3.4.4 \(\int \frac {1}{(d+e x) (b x+c x^2)^{3/2}} \, dx\)

Optimal. Leaf size=126 \[ \frac {e^2 \tanh ^{-1}\left (\frac {x (2 c d-b e)+b d}{2 \sqrt {d} \sqrt {b x+c x^2} \sqrt {c d-b e}}\right )}{d^{3/2} (c d-b e)^{3/2}}-\frac {2 (c x (2 c d-b e)+b (c d-b e))}{b^2 d \sqrt {b x+c x^2} (c d-b e)} \]

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Rubi [A]  time = 0.10, antiderivative size = 126, normalized size of antiderivative = 1.00, number of steps used = 4, number of rules used = 4, integrand size = 21, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.190, Rules used = {740, 12, 724, 206} \begin {gather*} \frac {e^2 \tanh ^{-1}\left (\frac {x (2 c d-b e)+b d}{2 \sqrt {d} \sqrt {b x+c x^2} \sqrt {c d-b e}}\right )}{d^{3/2} (c d-b e)^{3/2}}-\frac {2 (c x (2 c d-b e)+b (c d-b e))}{b^2 d \sqrt {b x+c x^2} (c d-b e)} \end {gather*}

Antiderivative was successfully verified.

[In]

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

[Out]

(-2*(b*(c*d - b*e) + c*(2*c*d - b*e)*x))/(b^2*d*(c*d - b*e)*Sqrt[b*x + c*x^2]) + (e^2*ArcTanh[(b*d + (2*c*d -
b*e)*x)/(2*Sqrt[d]*Sqrt[c*d - b*e]*Sqrt[b*x + c*x^2])])/(d^(3/2)*(c*d - b*e)^(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 724

Int[1/(((d_.) + (e_.)*(x_))*Sqrt[(a_.) + (b_.)*(x_) + (c_.)*(x_)^2]), x_Symbol] :> Dist[-2, Subst[Int[1/(4*c*d
^2 - 4*b*d*e + 4*a*e^2 - x^2), x], x, (2*a*e - b*d - (2*c*d - b*e)*x)/Sqrt[a + b*x + c*x^2]], x] /; FreeQ[{a,
b, c, d, e}, x] && NeQ[b^2 - 4*a*c, 0] && NeQ[2*c*d - b*e, 0]

Rule 740

Int[((d_.) + (e_.)*(x_))^(m_)*((a_.) + (b_.)*(x_) + (c_.)*(x_)^2)^(p_), x_Symbol] :> Simp[((d + e*x)^(m + 1)*(
b*c*d - b^2*e + 2*a*c*e + c*(2*c*d - b*e)*x)*(a + b*x + c*x^2)^(p + 1))/((p + 1)*(b^2 - 4*a*c)*(c*d^2 - b*d*e
+ a*e^2)), x] + Dist[1/((p + 1)*(b^2 - 4*a*c)*(c*d^2 - b*d*e + a*e^2)), Int[(d + e*x)^m*Simp[b*c*d*e*(2*p - m
+ 2) + b^2*e^2*(m + p + 2) - 2*c^2*d^2*(2*p + 3) - 2*a*c*e^2*(m + 2*p + 3) - c*e*(2*c*d - b*e)*(m + 2*p + 4)*x
, x]*(a + b*x + c*x^2)^(p + 1), x], x] /; FreeQ[{a, b, c, d, e, m}, x] && NeQ[b^2 - 4*a*c, 0] && NeQ[c*d^2 - b
*d*e + a*e^2, 0] && NeQ[2*c*d - b*e, 0] && LtQ[p, -1] && IntQuadraticQ[a, b, c, d, e, m, p, x]

Rubi steps

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

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Mathematica [A]  time = 0.11, size = 129, normalized size = 1.02 \begin {gather*} \frac {2 \left (\sqrt {d} \sqrt {c d-b e} \left (b^2 e-b c d+b c e x-2 c^2 d x\right )+b^2 e^2 \sqrt {x} \sqrt {b+c x} \tanh ^{-1}\left (\frac {\sqrt {x} \sqrt {c d-b e}}{\sqrt {d} \sqrt {b+c x}}\right )\right )}{b^2 d^{3/2} \sqrt {x (b+c x)} (c d-b e)^{3/2}} \end {gather*}

Antiderivative was successfully verified.

[In]

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

[Out]

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

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IntegrateAlgebraic [A]  time = 0.61, size = 170, normalized size = 1.35 \begin {gather*} \frac {2 e^2 \tanh ^{-1}\left (-\frac {e \sqrt {b x+c x^2}}{\sqrt {d} \sqrt {c d-b e}}+\frac {\sqrt {c} e x}{\sqrt {d} \sqrt {c d-b e}}+\frac {\sqrt {c} \sqrt {d}}{\sqrt {c d-b e}}\right )}{d^{3/2} (c d-b e)^{3/2}}-\frac {2 \sqrt {b x+c x^2} \left (b^2 e-b c d+b c e x-2 c^2 d x\right )}{b^2 d x (b+c x) (b e-c d)} \end {gather*}

Antiderivative was successfully verified.

[In]

IntegrateAlgebraic[1/((d + e*x)*(b*x + c*x^2)^(3/2)),x]

[Out]

(-2*(-(b*c*d) + b^2*e - 2*c^2*d*x + b*c*e*x)*Sqrt[b*x + c*x^2])/(b^2*d*(-(c*d) + b*e)*x*(b + c*x)) + (2*e^2*Ar
cTanh[(Sqrt[c]*Sqrt[d])/Sqrt[c*d - b*e] + (Sqrt[c]*e*x)/(Sqrt[d]*Sqrt[c*d - b*e]) - (e*Sqrt[b*x + c*x^2])/(Sqr
t[d]*Sqrt[c*d - b*e])])/(d^(3/2)*(c*d - b*e)^(3/2))

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fricas [A]  time = 0.43, size = 453, normalized size = 3.60 \begin {gather*} \left [-\frac {{\left (b^{2} c e^{2} x^{2} + b^{3} e^{2} x\right )} \sqrt {c d^{2} - b d e} \log \left (\frac {b d + {\left (2 \, c d - b e\right )} x - 2 \, \sqrt {c d^{2} - b d e} \sqrt {c x^{2} + b x}}{e x + d}\right ) + 2 \, {\left (b c^{2} d^{3} - 2 \, b^{2} c d^{2} e + b^{3} d e^{2} + {\left (2 \, c^{3} d^{3} - 3 \, b c^{2} d^{2} e + b^{2} c d e^{2}\right )} x\right )} \sqrt {c x^{2} + b x}}{{\left (b^{2} c^{3} d^{4} - 2 \, b^{3} c^{2} d^{3} e + b^{4} c d^{2} e^{2}\right )} x^{2} + {\left (b^{3} c^{2} d^{4} - 2 \, b^{4} c d^{3} e + b^{5} d^{2} e^{2}\right )} x}, \frac {2 \, {\left ({\left (b^{2} c e^{2} x^{2} + b^{3} e^{2} x\right )} \sqrt {-c d^{2} + b d e} \arctan \left (-\frac {\sqrt {-c d^{2} + b d e} \sqrt {c x^{2} + b x}}{{\left (c d - b e\right )} x}\right ) - {\left (b c^{2} d^{3} - 2 \, b^{2} c d^{2} e + b^{3} d e^{2} + {\left (2 \, c^{3} d^{3} - 3 \, b c^{2} d^{2} e + b^{2} c d e^{2}\right )} x\right )} \sqrt {c x^{2} + b x}\right )}}{{\left (b^{2} c^{3} d^{4} - 2 \, b^{3} c^{2} d^{3} e + b^{4} c d^{2} e^{2}\right )} x^{2} + {\left (b^{3} c^{2} d^{4} - 2 \, b^{4} c d^{3} e + b^{5} d^{2} e^{2}\right )} x}\right ] \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

[-((b^2*c*e^2*x^2 + b^3*e^2*x)*sqrt(c*d^2 - b*d*e)*log((b*d + (2*c*d - b*e)*x - 2*sqrt(c*d^2 - b*d*e)*sqrt(c*x
^2 + b*x))/(e*x + d)) + 2*(b*c^2*d^3 - 2*b^2*c*d^2*e + b^3*d*e^2 + (2*c^3*d^3 - 3*b*c^2*d^2*e + b^2*c*d*e^2)*x
)*sqrt(c*x^2 + b*x))/((b^2*c^3*d^4 - 2*b^3*c^2*d^3*e + b^4*c*d^2*e^2)*x^2 + (b^3*c^2*d^4 - 2*b^4*c*d^3*e + b^5
*d^2*e^2)*x), 2*((b^2*c*e^2*x^2 + b^3*e^2*x)*sqrt(-c*d^2 + b*d*e)*arctan(-sqrt(-c*d^2 + b*d*e)*sqrt(c*x^2 + b*
x)/((c*d - b*e)*x)) - (b*c^2*d^3 - 2*b^2*c*d^2*e + b^3*d*e^2 + (2*c^3*d^3 - 3*b*c^2*d^2*e + b^2*c*d*e^2)*x)*sq
rt(c*x^2 + b*x))/((b^2*c^3*d^4 - 2*b^3*c^2*d^3*e + b^4*c*d^2*e^2)*x^2 + (b^3*c^2*d^4 - 2*b^4*c*d^3*e + b^5*d^2
*e^2)*x)]

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giac [A]  time = 0.24, size = 167, normalized size = 1.33 \begin {gather*} -\frac {2 \, {\left (\frac {{\left (2 \, c^{2} d^{2} - b c d e\right )} x}{b^{2} c d^{3} - b^{3} d^{2} e} + \frac {b c d^{2} - b^{2} d e}{b^{2} c d^{3} - b^{3} d^{2} e}\right )}}{\sqrt {c x^{2} + b x}} - \frac {2 \, \arctan \left (\frac {{\left (\sqrt {c} x - \sqrt {c x^{2} + b x}\right )} e + \sqrt {c} d}{\sqrt {-c d^{2} + b d e}}\right ) e^{2}}{{\left (c d^{2} - b d e\right )} \sqrt {-c d^{2} + b d e}} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

-2*((2*c^2*d^2 - b*c*d*e)*x/(b^2*c*d^3 - b^3*d^2*e) + (b*c*d^2 - b^2*d*e)/(b^2*c*d^3 - b^3*d^2*e))/sqrt(c*x^2
+ b*x) - 2*arctan(((sqrt(c)*x - sqrt(c*x^2 + b*x))*e + sqrt(c)*d)/sqrt(-c*d^2 + b*d*e))*e^2/((c*d^2 - b*d*e)*s
qrt(-c*d^2 + b*d*e))

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maple [B]  time = 0.06, size = 403, normalized size = 3.20 \begin {gather*} -\frac {2 c e x}{\left (b e -c d \right ) \sqrt {\left (x +\frac {d}{e}\right )^{2} c -\frac {\left (b e -c d \right ) d}{e^{2}}+\frac {\left (b e -2 c d \right ) \left (x +\frac {d}{e}\right )}{e}}\, b d}+\frac {4 c^{2} x}{\left (b e -c d \right ) \sqrt {\left (x +\frac {d}{e}\right )^{2} c -\frac {\left (b e -c d \right ) d}{e^{2}}+\frac {\left (b e -2 c d \right ) \left (x +\frac {d}{e}\right )}{e}}\, b^{2}}+\frac {e \ln \left (\frac {-\frac {2 \left (b e -c d \right ) d}{e^{2}}+\frac {\left (b e -2 c d \right ) \left (x +\frac {d}{e}\right )}{e}+2 \sqrt {-\frac {\left (b e -c d \right ) d}{e^{2}}}\, \sqrt {\left (x +\frac {d}{e}\right )^{2} c -\frac {\left (b e -c d \right ) d}{e^{2}}+\frac {\left (b e -2 c d \right ) \left (x +\frac {d}{e}\right )}{e}}}{x +\frac {d}{e}}\right )}{\left (b e -c d \right ) \sqrt {-\frac {\left (b e -c d \right ) d}{e^{2}}}\, d}+\frac {2 c}{\left (b e -c d \right ) \sqrt {\left (x +\frac {d}{e}\right )^{2} c -\frac {\left (b e -c d \right ) d}{e^{2}}+\frac {\left (b e -2 c d \right ) \left (x +\frac {d}{e}\right )}{e}}\, b}-\frac {2 e}{\left (b e -c d \right ) \sqrt {\left (x +\frac {d}{e}\right )^{2} c -\frac {\left (b e -c d \right ) d}{e^{2}}+\frac {\left (b e -2 c d \right ) \left (x +\frac {d}{e}\right )}{e}}\, d} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

-2*e/(b*e-c*d)/d/((x+d/e)^2*c-(b*e-c*d)*d/e^2+(b*e-2*c*d)*(x+d/e)/e)^(1/2)-2*e/(b*e-c*d)/d/b/((x+d/e)^2*c-(b*e
-c*d)*d/e^2+(b*e-2*c*d)*(x+d/e)/e)^(1/2)*x*c+4/(b*e-c*d)/b^2/((x+d/e)^2*c-(b*e-c*d)*d/e^2+(b*e-2*c*d)*(x+d/e)/
e)^(1/2)*x*c^2+2/(b*e-c*d)/b/((x+d/e)^2*c-(b*e-c*d)*d/e^2+(b*e-2*c*d)*(x+d/e)/e)^(1/2)*c+e/(b*e-c*d)/d/(-(b*e-
c*d)*d/e^2)^(1/2)*ln((-2*(b*e-c*d)*d/e^2+(b*e-2*c*d)*(x+d/e)/e+2*(-(b*e-c*d)*d/e^2)^(1/2)*((x+d/e)^2*c-(b*e-c*
d)*d/e^2+(b*e-2*c*d)*(x+d/e)/e)^(1/2))/(x+d/e))

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maxima [F(-2)]  time = 0.00, size = 0, normalized size = 0.00 \begin {gather*} \text {Exception raised: ValueError} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

Exception raised: ValueError >> Computation failed since Maxima requested additional constraints; using the 'a
ssume' command before evaluation *may* help (example of legal syntax is 'assume(b*e-c*d>0)', see `assume?` for
more details)Is b*e-c*d positive, negative or zero?

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

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

int(1/((b*x + 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 {1}{\left (x \left (b + c x\right )\right )^{\frac {3}{2}} \left (d + e x\right )}\, dx \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

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

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