3.4.33 \(\int \frac {1}{x^3 (d+e x) \sqrt {a+c x^2}} \, dx\) [333]

Optimal. Leaf size=168 \[ -\frac {\sqrt {a+c x^2}}{2 a d x^2}+\frac {e \sqrt {a+c x^2}}{a d^2 x}+\frac {e^3 \tanh ^{-1}\left (\frac {a e-c d x}{\sqrt {c d^2+a e^2} \sqrt {a+c x^2}}\right )}{d^3 \sqrt {c d^2+a e^2}}+\frac {c \tanh ^{-1}\left (\frac {\sqrt {a+c x^2}}{\sqrt {a}}\right )}{2 a^{3/2} d}-\frac {e^2 \tanh ^{-1}\left (\frac {\sqrt {a+c x^2}}{\sqrt {a}}\right )}{\sqrt {a} d^3} \]

[Out]

1/2*c*arctanh((c*x^2+a)^(1/2)/a^(1/2))/a^(3/2)/d-e^2*arctanh((c*x^2+a)^(1/2)/a^(1/2))/d^3/a^(1/2)+e^3*arctanh(
(-c*d*x+a*e)/(a*e^2+c*d^2)^(1/2)/(c*x^2+a)^(1/2))/d^3/(a*e^2+c*d^2)^(1/2)-1/2*(c*x^2+a)^(1/2)/a/d/x^2+e*(c*x^2
+a)^(1/2)/a/d^2/x

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Rubi [A]
time = 0.10, antiderivative size = 168, normalized size of antiderivative = 1.00, number of steps used = 12, number of rules used = 8, integrand size = 22, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.364, Rules used = {975, 272, 44, 65, 214, 270, 739, 212} \begin {gather*} \frac {c \tanh ^{-1}\left (\frac {\sqrt {a+c x^2}}{\sqrt {a}}\right )}{2 a^{3/2} d}-\frac {e^2 \tanh ^{-1}\left (\frac {\sqrt {a+c x^2}}{\sqrt {a}}\right )}{\sqrt {a} d^3}+\frac {e \sqrt {a+c x^2}}{a d^2 x}+\frac {e^3 \tanh ^{-1}\left (\frac {a e-c d x}{\sqrt {a+c x^2} \sqrt {a e^2+c d^2}}\right )}{d^3 \sqrt {a e^2+c d^2}}-\frac {\sqrt {a+c x^2}}{2 a d x^2} \end {gather*}

Antiderivative was successfully verified.

[In]

Int[1/(x^3*(d + e*x)*Sqrt[a + c*x^2]),x]

[Out]

-1/2*Sqrt[a + c*x^2]/(a*d*x^2) + (e*Sqrt[a + c*x^2])/(a*d^2*x) + (e^3*ArcTanh[(a*e - c*d*x)/(Sqrt[c*d^2 + a*e^
2]*Sqrt[a + c*x^2])])/(d^3*Sqrt[c*d^2 + a*e^2]) + (c*ArcTanh[Sqrt[a + c*x^2]/Sqrt[a]])/(2*a^(3/2)*d) - (e^2*Ar
cTanh[Sqrt[a + c*x^2]/Sqrt[a]])/(Sqrt[a]*d^3)

Rule 44

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

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 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 270

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 272

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

Rule 739

Int[1/(((d_) + (e_.)*(x_))*Sqrt[(a_) + (c_.)*(x_)^2]), x_Symbol] :> -Subst[Int[1/(c*d^2 + a*e^2 - x^2), x], x,
(a*e - c*d*x)/Sqrt[a + c*x^2]] /; FreeQ[{a, c, d, e}, x]

Rule 975

Int[((d_.) + (e_.)*(x_))^(m_)*((f_.) + (g_.)*(x_))^(n_)*((a_) + (c_.)*(x_)^2)^(p_.), x_Symbol] :> Int[ExpandIn
tegrand[(d + e*x)^m*(f + g*x)^n*(a + c*x^2)^p, x], x] /; FreeQ[{a, c, d, e, f, g}, x] && NeQ[e*f - d*g, 0] &&
NeQ[c*d^2 + a*e^2, 0] && (IntegerQ[p] || (ILtQ[m, 0] && ILtQ[n, 0])) &&  !(IGtQ[m, 0] || IGtQ[n, 0])

Rubi steps

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

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Mathematica [A]
time = 0.51, size = 198, normalized size = 1.18 \begin {gather*} \frac {\sqrt {a} \left (d \left (c d^2+a e^2\right ) (-d+2 e x) \sqrt {a+c x^2}-4 a e^3 \sqrt {-c d^2-a e^2} x^2 \tan ^{-1}\left (\frac {\sqrt {c} (d+e x)-e \sqrt {a+c x^2}}{\sqrt {-c d^2-a e^2}}\right )\right )-2 \left (c^2 d^4-a c d^2 e^2-2 a^2 e^4\right ) x^2 \tanh ^{-1}\left (\frac {\sqrt {c} x-\sqrt {a+c x^2}}{\sqrt {a}}\right )}{2 a^{3/2} d^3 \left (c d^2+a e^2\right ) x^2} \end {gather*}

Antiderivative was successfully verified.

[In]

Integrate[1/(x^3*(d + e*x)*Sqrt[a + c*x^2]),x]

[Out]

(Sqrt[a]*(d*(c*d^2 + a*e^2)*(-d + 2*e*x)*Sqrt[a + c*x^2] - 4*a*e^3*Sqrt[-(c*d^2) - a*e^2]*x^2*ArcTan[(Sqrt[c]*
(d + e*x) - e*Sqrt[a + c*x^2])/Sqrt[-(c*d^2) - a*e^2]]) - 2*(c^2*d^4 - a*c*d^2*e^2 - 2*a^2*e^4)*x^2*ArcTanh[(S
qrt[c]*x - Sqrt[a + c*x^2])/Sqrt[a]])/(2*a^(3/2)*d^3*(c*d^2 + a*e^2)*x^2)

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Maple [A]
time = 0.08, size = 235, normalized size = 1.40

method	result	size
risch	\(-\frac {\sqrt {c \,x^{2}+a}\, \left (-2 e x +d \right )}{2 a \,d^{2} x^{2}}+\frac {e^{2} \ln \left (\frac {\frac {2 a \,e^{2}+2 c \,d^{2}}{e^{2}}-\frac {2 c d \left (x +\frac {d}{e}\right )}{e}+2 \sqrt {\frac {a \,e^{2}+c \,d^{2}}{e^{2}}}\, \sqrt {c \left (x +\frac {d}{e}\right )^{2}-\frac {2 c d \left (x +\frac {d}{e}\right )}{e}+\frac {a \,e^{2}+c \,d^{2}}{e^{2}}}}{x +\frac {d}{e}}\right )}{d^{3} \sqrt {\frac {a \,e^{2}+c \,d^{2}}{e^{2}}}}-\frac {e^{2} \ln \left (\frac {2 a +2 \sqrt {a}\, \sqrt {c \,x^{2}+a}}{x}\right )}{d^{3} \sqrt {a}}+\frac {\ln \left (\frac {2 a +2 \sqrt {a}\, \sqrt {c \,x^{2}+a}}{x}\right ) c}{2 a^{\frac {3}{2}} d}\)	\(222\)
default	\(\frac {e^{2} \ln \left (\frac {\frac {2 a \,e^{2}+2 c \,d^{2}}{e^{2}}-\frac {2 c d \left (x +\frac {d}{e}\right )}{e}+2 \sqrt {\frac {a \,e^{2}+c \,d^{2}}{e^{2}}}\, \sqrt {c \left (x +\frac {d}{e}\right )^{2}-\frac {2 c d \left (x +\frac {d}{e}\right )}{e}+\frac {a \,e^{2}+c \,d^{2}}{e^{2}}}}{x +\frac {d}{e}}\right )}{d^{3} \sqrt {\frac {a \,e^{2}+c \,d^{2}}{e^{2}}}}+\frac {-\frac {\sqrt {c \,x^{2}+a}}{2 a \,x^{2}}+\frac {c \ln \left (\frac {2 a +2 \sqrt {a}\, \sqrt {c \,x^{2}+a}}{x}\right )}{2 a^{\frac {3}{2}}}}{d}+\frac {e \sqrt {c \,x^{2}+a}}{a \,d^{2} x}-\frac {e^{2} \ln \left (\frac {2 a +2 \sqrt {a}\, \sqrt {c \,x^{2}+a}}{x}\right )}{d^{3} \sqrt {a}}\)	\(235\)

Verification of antiderivative is not currently implemented for this CAS.

[In]

int(1/x^3/(e*x+d)/(c*x^2+a)^(1/2),x,method=_RETURNVERBOSE)

[Out]

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

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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(1/x^3/(e*x+d)/(c*x^2+a)^(1/2),x, algorithm="maxima")

[Out]

integrate(1/(sqrt(c*x^2 + a)*(x*e + d)*x^3), x)

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Fricas [A]
time = 3.57, size = 969, normalized size = 5.77 \begin {gather*} \left [\frac {2 \, \sqrt {c d^{2} + a e^{2}} a^{2} x^{2} e^{3} \log \left (-\frac {2 \, c^{2} d^{2} x^{2} - 2 \, a c d x e + a c d^{2} - 2 \, \sqrt {c d^{2} + a e^{2}} {\left (c d x - a e\right )} \sqrt {c x^{2} + a} + {\left (a c x^{2} + 2 \, a^{2}\right )} e^{2}}{x^{2} e^{2} + 2 \, d x e + d^{2}}\right ) - {\left (c^{2} d^{4} x^{2} - a c d^{2} x^{2} e^{2} - 2 \, a^{2} x^{2} e^{4}\right )} \sqrt {a} \log \left (-\frac {c x^{2} - 2 \, \sqrt {c x^{2} + a} \sqrt {a} + 2 \, a}{x^{2}}\right ) + 2 \, {\left (2 \, a c d^{3} x e - a c d^{4} + 2 \, a^{2} d x e^{3} - a^{2} d^{2} e^{2}\right )} \sqrt {c x^{2} + a}}{4 \, {\left (a^{2} c d^{5} x^{2} + a^{3} d^{3} x^{2} e^{2}\right )}}, -\frac {4 \, \sqrt {-c d^{2} - a e^{2}} a^{2} x^{2} \arctan \left (-\frac {\sqrt {-c d^{2} - a e^{2}} {\left (c d x - a e\right )} \sqrt {c x^{2} + a}}{c^{2} d^{2} x^{2} + a c d^{2} + {\left (a c x^{2} + a^{2}\right )} e^{2}}\right ) e^{3} + {\left (c^{2} d^{4} x^{2} - a c d^{2} x^{2} e^{2} - 2 \, a^{2} x^{2} e^{4}\right )} \sqrt {a} \log \left (-\frac {c x^{2} - 2 \, \sqrt {c x^{2} + a} \sqrt {a} + 2 \, a}{x^{2}}\right ) - 2 \, {\left (2 \, a c d^{3} x e - a c d^{4} + 2 \, a^{2} d x e^{3} - a^{2} d^{2} e^{2}\right )} \sqrt {c x^{2} + a}}{4 \, {\left (a^{2} c d^{5} x^{2} + a^{3} d^{3} x^{2} e^{2}\right )}}, \frac {\sqrt {c d^{2} + a e^{2}} a^{2} x^{2} e^{3} \log \left (-\frac {2 \, c^{2} d^{2} x^{2} - 2 \, a c d x e + a c d^{2} - 2 \, \sqrt {c d^{2} + a e^{2}} {\left (c d x - a e\right )} \sqrt {c x^{2} + a} + {\left (a c x^{2} + 2 \, a^{2}\right )} e^{2}}{x^{2} e^{2} + 2 \, d x e + d^{2}}\right ) - {\left (c^{2} d^{4} x^{2} - a c d^{2} x^{2} e^{2} - 2 \, a^{2} x^{2} e^{4}\right )} \sqrt {-a} \arctan \left (\frac {\sqrt {-a}}{\sqrt {c x^{2} + a}}\right ) + {\left (2 \, a c d^{3} x e - a c d^{4} + 2 \, a^{2} d x e^{3} - a^{2} d^{2} e^{2}\right )} \sqrt {c x^{2} + a}}{2 \, {\left (a^{2} c d^{5} x^{2} + a^{3} d^{3} x^{2} e^{2}\right )}}, -\frac {2 \, \sqrt {-c d^{2} - a e^{2}} a^{2} x^{2} \arctan \left (-\frac {\sqrt {-c d^{2} - a e^{2}} {\left (c d x - a e\right )} \sqrt {c x^{2} + a}}{c^{2} d^{2} x^{2} + a c d^{2} + {\left (a c x^{2} + a^{2}\right )} e^{2}}\right ) e^{3} + {\left (c^{2} d^{4} x^{2} - a c d^{2} x^{2} e^{2} - 2 \, a^{2} x^{2} e^{4}\right )} \sqrt {-a} \arctan \left (\frac {\sqrt {-a}}{\sqrt {c x^{2} + a}}\right ) - {\left (2 \, a c d^{3} x e - a c d^{4} + 2 \, a^{2} d x e^{3} - a^{2} d^{2} e^{2}\right )} \sqrt {c x^{2} + a}}{2 \, {\left (a^{2} c d^{5} x^{2} + a^{3} d^{3} x^{2} e^{2}\right )}}\right ] \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate(1/x^3/(e*x+d)/(c*x^2+a)^(1/2),x, algorithm="fricas")

[Out]

[1/4*(2*sqrt(c*d^2 + a*e^2)*a^2*x^2*e^3*log(-(2*c^2*d^2*x^2 - 2*a*c*d*x*e + a*c*d^2 - 2*sqrt(c*d^2 + a*e^2)*(c
*d*x - a*e)*sqrt(c*x^2 + a) + (a*c*x^2 + 2*a^2)*e^2)/(x^2*e^2 + 2*d*x*e + d^2)) - (c^2*d^4*x^2 - a*c*d^2*x^2*e
^2 - 2*a^2*x^2*e^4)*sqrt(a)*log(-(c*x^2 - 2*sqrt(c*x^2 + a)*sqrt(a) + 2*a)/x^2) + 2*(2*a*c*d^3*x*e - a*c*d^4 +
2*a^2*d*x*e^3 - a^2*d^2*e^2)*sqrt(c*x^2 + a))/(a^2*c*d^5*x^2 + a^3*d^3*x^2*e^2), -1/4*(4*sqrt(-c*d^2 - a*e^2)
*a^2*x^2*arctan(-sqrt(-c*d^2 - a*e^2)*(c*d*x - a*e)*sqrt(c*x^2 + a)/(c^2*d^2*x^2 + a*c*d^2 + (a*c*x^2 + a^2)*e
^2))*e^3 + (c^2*d^4*x^2 - a*c*d^2*x^2*e^2 - 2*a^2*x^2*e^4)*sqrt(a)*log(-(c*x^2 - 2*sqrt(c*x^2 + a)*sqrt(a) + 2
*a)/x^2) - 2*(2*a*c*d^3*x*e - a*c*d^4 + 2*a^2*d*x*e^3 - a^2*d^2*e^2)*sqrt(c*x^2 + a))/(a^2*c*d^5*x^2 + a^3*d^3
*x^2*e^2), 1/2*(sqrt(c*d^2 + a*e^2)*a^2*x^2*e^3*log(-(2*c^2*d^2*x^2 - 2*a*c*d*x*e + a*c*d^2 - 2*sqrt(c*d^2 + a
*e^2)*(c*d*x - a*e)*sqrt(c*x^2 + a) + (a*c*x^2 + 2*a^2)*e^2)/(x^2*e^2 + 2*d*x*e + d^2)) - (c^2*d^4*x^2 - a*c*d
^2*x^2*e^2 - 2*a^2*x^2*e^4)*sqrt(-a)*arctan(sqrt(-a)/sqrt(c*x^2 + a)) + (2*a*c*d^3*x*e - a*c*d^4 + 2*a^2*d*x*e
^3 - a^2*d^2*e^2)*sqrt(c*x^2 + a))/(a^2*c*d^5*x^2 + a^3*d^3*x^2*e^2), -1/2*(2*sqrt(-c*d^2 - a*e^2)*a^2*x^2*arc
tan(-sqrt(-c*d^2 - a*e^2)*(c*d*x - a*e)*sqrt(c*x^2 + a)/(c^2*d^2*x^2 + a*c*d^2 + (a*c*x^2 + a^2)*e^2))*e^3 + (
c^2*d^4*x^2 - a*c*d^2*x^2*e^2 - 2*a^2*x^2*e^4)*sqrt(-a)*arctan(sqrt(-a)/sqrt(c*x^2 + a)) - (2*a*c*d^3*x*e - a*
c*d^4 + 2*a^2*d*x*e^3 - a^2*d^2*e^2)*sqrt(c*x^2 + a))/(a^2*c*d^5*x^2 + a^3*d^3*x^2*e^2)]

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Sympy [F]
time = 0.00, size = 0, normalized size = 0.00 \begin {gather*} \int \frac {1}{x^{3} \sqrt {a + c x^{2}} \left (d + e x\right )}\, dx \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate(1/x**3/(e*x+d)/(c*x**2+a)**(1/2),x)

[Out]

Integral(1/(x**3*sqrt(a + c*x**2)*(d + e*x)), x)

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Giac [A]
time = 1.98, size = 239, normalized size = 1.42 \begin {gather*} -c^{\frac {3}{2}} {\left (\frac {2 \, \arctan \left (-\frac {{\left (\sqrt {c} x - \sqrt {c x^{2} + a}\right )} e + \sqrt {c} d}{\sqrt {-c d^{2} - a e^{2}}}\right ) e^{3}}{\sqrt {-c d^{2} - a e^{2}} c^{\frac {3}{2}} d^{3}} + \frac {{\left (c d^{2} - 2 \, a e^{2}\right )} \arctan \left (-\frac {\sqrt {c} x - \sqrt {c x^{2} + a}}{\sqrt {-a}}\right )}{\sqrt {-a} a c^{\frac {3}{2}} d^{3}} - \frac {{\left (\sqrt {c} x - \sqrt {c x^{2} + a}\right )}^{3} \sqrt {c} d - 2 \, {\left (\sqrt {c} x - \sqrt {c x^{2} + a}\right )}^{2} a e + {\left (\sqrt {c} x - \sqrt {c x^{2} + a}\right )} a \sqrt {c} d + 2 \, a^{2} e}{{\left ({\left (\sqrt {c} x - \sqrt {c x^{2} + a}\right )}^{2} - a\right )}^{2} a c d^{2}}\right )} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate(1/x^3/(e*x+d)/(c*x^2+a)^(1/2),x, algorithm="giac")

[Out]

-c^(3/2)*(2*arctan(-((sqrt(c)*x - sqrt(c*x^2 + a))*e + sqrt(c)*d)/sqrt(-c*d^2 - a*e^2))*e^3/(sqrt(-c*d^2 - a*e
^2)*c^(3/2)*d^3) + (c*d^2 - 2*a*e^2)*arctan(-(sqrt(c)*x - sqrt(c*x^2 + a))/sqrt(-a))/(sqrt(-a)*a*c^(3/2)*d^3)
- ((sqrt(c)*x - sqrt(c*x^2 + a))^3*sqrt(c)*d - 2*(sqrt(c)*x - sqrt(c*x^2 + a))^2*a*e + (sqrt(c)*x - sqrt(c*x^2
+ a))*a*sqrt(c)*d + 2*a^2*e)/(((sqrt(c)*x - sqrt(c*x^2 + a))^2 - a)^2*a*c*d^2))

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

Verification of antiderivative is not currently implemented for this CAS.

[In]

int(1/(x^3*(a + c*x^2)^(1/2)*(d + e*x)),x)

[Out]

int(1/(x^3*(a + c*x^2)^(1/2)*(d + e*x)), x)

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