*d*e*x*(a + b*ArcSin[c*x]) + (e^2*x^3*(a + b*ArcSin[c*x]))/3 - b*c*d^2*ArcTanh[Sqrt[1 - c^2*x^2]]

Rule 214

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

Rule 276

Int[((c_.)*(x_))^(m_.)*((a_) + (b_.)*(x_)^(n_))^(p_.), x_Symbol] :> Int[ExpandIntegrand[(c*x)^m*(a + b*x^n)^p,
x], x] /; FreeQ[{a, b, c, m, n}, x] && IGtQ[p, 0]

Rule 911

Int[((d_.) + (e_.)*(x_))^(m_)*((f_.) + (g_.)*(x_))^(n_)*((a_.) + (b_.)*(x_) + (c_.)*(x_)^2)^(p_.), x_Symbol] :
> With[{q = Denominator[m]}, Dist[q/e, Subst[Int[x^(q*(m + 1) - 1)*((e*f - d*g)/e + g*(x^q/e))^n*((c*d^2 - b*d
*e + a*e^2)/e^2 - (2*c*d - b*e)*(x^q/e^2) + c*(x^(2*q)/e^2))^p, x], x, (d + e*x)^(1/q)], 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] && IntegersQ[n,
p] && FractionQ[m]

Rule 1167

Int[((d_) + (e_.)*(x_)^2)^(q_.)*((a_) + (b_.)*(x_)^2 + (c_.)*(x_)^4)^(p_.), x_Symbol] :> Int[ExpandIntegrand[(
d + e*x^2)^q*(a + b*x^2 + c*x^4)^p, x], x] /; FreeQ[{a, b, c, d, e}, x] && NeQ[b^2 - 4*a*c, 0] && NeQ[c*d^2 -
b*d*e + a*e^2, 0] && IGtQ[p, 0] && IGtQ[q, -2]

Rule 1265

Int[(x_)^(m_.)*((d_) + (e_.)*(x_)^2)^(q_.)*((a_) + (b_.)*(x_)^2 + (c_.)*(x_)^4)^(p_.), x_Symbol] :> Dist[1/2,
Subst[Int[x^((m - 1)/2)*(d + e*x)^q*(a + b*x + c*x^2)^p, x], x, x^2], x] /; FreeQ[{a, b, c, d, e, p, q}, x] &&
IntegerQ[(m - 1)/2]

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*} \int \frac {\left (d+e x^2\right )^2 \left (a+b \sin ^{-1}(c x)\right )}{x^2} \, dx &=-\frac {d^2 \left (a+b \sin ^{-1}(c x)\right )}{x}+2 d e x \left (a+b \sin ^{-1}(c x)\right )+\frac {1}{3} e^2 x^3 \left (a+b \sin ^{-1}(c x)\right )-(b c) \int \frac {-d^2+2 d e x^2+\frac {e^2 x^4}{3}}{x \sqrt {1-c^2 x^2}} \, dx\\ &=-\frac {d^2 \left (a+b \sin ^{-1}(c x)\right )}{x}+2 d e x \left (a+b \sin ^{-1}(c x)\right )+\frac {1}{3} e^2 x^3 \left (a+b \sin ^{-1}(c x)\right )-\frac {1}{2} (b c) \text {Subst}\left (\int \frac {-d^2+2 d e x+\frac {e^2 x^2}{3}}{x \sqrt {1-c^2 x}} \, dx,x,x^2\right )\\ &=-\frac {d^2 \left (a+b \sin ^{-1}(c x)\right )}{x}+2 d e x \left (a+b \sin ^{-1}(c x)\right )+\frac {1}{3} e^2 x^3 \left (a+b \sin ^{-1}(c x)\right )+\frac {b \text {Subst}\left (\int \frac {\frac {-c^4 d^2+2 c^2 d e+\frac {e^2}{3}}{c^4}-\frac {\left (2 c^2 d e+\frac {2 e^2}{3}\right ) x^2}{c^4}+\frac {e^2 x^4}{3 c^4}}{\frac {1}{c^2}-\frac {x^2}{c^2}} \, dx,x,\sqrt {1-c^2 x^2}\right )}{c}\\ &=-\frac {d^2 \left (a+b \sin ^{-1}(c x)\right )}{x}+2 d e x \left (a+b \sin ^{-1}(c x)\right )+\frac {1}{3} e^2 x^3 \left (a+b \sin ^{-1}(c x)\right )+\frac {b \text {Subst}\left (\int \left (\frac {1}{3} e \left (6 d+\frac {e}{c^2}\right )-\frac {e^2 x^2}{3 c^2}-\frac {d^2}{\frac {1}{c^2}-\frac {x^2}{c^2}}\right ) \, dx,x,\sqrt {1-c^2 x^2}\right )}{c}\\ &=\frac {b e \left (6 c^2 d+e\right ) \sqrt {1-c^2 x^2}}{3 c^3}-\frac {b e^2 \left (1-c^2 x^2\right )^{3/2}}{9 c^3}-\frac {d^2 \left (a+b \sin ^{-1}(c x)\right )}{x}+2 d e x \left (a+b \sin ^{-1}(c x)\right )+\frac {1}{3} e^2 x^3 \left (a+b \sin ^{-1}(c x)\right )-\frac {\left (b d^2\right ) \text {Subst}\left (\int \frac {1}{\frac {1}{c^2}-\frac {x^2}{c^2}} \, dx,x,\sqrt {1-c^2 x^2}\right )}{c}\\ &=\frac {b e \left (6 c^2 d+e\right ) \sqrt {1-c^2 x^2}}{3 c^3}-\frac {b e^2 \left (1-c^2 x^2\right )^{3/2}}{9 c^3}-\frac {d^2 \left (a+b \sin ^{-1}(c x)\right )}{x}+2 d e x \left (a+b \sin ^{-1}(c x)\right )+\frac {1}{3} e^2 x^3 \left (a+b \sin ^{-1}(c x)\right )-b c d^2 \tanh ^{-1}\left (\sqrt {1-c^2 x^2}\right )\\ \end {align*}

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Mathematica [A]
time = 0.10, size = 129, normalized size = 1.02 \begin {gather*} \frac {1}{9} \left (-\frac {9 a d^2}{x}+18 a d e x+3 a e^2 x^3+\frac {b e \sqrt {1-c^2 x^2} \left (2 e+c^2 \left (18 d+e x^2\right )\right )}{c^3}+\frac {3 b \left (-3 d^2+6 d e x^2+e^2 x^4\right ) \text {ArcSin}(c x)}{x}+9 b c d^2 \log (x)-9 b c d^2 \log \left (1+\sqrt {1-c^2 x^2}\right )\right ) \end {gather*}

Antiderivative was successfully verified.

[In]

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

[Out]

((-9*a*d^2)/x + 18*a*d*e*x + 3*a*e^2*x^3 + (b*e*Sqrt[1 - c^2*x^2]*(2*e + c^2*(18*d + e*x^2)))/c^3 + (3*b*(-3*d
^2 + 6*d*e*x^2 + e^2*x^4)*ArcSin[c*x])/x + 9*b*c*d^2*Log[x] - 9*b*c*d^2*Log[1 + Sqrt[1 - c^2*x^2]])/9

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Maple [A]
time = 0.11, size = 168, normalized size = 1.33

method	result	size
derivativedivides	\(c \left (\frac {a \left (2 c^{3} d e x +\frac {e^{2} c^{3} x^{3}}{3}-\frac {c^{3} d^{2}}{x}\right )}{c^{4}}+\frac {b \left (2 \arcsin \left (c x \right ) c^{3} d e x +\frac {\arcsin \left (c x \right ) e^{2} c^{3} x^{3}}{3}-\frac {\arcsin \left (c x \right ) c^{3} d^{2}}{x}+2 c^{2} d e \sqrt {-c^{2} x^{2}+1}-\frac {e^{2} \left (-\frac {c^{2} x^{2} \sqrt {-c^{2} x^{2}+1}}{3}-\frac {2 \sqrt {-c^{2} x^{2}+1}}{3}\right )}{3}-c^{4} d^{2} \arctanh \left (\frac {1}{\sqrt {-c^{2} x^{2}+1}}\right )\right )}{c^{4}}\right )\)	\(168\)
default	\(c \left (\frac {a \left (2 c^{3} d e x +\frac {e^{2} c^{3} x^{3}}{3}-\frac {c^{3} d^{2}}{x}\right )}{c^{4}}+\frac {b \left (2 \arcsin \left (c x \right ) c^{3} d e x +\frac {\arcsin \left (c x \right ) e^{2} c^{3} x^{3}}{3}-\frac {\arcsin \left (c x \right ) c^{3} d^{2}}{x}+2 c^{2} d e \sqrt {-c^{2} x^{2}+1}-\frac {e^{2} \left (-\frac {c^{2} x^{2} \sqrt {-c^{2} x^{2}+1}}{3}-\frac {2 \sqrt {-c^{2} x^{2}+1}}{3}\right )}{3}-c^{4} d^{2} \arctanh \left (\frac {1}{\sqrt {-c^{2} x^{2}+1}}\right )\right )}{c^{4}}\right )\)	\(168\)

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

c*(a/c^4*(2*c^3*d*e*x+1/3*e^2*c^3*x^3-c^3*d^2/x)+b/c^4*(2*arcsin(c*x)*c^3*d*e*x+1/3*arcsin(c*x)*e^2*c^3*x^3-ar
csin(c*x)*c^3*d^2/x+2*c^2*d*e*(-c^2*x^2+1)^(1/2)-1/3*e^2*(-1/3*c^2*x^2*(-c^2*x^2+1)^(1/2)-2/3*(-c^2*x^2+1)^(1/
2))-c^4*d^2*arctanh(1/(-c^2*x^2+1)^(1/2))))

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Maxima [A]
time = 0.48, size = 151, normalized size = 1.20 \begin {gather*} \frac {1}{3} \, a x^{3} e^{2} - {\left (c \log \left (\frac {2 \, \sqrt {-c^{2} x^{2} + 1}}{{\left | x \right |}} + \frac {2}{{\left | x \right |}}\right ) + \frac {\arcsin \left (c x\right )}{x}\right )} b d^{2} + 2 \, a d x e + \frac {1}{9} \, {\left (3 \, x^{3} \arcsin \left (c x\right ) + c {\left (\frac {\sqrt {-c^{2} x^{2} + 1} x^{2}}{c^{2}} + \frac {2 \, \sqrt {-c^{2} x^{2} + 1}}{c^{4}}\right )}\right )} b e^{2} + \frac {2 \, {\left (c x \arcsin \left (c x\right ) + \sqrt {-c^{2} x^{2} + 1}\right )} b d e}{c} - \frac {a d^{2}}{x} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

1/3*a*x^3*e^2 - (c*log(2*sqrt(-c^2*x^2 + 1)/abs(x) + 2/abs(x)) + arcsin(c*x)/x)*b*d^2 + 2*a*d*x*e + 1/9*(3*x^3
*arcsin(c*x) + c*(sqrt(-c^2*x^2 + 1)*x^2/c^2 + 2*sqrt(-c^2*x^2 + 1)/c^4))*b*e^2 + 2*(c*x*arcsin(c*x) + sqrt(-c
^2*x^2 + 1))*b*d*e/c - a*d^2/x

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Fricas [A]
time = 1.07, size = 170, normalized size = 1.35 \begin {gather*} \frac {6 \, a c^{3} x^{4} e^{2} - 9 \, b c^{4} d^{2} x \log \left (\sqrt {-c^{2} x^{2} + 1} + 1\right ) + 9 \, b c^{4} d^{2} x \log \left (\sqrt {-c^{2} x^{2} + 1} - 1\right ) + 36 \, a c^{3} d x^{2} e - 18 \, a c^{3} d^{2} + 6 \, {\left (b c^{3} x^{4} e^{2} + 6 \, b c^{3} d x^{2} e - 3 \, b c^{3} d^{2}\right )} \arcsin \left (c x\right ) + 2 \, {\left (18 \, b c^{2} d x e + {\left (b c^{2} x^{3} + 2 \, b x\right )} e^{2}\right )} \sqrt {-c^{2} x^{2} + 1}}{18 \, c^{3} x} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

1/18*(6*a*c^3*x^4*e^2 - 9*b*c^4*d^2*x*log(sqrt(-c^2*x^2 + 1) + 1) + 9*b*c^4*d^2*x*log(sqrt(-c^2*x^2 + 1) - 1)
+ 36*a*c^3*d*x^2*e - 18*a*c^3*d^2 + 6*(b*c^3*x^4*e^2 + 6*b*c^3*d*x^2*e - 3*b*c^3*d^2)*arcsin(c*x) + 2*(18*b*c^
2*d*x*e + (b*c^2*x^3 + 2*b*x)*e^2)*sqrt(-c^2*x^2 + 1))/(c^3*x)

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Sympy [A]
time = 3.42, size = 167, normalized size = 1.33 \begin {gather*} - \frac {a d^{2}}{x} + 2 a d e x + \frac {a e^{2} x^{3}}{3} + b c d^{2} \left (\begin {cases} - \operatorname {acosh}{\left (\frac {1}{c x} \right )} & \text {for}\: \frac {1}{\left |{c^{2} x^{2}}\right |} > 1 \\i \operatorname {asin}{\left (\frac {1}{c x} \right )} & \text {otherwise} \end {cases}\right ) - \frac {b c e^{2} \left (\begin {cases} - \frac {x^{2} \sqrt {- c^{2} x^{2} + 1}}{3 c^{2}} - \frac {2 \sqrt {- c^{2} x^{2} + 1}}{3 c^{4}} & \text {for}\: c \neq 0 \\\frac {x^{4}}{4} & \text {otherwise} \end {cases}\right )}{3} - \frac {b d^{2} \operatorname {asin}{\left (c x \right )}}{x} + 2 b d e \left (\begin {cases} 0 & \text {for}\: c = 0 \\x \operatorname {asin}{\left (c x \right )} + \frac {\sqrt {- c^{2} x^{2} + 1}}{c} & \text {otherwise} \end {cases}\right ) + \frac {b e^{2} x^{3} \operatorname {asin}{\left (c x \right )}}{3} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

-a*d**2/x + 2*a*d*e*x + a*e**2*x**3/3 + b*c*d**2*Piecewise((-acosh(1/(c*x)), 1/Abs(c**2*x**2) > 1), (I*asin(1/
(c*x)), True)) - b*c*e**2*Piecewise((-x**2*sqrt(-c**2*x**2 + 1)/(3*c**2) - 2*sqrt(-c**2*x**2 + 1)/(3*c**4), Ne
(c, 0)), (x**4/4, True))/3 - b*d**2*asin(c*x)/x + 2*b*d*e*Piecewise((0, Eq(c, 0)), (x*asin(c*x) + sqrt(-c**2*x
**2 + 1)/c, True)) + b*e**2*x**3*asin(c*x)/3

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Giac [B] Leaf count of result is larger than twice the leaf count of optimal. 4243 vs. \(2 (114) = 228\).
time = 1.92, size = 4243, normalized size = 33.67 \begin {gather*} \text {Too large to display} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

-1/2*b*c^12*d^2*x^8*arcsin(c*x)/((c^10*x^7/(sqrt(-c^2*x^2 + 1) + 1)^7 + 3*c^8*x^5/(sqrt(-c^2*x^2 + 1) + 1)^5 +
3*c^6*x^3/(sqrt(-c^2*x^2 + 1) + 1)^3 + c^4*x/(sqrt(-c^2*x^2 + 1) + 1))*(sqrt(-c^2*x^2 + 1) + 1)^8) - 1/2*a*c^
12*d^2*x^8/((c^10*x^7/(sqrt(-c^2*x^2 + 1) + 1)^7 + 3*c^8*x^5/(sqrt(-c^2*x^2 + 1) + 1)^5 + 3*c^6*x^3/(sqrt(-c^2
*x^2 + 1) + 1)^3 + c^4*x/(sqrt(-c^2*x^2 + 1) + 1))*(sqrt(-c^2*x^2 + 1) + 1)^8) + b*c^11*d^2*x^7*log(abs(c)*abs
(x))/((c^10*x^7/(sqrt(-c^2*x^2 + 1) + 1)^7 + 3*c^8*x^5/(sqrt(-c^2*x^2 + 1) + 1)^5 + 3*c^6*x^3/(sqrt(-c^2*x^2 +
1) + 1)^3 + c^4*x/(sqrt(-c^2*x^2 + 1) + 1))*(sqrt(-c^2*x^2 + 1) + 1)^7) - b*c^11*d^2*x^7*log(sqrt(-c^2*x^2 +
1) + 1)/((c^10*x^7/(sqrt(-c^2*x^2 + 1) + 1)^7 + 3*c^8*x^5/(sqrt(-c^2*x^2 + 1) + 1)^5 + 3*c^6*x^3/(sqrt(-c^2*x^
2 + 1) + 1)^3 + c^4*x/(sqrt(-c^2*x^2 + 1) + 1))*(sqrt(-c^2*x^2 + 1) + 1)^7) - 2*b*c^10*d^2*x^6*arcsin(c*x)/((c
^10*x^7/(sqrt(-c^2*x^2 + 1) + 1)^7 + 3*c^8*x^5/(sqrt(-c^2*x^2 + 1) + 1)^5 + 3*c^6*x^3/(sqrt(-c^2*x^2 + 1) + 1)
^3 + c^4*x/(sqrt(-c^2*x^2 + 1) + 1))*(sqrt(-c^2*x^2 + 1) + 1)^6) - 2*a*c^10*d^2*x^6/((c^10*x^7/(sqrt(-c^2*x^2
+ 1) + 1)^7 + 3*c^8*x^5/(sqrt(-c^2*x^2 + 1) + 1)^5 + 3*c^6*x^3/(sqrt(-c^2*x^2 + 1) + 1)^3 + c^4*x/(sqrt(-c^2*x
^2 + 1) + 1))*(sqrt(-c^2*x^2 + 1) + 1)^6) + 3*b*c^9*d^2*x^5*log(abs(c)*abs(x))/((c^10*x^7/(sqrt(-c^2*x^2 + 1)
+ 1)^7 + 3*c^8*x^5/(sqrt(-c^2*x^2 + 1) + 1)^5 + 3*c^6*x^3/(sqrt(-c^2*x^2 + 1) + 1)^3 + c^4*x/(sqrt(-c^2*x^2 +
1) + 1))*(sqrt(-c^2*x^2 + 1) + 1)^5) - 3*b*c^9*d^2*x^5*log(sqrt(-c^2*x^2 + 1) + 1)/((c^10*x^7/(sqrt(-c^2*x^2 +
1) + 1)^7 + 3*c^8*x^5/(sqrt(-c^2*x^2 + 1) + 1)^5 + 3*c^6*x^3/(sqrt(-c^2*x^2 + 1) + 1)^3 + c^4*x/(sqrt(-c^2*x^
2 + 1) + 1))*(sqrt(-c^2*x^2 + 1) + 1)^5) - 2*b*c^9*d*e*x^7/((c^10*x^7/(sqrt(-c^2*x^2 + 1) + 1)^7 + 3*c^8*x^5/(
sqrt(-c^2*x^2 + 1) + 1)^5 + 3*c^6*x^3/(sqrt(-c^2*x^2 + 1) + 1)^3 + c^4*x/(sqrt(-c^2*x^2 + 1) + 1))*(sqrt(-c^2*
x^2 + 1) + 1)^7) + 4*b*c^8*d*e*x^6*arcsin(c*x)/((c^10*x^7/(sqrt(-c^2*x^2 + 1) + 1)^7 + 3*c^8*x^5/(sqrt(-c^2*x^
2 + 1) + 1)^5 + 3*c^6*x^3/(sqrt(-c^2*x^2 + 1) + 1)^3 + c^4*x/(sqrt(-c^2*x^2 + 1) + 1))*(sqrt(-c^2*x^2 + 1) + 1
)^6) - 3*b*c^8*d^2*x^4*arcsin(c*x)/((c^10*x^7/(sqrt(-c^2*x^2 + 1) + 1)^7 + 3*c^8*x^5/(sqrt(-c^2*x^2 + 1) + 1)^
5 + 3*c^6*x^3/(sqrt(-c^2*x^2 + 1) + 1)^3 + c^4*x/(sqrt(-c^2*x^2 + 1) + 1))*(sqrt(-c^2*x^2 + 1) + 1)^4) + 4*a*c
^8*d*e*x^6/((c^10*x^7/(sqrt(-c^2*x^2 + 1) + 1)^7 + 3*c^8*x^5/(sqrt(-c^2*x^2 + 1) + 1)^5 + 3*c^6*x^3/(sqrt(-c^2
*x^2 + 1) + 1)^3 + c^4*x/(sqrt(-c^2*x^2 + 1) + 1))*(sqrt(-c^2*x^2 + 1) + 1)^6) - 3*a*c^8*d^2*x^4/((c^10*x^7/(s
qrt(-c^2*x^2 + 1) + 1)^7 + 3*c^8*x^5/(sqrt(-c^2*x^2 + 1) + 1)^5 + 3*c^6*x^3/(sqrt(-c^2*x^2 + 1) + 1)^3 + c^4*x
/(sqrt(-c^2*x^2 + 1) + 1))*(sqrt(-c^2*x^2 + 1) + 1)^4) + 3*b*c^7*d^2*x^3*log(abs(c)*abs(x))/((c^10*x^7/(sqrt(-
c^2*x^2 + 1) + 1)^7 + 3*c^8*x^5/(sqrt(-c^2*x^2 + 1) + 1)^5 + 3*c^6*x^3/(sqrt(-c^2*x^2 + 1) + 1)^3 + c^4*x/(sqr
t(-c^2*x^2 + 1) + 1))*(sqrt(-c^2*x^2 + 1) + 1)^3) - 3*b*c^7*d^2*x^3*log(sqrt(-c^2*x^2 + 1) + 1)/((c^10*x^7/(sq
rt(-c^2*x^2 + 1) + 1)^7 + 3*c^8*x^5/(sqrt(-c^2*x^2 + 1) + 1)^5 + 3*c^6*x^3/(sqrt(-c^2*x^2 + 1) + 1)^3 + c^4*x/
(sqrt(-c^2*x^2 + 1) + 1))*(sqrt(-c^2*x^2 + 1) + 1)^3) - 2/9*b*c^7*e^2*x^7/((c^10*x^7/(sqrt(-c^2*x^2 + 1) + 1)^
7 + 3*c^8*x^5/(sqrt(-c^2*x^2 + 1) + 1)^5 + 3*c^6*x^3/(sqrt(-c^2*x^2 + 1) + 1)^3 + c^4*x/(sqrt(-c^2*x^2 + 1) +
1))*(sqrt(-c^2*x^2 + 1) + 1)^7) - 2*b*c^7*d*e*x^5/((c^10*x^7/(sqrt(-c^2*x^2 + 1) + 1)^7 + 3*c^8*x^5/(sqrt(-c^2
*x^2 + 1) + 1)^5 + 3*c^6*x^3/(sqrt(-c^2*x^2 + 1) + 1)^3 + c^4*x/(sqrt(-c^2*x^2 + 1) + 1))*(sqrt(-c^2*x^2 + 1)
+ 1)^5) + 8*b*c^6*d*e*x^4*arcsin(c*x)/((c^10*x^7/(sqrt(-c^2*x^2 + 1) + 1)^7 + 3*c^8*x^5/(sqrt(-c^2*x^2 + 1) +
1)^5 + 3*c^6*x^3/(sqrt(-c^2*x^2 + 1) + 1)^3 + c^4*x/(sqrt(-c^2*x^2 + 1) + 1))*(sqrt(-c^2*x^2 + 1) + 1)^4) - 2*
b*c^6*d^2*x^2*arcsin(c*x)/((c^10*x^7/(sqrt(-c^2*x^2 + 1) + 1)^7 + 3*c^8*x^5/(sqrt(-c^2*x^2 + 1) + 1)^5 + 3*c^6
*x^3/(sqrt(-c^2*x^2 + 1) + 1)^3 + c^4*x/(sqrt(-c^2*x^2 + 1) + 1))*(sqrt(-c^2*x^2 + 1) + 1)^2) + 8*a*c^6*d*e*x^
4/((c^10*x^7/(sqrt(-c^2*x^2 + 1) + 1)^7 + 3*c^8*x^5/(sqrt(-c^2*x^2 + 1) + 1)^5 + 3*c^6*x^3/(sqrt(-c^2*x^2 + 1)
+ 1)^3 + c^4*x/(sqrt(-c^2*x^2 + 1) + 1))*(sqrt(-c^2*x^2 + 1) + 1)^4) - 2*a*c^6*d^2*x^2/((c^10*x^7/(sqrt(-c^2*
x^2 + 1) + 1)^7 + 3*c^8*x^5/(sqrt(-c^2*x^2 + 1) + 1)^5 + 3*c^6*x^3/(sqrt(-c^2*x^2 + 1) + 1)^3 + c^4*x/(sqrt(-c
^2*x^2 + 1) + 1))*(sqrt(-c^2*x^2 + 1) + 1)^2) + b*c^5*d^2*x*log(abs(c)*abs(x))/((c^10*x^7/(sqrt(-c^2*x^2 + 1)
+ 1)^7 + 3*c^8*x^5/(sqrt(-c^2*x^2 + 1) + 1)^5 + 3*c^6*x^3/(sqrt(-c^2*x^2 + 1) + 1)^3 + c^4*x/(sqrt(-c^2*x^2 +
1) + 1))*(sqrt(-c^2*x^2 + 1) + 1)) - b*c^5*d^2*x*log(sqrt(-c^2*x^2 + 1) + 1)/((c^10*x^7/(sqrt(-c^2*x^2 + 1) +
1)^7 + 3*c^8*x^5/(sqrt(-c^2*x^2 + 1) + 1)^5 + 3*c^6*x^3/(sqrt(-c^2*x^2 + 1) + 1)^3 + c^4*x/(sqrt(-c^2*x^2 + 1)
+ 1))*(sqrt(-c^2*x^2 + 1) + 1)) - 2/3*b*c^5*e^2*x^5/((c^10*x^7/(sqrt(-c^2*x^2 + 1) + 1)^7 + 3*c^8*x^5/(sqrt(-
c^2*x^2 + 1) + 1)^5 + 3*c^6*x^3/(sqrt(-c^2*x^2 + 1) + 1)^3 + c^4*x/(sqrt(-c^2*x^2 + 1) + 1))*(sqrt(-c^2*x^2 +
1) + 1)^5) + 2*b*c^5*d*e*x^3/((c^10*x^7/(sqrt(-...

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Mupad [F]
time = 0.00, size = -1, normalized size = -0.01 \begin {gather*} \left \{\begin {array}{cl} \frac {a\,\left (-3\,d^2+6\,d\,e\,x^2+e^2\,x^4\right )}{3\,x}+b\,e^2\,\left (\frac {\sqrt {\frac {1}{c^2}-x^2}\,\left (\frac {2}{c^2}+x^2\right )}{9}+\frac {x^3\,\mathrm {asin}\left (c\,x\right )}{3}\right )-b\,c\,d^2\,\mathrm {atanh}\left (\frac {1}{\sqrt {1-c^2\,x^2}}\right )-\frac {b\,d^2\,\mathrm {asin}\left (c\,x\right )}{x}+\frac {2\,b\,d\,e\,\left (\sqrt {1-c^2\,x^2}+c\,x\,\mathrm {asin}\left (c\,x\right )\right )}{c} & \text {\ if\ \ }0<c\\ \int \frac {\left (a+b\,\mathrm {asin}\left (c\,x\right )\right )\,{\left (e\,x^2+d\right )}^2}{x^2} \,d x & \text {\ if\ \ }\neg 0<c \end {array}\right . \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

piecewise(0 < c, (a*(- 3*d^2 + e^2*x^4 + 6*d*e*x^2))/(3*x) + b*e^2*(((1/c^2 - x^2)^(1/2)*(2/c^2 + x^2))/9 + (x
^3*asin(c*x))/3) - b*c*d^2*atanh(1/(- c^2*x^2 + 1)^(1/2)) - (b*d^2*asin(c*x))/x + (2*b*d*e*((- c^2*x^2 + 1)^(1
/2) + c*x*asin(c*x)))/c, ~0 < c, int(((a + b*asin(c*x))*(d + e*x^2)^2)/x^2, x))

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