3.5.17 \(\int \frac {(c+d x^3)^{3/2}}{x^7 (8 c-d x^3)^2} \, dx\) [417]

Optimal. Leaf size=161 \[ \frac {7 d^2 \sqrt {c+d x^3}}{512 c^2 \left (8 c-d x^3\right )}-\frac {\sqrt {c+d x^3}}{48 x^6 \left (8 c-d x^3\right )}-\frac {23 d \sqrt {c+d x^3}}{384 c x^3 \left (8 c-d x^3\right )}+\frac {15 d^2 \tanh ^{-1}\left (\frac {\sqrt {c+d x^3}}{3 \sqrt {c}}\right )}{2048 c^{5/2}}-\frac {17 d^2 \tanh ^{-1}\left (\frac {\sqrt {c+d x^3}}{\sqrt {c}}\right )}{2048 c^{5/2}} \]

[Out]

15/2048*d^2*arctanh(1/3*(d*x^3+c)^(1/2)/c^(1/2))/c^(5/2)-17/2048*d^2*arctanh((d*x^3+c)^(1/2)/c^(1/2))/c^(5/2)+
7/512*d^2*(d*x^3+c)^(1/2)/c^2/(-d*x^3+8*c)-1/48*(d*x^3+c)^(1/2)/x^6/(-d*x^3+8*c)-23/384*d*(d*x^3+c)^(1/2)/c/x^
3/(-d*x^3+8*c)

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Rubi [A]
time = 0.09, antiderivative size = 161, normalized size of antiderivative = 1.00, number of steps used = 9, number of rules used = 7, integrand size = 27, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.259, Rules used = {457, 100, 156, 162, 65, 214, 212} \begin {gather*} \frac {15 d^2 \tanh ^{-1}\left (\frac {\sqrt {c+d x^3}}{3 \sqrt {c}}\right )}{2048 c^{5/2}}-\frac {17 d^2 \tanh ^{-1}\left (\frac {\sqrt {c+d x^3}}{\sqrt {c}}\right )}{2048 c^{5/2}}+\frac {7 d^2 \sqrt {c+d x^3}}{512 c^2 \left (8 c-d x^3\right )}-\frac {23 d \sqrt {c+d x^3}}{384 c x^3 \left (8 c-d x^3\right )}-\frac {\sqrt {c+d x^3}}{48 x^6 \left (8 c-d x^3\right )} \end {gather*}

Antiderivative was successfully verified.

[In]

Int[(c + d*x^3)^(3/2)/(x^7*(8*c - d*x^3)^2),x]

[Out]

(7*d^2*Sqrt[c + d*x^3])/(512*c^2*(8*c - d*x^3)) - Sqrt[c + d*x^3]/(48*x^6*(8*c - d*x^3)) - (23*d*Sqrt[c + d*x^
3])/(384*c*x^3*(8*c - d*x^3)) + (15*d^2*ArcTanh[Sqrt[c + d*x^3]/(3*Sqrt[c])])/(2048*c^(5/2)) - (17*d^2*ArcTanh
[Sqrt[c + d*x^3]/Sqrt[c]])/(2048*c^(5/2))

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 100

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

Rule 156

Int[((a_.) + (b_.)*(x_))^(m_)*((c_.) + (d_.)*(x_))^(n_)*((e_.) + (f_.)*(x_))^(p_)*((g_.) + (h_.)*(x_)), x_Symb
ol] :> Simp[(b*g - a*h)*(a + b*x)^(m + 1)*(c + d*x)^(n + 1)*((e + f*x)^(p + 1)/((m + 1)*(b*c - a*d)*(b*e - a*f
))), x] + Dist[1/((m + 1)*(b*c - a*d)*(b*e - a*f)), Int[(a + b*x)^(m + 1)*(c + d*x)^n*(e + f*x)^p*Simp[(a*d*f*
g - b*(d*e + c*f)*g + b*c*e*h)*(m + 1) - (b*g - a*h)*(d*e*(n + 1) + c*f*(p + 1)) - d*f*(b*g - a*h)*(m + n + p
+ 3)*x, x], x], x] /; FreeQ[{a, b, c, d, e, f, g, h, n, p}, x] && ILtQ[m, -1]

Rule 162

Int[(((e_.) + (f_.)*(x_))^(p_)*((g_.) + (h_.)*(x_)))/(((a_.) + (b_.)*(x_))*((c_.) + (d_.)*(x_))), x_Symbol] :>
Dist[(b*g - a*h)/(b*c - a*d), Int[(e + f*x)^p/(a + b*x), x], x] - Dist[(d*g - c*h)/(b*c - a*d), Int[(e + f*x)
^p/(c + d*x), x], x] /; FreeQ[{a, b, c, d, e, f, g, h}, x]

Rule 212

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

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 457

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 (c+d x^3\right )^{3/2}}{x^7 \left (8 c-d x^3\right )^2} \, dx &=\frac {1}{3} \text {Subst}\left (\int \frac {(c+d x)^{3/2}}{x^3 (8 c-d x)^2} \, dx,x,x^3\right )\\ &=-\frac {\sqrt {c+d x^3}}{48 x^6 \left (8 c-d x^3\right )}-\frac {\text {Subst}\left (\int \frac {-23 c^2 d-\frac {37}{2} c d^2 x}{x^2 (8 c-d x)^2 \sqrt {c+d x}} \, dx,x,x^3\right )}{48 c}\\ &=-\frac {\sqrt {c+d x^3}}{48 x^6 \left (8 c-d x^3\right )}-\frac {23 d \sqrt {c+d x^3}}{384 c x^3 \left (8 c-d x^3\right )}+\frac {\text {Subst}\left (\int \frac {102 c^3 d^2+\frac {69}{2} c^2 d^3 x}{x (8 c-d x)^2 \sqrt {c+d x}} \, dx,x,x^3\right )}{384 c^3}\\ &=\frac {7 d^2 \sqrt {c+d x^3}}{512 c^2 \left (8 c-d x^3\right )}-\frac {\sqrt {c+d x^3}}{48 x^6 \left (8 c-d x^3\right )}-\frac {23 d \sqrt {c+d x^3}}{384 c x^3 \left (8 c-d x^3\right )}-\frac {\text {Subst}\left (\int \frac {-918 c^4 d^3-189 c^3 d^4 x}{x (8 c-d x) \sqrt {c+d x}} \, dx,x,x^3\right )}{27648 c^5 d}\\ &=\frac {7 d^2 \sqrt {c+d x^3}}{512 c^2 \left (8 c-d x^3\right )}-\frac {\sqrt {c+d x^3}}{48 x^6 \left (8 c-d x^3\right )}-\frac {23 d \sqrt {c+d x^3}}{384 c x^3 \left (8 c-d x^3\right )}+\frac {\left (17 d^2\right ) \text {Subst}\left (\int \frac {1}{x \sqrt {c+d x}} \, dx,x,x^3\right )}{4096 c^2}+\frac {\left (45 d^3\right ) \text {Subst}\left (\int \frac {1}{(8 c-d x) \sqrt {c+d x}} \, dx,x,x^3\right )}{4096 c^2}\\ &=\frac {7 d^2 \sqrt {c+d x^3}}{512 c^2 \left (8 c-d x^3\right )}-\frac {\sqrt {c+d x^3}}{48 x^6 \left (8 c-d x^3\right )}-\frac {23 d \sqrt {c+d x^3}}{384 c x^3 \left (8 c-d x^3\right )}+\frac {(17 d) \text {Subst}\left (\int \frac {1}{-\frac {c}{d}+\frac {x^2}{d}} \, dx,x,\sqrt {c+d x^3}\right )}{2048 c^2}+\frac {\left (45 d^2\right ) \text {Subst}\left (\int \frac {1}{9 c-x^2} \, dx,x,\sqrt {c+d x^3}\right )}{2048 c^2}\\ &=\frac {7 d^2 \sqrt {c+d x^3}}{512 c^2 \left (8 c-d x^3\right )}-\frac {\sqrt {c+d x^3}}{48 x^6 \left (8 c-d x^3\right )}-\frac {23 d \sqrt {c+d x^3}}{384 c x^3 \left (8 c-d x^3\right )}+\frac {15 d^2 \tanh ^{-1}\left (\frac {\sqrt {c+d x^3}}{3 \sqrt {c}}\right )}{2048 c^{5/2}}-\frac {17 d^2 \tanh ^{-1}\left (\frac {\sqrt {c+d x^3}}{\sqrt {c}}\right )}{2048 c^{5/2}}\\ \end {align*}

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Mathematica [A]
time = 0.22, size = 112, normalized size = 0.70 \begin {gather*} \frac {\frac {4 \sqrt {c} \sqrt {c+d x^3} \left (32 c^2+92 c d x^3-21 d^2 x^6\right )}{-8 c x^6+d x^9}+45 d^2 \tanh ^{-1}\left (\frac {\sqrt {c+d x^3}}{3 \sqrt {c}}\right )-51 d^2 \tanh ^{-1}\left (\frac {\sqrt {c+d x^3}}{\sqrt {c}}\right )}{6144 c^{5/2}} \end {gather*}

Antiderivative was successfully verified.

[In]

Integrate[(c + d*x^3)^(3/2)/(x^7*(8*c - d*x^3)^2),x]

[Out]

((4*Sqrt[c]*Sqrt[c + d*x^3]*(32*c^2 + 92*c*d*x^3 - 21*d^2*x^6))/(-8*c*x^6 + d*x^9) + 45*d^2*ArcTanh[Sqrt[c + d
*x^3]/(3*Sqrt[c])] - 51*d^2*ArcTanh[Sqrt[c + d*x^3]/Sqrt[c]])/(6144*c^(5/2))

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Maple [C] Result contains higher order function than in optimal. Order 9 vs. order 3.
time = 0.44, size = 1076, normalized size = 6.68

method	result	size
risch	\(\text {Expression too large to display}\)	\(912\)
default	\(\text {Expression too large to display}\)	\(1076\)
elliptic	\(\text {Expression too large to display}\)	\(1580\)

Verification of antiderivative is not currently implemented for this CAS.

[In]

int((d*x^3+c)^(3/2)/x^7/(-d*x^3+8*c)^2,x,method=_RETURNVERBOSE)

[Out]

1/64/c^2*(-1/6*c*(d*x^3+c)^(1/2)/x^6-5/12*d*(d*x^3+c)^(1/2)/x^3-1/4*d^2*arctanh((d*x^3+c)^(1/2)/c^(1/2))/c^(1/
2))-3/4096*d^3/c^4*(2/9*x^3*(d*x^3+c)^(1/2)+56/9*c*(d*x^3+c)^(1/2)/d+3*I*c/d^3*2^(1/2)*sum((-c*d^2)^(1/3)*(1/2
*I*d*(2*x+1/d*(-I*3^(1/2)*(-c*d^2)^(1/3)+(-c*d^2)^(1/3)))/(-c*d^2)^(1/3))^(1/2)*(d*(x-1/d*(-c*d^2)^(1/3))/(-3*
(-c*d^2)^(1/3)+I*3^(1/2)*(-c*d^2)^(1/3)))^(1/2)*(-1/2*I*d*(2*x+1/d*(I*3^(1/2)*(-c*d^2)^(1/3)+(-c*d^2)^(1/3)))/
(-c*d^2)^(1/3))^(1/2)/(d*x^3+c)^(1/2)*(I*(-c*d^2)^(1/3)*_alpha*3^(1/2)*d-I*3^(1/2)*(-c*d^2)^(2/3)+2*_alpha^2*d
^2-(-c*d^2)^(1/3)*_alpha*d-(-c*d^2)^(2/3))*EllipticPi(1/3*3^(1/2)*(I*(x+1/2/d*(-c*d^2)^(1/3)-1/2*I*3^(1/2)/d*(
-c*d^2)^(1/3))*3^(1/2)*d/(-c*d^2)^(1/3))^(1/2),-1/18/d*(2*I*(-c*d^2)^(1/3)*3^(1/2)*_alpha^2*d-I*(-c*d^2)^(2/3)
*3^(1/2)*_alpha+I*3^(1/2)*c*d-3*(-c*d^2)^(2/3)*_alpha-3*c*d)/c,(I*3^(1/2)/d*(-c*d^2)^(1/3)/(-3/2/d*(-c*d^2)^(1
/3)+1/2*I*3^(1/2)/d*(-c*d^2)^(1/3)))^(1/2)),_alpha=RootOf(_Z^3*d-8*c)))+1/256/c^3*d*(-1/3*c*(d*x^3+c)^(1/2)/x^
3+2/3*d*(d*x^3+c)^(1/2)-c^(1/2)*d*arctanh((d*x^3+c)^(1/2)/c^(1/2)))+3/4096*d^2/c^4*(2/9*d*x^3*(d*x^3+c)^(1/2)+
8/9*c*(d*x^3+c)^(1/2)-2/3*c^(3/2)*arctanh((d*x^3+c)^(1/2)/c^(1/2)))+1/512/c^3*d^3*(3*c/d*(d*x^3+c)^(1/2)/(-d*x
^3+8*c)+2/3*(d*x^3+c)^(1/2)/d+1/2*I/d^3*2^(1/2)*sum((-c*d^2)^(1/3)*(1/2*I*d*(2*x+1/d*(-I*3^(1/2)*(-c*d^2)^(1/3
)+(-c*d^2)^(1/3)))/(-c*d^2)^(1/3))^(1/2)*(d*(x-1/d*(-c*d^2)^(1/3))/(-3*(-c*d^2)^(1/3)+I*3^(1/2)*(-c*d^2)^(1/3)
))^(1/2)*(-1/2*I*d*(2*x+1/d*(I*3^(1/2)*(-c*d^2)^(1/3)+(-c*d^2)^(1/3)))/(-c*d^2)^(1/3))^(1/2)/(d*x^3+c)^(1/2)*(
I*(-c*d^2)^(1/3)*_alpha*3^(1/2)*d-I*3^(1/2)*(-c*d^2)^(2/3)+2*_alpha^2*d^2-(-c*d^2)^(1/3)*_alpha*d-(-c*d^2)^(2/
3))*EllipticPi(1/3*3^(1/2)*(I*(x+1/2/d*(-c*d^2)^(1/3)-1/2*I*3^(1/2)/d*(-c*d^2)^(1/3))*3^(1/2)*d/(-c*d^2)^(1/3)
)^(1/2),-1/18/d*(2*I*(-c*d^2)^(1/3)*3^(1/2)*_alpha^2*d-I*(-c*d^2)^(2/3)*3^(1/2)*_alpha+I*3^(1/2)*c*d-3*(-c*d^2
)^(2/3)*_alpha-3*c*d)/c,(I*3^(1/2)/d*(-c*d^2)^(1/3)/(-3/2/d*(-c*d^2)^(1/3)+1/2*I*3^(1/2)/d*(-c*d^2)^(1/3)))^(1
/2)),_alpha=RootOf(_Z^3*d-8*c)))

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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((d*x^3+c)^(3/2)/x^7/(-d*x^3+8*c)^2,x, algorithm="maxima")

[Out]

integrate((d*x^3 + c)^(3/2)/((d*x^3 - 8*c)^2*x^7), x)

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Fricas [A]
time = 2.65, size = 310, normalized size = 1.93 \begin {gather*} \left [\frac {45 \, {\left (d^{3} x^{9} - 8 \, c d^{2} x^{6}\right )} \sqrt {c} \log \left (\frac {d x^{3} + 6 \, \sqrt {d x^{3} + c} \sqrt {c} + 10 \, c}{d x^{3} - 8 \, c}\right ) + 51 \, {\left (d^{3} x^{9} - 8 \, c d^{2} x^{6}\right )} \sqrt {c} \log \left (\frac {d x^{3} - 2 \, \sqrt {d x^{3} + c} \sqrt {c} + 2 \, c}{x^{3}}\right ) - 8 \, {\left (21 \, c d^{2} x^{6} - 92 \, c^{2} d x^{3} - 32 \, c^{3}\right )} \sqrt {d x^{3} + c}}{12288 \, {\left (c^{3} d x^{9} - 8 \, c^{4} x^{6}\right )}}, \frac {51 \, {\left (d^{3} x^{9} - 8 \, c d^{2} x^{6}\right )} \sqrt {-c} \arctan \left (\frac {\sqrt {d x^{3} + c} \sqrt {-c}}{c}\right ) - 45 \, {\left (d^{3} x^{9} - 8 \, c d^{2} x^{6}\right )} \sqrt {-c} \arctan \left (\frac {\sqrt {d x^{3} + c} \sqrt {-c}}{3 \, c}\right ) - 4 \, {\left (21 \, c d^{2} x^{6} - 92 \, c^{2} d x^{3} - 32 \, c^{3}\right )} \sqrt {d x^{3} + c}}{6144 \, {\left (c^{3} d x^{9} - 8 \, c^{4} x^{6}\right )}}\right ] \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate((d*x^3+c)^(3/2)/x^7/(-d*x^3+8*c)^2,x, algorithm="fricas")

[Out]

[1/12288*(45*(d^3*x^9 - 8*c*d^2*x^6)*sqrt(c)*log((d*x^3 + 6*sqrt(d*x^3 + c)*sqrt(c) + 10*c)/(d*x^3 - 8*c)) + 5
1*(d^3*x^9 - 8*c*d^2*x^6)*sqrt(c)*log((d*x^3 - 2*sqrt(d*x^3 + c)*sqrt(c) + 2*c)/x^3) - 8*(21*c*d^2*x^6 - 92*c^
2*d*x^3 - 32*c^3)*sqrt(d*x^3 + c))/(c^3*d*x^9 - 8*c^4*x^6), 1/6144*(51*(d^3*x^9 - 8*c*d^2*x^6)*sqrt(-c)*arctan
(sqrt(d*x^3 + c)*sqrt(-c)/c) - 45*(d^3*x^9 - 8*c*d^2*x^6)*sqrt(-c)*arctan(1/3*sqrt(d*x^3 + c)*sqrt(-c)/c) - 4*
(21*c*d^2*x^6 - 92*c^2*d*x^3 - 32*c^3)*sqrt(d*x^3 + c))/(c^3*d*x^9 - 8*c^4*x^6)]

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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((d*x**3+c)**(3/2)/x**7/(-d*x**3+8*c)**2,x)

[Out]

Timed out

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Giac [A]
time = 0.83, size = 129, normalized size = 0.80 \begin {gather*} \frac {17 \, d^{2} \arctan \left (\frac {\sqrt {d x^{3} + c}}{\sqrt {-c}}\right )}{2048 \, \sqrt {-c} c^{2}} - \frac {15 \, d^{2} \arctan \left (\frac {\sqrt {d x^{3} + c}}{3 \, \sqrt {-c}}\right )}{2048 \, \sqrt {-c} c^{2}} - \frac {3 \, \sqrt {d x^{3} + c} d^{2}}{512 \, {\left (d x^{3} - 8 \, c\right )} c^{2}} - \frac {3 \, {\left (d x^{3} + c\right )}^{\frac {3}{2}} d^{2} - 2 \, \sqrt {d x^{3} + c} c d^{2}}{384 \, c^{2} d^{2} x^{6}} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate((d*x^3+c)^(3/2)/x^7/(-d*x^3+8*c)^2,x, algorithm="giac")

[Out]

17/2048*d^2*arctan(sqrt(d*x^3 + c)/sqrt(-c))/(sqrt(-c)*c^2) - 15/2048*d^2*arctan(1/3*sqrt(d*x^3 + c)/sqrt(-c))
/(sqrt(-c)*c^2) - 3/512*sqrt(d*x^3 + c)*d^2/((d*x^3 - 8*c)*c^2) - 1/384*(3*(d*x^3 + c)^(3/2)*d^2 - 2*sqrt(d*x^
3 + c)*c*d^2)/(c^2*d^2*x^6)

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Mupad [B]
time = 4.62, size = 151, normalized size = 0.94 \begin {gather*} \frac {\frac {81\,d^2\,\sqrt {d\,x^3+c}}{512}-\frac {67\,d^2\,{\left (d\,x^3+c\right )}^{3/2}}{256\,c}+\frac {21\,d^2\,{\left (d\,x^3+c\right )}^{5/2}}{512\,c^2}}{33\,c\,{\left (d\,x^3+c\right )}^2-57\,c^2\,\left (d\,x^3+c\right )-3\,{\left (d\,x^3+c\right )}^3+27\,c^3}+\frac {d^2\,\left (\mathrm {atanh}\left (\frac {c^2\,\sqrt {d\,x^3+c}}{\sqrt {c^5}}\right )\,1{}\mathrm {i}-\frac {\mathrm {atanh}\left (\frac {c^2\,\sqrt {d\,x^3+c}}{3\,\sqrt {c^5}}\right )\,15{}\mathrm {i}}{17}\right )\,17{}\mathrm {i}}{2048\,\sqrt {c^5}} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

int((c + d*x^3)^(3/2)/(x^7*(8*c - d*x^3)^2),x)

[Out]

((81*d^2*(c + d*x^3)^(1/2))/512 - (67*d^2*(c + d*x^3)^(3/2))/(256*c) + (21*d^2*(c + d*x^3)^(5/2))/(512*c^2))/(
33*c*(c + d*x^3)^2 - 57*c^2*(c + d*x^3) - 3*(c + d*x^3)^3 + 27*c^3) + (d^2*(atanh((c^2*(c + d*x^3)^(1/2))/(c^5
)^(1/2))*1i - (atanh((c^2*(c + d*x^3)^(1/2))/(3*(c^5)^(1/2)))*15i)/17)*17i)/(2048*(c^5)^(1/2))

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