3.1185 \(\int x (d+e x^2)^{3/2} (a+b \tan ^{-1}(c x)) \, dx\)

Optimal. Leaf size=181 \[ \frac{\left (d+e x^2\right )^{5/2} \left (a+b \tan ^{-1}(c x)\right )}{5 e}-\frac{b \left (15 c^4 d^2-20 c^2 d e+8 e^2\right ) \tanh ^{-1}\left (\frac{\sqrt{e} x}{\sqrt{d+e x^2}}\right )}{40 c^5 \sqrt{e}}-\frac{b x \left (7 c^2 d-4 e\right ) \sqrt{d+e x^2}}{40 c^3}-\frac{b \left (c^2 d-e\right )^{5/2} \tan ^{-1}\left (\frac{x \sqrt{c^2 d-e}}{\sqrt{d+e x^2}}\right )}{5 c^5 e}-\frac{b x \left (d+e x^2\right )^{3/2}}{20 c} \]

[Out]

-(b*(7*c^2*d - 4*e)*x*Sqrt[d + e*x^2])/(40*c^3) - (b*x*(d + e*x^2)^(3/2))/(20*c) + ((d + e*x^2)^(5/2)*(a + b*A
rcTan[c*x]))/(5*e) - (b*(c^2*d - e)^(5/2)*ArcTan[(Sqrt[c^2*d - e]*x)/Sqrt[d + e*x^2]])/(5*c^5*e) - (b*(15*c^4*
d^2 - 20*c^2*d*e + 8*e^2)*ArcTanh[(Sqrt[e]*x)/Sqrt[d + e*x^2]])/(40*c^5*Sqrt[e])

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Rubi [A]  time = 0.233115, antiderivative size = 181, normalized size of antiderivative = 1., number of steps used = 8, number of rules used = 8, integrand size = 21, \(\frac{\text{number of rules}}{\text{integrand size}}\) = 0.381, Rules used = {4974, 416, 528, 523, 217, 206, 377, 203} \[ \frac{\left (d+e x^2\right )^{5/2} \left (a+b \tan ^{-1}(c x)\right )}{5 e}-\frac{b \left (15 c^4 d^2-20 c^2 d e+8 e^2\right ) \tanh ^{-1}\left (\frac{\sqrt{e} x}{\sqrt{d+e x^2}}\right )}{40 c^5 \sqrt{e}}-\frac{b x \left (7 c^2 d-4 e\right ) \sqrt{d+e x^2}}{40 c^3}-\frac{b \left (c^2 d-e\right )^{5/2} \tan ^{-1}\left (\frac{x \sqrt{c^2 d-e}}{\sqrt{d+e x^2}}\right )}{5 c^5 e}-\frac{b x \left (d+e x^2\right )^{3/2}}{20 c} \]

Antiderivative was successfully verified.

[In]

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

[Out]

-(b*(7*c^2*d - 4*e)*x*Sqrt[d + e*x^2])/(40*c^3) - (b*x*(d + e*x^2)^(3/2))/(20*c) + ((d + e*x^2)^(5/2)*(a + b*A
rcTan[c*x]))/(5*e) - (b*(c^2*d - e)^(5/2)*ArcTan[(Sqrt[c^2*d - e]*x)/Sqrt[d + e*x^2]])/(5*c^5*e) - (b*(15*c^4*
d^2 - 20*c^2*d*e + 8*e^2)*ArcTanh[(Sqrt[e]*x)/Sqrt[d + e*x^2]])/(40*c^5*Sqrt[e])

Rule 4974

Int[((a_.) + ArcTan[(c_.)*(x_)]*(b_.))*(x_)*((d_.) + (e_.)*(x_)^2)^(q_.), x_Symbol] :> Simp[((d + e*x^2)^(q +
1)*(a + b*ArcTan[c*x]))/(2*e*(q + 1)), x] - Dist[(b*c)/(2*e*(q + 1)), Int[(d + e*x^2)^(q + 1)/(1 + c^2*x^2), x
], x] /; FreeQ[{a, b, c, d, e, q}, x] && NeQ[q, -1]

Rule 416

Int[((a_) + (b_.)*(x_)^(n_))^(p_)*((c_) + (d_.)*(x_)^(n_))^(q_), x_Symbol] :> Simp[(d*x*(a + b*x^n)^(p + 1)*(c
+ d*x^n)^(q - 1))/(b*(n*(p + q) + 1)), x] + Dist[1/(b*(n*(p + q) + 1)), Int[(a + b*x^n)^p*(c + d*x^n)^(q - 2)
*Simp[c*(b*c*(n*(p + q) + 1) - a*d) + d*(b*c*(n*(p + 2*q - 1) + 1) - a*d*(n*(q - 1) + 1))*x^n, x], x], x] /; F
reeQ[{a, b, c, d, n, p}, x] && NeQ[b*c - a*d, 0] && GtQ[q, 1] && NeQ[n*(p + q) + 1, 0] &&  !IGtQ[p, 1] && IntB
inomialQ[a, b, c, d, n, p, q, x]

Rule 528

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

Rule 523

Int[((e_) + (f_.)*(x_)^(n_))/(((a_) + (b_.)*(x_)^(n_))*Sqrt[(c_) + (d_.)*(x_)^(n_)]), x_Symbol] :> Dist[f/b, I
nt[1/Sqrt[c + d*x^n], x], x] + Dist[(b*e - a*f)/b, Int[1/((a + b*x^n)*Sqrt[c + d*x^n]), x], x] /; FreeQ[{a, b,
c, d, e, f, n}, x]

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 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 377

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

Rule 203

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

Rubi steps

\begin{align*} \int x \left (d+e x^2\right )^{3/2} \left (a+b \tan ^{-1}(c x)\right ) \, dx &=\frac{\left (d+e x^2\right )^{5/2} \left (a+b \tan ^{-1}(c x)\right )}{5 e}-\frac{(b c) \int \frac{\left (d+e x^2\right )^{5/2}}{1+c^2 x^2} \, dx}{5 e}\\ &=-\frac{b x \left (d+e x^2\right )^{3/2}}{20 c}+\frac{\left (d+e x^2\right )^{5/2} \left (a+b \tan ^{-1}(c x)\right )}{5 e}-\frac{b \int \frac{\sqrt{d+e x^2} \left (d \left (4 c^2 d-e\right )+\left (7 c^2 d-4 e\right ) e x^2\right )}{1+c^2 x^2} \, dx}{20 c e}\\ &=-\frac{b \left (7 c^2 d-4 e\right ) x \sqrt{d+e x^2}}{40 c^3}-\frac{b x \left (d+e x^2\right )^{3/2}}{20 c}+\frac{\left (d+e x^2\right )^{5/2} \left (a+b \tan ^{-1}(c x)\right )}{5 e}-\frac{b \int \frac{d \left (8 c^4 d^2-9 c^2 d e+4 e^2\right )+e \left (15 c^4 d^2-20 c^2 d e+8 e^2\right ) x^2}{\left (1+c^2 x^2\right ) \sqrt{d+e x^2}} \, dx}{40 c^3 e}\\ &=-\frac{b \left (7 c^2 d-4 e\right ) x \sqrt{d+e x^2}}{40 c^3}-\frac{b x \left (d+e x^2\right )^{3/2}}{20 c}+\frac{\left (d+e x^2\right )^{5/2} \left (a+b \tan ^{-1}(c x)\right )}{5 e}-\frac{\left (b \left (c^2 d-e\right )^3\right ) \int \frac{1}{\left (1+c^2 x^2\right ) \sqrt{d+e x^2}} \, dx}{5 c^5 e}-\frac{\left (b \left (15 c^4 d^2-20 c^2 d e+8 e^2\right )\right ) \int \frac{1}{\sqrt{d+e x^2}} \, dx}{40 c^5}\\ &=-\frac{b \left (7 c^2 d-4 e\right ) x \sqrt{d+e x^2}}{40 c^3}-\frac{b x \left (d+e x^2\right )^{3/2}}{20 c}+\frac{\left (d+e x^2\right )^{5/2} \left (a+b \tan ^{-1}(c x)\right )}{5 e}-\frac{\left (b \left (c^2 d-e\right )^3\right ) \operatorname{Subst}\left (\int \frac{1}{1-\left (-c^2 d+e\right ) x^2} \, dx,x,\frac{x}{\sqrt{d+e x^2}}\right )}{5 c^5 e}-\frac{\left (b \left (15 c^4 d^2-20 c^2 d e+8 e^2\right )\right ) \operatorname{Subst}\left (\int \frac{1}{1-e x^2} \, dx,x,\frac{x}{\sqrt{d+e x^2}}\right )}{40 c^5}\\ &=-\frac{b \left (7 c^2 d-4 e\right ) x \sqrt{d+e x^2}}{40 c^3}-\frac{b x \left (d+e x^2\right )^{3/2}}{20 c}+\frac{\left (d+e x^2\right )^{5/2} \left (a+b \tan ^{-1}(c x)\right )}{5 e}-\frac{b \left (c^2 d-e\right )^{5/2} \tan ^{-1}\left (\frac{\sqrt{c^2 d-e} x}{\sqrt{d+e x^2}}\right )}{5 c^5 e}-\frac{b \left (15 c^4 d^2-20 c^2 d e+8 e^2\right ) \tanh ^{-1}\left (\frac{\sqrt{e} x}{\sqrt{d+e x^2}}\right )}{40 c^5 \sqrt{e}}\\ \end{align*}

Mathematica [C]  time = 0.425465, size = 313, normalized size = 1.73 \[ \frac{c^2 \sqrt{d+e x^2} \left (8 a c^3 \left (d+e x^2\right )^2+b e x \left (4 e-c^2 \left (9 d+2 e x^2\right )\right )\right )-b \sqrt{e} \left (15 c^4 d^2-20 c^2 d e+8 e^2\right ) \log \left (\sqrt{e} \sqrt{d+e x^2}+e x\right )-4 i b \left (c^2 d-e\right )^{5/2} \log \left (\frac{20 c^6 e \left (-i \sqrt{c^2 d-e} \sqrt{d+e x^2}-i c d+e x\right )}{b (c x-i) \left (c^2 d-e\right )^{7/2}}\right )+4 i b \left (c^2 d-e\right )^{5/2} \log \left (\frac{20 c^6 e \left (i \sqrt{c^2 d-e} \sqrt{d+e x^2}+i c d+e x\right )}{b (c x+i) \left (c^2 d-e\right )^{7/2}}\right )+8 b c^5 \tan ^{-1}(c x) \left (d+e x^2\right )^{5/2}}{40 c^5 e} \]

Antiderivative was successfully verified.

[In]

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

[Out]

(c^2*Sqrt[d + e*x^2]*(8*a*c^3*(d + e*x^2)^2 + b*e*x*(4*e - c^2*(9*d + 2*e*x^2))) + 8*b*c^5*(d + e*x^2)^(5/2)*A
rcTan[c*x] - (4*I)*b*(c^2*d - e)^(5/2)*Log[(20*c^6*e*((-I)*c*d + e*x - I*Sqrt[c^2*d - e]*Sqrt[d + e*x^2]))/(b*
(c^2*d - e)^(7/2)*(-I + c*x))] + (4*I)*b*(c^2*d - e)^(5/2)*Log[(20*c^6*e*(I*c*d + e*x + I*Sqrt[c^2*d - e]*Sqrt
[d + e*x^2]))/(b*(c^2*d - e)^(7/2)*(I + c*x))] - b*Sqrt[e]*(15*c^4*d^2 - 20*c^2*d*e + 8*e^2)*Log[e*x + Sqrt[e]
*Sqrt[d + e*x^2]])/(40*c^5*e)

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Maple [F]  time = 0.619, size = 0, normalized size = 0. \begin{align*} \int x \left ( e{x}^{2}+d \right ) ^{{\frac{3}{2}}} \left ( a+b\arctan \left ( cx \right ) \right ) \, dx \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

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

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Maxima [F(-2)]  time = 0., size = 0, normalized size = 0. \begin{align*} \text{Exception raised: ValueError} \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

Exception raised: ValueError

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Fricas [A]  time = 29.7106, size = 2603, normalized size = 14.38 \begin{align*} \text{result too large to display} \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

[1/80*((15*b*c^4*d^2 - 20*b*c^2*d*e + 8*b*e^2)*sqrt(e)*log(-2*e*x^2 + 2*sqrt(e*x^2 + d)*sqrt(e)*x - d) + 4*(b*
c^4*d^2 - 2*b*c^2*d*e + b*e^2)*sqrt(-c^2*d + e)*log(((c^4*d^2 - 8*c^2*d*e + 8*e^2)*x^4 - 2*(3*c^2*d^2 - 4*d*e)
*x^2 - 4*((c^2*d - 2*e)*x^3 - d*x)*sqrt(-c^2*d + e)*sqrt(e*x^2 + d) + d^2)/(c^4*x^4 + 2*c^2*x^2 + 1)) + 2*(8*a
*c^5*e^2*x^4 + 16*a*c^5*d*e*x^2 - 2*b*c^4*e^2*x^3 + 8*a*c^5*d^2 - (9*b*c^4*d*e - 4*b*c^2*e^2)*x + 8*(b*c^5*e^2
*x^4 + 2*b*c^5*d*e*x^2 + b*c^5*d^2)*arctan(c*x))*sqrt(e*x^2 + d))/(c^5*e), -1/80*(8*(b*c^4*d^2 - 2*b*c^2*d*e +
b*e^2)*sqrt(c^2*d - e)*arctan(1/2*sqrt(c^2*d - e)*((c^2*d - 2*e)*x^2 - d)*sqrt(e*x^2 + d)/((c^2*d*e - e^2)*x^
3 + (c^2*d^2 - d*e)*x)) - (15*b*c^4*d^2 - 20*b*c^2*d*e + 8*b*e^2)*sqrt(e)*log(-2*e*x^2 + 2*sqrt(e*x^2 + d)*sqr
t(e)*x - d) - 2*(8*a*c^5*e^2*x^4 + 16*a*c^5*d*e*x^2 - 2*b*c^4*e^2*x^3 + 8*a*c^5*d^2 - (9*b*c^4*d*e - 4*b*c^2*e
^2)*x + 8*(b*c^5*e^2*x^4 + 2*b*c^5*d*e*x^2 + b*c^5*d^2)*arctan(c*x))*sqrt(e*x^2 + d))/(c^5*e), 1/40*((15*b*c^4
*d^2 - 20*b*c^2*d*e + 8*b*e^2)*sqrt(-e)*arctan(sqrt(-e)*x/sqrt(e*x^2 + d)) + 2*(b*c^4*d^2 - 2*b*c^2*d*e + b*e^
2)*sqrt(-c^2*d + e)*log(((c^4*d^2 - 8*c^2*d*e + 8*e^2)*x^4 - 2*(3*c^2*d^2 - 4*d*e)*x^2 - 4*((c^2*d - 2*e)*x^3
- d*x)*sqrt(-c^2*d + e)*sqrt(e*x^2 + d) + d^2)/(c^4*x^4 + 2*c^2*x^2 + 1)) + (8*a*c^5*e^2*x^4 + 16*a*c^5*d*e*x^
2 - 2*b*c^4*e^2*x^3 + 8*a*c^5*d^2 - (9*b*c^4*d*e - 4*b*c^2*e^2)*x + 8*(b*c^5*e^2*x^4 + 2*b*c^5*d*e*x^2 + b*c^5
*d^2)*arctan(c*x))*sqrt(e*x^2 + d))/(c^5*e), -1/40*(4*(b*c^4*d^2 - 2*b*c^2*d*e + b*e^2)*sqrt(c^2*d - e)*arctan
(1/2*sqrt(c^2*d - e)*((c^2*d - 2*e)*x^2 - d)*sqrt(e*x^2 + d)/((c^2*d*e - e^2)*x^3 + (c^2*d^2 - d*e)*x)) - (15*
b*c^4*d^2 - 20*b*c^2*d*e + 8*b*e^2)*sqrt(-e)*arctan(sqrt(-e)*x/sqrt(e*x^2 + d)) - (8*a*c^5*e^2*x^4 + 16*a*c^5*
d*e*x^2 - 2*b*c^4*e^2*x^3 + 8*a*c^5*d^2 - (9*b*c^4*d*e - 4*b*c^2*e^2)*x + 8*(b*c^5*e^2*x^4 + 2*b*c^5*d*e*x^2 +
b*c^5*d^2)*arctan(c*x))*sqrt(e*x^2 + d))/(c^5*e)]

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Sympy [F]  time = 0., size = 0, normalized size = 0. \begin{align*} \int x \left (a + b \operatorname{atan}{\left (c x \right )}\right ) \left (d + e x^{2}\right )^{\frac{3}{2}}\, dx \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

Integral(x*(a + b*atan(c*x))*(d + e*x**2)**(3/2), x)

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Giac [B]  time = 1.50523, size = 620, normalized size = 3.43 \begin{align*} \frac{1}{3} \,{\left (x^{2} e + d\right )}^{\frac{3}{2}} a d e^{\left (-1\right )} + \frac{1}{12} \,{\left (4 \,{\left (x^{2} e + d\right )}^{\frac{3}{2}} \arctan \left (c x\right ) e^{\left (-1\right )} - c{\left (\frac{2 \, \sqrt{x^{2} e + d} x}{c^{2}} - \frac{{\left (3 \, c^{2} d - 2 \, e\right )} e^{\left (-\frac{1}{2}\right )} \log \left ({\left (x e^{\frac{1}{2}} - \sqrt{x^{2} e + d}\right )}^{2}\right )}{c^{4}} - \frac{4 \,{\left (c^{4} d^{2} e^{\frac{1}{2}} - 2 \, c^{2} d e^{\frac{3}{2}} + e^{\frac{5}{2}}\right )} \arctan \left (\frac{{\left ({\left (x e^{\frac{1}{2}} - \sqrt{x^{2} e + d}\right )}^{2} c^{2} - c^{2} d + 2 \, e\right )} e^{\left (-\frac{1}{2}\right )}}{2 \, \sqrt{c^{2} d - e}}\right ) e^{\left (-\frac{3}{2}\right )}}{\sqrt{c^{2} d - e} c^{4}}\right )}\right )} b d + \frac{1}{240} \,{\left (16 \,{\left (3 \,{\left (x^{2} e + d\right )}^{\frac{5}{2}} - 5 \,{\left (x^{2} e + d\right )}^{\frac{3}{2}} d\right )} a e^{\left (-2\right )} +{\left (16 \,{\left (3 \,{\left (x^{2} e + d\right )}^{\frac{5}{2}} - 5 \,{\left (x^{2} e + d\right )}^{\frac{3}{2}} d\right )} \arctan \left (c x\right ) e^{\left (-2\right )} -{\left (2 \, \sqrt{x^{2} e + d} x{\left (\frac{6 \, x^{2}}{c^{2}} + \frac{{\left (7 \, c^{10} d e^{2} - 12 \, c^{8} e^{3}\right )} e^{\left (-3\right )}}{c^{12}}\right )} + \frac{{\left (15 \, c^{4} d^{2} + 20 \, c^{2} d e - 24 \, e^{2}\right )} e^{\left (-\frac{3}{2}\right )} \log \left ({\left (x e^{\frac{1}{2}} - \sqrt{x^{2} e + d}\right )}^{2}\right )}{c^{6}} + \frac{16 \,{\left (2 \, c^{6} d^{3} e^{\frac{1}{2}} - c^{4} d^{2} e^{\frac{3}{2}} - 4 \, c^{2} d e^{\frac{5}{2}} + 3 \, e^{\frac{7}{2}}\right )} \arctan \left (\frac{{\left ({\left (x e^{\frac{1}{2}} - \sqrt{x^{2} e + d}\right )}^{2} c^{2} - c^{2} d + 2 \, e\right )} e^{\left (-\frac{1}{2}\right )}}{2 \, \sqrt{c^{2} d - e}}\right ) e^{\left (-\frac{5}{2}\right )}}{\sqrt{c^{2} d - e} c^{6}}\right )} c\right )} b\right )} e \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

1/3*(x^2*e + d)^(3/2)*a*d*e^(-1) + 1/12*(4*(x^2*e + d)^(3/2)*arctan(c*x)*e^(-1) - c*(2*sqrt(x^2*e + d)*x/c^2 -
(3*c^2*d - 2*e)*e^(-1/2)*log((x*e^(1/2) - sqrt(x^2*e + d))^2)/c^4 - 4*(c^4*d^2*e^(1/2) - 2*c^2*d*e^(3/2) + e^
(5/2))*arctan(1/2*((x*e^(1/2) - sqrt(x^2*e + d))^2*c^2 - c^2*d + 2*e)*e^(-1/2)/sqrt(c^2*d - e))*e^(-3/2)/(sqrt
(c^2*d - e)*c^4)))*b*d + 1/240*(16*(3*(x^2*e + d)^(5/2) - 5*(x^2*e + d)^(3/2)*d)*a*e^(-2) + (16*(3*(x^2*e + d)
^(5/2) - 5*(x^2*e + d)^(3/2)*d)*arctan(c*x)*e^(-2) - (2*sqrt(x^2*e + d)*x*(6*x^2/c^2 + (7*c^10*d*e^2 - 12*c^8*
e^3)*e^(-3)/c^12) + (15*c^4*d^2 + 20*c^2*d*e - 24*e^2)*e^(-3/2)*log((x*e^(1/2) - sqrt(x^2*e + d))^2)/c^6 + 16*
(2*c^6*d^3*e^(1/2) - c^4*d^2*e^(3/2) - 4*c^2*d*e^(5/2) + 3*e^(7/2))*arctan(1/2*((x*e^(1/2) - sqrt(x^2*e + d))^
2*c^2 - c^2*d + 2*e)*e^(-1/2)/sqrt(c^2*d - e))*e^(-5/2)/(sqrt(c^2*d - e)*c^6))*c)*b)*e