3.7.3 \(\int \frac {(a+b x^2)^2 (c+d x^2)^{3/2}}{x^4} \, dx\)

Optimal. Leaf size=184 \[ -\frac {a^2 \left (c+d x^2\right )^{5/2}}{3 c x^3}+\frac {x \left (c+d x^2\right )^{3/2} \left (8 a d (a d+3 b c)+3 b^2 c^2\right )}{12 c^2}+\frac {x \sqrt {c+d x^2} \left (8 a d (a d+3 b c)+3 b^2 c^2\right )}{8 c}+\frac {\left (8 a d (a d+3 b c)+3 b^2 c^2\right ) \tanh ^{-1}\left (\frac {\sqrt {d} x}{\sqrt {c+d x^2}}\right )}{8 \sqrt {d}}-\frac {2 a \left (c+d x^2\right )^{5/2} (a d+3 b c)}{3 c^2 x} \]

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Rubi [A]  time = 0.13, antiderivative size = 181, normalized size of antiderivative = 0.98, number of steps used = 6, number of rules used = 5, integrand size = 24, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.208, Rules used = {462, 453, 195, 217, 206} \begin {gather*} -\frac {a^2 \left (c+d x^2\right )^{5/2}}{3 c x^3}+\frac {1}{12} x \left (c+d x^2\right )^{3/2} \left (\frac {8 a d (a d+3 b c)}{c^2}+3 b^2\right )+\frac {x \sqrt {c+d x^2} \left (8 a d (a d+3 b c)+3 b^2 c^2\right )}{8 c}+\frac {\left (8 a d (a d+3 b c)+3 b^2 c^2\right ) \tanh ^{-1}\left (\frac {\sqrt {d} x}{\sqrt {c+d x^2}}\right )}{8 \sqrt {d}}-\frac {2 a \left (c+d x^2\right )^{5/2} (a d+3 b c)}{3 c^2 x} \end {gather*}

Antiderivative was successfully verified.

[In]

Int[((a + b*x^2)^2*(c + d*x^2)^(3/2))/x^4,x]

[Out]

((3*b^2*c^2 + 8*a*d*(3*b*c + a*d))*x*Sqrt[c + d*x^2])/(8*c) + ((3*b^2 + (8*a*d*(3*b*c + a*d))/c^2)*x*(c + d*x^
2)^(3/2))/12 - (a^2*(c + d*x^2)^(5/2))/(3*c*x^3) - (2*a*(3*b*c + a*d)*(c + d*x^2)^(5/2))/(3*c^2*x) + ((3*b^2*c
^2 + 8*a*d*(3*b*c + a*d))*ArcTanh[(Sqrt[d]*x)/Sqrt[c + d*x^2]])/(8*Sqrt[d])

Rule 195

Int[((a_) + (b_.)*(x_)^(n_))^(p_), x_Symbol] :> Simp[(x*(a + b*x^n)^p)/(n*p + 1), x] + Dist[(a*n*p)/(n*p + 1),
Int[(a + b*x^n)^(p - 1), x], x] /; FreeQ[{a, b}, x] && IGtQ[n, 0] && GtQ[p, 0] && (IntegerQ[2*p] || (EqQ[n, 2
] && IntegerQ[4*p]) || (EqQ[n, 2] && IntegerQ[3*p]) || LtQ[Denominator[p + 1/n], Denominator[p]])

Rule 206

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

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 453

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

Rule 462

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

Rubi steps

\begin {align*} \int \frac {\left (a+b x^2\right )^2 \left (c+d x^2\right )^{3/2}}{x^4} \, dx &=-\frac {a^2 \left (c+d x^2\right )^{5/2}}{3 c x^3}+\frac {\int \frac {\left (2 a (3 b c+a d)+3 b^2 c x^2\right ) \left (c+d x^2\right )^{3/2}}{x^2} \, dx}{3 c}\\ &=-\frac {a^2 \left (c+d x^2\right )^{5/2}}{3 c x^3}-\frac {2 a (3 b c+a d) \left (c+d x^2\right )^{5/2}}{3 c^2 x}-\frac {1}{3} \left (-3 b^2-\frac {8 a d (3 b c+a d)}{c^2}\right ) \int \left (c+d x^2\right )^{3/2} \, dx\\ &=\frac {1}{12} \left (3 b^2+\frac {8 a d (3 b c+a d)}{c^2}\right ) x \left (c+d x^2\right )^{3/2}-\frac {a^2 \left (c+d x^2\right )^{5/2}}{3 c x^3}-\frac {2 a (3 b c+a d) \left (c+d x^2\right )^{5/2}}{3 c^2 x}-\frac {1}{4} \left (c \left (-3 b^2-\frac {8 a d (3 b c+a d)}{c^2}\right )\right ) \int \sqrt {c+d x^2} \, dx\\ &=\frac {1}{8} c \left (3 b^2+\frac {8 a d (3 b c+a d)}{c^2}\right ) x \sqrt {c+d x^2}+\frac {1}{12} \left (3 b^2+\frac {8 a d (3 b c+a d)}{c^2}\right ) x \left (c+d x^2\right )^{3/2}-\frac {a^2 \left (c+d x^2\right )^{5/2}}{3 c x^3}-\frac {2 a (3 b c+a d) \left (c+d x^2\right )^{5/2}}{3 c^2 x}-\frac {1}{8} \left (-3 b^2 c^2-24 a b c d-8 a^2 d^2\right ) \int \frac {1}{\sqrt {c+d x^2}} \, dx\\ &=\frac {1}{8} c \left (3 b^2+\frac {8 a d (3 b c+a d)}{c^2}\right ) x \sqrt {c+d x^2}+\frac {1}{12} \left (3 b^2+\frac {8 a d (3 b c+a d)}{c^2}\right ) x \left (c+d x^2\right )^{3/2}-\frac {a^2 \left (c+d x^2\right )^{5/2}}{3 c x^3}-\frac {2 a (3 b c+a d) \left (c+d x^2\right )^{5/2}}{3 c^2 x}-\frac {1}{8} \left (-3 b^2 c^2-24 a b c d-8 a^2 d^2\right ) \operatorname {Subst}\left (\int \frac {1}{1-d x^2} \, dx,x,\frac {x}{\sqrt {c+d x^2}}\right )\\ &=\frac {1}{8} c \left (3 b^2+\frac {8 a d (3 b c+a d)}{c^2}\right ) x \sqrt {c+d x^2}+\frac {1}{12} \left (3 b^2+\frac {8 a d (3 b c+a d)}{c^2}\right ) x \left (c+d x^2\right )^{3/2}-\frac {a^2 \left (c+d x^2\right )^{5/2}}{3 c x^3}-\frac {2 a (3 b c+a d) \left (c+d x^2\right )^{5/2}}{3 c^2 x}+\frac {\left (3 b^2 c^2+24 a b c d+8 a^2 d^2\right ) \tanh ^{-1}\left (\frac {\sqrt {d} x}{\sqrt {c+d x^2}}\right )}{8 \sqrt {d}}\\ \end {align*}

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Mathematica [A]  time = 0.10, size = 118, normalized size = 0.64 \begin {gather*} \frac {1}{24} \left (\frac {3 \left (8 a^2 d^2+24 a b c d+3 b^2 c^2\right ) \log \left (\sqrt {d} \sqrt {c+d x^2}+d x\right )}{\sqrt {d}}+\frac {\sqrt {c+d x^2} \left (-8 a^2 c+3 b x^4 (8 a d+5 b c)-16 a x^2 (2 a d+3 b c)+6 b^2 d x^6\right )}{x^3}\right ) \end {gather*}

Antiderivative was successfully verified.

[In]

Integrate[((a + b*x^2)^2*(c + d*x^2)^(3/2))/x^4,x]

[Out]

((Sqrt[c + d*x^2]*(-8*a^2*c - 16*a*(3*b*c + 2*a*d)*x^2 + 3*b*(5*b*c + 8*a*d)*x^4 + 6*b^2*d*x^6))/x^3 + (3*(3*b
^2*c^2 + 24*a*b*c*d + 8*a^2*d^2)*Log[d*x + Sqrt[d]*Sqrt[c + d*x^2]])/Sqrt[d])/24

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IntegrateAlgebraic [A]  time = 0.32, size = 122, normalized size = 0.66 \begin {gather*} \frac {\left (-8 a^2 d^2-24 a b c d-3 b^2 c^2\right ) \log \left (\sqrt {c+d x^2}-\sqrt {d} x\right )}{8 \sqrt {d}}+\frac {\sqrt {c+d x^2} \left (-8 a^2 c-32 a^2 d x^2-48 a b c x^2+24 a b d x^4+15 b^2 c x^4+6 b^2 d x^6\right )}{24 x^3} \end {gather*}

Antiderivative was successfully verified.

[In]

IntegrateAlgebraic[((a + b*x^2)^2*(c + d*x^2)^(3/2))/x^4,x]

[Out]

(Sqrt[c + d*x^2]*(-8*a^2*c - 48*a*b*c*x^2 - 32*a^2*d*x^2 + 15*b^2*c*x^4 + 24*a*b*d*x^4 + 6*b^2*d*x^6))/(24*x^3
) + ((-3*b^2*c^2 - 24*a*b*c*d - 8*a^2*d^2)*Log[-(Sqrt[d]*x) + Sqrt[c + d*x^2]])/(8*Sqrt[d])

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fricas [A]  time = 1.58, size = 266, normalized size = 1.45 \begin {gather*} \left [\frac {3 \, {\left (3 \, b^{2} c^{2} + 24 \, a b c d + 8 \, a^{2} d^{2}\right )} \sqrt {d} x^{3} \log \left (-2 \, d x^{2} - 2 \, \sqrt {d x^{2} + c} \sqrt {d} x - c\right ) + 2 \, {\left (6 \, b^{2} d^{2} x^{6} + 3 \, {\left (5 \, b^{2} c d + 8 \, a b d^{2}\right )} x^{4} - 8 \, a^{2} c d - 16 \, {\left (3 \, a b c d + 2 \, a^{2} d^{2}\right )} x^{2}\right )} \sqrt {d x^{2} + c}}{48 \, d x^{3}}, -\frac {3 \, {\left (3 \, b^{2} c^{2} + 24 \, a b c d + 8 \, a^{2} d^{2}\right )} \sqrt {-d} x^{3} \arctan \left (\frac {\sqrt {-d} x}{\sqrt {d x^{2} + c}}\right ) - {\left (6 \, b^{2} d^{2} x^{6} + 3 \, {\left (5 \, b^{2} c d + 8 \, a b d^{2}\right )} x^{4} - 8 \, a^{2} c d - 16 \, {\left (3 \, a b c d + 2 \, a^{2} d^{2}\right )} x^{2}\right )} \sqrt {d x^{2} + c}}{24 \, d x^{3}}\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^4,x, algorithm="fricas")

[Out]

[1/48*(3*(3*b^2*c^2 + 24*a*b*c*d + 8*a^2*d^2)*sqrt(d)*x^3*log(-2*d*x^2 - 2*sqrt(d*x^2 + c)*sqrt(d)*x - c) + 2*
(6*b^2*d^2*x^6 + 3*(5*b^2*c*d + 8*a*b*d^2)*x^4 - 8*a^2*c*d - 16*(3*a*b*c*d + 2*a^2*d^2)*x^2)*sqrt(d*x^2 + c))/
(d*x^3), -1/24*(3*(3*b^2*c^2 + 24*a*b*c*d + 8*a^2*d^2)*sqrt(-d)*x^3*arctan(sqrt(-d)*x/sqrt(d*x^2 + c)) - (6*b^
2*d^2*x^6 + 3*(5*b^2*c*d + 8*a*b*d^2)*x^4 - 8*a^2*c*d - 16*(3*a*b*c*d + 2*a^2*d^2)*x^2)*sqrt(d*x^2 + c))/(d*x^
3)]

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giac [A]  time = 0.46, size = 262, normalized size = 1.42 \begin {gather*} \frac {1}{8} \, {\left (2 \, b^{2} d x^{2} + \frac {5 \, b^{2} c d^{2} + 8 \, a b d^{3}}{d^{2}}\right )} \sqrt {d x^{2} + c} x - \frac {{\left (3 \, b^{2} c^{2} \sqrt {d} + 24 \, a b c d^{\frac {3}{2}} + 8 \, a^{2} d^{\frac {5}{2}}\right )} \log \left ({\left (\sqrt {d} x - \sqrt {d x^{2} + c}\right )}^{2}\right )}{16 \, d} + \frac {4 \, {\left (3 \, {\left (\sqrt {d} x - \sqrt {d x^{2} + c}\right )}^{4} a b c^{2} \sqrt {d} + 3 \, {\left (\sqrt {d} x - \sqrt {d x^{2} + c}\right )}^{4} a^{2} c d^{\frac {3}{2}} - 6 \, {\left (\sqrt {d} x - \sqrt {d x^{2} + c}\right )}^{2} a b c^{3} \sqrt {d} - 3 \, {\left (\sqrt {d} x - \sqrt {d x^{2} + c}\right )}^{2} a^{2} c^{2} d^{\frac {3}{2}} + 3 \, a b c^{4} \sqrt {d} + 2 \, a^{2} c^{3} d^{\frac {3}{2}}\right )}}{3 \, {\left ({\left (\sqrt {d} x - \sqrt {d x^{2} + c}\right )}^{2} - c\right )}^{3}} \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^4,x, algorithm="giac")

[Out]

1/8*(2*b^2*d*x^2 + (5*b^2*c*d^2 + 8*a*b*d^3)/d^2)*sqrt(d*x^2 + c)*x - 1/16*(3*b^2*c^2*sqrt(d) + 24*a*b*c*d^(3/
2) + 8*a^2*d^(5/2))*log((sqrt(d)*x - sqrt(d*x^2 + c))^2)/d + 4/3*(3*(sqrt(d)*x - sqrt(d*x^2 + c))^4*a*b*c^2*sq
rt(d) + 3*(sqrt(d)*x - sqrt(d*x^2 + c))^4*a^2*c*d^(3/2) - 6*(sqrt(d)*x - sqrt(d*x^2 + c))^2*a*b*c^3*sqrt(d) -
3*(sqrt(d)*x - sqrt(d*x^2 + c))^2*a^2*c^2*d^(3/2) + 3*a*b*c^4*sqrt(d) + 2*a^2*c^3*d^(3/2))/((sqrt(d)*x - sqrt(
d*x^2 + c))^2 - c)^3

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maple [A]  time = 0.02, size = 241, normalized size = 1.31 \begin {gather*} a^{2} d^{\frac {3}{2}} \ln \left (\sqrt {d}\, x +\sqrt {d \,x^{2}+c}\right )+3 a b c \sqrt {d}\, \ln \left (\sqrt {d}\, x +\sqrt {d \,x^{2}+c}\right )+\frac {3 b^{2} c^{2} \ln \left (\sqrt {d}\, x +\sqrt {d \,x^{2}+c}\right )}{8 \sqrt {d}}+\frac {\sqrt {d \,x^{2}+c}\, a^{2} d^{2} x}{c}+3 \sqrt {d \,x^{2}+c}\, a b d x +\frac {3 \sqrt {d \,x^{2}+c}\, b^{2} c x}{8}+\frac {2 \left (d \,x^{2}+c \right )^{\frac {3}{2}} a^{2} d^{2} x}{3 c^{2}}+\frac {2 \left (d \,x^{2}+c \right )^{\frac {3}{2}} a b d x}{c}+\frac {\left (d \,x^{2}+c \right )^{\frac {3}{2}} b^{2} x}{4}-\frac {2 \left (d \,x^{2}+c \right )^{\frac {5}{2}} a^{2} d}{3 c^{2} x}-\frac {2 \left (d \,x^{2}+c \right )^{\frac {5}{2}} a b}{c x}-\frac {\left (d \,x^{2}+c \right )^{\frac {5}{2}} a^{2}}{3 c \,x^{3}} \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^4,x)

[Out]

1/4*x*b^2*(d*x^2+c)^(3/2)+3/8*b^2*c*x*(d*x^2+c)^(1/2)+3/8*b^2*c^2/d^(1/2)*ln(d^(1/2)*x+(d*x^2+c)^(1/2))-2*a*b/
c/x*(d*x^2+c)^(5/2)+2*a*b*d/c*x*(d*x^2+c)^(3/2)+3*a*b*d*x*(d*x^2+c)^(1/2)+3*a*b*d^(1/2)*c*ln(d^(1/2)*x+(d*x^2+
c)^(1/2))-1/3*a^2*(d*x^2+c)^(5/2)/c/x^3-2/3*a^2*d/c^2/x*(d*x^2+c)^(5/2)+2/3*a^2*d^2/c^2*x*(d*x^2+c)^(3/2)+a^2*
d^2/c*x*(d*x^2+c)^(1/2)+a^2*d^(3/2)*ln(d^(1/2)*x+(d*x^2+c)^(1/2))

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maxima [A]  time = 1.13, size = 177, normalized size = 0.96 \begin {gather*} \frac {1}{4} \, {\left (d x^{2} + c\right )}^{\frac {3}{2}} b^{2} x + \frac {3}{8} \, \sqrt {d x^{2} + c} b^{2} c x + 3 \, \sqrt {d x^{2} + c} a b d x + \frac {\sqrt {d x^{2} + c} a^{2} d^{2} x}{c} + \frac {3 \, b^{2} c^{2} \operatorname {arsinh}\left (\frac {d x}{\sqrt {c d}}\right )}{8 \, \sqrt {d}} + 3 \, a b c \sqrt {d} \operatorname {arsinh}\left (\frac {d x}{\sqrt {c d}}\right ) + a^{2} d^{\frac {3}{2}} \operatorname {arsinh}\left (\frac {d x}{\sqrt {c d}}\right ) - \frac {2 \, {\left (d x^{2} + c\right )}^{\frac {3}{2}} a b}{x} - \frac {2 \, {\left (d x^{2} + c\right )}^{\frac {3}{2}} a^{2} d}{3 \, c x} - \frac {{\left (d x^{2} + c\right )}^{\frac {5}{2}} a^{2}}{3 \, c x^{3}} \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^4,x, algorithm="maxima")

[Out]

1/4*(d*x^2 + c)^(3/2)*b^2*x + 3/8*sqrt(d*x^2 + c)*b^2*c*x + 3*sqrt(d*x^2 + c)*a*b*d*x + sqrt(d*x^2 + c)*a^2*d^
2*x/c + 3/8*b^2*c^2*arcsinh(d*x/sqrt(c*d))/sqrt(d) + 3*a*b*c*sqrt(d)*arcsinh(d*x/sqrt(c*d)) + a^2*d^(3/2)*arcs
inh(d*x/sqrt(c*d)) - 2*(d*x^2 + c)^(3/2)*a*b/x - 2/3*(d*x^2 + c)^(3/2)*a^2*d/(c*x) - 1/3*(d*x^2 + c)^(5/2)*a^2
/(c*x^3)

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mupad [F]  time = 0.00, size = -1, normalized size = -0.01 \begin {gather*} \int \frac {{\left (b\,x^2+a\right )}^2\,{\left (d\,x^2+c\right )}^{3/2}}{x^4} \,d x \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^4,x)

[Out]

int(((a + b*x^2)^2*(c + d*x^2)^(3/2))/x^4, x)

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sympy [B]  time = 14.13, size = 352, normalized size = 1.91 \begin {gather*} - \frac {a^{2} \sqrt {c} d}{x \sqrt {1 + \frac {d x^{2}}{c}}} - \frac {a^{2} c \sqrt {d} \sqrt {\frac {c}{d x^{2}} + 1}}{3 x^{2}} - \frac {a^{2} d^{\frac {3}{2}} \sqrt {\frac {c}{d x^{2}} + 1}}{3} + a^{2} d^{\frac {3}{2}} \operatorname {asinh}{\left (\frac {\sqrt {d} x}{\sqrt {c}} \right )} - \frac {a^{2} d^{2} x}{\sqrt {c} \sqrt {1 + \frac {d x^{2}}{c}}} - \frac {2 a b c^{\frac {3}{2}}}{x \sqrt {1 + \frac {d x^{2}}{c}}} + a b \sqrt {c} d x \sqrt {1 + \frac {d x^{2}}{c}} - \frac {2 a b \sqrt {c} d x}{\sqrt {1 + \frac {d x^{2}}{c}}} + 3 a b c \sqrt {d} \operatorname {asinh}{\left (\frac {\sqrt {d} x}{\sqrt {c}} \right )} + \frac {b^{2} c^{\frac {3}{2}} x \sqrt {1 + \frac {d x^{2}}{c}}}{2} + \frac {b^{2} c^{\frac {3}{2}} x}{8 \sqrt {1 + \frac {d x^{2}}{c}}} + \frac {3 b^{2} \sqrt {c} d x^{3}}{8 \sqrt {1 + \frac {d x^{2}}{c}}} + \frac {3 b^{2} c^{2} \operatorname {asinh}{\left (\frac {\sqrt {d} x}{\sqrt {c}} \right )}}{8 \sqrt {d}} + \frac {b^{2} d^{2} x^{5}}{4 \sqrt {c} \sqrt {1 + \frac {d x^{2}}{c}}} \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**4,x)

[Out]

-a**2*sqrt(c)*d/(x*sqrt(1 + d*x**2/c)) - a**2*c*sqrt(d)*sqrt(c/(d*x**2) + 1)/(3*x**2) - a**2*d**(3/2)*sqrt(c/(
d*x**2) + 1)/3 + a**2*d**(3/2)*asinh(sqrt(d)*x/sqrt(c)) - a**2*d**2*x/(sqrt(c)*sqrt(1 + d*x**2/c)) - 2*a*b*c**
(3/2)/(x*sqrt(1 + d*x**2/c)) + a*b*sqrt(c)*d*x*sqrt(1 + d*x**2/c) - 2*a*b*sqrt(c)*d*x/sqrt(1 + d*x**2/c) + 3*a
*b*c*sqrt(d)*asinh(sqrt(d)*x/sqrt(c)) + b**2*c**(3/2)*x*sqrt(1 + d*x**2/c)/2 + b**2*c**(3/2)*x/(8*sqrt(1 + d*x
**2/c)) + 3*b**2*sqrt(c)*d*x**3/(8*sqrt(1 + d*x**2/c)) + 3*b**2*c**2*asinh(sqrt(d)*x/sqrt(c))/(8*sqrt(d)) + b*
*2*d**2*x**5/(4*sqrt(c)*sqrt(1 + d*x**2/c))

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