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

Optimal. Leaf size=101 \[ \frac {2 e \sqrt {b x+c x^2} (2 c d-b e)}{b^2 c}-\frac {2 (d+e x) (x (2 c d-b e)+b d)}{b^2 \sqrt {b x+c x^2}}+\frac {2 e^2 \tanh ^{-1}\left (\frac {\sqrt {c} x}{\sqrt {b x+c x^2}}\right )}{c^{3/2}} \]

________________________________________________________________________________________

Rubi [A]  time = 0.06, antiderivative size = 101, 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 = {738, 640, 620, 206} \begin {gather*} \frac {2 e \sqrt {b x+c x^2} (2 c d-b e)}{b^2 c}-\frac {2 (d+e x) (x (2 c d-b e)+b d)}{b^2 \sqrt {b x+c x^2}}+\frac {2 e^2 \tanh ^{-1}\left (\frac {\sqrt {c} x}{\sqrt {b x+c x^2}}\right )}{c^{3/2}} \end {gather*}

Antiderivative was successfully verified.

[In]

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

[Out]

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

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 620

Int[1/Sqrt[(b_.)*(x_) + (c_.)*(x_)^2], x_Symbol] :> Dist[2, Subst[Int[1/(1 - c*x^2), x], x, x/Sqrt[b*x + c*x^2
]], x] /; FreeQ[{b, c}, x]

Rule 640

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

Rule 738

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

Rubi steps

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

________________________________________________________________________________________

Mathematica [A]  time = 0.09, size = 100, normalized size = 0.99 \begin {gather*} \frac {2 b^{5/2} e^2 \sqrt {x} \sqrt {\frac {c x}{b}+1} \sinh ^{-1}\left (\frac {\sqrt {c} \sqrt {x}}{\sqrt {b}}\right )-2 \sqrt {c} \left (b^2 e^2 x+b c d (d-2 e x)+2 c^2 d^2 x\right )}{b^2 c^{3/2} \sqrt {x (b+c x)}} \end {gather*}

Antiderivative was successfully verified.

[In]

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

[Out]

(-2*Sqrt[c]*(2*c^2*d^2*x + b^2*e^2*x + b*c*d*(d - 2*e*x)) + 2*b^(5/2)*e^2*Sqrt[x]*Sqrt[1 + (c*x)/b]*ArcSinh[(S
qrt[c]*Sqrt[x])/Sqrt[b]])/(b^2*c^(3/2)*Sqrt[x*(b + c*x)])

________________________________________________________________________________________

IntegrateAlgebraic [A]  time = 0.41, size = 104, normalized size = 1.03 \begin {gather*} -\frac {2 \sqrt {b x+c x^2} \left (b^2 e^2 x+b c d^2-2 b c d e x+2 c^2 d^2 x\right )}{b^2 c x (b+c x)}-\frac {e^2 \log \left (-2 c^{3/2} \sqrt {b x+c x^2}+b c+2 c^2 x\right )}{c^{3/2}} \end {gather*}

Antiderivative was successfully verified.

[In]

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

[Out]

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

________________________________________________________________________________________

fricas [A]  time = 0.42, size = 242, normalized size = 2.40 \begin {gather*} \left [\frac {{\left (b^{2} c e^{2} x^{2} + b^{3} e^{2} x\right )} \sqrt {c} \log \left (2 \, c x + b + 2 \, \sqrt {c x^{2} + b x} \sqrt {c}\right ) - 2 \, {\left (b c^{2} d^{2} + {\left (2 \, c^{3} d^{2} - 2 \, b c^{2} d e + b^{2} c e^{2}\right )} x\right )} \sqrt {c x^{2} + b x}}{b^{2} c^{3} x^{2} + b^{3} c^{2} x}, -\frac {2 \, {\left ({\left (b^{2} c e^{2} x^{2} + b^{3} e^{2} x\right )} \sqrt {-c} \arctan \left (\frac {\sqrt {c x^{2} + b x} \sqrt {-c}}{c x}\right ) + {\left (b c^{2} d^{2} + {\left (2 \, c^{3} d^{2} - 2 \, b c^{2} d e + b^{2} c e^{2}\right )} x\right )} \sqrt {c x^{2} + b x}\right )}}{b^{2} c^{3} x^{2} + b^{3} c^{2} x}\right ] \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

________________________________________________________________________________________

giac [A]  time = 0.40, size = 89, normalized size = 0.88 \begin {gather*} -\frac {2 \, {\left (\frac {d^{2}}{b} + \frac {{\left (2 \, c^{2} d^{2} - 2 \, b c d e + b^{2} e^{2}\right )} x}{b^{2} c}\right )}}{\sqrt {c x^{2} + b x}} - \frac {e^{2} \log \left ({\left | -2 \, {\left (\sqrt {c} x - \sqrt {c x^{2} + b x}\right )} \sqrt {c} - b \right |}\right )}{c^{\frac {3}{2}}} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

-2*(d^2/b + (2*c^2*d^2 - 2*b*c*d*e + b^2*e^2)*x/(b^2*c))/sqrt(c*x^2 + b*x) - e^2*log(abs(-2*(sqrt(c)*x - sqrt(
c*x^2 + b*x))*sqrt(c) - b))/c^(3/2)

________________________________________________________________________________________

maple [A]  time = 0.05, size = 97, normalized size = 0.96 \begin {gather*} \frac {4 d e x}{\sqrt {c \,x^{2}+b x}\, b}-\frac {2 e^{2} x}{\sqrt {c \,x^{2}+b x}\, c}+\frac {e^{2} \ln \left (\frac {c x +\frac {b}{2}}{\sqrt {c}}+\sqrt {c \,x^{2}+b x}\right )}{c^{\frac {3}{2}}}-\frac {2 \left (2 c x +b \right ) d^{2}}{\sqrt {c \,x^{2}+b x}\, b^{2}} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

-2*e^2/c/(c*x^2+b*x)^(1/2)*x+e^2/c^(3/2)*ln((c*x+1/2*b)/c^(1/2)+(c*x^2+b*x)^(1/2))+4*d*e/b/(c*x^2+b*x)^(1/2)*x
-2*d^2*(2*c*x+b)/b^2/(c*x^2+b*x)^(1/2)

________________________________________________________________________________________

maxima [A]  time = 1.33, size = 110, normalized size = 1.09 \begin {gather*} -\frac {4 \, c d^{2} x}{\sqrt {c x^{2} + b x} b^{2}} + \frac {4 \, d e x}{\sqrt {c x^{2} + b x} b} - \frac {2 \, e^{2} x}{\sqrt {c x^{2} + b x} c} + \frac {e^{2} \log \left (2 \, c x + b + 2 \, \sqrt {c x^{2} + b x} \sqrt {c}\right )}{c^{\frac {3}{2}}} - \frac {2 \, d^{2}}{\sqrt {c x^{2} + b x} b} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

-4*c*d^2*x/(sqrt(c*x^2 + b*x)*b^2) + 4*d*e*x/(sqrt(c*x^2 + b*x)*b) - 2*e^2*x/(sqrt(c*x^2 + b*x)*c) + e^2*log(2
*c*x + b + 2*sqrt(c*x^2 + b*x)*sqrt(c))/c^(3/2) - 2*d^2/(sqrt(c*x^2 + b*x)*b)

________________________________________________________________________________________

mupad [B]  time = 0.59, size = 96, normalized size = 0.95 \begin {gather*} \frac {e^2\,\ln \left (\frac {\frac {b}{2}+c\,x}{\sqrt {c}}+\sqrt {c\,x^2+b\,x}\right )}{c^{3/2}}-\frac {d^2\,\left (2\,b+4\,c\,x\right )}{b^2\,\sqrt {c\,x^2+b\,x}}-\frac {2\,e^2\,x}{c\,\sqrt {c\,x^2+b\,x}}+\frac {4\,d\,e\,x}{b\,\sqrt {x\,\left (b+c\,x\right )}} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

(e^2*log((b/2 + c*x)/c^(1/2) + (b*x + c*x^2)^(1/2)))/c^(3/2) - (d^2*(2*b + 4*c*x))/(b^2*(b*x + c*x^2)^(1/2)) -
(2*e^2*x)/(c*(b*x + c*x^2)^(1/2)) + (4*d*e*x)/(b*(x*(b + c*x))^(1/2))

________________________________________________________________________________________

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

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

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

________________________________________________________________________________________