3.4.23 \(\int \frac {(d+e x)^2}{(b x+c x^2)^{3/2}} \, dx\) [323]

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

[Out]

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

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Rubi [A]
time = 0.04, 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 = {752, 654, 634, 212} \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 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 634

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 654

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 752

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 ) \text {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*}

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Mathematica [A]
time = 0.16, size = 100, normalized size = 0.99 \begin {gather*} -\frac {2 \left (\sqrt {c} \left (2 c^2 d^2 x+b^2 e^2 x+b c d (d-2 e x)\right )+b^2 e^2 \sqrt {x} \sqrt {b+c x} \log \left (-\sqrt {c} \sqrt {x}+\sqrt {b+c x}\right )\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)) + b^2*e^2*Sqrt[x]*Sqrt[b + c*x]*Log[-(Sqrt[c]*Sqrt[
x]) + Sqrt[b + c*x]]))/(b^2*c^(3/2)*Sqrt[x*(b + c*x)])

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Maple [A]
time = 0.45, size = 169, normalized size = 1.67

method	result	size
default	\(e^{2} \left (-\frac {x}{c \sqrt {c \,x^{2}+b x}}-\frac {b \left (-\frac {1}{c \sqrt {c \,x^{2}+b x}}+\frac {2 c x +b}{b c \sqrt {c \,x^{2}+b x}}\right )}{2 c}+\frac {\ln \left (\frac {\frac {b}{2}+c x}{\sqrt {c}}+\sqrt {c \,x^{2}+b x}\right )}{c^{\frac {3}{2}}}\right )+2 d e \left (-\frac {1}{c \sqrt {c \,x^{2}+b x}}+\frac {2 c x +b}{b c \sqrt {c \,x^{2}+b x}}\right )-\frac {2 d^{2} \left (2 c x +b \right )}{b^{2} \sqrt {c \,x^{2}+b x}}\)	\(169\)
risch	\(-\frac {2 d^{2} \left (c x +b \right )}{b^{2} \sqrt {x \left (c x +b \right )}}+\frac {e^{2} \ln \left (\frac {\frac {b}{2}+c x}{\sqrt {c}}+\sqrt {c \,x^{2}+b x}\right )}{c^{\frac {3}{2}}}-\frac {2 \sqrt {c \left (\frac {b}{c}+x \right )^{2}-\left (\frac {b}{c}+x \right ) b}\, e^{2}}{c^{2} \left (\frac {b}{c}+x \right )}+\frac {4 \sqrt {c \left (\frac {b}{c}+x \right )^{2}-\left (\frac {b}{c}+x \right ) b}\, d e}{b c \left (\frac {b}{c}+x \right )}-\frac {2 \sqrt {c \left (\frac {b}{c}+x \right )^{2}-\left (\frac {b}{c}+x \right ) b}\, d^{2}}{b^{2} \left (\frac {b}{c}+x \right )}\)	\(180\)

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

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

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Maxima [A]
time = 0.27, size = 109, normalized size = 1.08 \begin {gather*} -\frac {4 \, c d^{2} x}{\sqrt {c x^{2} + b x} b^{2}} + \frac {4 \, d x e}{\sqrt {c x^{2} + b x} b} - \frac {2 \, d^{2}}{\sqrt {c x^{2} + b x} b} - \frac {2 \, x e^{2}}{\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}}} \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*x*e/(sqrt(c*x^2 + b*x)*b) - 2*d^2/(sqrt(c*x^2 + b*x)*b) - 2*x*e^2/(sq
rt(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)

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Fricas [A]
time = 1.35, size = 234, normalized size = 2.32 \begin {gather*} \left [\frac {{\left (b^{2} c x^{2} + b^{3} x\right )} \sqrt {c} e^{2} \log \left (2 \, c x + b + 2 \, \sqrt {c x^{2} + b x} \sqrt {c}\right ) - 2 \, {\left (2 \, c^{3} d^{2} x - 2 \, b c^{2} d x e + b c^{2} d^{2} + b^{2} c x e^{2}\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 x^{2} + b^{3} x\right )} \sqrt {-c} \arctan \left (\frac {\sqrt {c x^{2} + b x} \sqrt {-c}}{c x}\right ) e^{2} + {\left (2 \, c^{3} d^{2} x - 2 \, b c^{2} d x e + b c^{2} d^{2} + b^{2} c x e^{2}\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*x^2 + b^3*x)*sqrt(c)*e^2*log(2*c*x + b + 2*sqrt(c*x^2 + b*x)*sqrt(c)) - 2*(2*c^3*d^2*x - 2*b*c^2*d*x*
e + b*c^2*d^2 + b^2*c*x*e^2)*sqrt(c*x^2 + b*x))/(b^2*c^3*x^2 + b^3*c^2*x), -2*((b^2*c*x^2 + b^3*x)*sqrt(-c)*ar
ctan(sqrt(c*x^2 + b*x)*sqrt(-c)/(c*x))*e^2 + (2*c^3*d^2*x - 2*b*c^2*d*x*e + b*c^2*d^2 + b^2*c*x*e^2)*sqrt(c*x^
2 + b*x))/(b^2*c^3*x^2 + b^3*c^2*x)]

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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)

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Giac [A]
time = 1.31, 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)

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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))

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