3.3.48 \(\int \frac {x}{(d+e x) (a+c x^2)^{3/2}} \, dx\)

Optimal. Leaf size=88 \[ \frac {d e \tanh ^{-1}\left (\frac {a e-c d x}{\sqrt {a+c x^2} \sqrt {a e^2+c d^2}}\right )}{\left (a e^2+c d^2\right )^{3/2}}-\frac {d-e x}{\sqrt {a+c x^2} \left (a e^2+c d^2\right )} \]

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Rubi [A]  time = 0.05, antiderivative size = 88, normalized size of antiderivative = 1.00, number of steps used = 4, number of rules used = 4, integrand size = 20, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.200, Rules used = {823, 12, 725, 206} \begin {gather*} \frac {d e \tanh ^{-1}\left (\frac {a e-c d x}{\sqrt {a+c x^2} \sqrt {a e^2+c d^2}}\right )}{\left (a e^2+c d^2\right )^{3/2}}-\frac {d-e x}{\sqrt {a+c x^2} \left (a e^2+c d^2\right )} \end {gather*}

Antiderivative was successfully verified.

[In]

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

[Out]

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

Rule 12

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

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 725

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 823

Int[((d_.) + (e_.)*(x_))^(m_)*((f_.) + (g_.)*(x_))*((a_) + (c_.)*(x_)^2)^(p_), x_Symbol] :> -Simp[((d + e*x)^(
m + 1)*(f*a*c*e - a*g*c*d + c*(c*d*f + a*e*g)*x)*(a + c*x^2)^(p + 1))/(2*a*c*(p + 1)*(c*d^2 + a*e^2)), x] + Di
st[1/(2*a*c*(p + 1)*(c*d^2 + a*e^2)), Int[(d + e*x)^m*(a + c*x^2)^(p + 1)*Simp[f*(c^2*d^2*(2*p + 3) + a*c*e^2*
(m + 2*p + 3)) - a*c*d*e*g*m + c*e*(c*d*f + a*e*g)*(m + 2*p + 4)*x, x], x], x] /; FreeQ[{a, c, d, e, f, g}, x]
&& NeQ[c*d^2 + a*e^2, 0] && LtQ[p, -1] && (IntegerQ[m] || IntegerQ[p] || IntegersQ[2*m, 2*p])

Rubi steps

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

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Mathematica [A]  time = 0.05, size = 88, normalized size = 1.00 \begin {gather*} \frac {e x-d}{\sqrt {a+c x^2} \left (a e^2+c d^2\right )}+\frac {d e \tanh ^{-1}\left (\frac {a e-c d x}{\sqrt {a+c x^2} \sqrt {a e^2+c d^2}}\right )}{\left (a e^2+c d^2\right )^{3/2}} \end {gather*}

Antiderivative was successfully verified.

[In]

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

[Out]

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

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IntegrateAlgebraic [A]  time = 0.53, size = 149, normalized size = 1.69 \begin {gather*} \frac {e x-d}{\sqrt {a+c x^2} \left (a e^2+c d^2\right )}-\frac {2 d e \sqrt {-a e^2-c d^2} \tan ^{-1}\left (-\frac {e \sqrt {a+c x^2}}{\sqrt {-a e^2-c d^2}}+\frac {\sqrt {c} e x}{\sqrt {-a e^2-c d^2}}+\frac {\sqrt {c} d}{\sqrt {-a e^2-c d^2}}\right )}{\left (a e^2+c d^2\right )^2} \end {gather*}

Antiderivative was successfully verified.

[In]

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

[Out]

(-d + e*x)/((c*d^2 + a*e^2)*Sqrt[a + c*x^2]) - (2*d*e*Sqrt[-(c*d^2) - a*e^2]*ArcTan[(Sqrt[c]*d)/Sqrt[-(c*d^2)
- a*e^2] + (Sqrt[c]*e*x)/Sqrt[-(c*d^2) - a*e^2] - (e*Sqrt[a + c*x^2])/Sqrt[-(c*d^2) - a*e^2]])/(c*d^2 + a*e^2)
^2

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fricas [B]  time = 0.48, size = 425, normalized size = 4.83 \begin {gather*} \left [\frac {{\left (c d e x^{2} + a d e\right )} \sqrt {c d^{2} + a e^{2}} \log \left (\frac {2 \, a c d e x - a c d^{2} - 2 \, a^{2} e^{2} - {\left (2 \, c^{2} d^{2} + a c e^{2}\right )} x^{2} + 2 \, \sqrt {c d^{2} + a e^{2}} {\left (c d x - a e\right )} \sqrt {c x^{2} + a}}{e^{2} x^{2} + 2 \, d e x + d^{2}}\right ) - 2 \, {\left (c d^{3} + a d e^{2} - {\left (c d^{2} e + a e^{3}\right )} x\right )} \sqrt {c x^{2} + a}}{2 \, {\left (a c^{2} d^{4} + 2 \, a^{2} c d^{2} e^{2} + a^{3} e^{4} + {\left (c^{3} d^{4} + 2 \, a c^{2} d^{2} e^{2} + a^{2} c e^{4}\right )} x^{2}\right )}}, \frac {{\left (c d e x^{2} + a d e\right )} \sqrt {-c d^{2} - a e^{2}} \arctan \left (\frac {\sqrt {-c d^{2} - a e^{2}} {\left (c d x - a e\right )} \sqrt {c x^{2} + a}}{a c d^{2} + a^{2} e^{2} + {\left (c^{2} d^{2} + a c e^{2}\right )} x^{2}}\right ) - {\left (c d^{3} + a d e^{2} - {\left (c d^{2} e + a e^{3}\right )} x\right )} \sqrt {c x^{2} + a}}{a c^{2} d^{4} + 2 \, a^{2} c d^{2} e^{2} + a^{3} e^{4} + {\left (c^{3} d^{4} + 2 \, a c^{2} d^{2} e^{2} + a^{2} c e^{4}\right )} x^{2}}\right ] \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

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

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giac [A]  time = 0.21, size = 162, normalized size = 1.84 \begin {gather*} \frac {2 \, d \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}{{\left (c d^{2} + a e^{2}\right )} \sqrt {-c d^{2} - a e^{2}}} + \frac {\frac {{\left (c d^{2} e + a e^{3}\right )} x}{c^{2} d^{4} + 2 \, a c d^{2} e^{2} + a^{2} e^{4}} - \frac {c d^{3} + a d e^{2}}{c^{2} d^{4} + 2 \, a c d^{2} e^{2} + a^{2} e^{4}}}{\sqrt {c x^{2} + a}} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

2*d*arctan(((sqrt(c)*x - sqrt(c*x^2 + a))*e + sqrt(c)*d)/sqrt(-c*d^2 - a*e^2))*e/((c*d^2 + a*e^2)*sqrt(-c*d^2
- a*e^2)) + ((c*d^2*e + a*e^3)*x/(c^2*d^4 + 2*a*c*d^2*e^2 + a^2*e^4) - (c*d^3 + a*d*e^2)/(c^2*d^4 + 2*a*c*d^2*
e^2 + a^2*e^4))/sqrt(c*x^2 + a)

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maple [B]  time = 0.01, size = 283, normalized size = 3.22 \begin {gather*} -\frac {c \,d^{2} x}{\left (a \,e^{2}+c \,d^{2}\right ) \sqrt {-\frac {2 \left (x +\frac {d}{e}\right ) c d}{e}+\left (x +\frac {d}{e}\right )^{2} c +\frac {a \,e^{2}+c \,d^{2}}{e^{2}}}\, a e}+\frac {d \ln \left (\frac {-\frac {2 \left (x +\frac {d}{e}\right ) c d}{e}+\frac {2 a \,e^{2}+2 c \,d^{2}}{e^{2}}+2 \sqrt {\frac {a \,e^{2}+c \,d^{2}}{e^{2}}}\, \sqrt {-\frac {2 \left (x +\frac {d}{e}\right ) c d}{e}+\left (x +\frac {d}{e}\right )^{2} c +\frac {a \,e^{2}+c \,d^{2}}{e^{2}}}}{x +\frac {d}{e}}\right )}{\left (a \,e^{2}+c \,d^{2}\right ) \sqrt {\frac {a \,e^{2}+c \,d^{2}}{e^{2}}}}-\frac {d}{\left (a \,e^{2}+c \,d^{2}\right ) \sqrt {-\frac {2 \left (x +\frac {d}{e}\right ) c d}{e}+\left (x +\frac {d}{e}\right )^{2} c +\frac {a \,e^{2}+c \,d^{2}}{e^{2}}}}+\frac {x}{\sqrt {c \,x^{2}+a}\, a e} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

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

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maxima [A]  time = 0.54, size = 148, normalized size = 1.68 \begin {gather*} -\frac {c d^{2} x}{\sqrt {c x^{2} + a} a c d^{2} e + \sqrt {c x^{2} + a} a^{2} e^{3}} - \frac {d}{\sqrt {c x^{2} + a} c d^{2} + \sqrt {c x^{2} + a} a e^{2}} + \frac {x}{\sqrt {c x^{2} + a} a e} - \frac {d \operatorname {arsinh}\left (\frac {c d x}{\sqrt {a c} {\left | e x + d \right |}} - \frac {a e}{\sqrt {a c} {\left | e x + d \right |}}\right )}{{\left (a + \frac {c d^{2}}{e^{2}}\right )}^{\frac {3}{2}} e^{2}} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

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

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

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

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

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

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

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

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