3.7.2 \(\int \frac {(a+b x^2)^2 (c+d x^2)^{3/2}}{x^3} \, dx\)

Optimal. Leaf size=136 \[ -\frac {a^2 \left (c+d x^2\right )^{5/2}}{2 c x^2}+\frac {a \left (c+d x^2\right )^{3/2} (3 a d+4 b c)}{6 c}+\frac {1}{2} a \sqrt {c+d x^2} (3 a d+4 b c)-\frac {1}{2} a \sqrt {c} (3 a d+4 b c) \tanh ^{-1}\left (\frac {\sqrt {c+d x^2}}{\sqrt {c}}\right )+\frac {b^2 \left (c+d x^2\right )^{5/2}}{5 d} \]

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Rubi [A]  time = 0.11, antiderivative size = 136, normalized size of antiderivative = 1.00, number of steps used = 7, number of rules used = 6, integrand size = 24, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.250, Rules used = {446, 89, 80, 50, 63, 208} \begin {gather*} -\frac {a^2 \left (c+d x^2\right )^{5/2}}{2 c x^2}+\frac {a \left (c+d x^2\right )^{3/2} (3 a d+4 b c)}{6 c}+\frac {1}{2} a \sqrt {c+d x^2} (3 a d+4 b c)-\frac {1}{2} a \sqrt {c} (3 a d+4 b c) \tanh ^{-1}\left (\frac {\sqrt {c+d x^2}}{\sqrt {c}}\right )+\frac {b^2 \left (c+d x^2\right )^{5/2}}{5 d} \end {gather*}

Antiderivative was successfully verified.

[In]

Int[((a + b*x^2)^2*(c + d*x^2)^(3/2))/x^3,x]

[Out]

(a*(4*b*c + 3*a*d)*Sqrt[c + d*x^2])/2 + (a*(4*b*c + 3*a*d)*(c + d*x^2)^(3/2))/(6*c) + (b^2*(c + d*x^2)^(5/2))/
(5*d) - (a^2*(c + d*x^2)^(5/2))/(2*c*x^2) - (a*Sqrt[c]*(4*b*c + 3*a*d)*ArcTanh[Sqrt[c + d*x^2]/Sqrt[c]])/2

Rule 50

Int[((a_.) + (b_.)*(x_))^(m_)*((c_.) + (d_.)*(x_))^(n_), x_Symbol] :> Simp[((a + b*x)^(m + 1)*(c + d*x)^n)/(b*
(m + n + 1)), x] + Dist[(n*(b*c - a*d))/(b*(m + n + 1)), Int[(a + b*x)^m*(c + d*x)^(n - 1), x], x] /; FreeQ[{a
, b, c, d}, x] && NeQ[b*c - a*d, 0] && GtQ[n, 0] && NeQ[m + n + 1, 0] &&  !(IGtQ[m, 0] && ( !IntegerQ[n] || (G
tQ[m, 0] && LtQ[m - n, 0]))) &&  !ILtQ[m + n + 2, 0] && IntLinearQ[a, b, c, d, m, n, x]

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 80

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

Rule 89

Int[((a_.) + (b_.)*(x_))^2*((c_.) + (d_.)*(x_))^(n_.)*((e_.) + (f_.)*(x_))^(p_.), x_Symbol] :> Simp[((b*c - a*
d)^2*(c + d*x)^(n + 1)*(e + f*x)^(p + 1))/(d^2*(d*e - c*f)*(n + 1)), x] - Dist[1/(d^2*(d*e - c*f)*(n + 1)), In
t[(c + d*x)^(n + 1)*(e + f*x)^p*Simp[a^2*d^2*f*(n + p + 2) + b^2*c*(d*e*(n + 1) + c*f*(p + 1)) - 2*a*b*d*(d*e*
(n + 1) + c*f*(p + 1)) - b^2*d*(d*e - c*f)*(n + 1)*x, x], x], x] /; FreeQ[{a, b, c, d, e, f, n, p}, x] && (LtQ
[n, -1] || (EqQ[n + p + 3, 0] && NeQ[n, -1] && (SumSimplerQ[n, 1] ||  !SumSimplerQ[p, 1])))

Rule 208

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

Rule 446

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

Rubi steps

\begin {align*} \int \frac {\left (a+b x^2\right )^2 \left (c+d x^2\right )^{3/2}}{x^3} \, dx &=\frac {1}{2} \operatorname {Subst}\left (\int \frac {(a+b x)^2 (c+d x)^{3/2}}{x^2} \, dx,x,x^2\right )\\ &=-\frac {a^2 \left (c+d x^2\right )^{5/2}}{2 c x^2}+\frac {\operatorname {Subst}\left (\int \frac {\left (\frac {1}{2} a (4 b c+3 a d)+b^2 c x\right ) (c+d x)^{3/2}}{x} \, dx,x,x^2\right )}{2 c}\\ &=\frac {b^2 \left (c+d x^2\right )^{5/2}}{5 d}-\frac {a^2 \left (c+d x^2\right )^{5/2}}{2 c x^2}+\frac {(a (4 b c+3 a d)) \operatorname {Subst}\left (\int \frac {(c+d x)^{3/2}}{x} \, dx,x,x^2\right )}{4 c}\\ &=\frac {a (4 b c+3 a d) \left (c+d x^2\right )^{3/2}}{6 c}+\frac {b^2 \left (c+d x^2\right )^{5/2}}{5 d}-\frac {a^2 \left (c+d x^2\right )^{5/2}}{2 c x^2}+\frac {1}{4} (a (4 b c+3 a d)) \operatorname {Subst}\left (\int \frac {\sqrt {c+d x}}{x} \, dx,x,x^2\right )\\ &=\frac {1}{2} a (4 b c+3 a d) \sqrt {c+d x^2}+\frac {a (4 b c+3 a d) \left (c+d x^2\right )^{3/2}}{6 c}+\frac {b^2 \left (c+d x^2\right )^{5/2}}{5 d}-\frac {a^2 \left (c+d x^2\right )^{5/2}}{2 c x^2}+\frac {1}{4} (a c (4 b c+3 a d)) \operatorname {Subst}\left (\int \frac {1}{x \sqrt {c+d x}} \, dx,x,x^2\right )\\ &=\frac {1}{2} a (4 b c+3 a d) \sqrt {c+d x^2}+\frac {a (4 b c+3 a d) \left (c+d x^2\right )^{3/2}}{6 c}+\frac {b^2 \left (c+d x^2\right )^{5/2}}{5 d}-\frac {a^2 \left (c+d x^2\right )^{5/2}}{2 c x^2}+\frac {(a c (4 b c+3 a d)) \operatorname {Subst}\left (\int \frac {1}{-\frac {c}{d}+\frac {x^2}{d}} \, dx,x,\sqrt {c+d x^2}\right )}{2 d}\\ &=\frac {1}{2} a (4 b c+3 a d) \sqrt {c+d x^2}+\frac {a (4 b c+3 a d) \left (c+d x^2\right )^{3/2}}{6 c}+\frac {b^2 \left (c+d x^2\right )^{5/2}}{5 d}-\frac {a^2 \left (c+d x^2\right )^{5/2}}{2 c x^2}-\frac {1}{2} a \sqrt {c} (4 b c+3 a d) \tanh ^{-1}\left (\frac {\sqrt {c+d x^2}}{\sqrt {c}}\right )\\ \end {align*}

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Mathematica [A]  time = 0.09, size = 108, normalized size = 0.79 \begin {gather*} \frac {\sqrt {c+d x^2} \left (-15 a^2 d \left (c-2 d x^2\right )+20 a b d x^2 \left (4 c+d x^2\right )+6 b^2 x^2 \left (c+d x^2\right )^2\right )}{30 d x^2}-\frac {1}{2} a \sqrt {c} (3 a d+4 b c) \tanh ^{-1}\left (\frac {\sqrt {c+d x^2}}{\sqrt {c}}\right ) \end {gather*}

Antiderivative was successfully verified.

[In]

Integrate[((a + b*x^2)^2*(c + d*x^2)^(3/2))/x^3,x]

[Out]

(Sqrt[c + d*x^2]*(-15*a^2*d*(c - 2*d*x^2) + 6*b^2*x^2*(c + d*x^2)^2 + 20*a*b*d*x^2*(4*c + d*x^2)))/(30*d*x^2)
- (a*Sqrt[c]*(4*b*c + 3*a*d)*ArcTanh[Sqrt[c + d*x^2]/Sqrt[c]])/2

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IntegrateAlgebraic [A]  time = 0.17, size = 135, normalized size = 0.99 \begin {gather*} \frac {\sqrt {c+d x^2} \left (-15 a^2 c d+30 a^2 d^2 x^2+80 a b c d x^2+20 a b d^2 x^4+6 b^2 c^2 x^2+12 b^2 c d x^4+6 b^2 d^2 x^6\right )}{30 d x^2}+\frac {1}{2} \left (-3 a^2 \sqrt {c} d-4 a b c^{3/2}\right ) \tanh ^{-1}\left (\frac {\sqrt {c+d x^2}}{\sqrt {c}}\right ) \end {gather*}

Antiderivative was successfully verified.

[In]

IntegrateAlgebraic[((a + b*x^2)^2*(c + d*x^2)^(3/2))/x^3,x]

[Out]

(Sqrt[c + d*x^2]*(-15*a^2*c*d + 6*b^2*c^2*x^2 + 80*a*b*c*d*x^2 + 30*a^2*d^2*x^2 + 12*b^2*c*d*x^4 + 20*a*b*d^2*
x^4 + 6*b^2*d^2*x^6))/(30*d*x^2) + ((-4*a*b*c^(3/2) - 3*a^2*Sqrt[c]*d)*ArcTanh[Sqrt[c + d*x^2]/Sqrt[c]])/2

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fricas [A]  time = 1.63, size = 267, normalized size = 1.96 \begin {gather*} \left [\frac {15 \, {\left (4 \, a b c d + 3 \, a^{2} d^{2}\right )} \sqrt {c} x^{2} \log \left (-\frac {d x^{2} - 2 \, \sqrt {d x^{2} + c} \sqrt {c} + 2 \, c}{x^{2}}\right ) + 2 \, {\left (6 \, b^{2} d^{2} x^{6} + 4 \, {\left (3 \, b^{2} c d + 5 \, a b d^{2}\right )} x^{4} - 15 \, a^{2} c d + 2 \, {\left (3 \, b^{2} c^{2} + 40 \, a b c d + 15 \, a^{2} d^{2}\right )} x^{2}\right )} \sqrt {d x^{2} + c}}{60 \, d x^{2}}, \frac {15 \, {\left (4 \, a b c d + 3 \, a^{2} d^{2}\right )} \sqrt {-c} x^{2} \arctan \left (\frac {\sqrt {-c}}{\sqrt {d x^{2} + c}}\right ) + {\left (6 \, b^{2} d^{2} x^{6} + 4 \, {\left (3 \, b^{2} c d + 5 \, a b d^{2}\right )} x^{4} - 15 \, a^{2} c d + 2 \, {\left (3 \, b^{2} c^{2} + 40 \, a b c d + 15 \, a^{2} d^{2}\right )} x^{2}\right )} \sqrt {d x^{2} + c}}{30 \, d x^{2}}\right ] \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate((b*x^2+a)^2*(d*x^2+c)^(3/2)/x^3,x, algorithm="fricas")

[Out]

[1/60*(15*(4*a*b*c*d + 3*a^2*d^2)*sqrt(c)*x^2*log(-(d*x^2 - 2*sqrt(d*x^2 + c)*sqrt(c) + 2*c)/x^2) + 2*(6*b^2*d
^2*x^6 + 4*(3*b^2*c*d + 5*a*b*d^2)*x^4 - 15*a^2*c*d + 2*(3*b^2*c^2 + 40*a*b*c*d + 15*a^2*d^2)*x^2)*sqrt(d*x^2
+ c))/(d*x^2), 1/30*(15*(4*a*b*c*d + 3*a^2*d^2)*sqrt(-c)*x^2*arctan(sqrt(-c)/sqrt(d*x^2 + c)) + (6*b^2*d^2*x^6
+ 4*(3*b^2*c*d + 5*a*b*d^2)*x^4 - 15*a^2*c*d + 2*(3*b^2*c^2 + 40*a*b*c*d + 15*a^2*d^2)*x^2)*sqrt(d*x^2 + c))/
(d*x^2)]

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giac [A]  time = 0.37, size = 126, normalized size = 0.93 \begin {gather*} \frac {6 \, {\left (d x^{2} + c\right )}^{\frac {5}{2}} b^{2} + 20 \, {\left (d x^{2} + c\right )}^{\frac {3}{2}} a b d + 60 \, \sqrt {d x^{2} + c} a b c d + 30 \, \sqrt {d x^{2} + c} a^{2} d^{2} - \frac {15 \, \sqrt {d x^{2} + c} a^{2} c d}{x^{2}} + \frac {15 \, {\left (4 \, a b c^{2} d + 3 \, a^{2} c d^{2}\right )} \arctan \left (\frac {\sqrt {d x^{2} + c}}{\sqrt {-c}}\right )}{\sqrt {-c}}}{30 \, d} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate((b*x^2+a)^2*(d*x^2+c)^(3/2)/x^3,x, algorithm="giac")

[Out]

1/30*(6*(d*x^2 + c)^(5/2)*b^2 + 20*(d*x^2 + c)^(3/2)*a*b*d + 60*sqrt(d*x^2 + c)*a*b*c*d + 30*sqrt(d*x^2 + c)*a
^2*d^2 - 15*sqrt(d*x^2 + c)*a^2*c*d/x^2 + 15*(4*a*b*c^2*d + 3*a^2*c*d^2)*arctan(sqrt(d*x^2 + c)/sqrt(-c))/sqrt
(-c))/d

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maple [A]  time = 0.01, size = 161, normalized size = 1.18 \begin {gather*} -\frac {3 a^{2} \sqrt {c}\, d \ln \left (\frac {2 c +2 \sqrt {d \,x^{2}+c}\, \sqrt {c}}{x}\right )}{2}-2 a b \,c^{\frac {3}{2}} \ln \left (\frac {2 c +2 \sqrt {d \,x^{2}+c}\, \sqrt {c}}{x}\right )+\frac {3 \sqrt {d \,x^{2}+c}\, a^{2} d}{2}+2 \sqrt {d \,x^{2}+c}\, a b c +\frac {\left (d \,x^{2}+c \right )^{\frac {3}{2}} a^{2} d}{2 c}+\frac {2 \left (d \,x^{2}+c \right )^{\frac {3}{2}} a b}{3}+\frac {\left (d \,x^{2}+c \right )^{\frac {5}{2}} b^{2}}{5 d}-\frac {\left (d \,x^{2}+c \right )^{\frac {5}{2}} a^{2}}{2 c \,x^{2}} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

int((b*x^2+a)^2*(d*x^2+c)^(3/2)/x^3,x)

[Out]

1/5*b^2*(d*x^2+c)^(5/2)/d-1/2*a^2*(d*x^2+c)^(5/2)/c/x^2+1/2*a^2*d/c*(d*x^2+c)^(3/2)-3/2*a^2*d*c^(1/2)*ln((2*c+
2*(d*x^2+c)^(1/2)*c^(1/2))/x)+3/2*a^2*d*(d*x^2+c)^(1/2)+2/3*a*b*(d*x^2+c)^(3/2)-2*a*b*c^(3/2)*ln((2*c+2*(d*x^2
+c)^(1/2)*c^(1/2))/x)+2*a*b*(d*x^2+c)^(1/2)*c

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maxima [A]  time = 1.15, size = 138, normalized size = 1.01 \begin {gather*} -2 \, a b c^{\frac {3}{2}} \operatorname {arsinh}\left (\frac {c}{\sqrt {c d} {\left | x \right |}}\right ) - \frac {3}{2} \, a^{2} \sqrt {c} d \operatorname {arsinh}\left (\frac {c}{\sqrt {c d} {\left | x \right |}}\right ) + \frac {2}{3} \, {\left (d x^{2} + c\right )}^{\frac {3}{2}} a b + 2 \, \sqrt {d x^{2} + c} a b c + \frac {{\left (d x^{2} + c\right )}^{\frac {5}{2}} b^{2}}{5 \, d} + \frac {3}{2} \, \sqrt {d x^{2} + c} a^{2} d + \frac {{\left (d x^{2} + c\right )}^{\frac {3}{2}} a^{2} d}{2 \, c} - \frac {{\left (d x^{2} + c\right )}^{\frac {5}{2}} a^{2}}{2 \, c x^{2}} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate((b*x^2+a)^2*(d*x^2+c)^(3/2)/x^3,x, algorithm="maxima")

[Out]

-2*a*b*c^(3/2)*arcsinh(c/(sqrt(c*d)*abs(x))) - 3/2*a^2*sqrt(c)*d*arcsinh(c/(sqrt(c*d)*abs(x))) + 2/3*(d*x^2 +
c)^(3/2)*a*b + 2*sqrt(d*x^2 + c)*a*b*c + 1/5*(d*x^2 + c)^(5/2)*b^2/d + 3/2*sqrt(d*x^2 + c)*a^2*d + 1/2*(d*x^2
+ c)^(3/2)*a^2*d/c - 1/2*(d*x^2 + c)^(5/2)*a^2/(c*x^2)

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mupad [B]  time = 1.17, size = 201, normalized size = 1.48 \begin {gather*} \frac {b^2\,{\left (d\,x^2+c\right )}^{5/2}}{5\,d}-\left (\frac {2\,b^2\,c-2\,a\,b\,d}{3\,d}-\frac {2\,b^2\,c}{3\,d}\right )\,{\left (d\,x^2+c\right )}^{3/2}-\sqrt {d\,x^2+c}\,\left (2\,c\,\left (\frac {2\,b^2\,c-2\,a\,b\,d}{d}-\frac {2\,b^2\,c}{d}\right )-\frac {{\left (a\,d-b\,c\right )}^2}{d}+\frac {b^2\,c^2}{d}\right )-\frac {a^2\,c\,\sqrt {d\,x^2+c}}{2\,x^2}+2\,a\,\mathrm {atan}\left (\frac {2\,a\,\sqrt {d\,x^2+c}\,\left (3\,a\,d+4\,b\,c\right )\,\sqrt {-\frac {c}{16}}}{\frac {3\,d\,a^2\,c}{2}+2\,b\,a\,c^2}\right )\,\left (3\,a\,d+4\,b\,c\right )\,\sqrt {-\frac {c}{16}} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

int(((a + b*x^2)^2*(c + d*x^2)^(3/2))/x^3,x)

[Out]

(b^2*(c + d*x^2)^(5/2))/(5*d) - ((2*b^2*c - 2*a*b*d)/(3*d) - (2*b^2*c)/(3*d))*(c + d*x^2)^(3/2) - (c + d*x^2)^
(1/2)*(2*c*((2*b^2*c - 2*a*b*d)/d - (2*b^2*c)/d) - (a*d - b*c)^2/d + (b^2*c^2)/d) - (a^2*c*(c + d*x^2)^(1/2))/
(2*x^2) + 2*a*atan((2*a*(c + d*x^2)^(1/2)*(3*a*d + 4*b*c)*(-c/16)^(1/2))/(2*a*b*c^2 + (3*a^2*c*d)/2))*(3*a*d +
4*b*c)*(-c/16)^(1/2)

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sympy [A]  time = 88.86, size = 303, normalized size = 2.23 \begin {gather*} - \frac {3 a^{2} \sqrt {c} d \operatorname {asinh}{\left (\frac {\sqrt {c}}{\sqrt {d} x} \right )}}{2} - \frac {a^{2} c \sqrt {d} \sqrt {\frac {c}{d x^{2}} + 1}}{2 x} + \frac {a^{2} c \sqrt {d}}{x \sqrt {\frac {c}{d x^{2}} + 1}} + \frac {a^{2} d^{\frac {3}{2}} x}{\sqrt {\frac {c}{d x^{2}} + 1}} - 2 a b c^{\frac {3}{2}} \operatorname {asinh}{\left (\frac {\sqrt {c}}{\sqrt {d} x} \right )} + \frac {2 a b c^{2}}{\sqrt {d} x \sqrt {\frac {c}{d x^{2}} + 1}} + \frac {2 a b c \sqrt {d} x}{\sqrt {\frac {c}{d x^{2}} + 1}} + 2 a b d \left (\begin {cases} \frac {\sqrt {c} x^{2}}{2} & \text {for}\: d = 0 \\\frac {\left (c + d x^{2}\right )^{\frac {3}{2}}}{3 d} & \text {otherwise} \end {cases}\right ) + b^{2} c \left (\begin {cases} \frac {\sqrt {c} x^{2}}{2} & \text {for}\: d = 0 \\\frac {\left (c + d x^{2}\right )^{\frac {3}{2}}}{3 d} & \text {otherwise} \end {cases}\right ) + b^{2} d \left (\begin {cases} - \frac {2 c^{2} \sqrt {c + d x^{2}}}{15 d^{2}} + \frac {c x^{2} \sqrt {c + d x^{2}}}{15 d} + \frac {x^{4} \sqrt {c + d x^{2}}}{5} & \text {for}\: d \neq 0 \\\frac {\sqrt {c} x^{4}}{4} & \text {otherwise} \end {cases}\right ) \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate((b*x**2+a)**2*(d*x**2+c)**(3/2)/x**3,x)

[Out]

-3*a**2*sqrt(c)*d*asinh(sqrt(c)/(sqrt(d)*x))/2 - a**2*c*sqrt(d)*sqrt(c/(d*x**2) + 1)/(2*x) + a**2*c*sqrt(d)/(x
*sqrt(c/(d*x**2) + 1)) + a**2*d**(3/2)*x/sqrt(c/(d*x**2) + 1) - 2*a*b*c**(3/2)*asinh(sqrt(c)/(sqrt(d)*x)) + 2*
a*b*c**2/(sqrt(d)*x*sqrt(c/(d*x**2) + 1)) + 2*a*b*c*sqrt(d)*x/sqrt(c/(d*x**2) + 1) + 2*a*b*d*Piecewise((sqrt(c
)*x**2/2, Eq(d, 0)), ((c + d*x**2)**(3/2)/(3*d), True)) + b**2*c*Piecewise((sqrt(c)*x**2/2, Eq(d, 0)), ((c + d
*x**2)**(3/2)/(3*d), True)) + b**2*d*Piecewise((-2*c**2*sqrt(c + d*x**2)/(15*d**2) + c*x**2*sqrt(c + d*x**2)/(
15*d) + x**4*sqrt(c + d*x**2)/5, Ne(d, 0)), (sqrt(c)*x**4/4, True))

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