3.1180 \(\int \frac{\sqrt{d+e x^2} (a+b \tan ^{-1}(c x))}{x^4} \, dx\)

Optimal. Leaf size=137 \[ -\frac{\left (d+e x^2\right )^{3/2} \left (a+b \tan ^{-1}(c x)\right )}{3 d x^3}-\frac{b \left (c^2 d-e\right )^{3/2} \tanh ^{-1}\left (\frac{c \sqrt{d+e x^2}}{\sqrt{c^2 d-e}}\right )}{3 d}+\frac{b c \left (2 c^2 d-3 e\right ) \tanh ^{-1}\left (\frac{\sqrt{d+e x^2}}{\sqrt{d}}\right )}{6 \sqrt{d}}-\frac{b c \sqrt{d+e x^2}}{6 x^2} \]

[Out]

-(b*c*Sqrt[d + e*x^2])/(6*x^2) - ((d + e*x^2)^(3/2)*(a + b*ArcTan[c*x]))/(3*d*x^3) + (b*c*(2*c^2*d - 3*e)*ArcT
anh[Sqrt[d + e*x^2]/Sqrt[d]])/(6*Sqrt[d]) - (b*(c^2*d - e)^(3/2)*ArcTanh[(c*Sqrt[d + e*x^2])/Sqrt[c^2*d - e]])
/(3*d)

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Rubi [A]  time = 0.279043, antiderivative size = 137, normalized size of antiderivative = 1., number of steps used = 9, number of rules used = 8, integrand size = 23, \(\frac{\text{number of rules}}{\text{integrand size}}\) = 0.348, Rules used = {264, 4976, 12, 446, 98, 156, 63, 208} \[ -\frac{\left (d+e x^2\right )^{3/2} \left (a+b \tan ^{-1}(c x)\right )}{3 d x^3}-\frac{b \left (c^2 d-e\right )^{3/2} \tanh ^{-1}\left (\frac{c \sqrt{d+e x^2}}{\sqrt{c^2 d-e}}\right )}{3 d}+\frac{b c \left (2 c^2 d-3 e\right ) \tanh ^{-1}\left (\frac{\sqrt{d+e x^2}}{\sqrt{d}}\right )}{6 \sqrt{d}}-\frac{b c \sqrt{d+e x^2}}{6 x^2} \]

Antiderivative was successfully verified.

[In]

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

[Out]

-(b*c*Sqrt[d + e*x^2])/(6*x^2) - ((d + e*x^2)^(3/2)*(a + b*ArcTan[c*x]))/(3*d*x^3) + (b*c*(2*c^2*d - 3*e)*ArcT
anh[Sqrt[d + e*x^2]/Sqrt[d]])/(6*Sqrt[d]) - (b*(c^2*d - e)^(3/2)*ArcTanh[(c*Sqrt[d + e*x^2])/Sqrt[c^2*d - e]])
/(3*d)

Rule 264

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

Rule 4976

Int[((a_.) + ArcTan[(c_.)*(x_)]*(b_.))*((f_.)*(x_))^(m_.)*((d_.) + (e_.)*(x_)^2)^(q_.), x_Symbol] :> With[{u =
IntHide[(f*x)^m*(d + e*x^2)^q, x]}, Dist[a + b*ArcTan[c*x], u, x] - Dist[b*c, Int[SimplifyIntegrand[u/(1 + c^
2*x^2), x], x], x]] /; FreeQ[{a, b, c, d, e, f, m, q}, x] && ((IGtQ[q, 0] &&  !(ILtQ[(m - 1)/2, 0] && GtQ[m +
2*q + 3, 0])) || (IGtQ[(m + 1)/2, 0] &&  !(ILtQ[q, 0] && GtQ[m + 2*q + 3, 0])) || (ILtQ[(m + 2*q + 1)/2, 0] &&
!ILtQ[(m - 1)/2, 0]))

Rule 12

Int[(a_)*(u_), x_Symbol] :> Dist[a, Int[u, x], x] /; FreeQ[a, x] &&  !MatchQ[u, (b_)*(v_) /; FreeQ[b, x]]

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]]

Rule 98

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[(((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 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 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]

Rubi steps

\begin{align*} \int \frac{\sqrt{d+e x^2} \left (a+b \tan ^{-1}(c x)\right )}{x^4} \, dx &=-\frac{\left (d+e x^2\right )^{3/2} \left (a+b \tan ^{-1}(c x)\right )}{3 d x^3}-(b c) \int \frac{\left (d+e x^2\right )^{3/2}}{3 x^3 \left (-d-c^2 d x^2\right )} \, dx\\ &=-\frac{\left (d+e x^2\right )^{3/2} \left (a+b \tan ^{-1}(c x)\right )}{3 d x^3}-\frac{1}{3} (b c) \int \frac{\left (d+e x^2\right )^{3/2}}{x^3 \left (-d-c^2 d x^2\right )} \, dx\\ &=-\frac{\left (d+e x^2\right )^{3/2} \left (a+b \tan ^{-1}(c x)\right )}{3 d x^3}-\frac{1}{6} (b c) \operatorname{Subst}\left (\int \frac{(d+e x)^{3/2}}{x^2 \left (-d-c^2 d x\right )} \, dx,x,x^2\right )\\ &=-\frac{b c \sqrt{d+e x^2}}{6 x^2}-\frac{\left (d+e x^2\right )^{3/2} \left (a+b \tan ^{-1}(c x)\right )}{3 d x^3}-\frac{(b c) \operatorname{Subst}\left (\int \frac{-\frac{1}{2} d^2 \left (2 c^2 d-3 e\right )-\frac{1}{2} d \left (c^2 d-2 e\right ) e x}{x \left (-d-c^2 d x\right ) \sqrt{d+e x}} \, dx,x,x^2\right )}{6 d}\\ &=-\frac{b c \sqrt{d+e x^2}}{6 x^2}-\frac{\left (d+e x^2\right )^{3/2} \left (a+b \tan ^{-1}(c x)\right )}{3 d x^3}-\frac{1}{12} \left (b c \left (2 c^2 d-3 e\right )\right ) \operatorname{Subst}\left (\int \frac{1}{x \sqrt{d+e x}} \, dx,x,x^2\right )-\frac{1}{6} \left (b c \left (c^2 d-e\right )^2\right ) \operatorname{Subst}\left (\int \frac{1}{\left (-d-c^2 d x\right ) \sqrt{d+e x}} \, dx,x,x^2\right )\\ &=-\frac{b c \sqrt{d+e x^2}}{6 x^2}-\frac{\left (d+e x^2\right )^{3/2} \left (a+b \tan ^{-1}(c x)\right )}{3 d x^3}-\frac{\left (b c \left (2 c^2 d-3 e\right )\right ) \operatorname{Subst}\left (\int \frac{1}{-\frac{d}{e}+\frac{x^2}{e}} \, dx,x,\sqrt{d+e x^2}\right )}{6 e}-\frac{\left (b c \left (c^2 d-e\right )^2\right ) \operatorname{Subst}\left (\int \frac{1}{-d+\frac{c^2 d^2}{e}-\frac{c^2 d x^2}{e}} \, dx,x,\sqrt{d+e x^2}\right )}{3 e}\\ &=-\frac{b c \sqrt{d+e x^2}}{6 x^2}-\frac{\left (d+e x^2\right )^{3/2} \left (a+b \tan ^{-1}(c x)\right )}{3 d x^3}+\frac{b c \left (2 c^2 d-3 e\right ) \tanh ^{-1}\left (\frac{\sqrt{d+e x^2}}{\sqrt{d}}\right )}{6 \sqrt{d}}-\frac{b \left (c^2 d-e\right )^{3/2} \tanh ^{-1}\left (\frac{c \sqrt{d+e x^2}}{\sqrt{c^2 d-e}}\right )}{3 d}\\ \end{align*}

Mathematica [C]  time = 0.637386, size = 288, normalized size = 2.1 \[ -\frac{\sqrt{d+e x^2} \left (2 a \left (d+e x^2\right )+b c d x\right )+b c \sqrt{d} x^3 \log (x) \left (2 c^2 d-3 e\right )-b c \sqrt{d} x^3 \left (2 c^2 d-3 e\right ) \log \left (\sqrt{d} \sqrt{d+e x^2}+d\right )+b x^3 \left (c^2 d-e\right )^{3/2} \log \left (\frac{12 c d \left (\sqrt{c^2 d-e} \sqrt{d+e x^2}+c d-i e x\right )}{b (c x+i) \left (c^2 d-e\right )^{5/2}}\right )+b x^3 \left (c^2 d-e\right )^{3/2} \log \left (\frac{12 c d \left (\sqrt{c^2 d-e} \sqrt{d+e x^2}+c d+i e x\right )}{b (c x-i) \left (c^2 d-e\right )^{5/2}}\right )+2 b \tan ^{-1}(c x) \left (d+e x^2\right )^{3/2}}{6 d x^3} \]

Antiderivative was successfully verified.

[In]

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

[Out]

-(Sqrt[d + e*x^2]*(b*c*d*x + 2*a*(d + e*x^2)) + 2*b*(d + e*x^2)^(3/2)*ArcTan[c*x] + b*c*Sqrt[d]*(2*c^2*d - 3*e
)*x^3*Log[x] - b*c*Sqrt[d]*(2*c^2*d - 3*e)*x^3*Log[d + Sqrt[d]*Sqrt[d + e*x^2]] + b*(c^2*d - e)^(3/2)*x^3*Log[
(12*c*d*(c*d - I*e*x + Sqrt[c^2*d - e]*Sqrt[d + e*x^2]))/(b*(c^2*d - e)^(5/2)*(I + c*x))] + b*(c^2*d - e)^(3/2
)*x^3*Log[(12*c*d*(c*d + I*e*x + Sqrt[c^2*d - e]*Sqrt[d + e*x^2]))/(b*(c^2*d - e)^(5/2)*(-I + c*x))])/(6*d*x^3
)

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Maple [F]  time = 0.829, size = 0, normalized size = 0. \begin{align*} \int{\frac{a+b\arctan \left ( cx \right ) }{{x}^{4}}\sqrt{e{x}^{2}+d}}\, dx \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

int((e*x^2+d)^(1/2)*(a+b*arctan(c*x))/x^4,x)

[Out]

int((e*x^2+d)^(1/2)*(a+b*arctan(c*x))/x^4,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((e*x^2+d)^(1/2)*(a+b*arctan(c*x))/x^4,x, algorithm="maxima")

[Out]

Exception raised: ValueError

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Fricas [A]  time = 3.74001, size = 1940, normalized size = 14.16 \begin{align*} \text{result too large to display} \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate((e*x^2+d)^(1/2)*(a+b*arctan(c*x))/x^4,x, algorithm="fricas")

[Out]

[-1/12*((b*c^2*d - b*e)*sqrt(c^2*d - e)*x^3*log((c^4*e^2*x^4 + 8*c^4*d^2 - 8*c^2*d*e + 2*(4*c^4*d*e - 3*c^2*e^
2)*x^2 + 4*(c^3*e*x^2 + 2*c^3*d - c*e)*sqrt(c^2*d - e)*sqrt(e*x^2 + d) + e^2)/(c^4*x^4 + 2*c^2*x^2 + 1)) + (2*
b*c^3*d - 3*b*c*e)*sqrt(d)*x^3*log(-(e*x^2 - 2*sqrt(e*x^2 + d)*sqrt(d) + 2*d)/x^2) + 2*(b*c*d*x + 2*a*e*x^2 +
2*a*d + 2*(b*e*x^2 + b*d)*arctan(c*x))*sqrt(e*x^2 + d))/(d*x^3), -1/12*(2*(b*c^2*d - b*e)*sqrt(-c^2*d + e)*x^3
*arctan(-1/2*(c^2*e*x^2 + 2*c^2*d - e)*sqrt(-c^2*d + e)*sqrt(e*x^2 + d)/(c^3*d^2 - c*d*e + (c^3*d*e - c*e^2)*x
^2)) + (2*b*c^3*d - 3*b*c*e)*sqrt(d)*x^3*log(-(e*x^2 - 2*sqrt(e*x^2 + d)*sqrt(d) + 2*d)/x^2) + 2*(b*c*d*x + 2*
a*e*x^2 + 2*a*d + 2*(b*e*x^2 + b*d)*arctan(c*x))*sqrt(e*x^2 + d))/(d*x^3), -1/12*(2*(2*b*c^3*d - 3*b*c*e)*sqrt
(-d)*x^3*arctan(sqrt(-d)/sqrt(e*x^2 + d)) + (b*c^2*d - b*e)*sqrt(c^2*d - e)*x^3*log((c^4*e^2*x^4 + 8*c^4*d^2 -
8*c^2*d*e + 2*(4*c^4*d*e - 3*c^2*e^2)*x^2 + 4*(c^3*e*x^2 + 2*c^3*d - c*e)*sqrt(c^2*d - e)*sqrt(e*x^2 + d) + e
^2)/(c^4*x^4 + 2*c^2*x^2 + 1)) + 2*(b*c*d*x + 2*a*e*x^2 + 2*a*d + 2*(b*e*x^2 + b*d)*arctan(c*x))*sqrt(e*x^2 +
d))/(d*x^3), -1/6*((b*c^2*d - b*e)*sqrt(-c^2*d + e)*x^3*arctan(-1/2*(c^2*e*x^2 + 2*c^2*d - e)*sqrt(-c^2*d + e)
*sqrt(e*x^2 + d)/(c^3*d^2 - c*d*e + (c^3*d*e - c*e^2)*x^2)) + (2*b*c^3*d - 3*b*c*e)*sqrt(-d)*x^3*arctan(sqrt(-
d)/sqrt(e*x^2 + d)) + (b*c*d*x + 2*a*e*x^2 + 2*a*d + 2*(b*e*x^2 + b*d)*arctan(c*x))*sqrt(e*x^2 + d))/(d*x^3)]

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Sympy [F]  time = 0., size = 0, normalized size = 0. \begin{align*} \int \frac{\left (a + b \operatorname{atan}{\left (c x \right )}\right ) \sqrt{d + e x^{2}}}{x^{4}}\, dx \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate((e*x**2+d)**(1/2)*(a+b*atan(c*x))/x**4,x)

[Out]

Integral((a + b*atan(c*x))*sqrt(d + e*x**2)/x**4, x)

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Giac [F]  time = 0., size = 0, normalized size = 0. \begin{align*} \int \frac{\sqrt{e x^{2} + d}{\left (b \arctan \left (c x\right ) + a\right )}}{x^{4}}\,{d x} \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate((e*x^2+d)^(1/2)*(a+b*arctan(c*x))/x^4,x, algorithm="giac")

[Out]

integrate(sqrt(e*x^2 + d)*(b*arctan(c*x) + a)/x^4, x)