3.4.22 \(\int \frac {(d+e x)^3}{(b x+c x^2)^{3/2}} \, dx\) [322]

Optimal. Leaf size=139 \[ -\frac {2 (d+e x)^2 (b d+(2 c d-b e) x)}{b^2 \sqrt {b x+c x^2}}+\frac {e \left (8 c^2 d^2-6 b c d e+3 b^2 e^2+2 c e (2 c d-b e) x\right ) \sqrt {b x+c x^2}}{b^2 c^2}+\frac {3 e^2 (2 c d-b e) \tanh ^{-1}\left (\frac {\sqrt {c} x}{\sqrt {b x+c x^2}}\right )}{c^{5/2}} \]

[Out]

3*e^2*(-b*e+2*c*d)*arctanh(x*c^(1/2)/(c*x^2+b*x)^(1/2))/c^(5/2)-2*(e*x+d)^2*(b*d+(-b*e+2*c*d)*x)/b^2/(c*x^2+b*
x)^(1/2)+e*(8*c^2*d^2-6*b*c*d*e+3*b^2*e^2+2*c*e*(-b*e+2*c*d)*x)*(c*x^2+b*x)^(1/2)/b^2/c^2

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Rubi [A]
time = 0.07, antiderivative size = 139, 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, 793, 634, 212} \begin {gather*} \frac {e \sqrt {b x+c x^2} \left (3 b^2 e^2+2 c e x (2 c d-b e)-6 b c d e+8 c^2 d^2\right )}{b^2 c^2}-\frac {2 (d+e x)^2 (x (2 c d-b e)+b d)}{b^2 \sqrt {b x+c x^2}}+\frac {3 e^2 (2 c d-b e) \tanh ^{-1}\left (\frac {\sqrt {c} x}{\sqrt {b x+c x^2}}\right )}{c^{5/2}} \end {gather*}

Antiderivative was successfully verified.

[In]

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

[Out]

(-2*(d + e*x)^2*(b*d + (2*c*d - b*e)*x))/(b^2*Sqrt[b*x + c*x^2]) + (e*(8*c^2*d^2 - 6*b*c*d*e + 3*b^2*e^2 + 2*c
*e*(2*c*d - b*e)*x)*Sqrt[b*x + c*x^2])/(b^2*c^2) + (3*e^2*(2*c*d - b*e)*ArcTanh[(Sqrt[c]*x)/Sqrt[b*x + c*x^2]]
)/c^(5/2)

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 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]

Rule 793

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

Rubi steps

\begin {align*} \int \frac {(d+e x)^3}{\left (b x+c x^2\right )^{3/2}} \, dx &=-\frac {2 (d+e x)^2 (b d+(2 c d-b e) x)}{b^2 \sqrt {b x+c x^2}}-\frac {2 \int \frac {(d+e x) (-2 b d e-2 e (2 c d-b e) x)}{\sqrt {b x+c x^2}} \, dx}{b^2}\\ &=-\frac {2 (d+e x)^2 (b d+(2 c d-b e) x)}{b^2 \sqrt {b x+c x^2}}+\frac {e \left (8 c^2 d^2-6 b c d e+3 b^2 e^2+2 c e (2 c d-b e) x\right ) \sqrt {b x+c x^2}}{b^2 c^2}+\frac {\left (3 e^2 (2 c d-b e)\right ) \int \frac {1}{\sqrt {b x+c x^2}} \, dx}{2 c^2}\\ &=-\frac {2 (d+e x)^2 (b d+(2 c d-b e) x)}{b^2 \sqrt {b x+c x^2}}+\frac {e \left (8 c^2 d^2-6 b c d e+3 b^2 e^2+2 c e (2 c d-b e) x\right ) \sqrt {b x+c x^2}}{b^2 c^2}+\frac {\left (3 e^2 (2 c d-b e)\right ) \text {Subst}\left (\int \frac {1}{1-c x^2} \, dx,x,\frac {x}{\sqrt {b x+c x^2}}\right )}{c^2}\\ &=-\frac {2 (d+e x)^2 (b d+(2 c d-b e) x)}{b^2 \sqrt {b x+c x^2}}+\frac {e \left (8 c^2 d^2-6 b c d e+3 b^2 e^2+2 c e (2 c d-b e) x\right ) \sqrt {b x+c x^2}}{b^2 c^2}+\frac {3 e^2 (2 c d-b e) \tanh ^{-1}\left (\frac {\sqrt {c} x}{\sqrt {b x+c x^2}}\right )}{c^{5/2}}\\ \end {align*}

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Mathematica [A]
time = 0.24, size = 130, normalized size = 0.94 \begin {gather*} \frac {\sqrt {c} \left (-4 c^3 d^3 x+3 b^3 e^3 x-2 b c^2 d^2 (d-3 e x)+b^2 c e^2 x (-6 d+e x)\right )+3 b^2 e^2 (-2 c d+b e) \sqrt {x} \sqrt {b+c x} \log \left (-\sqrt {c} \sqrt {x}+\sqrt {b+c x}\right )}{b^2 c^{5/2} \sqrt {x (b+c x)}} \end {gather*}

Antiderivative was successfully verified.

[In]

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

[Out]

(Sqrt[c]*(-4*c^3*d^3*x + 3*b^3*e^3*x - 2*b*c^2*d^2*(d - 3*e*x) + b^2*c*e^2*x*(-6*d + e*x)) + 3*b^2*e^2*(-2*c*d
+ b*e)*Sqrt[x]*Sqrt[b + c*x]*Log[-(Sqrt[c]*Sqrt[x]) + Sqrt[b + c*x]])/(b^2*c^(5/2)*Sqrt[x*(b + c*x)])

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Maple [B] Leaf count of result is larger than twice the leaf count of optimal. \(294\) vs. \(2(129)=258\).
time = 0.47, size = 295, normalized size = 2.12

method	result	size
risch	\(\frac {\left (c x +b \right ) \left (e^{3} x \,b^{2}-2 c^{2} d^{3}\right )}{b^{2} \sqrt {x \left (c x +b \right )}\, c^{2}}-\frac {3 b \,e^{3} \ln \left (\frac {\frac {b}{2}+c x}{\sqrt {c}}+\sqrt {c \,x^{2}+b x}\right )}{2 c^{\frac {5}{2}}}+\frac {3 d \,e^{2} \ln \left (\frac {\frac {b}{2}+c x}{\sqrt {c}}+\sqrt {c \,x^{2}+b x}\right )}{c^{\frac {3}{2}}}+\frac {2 b \sqrt {c \left (\frac {b}{c}+x \right )^{2}-\left (\frac {b}{c}+x \right ) b}\, e^{3}}{c^{3} \left (\frac {b}{c}+x \right )}-\frac {6 \sqrt {c \left (\frac {b}{c}+x \right )^{2}-\left (\frac {b}{c}+x \right ) b}\, d \,e^{2}}{c^{2} \left (\frac {b}{c}+x \right )}+\frac {6 \sqrt {c \left (\frac {b}{c}+x \right )^{2}-\left (\frac {b}{c}+x \right ) b}\, d^{2} e}{b c \left (\frac {b}{c}+x \right )}-\frac {2 \sqrt {c \left (\frac {b}{c}+x \right )^{2}-\left (\frac {b}{c}+x \right ) b}\, d^{3}}{b^{2} \left (\frac {b}{c}+x \right )}\)	\(276\)
default	\(e^{3} \left (\frac {x^{2}}{c \sqrt {c \,x^{2}+b x}}-\frac {3 b \left (-\frac {x}{c \sqrt {c \,x^{2}+b x}}-\frac {b \left (-\frac {1}{c \sqrt {c \,x^{2}+b x}}+\frac {2 c x +b}{b c \sqrt {c \,x^{2}+b x}}\right )}{2 c}+\frac {\ln \left (\frac {\frac {b}{2}+c x}{\sqrt {c}}+\sqrt {c \,x^{2}+b x}\right )}{c^{\frac {3}{2}}}\right )}{2 c}\right )+3 d \,e^{2} \left (-\frac {x}{c \sqrt {c \,x^{2}+b x}}-\frac {b \left (-\frac {1}{c \sqrt {c \,x^{2}+b x}}+\frac {2 c x +b}{b c \sqrt {c \,x^{2}+b x}}\right )}{2 c}+\frac {\ln \left (\frac {\frac {b}{2}+c x}{\sqrt {c}}+\sqrt {c \,x^{2}+b x}\right )}{c^{\frac {3}{2}}}\right )+3 d^{2} e \left (-\frac {1}{c \sqrt {c \,x^{2}+b x}}+\frac {2 c x +b}{b c \sqrt {c \,x^{2}+b x}}\right )-\frac {2 d^{3} \left (2 c x +b \right )}{b^{2} \sqrt {c \,x^{2}+b x}}\)	\(295\)

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

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

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Maxima [A]
time = 0.28, size = 185, normalized size = 1.33 \begin {gather*} -\frac {4 \, c d^{3} x}{\sqrt {c x^{2} + b x} b^{2}} + \frac {6 \, d^{2} x e}{\sqrt {c x^{2} + b x} b} - \frac {2 \, d^{3}}{\sqrt {c x^{2} + b x} b} + \frac {x^{2} e^{3}}{\sqrt {c x^{2} + b x} c} - \frac {6 \, d x e^{2}}{\sqrt {c x^{2} + b x} c} + \frac {3 \, d e^{2} \log \left (2 \, c x + b + 2 \, \sqrt {c x^{2} + b x} \sqrt {c}\right )}{c^{\frac {3}{2}}} + \frac {3 \, b x e^{3}}{\sqrt {c x^{2} + b x} c^{2}} - \frac {3 \, b e^{3} \log \left (2 \, c x + b + 2 \, \sqrt {c x^{2} + b x} \sqrt {c}\right )}{2 \, c^{\frac {5}{2}}} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

-4*c*d^3*x/(sqrt(c*x^2 + b*x)*b^2) + 6*d^2*x*e/(sqrt(c*x^2 + b*x)*b) - 2*d^3/(sqrt(c*x^2 + b*x)*b) + x^2*e^3/(
sqrt(c*x^2 + b*x)*c) - 6*d*x*e^2/(sqrt(c*x^2 + b*x)*c) + 3*d*e^2*log(2*c*x + b + 2*sqrt(c*x^2 + b*x)*sqrt(c))/
c^(3/2) + 3*b*x*e^3/(sqrt(c*x^2 + b*x)*c^2) - 3/2*b*e^3*log(2*c*x + b + 2*sqrt(c*x^2 + b*x)*sqrt(c))/c^(5/2)

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Fricas [A]
time = 1.61, size = 345, normalized size = 2.48 \begin {gather*} \left [\frac {3 \, {\left ({\left (b^{3} c x^{2} + b^{4} x\right )} e^{3} - 2 \, {\left (b^{2} c^{2} d x^{2} + b^{3} c d x\right )} e^{2}\right )} \sqrt {c} \log \left (2 \, c x + b - 2 \, \sqrt {c x^{2} + b x} \sqrt {c}\right ) - 2 \, {\left (4 \, c^{4} d^{3} x - 6 \, b c^{3} d^{2} x e + 2 \, b c^{3} d^{3} + 6 \, b^{2} c^{2} d x e^{2} - {\left (b^{2} c^{2} x^{2} + 3 \, b^{3} c x\right )} e^{3}\right )} \sqrt {c x^{2} + b x}}{2 \, {\left (b^{2} c^{4} x^{2} + b^{3} c^{3} x\right )}}, \frac {3 \, {\left ({\left (b^{3} c x^{2} + b^{4} x\right )} e^{3} - 2 \, {\left (b^{2} c^{2} d x^{2} + b^{3} c d x\right )} e^{2}\right )} \sqrt {-c} \arctan \left (\frac {\sqrt {c x^{2} + b x} \sqrt {-c}}{c x}\right ) - {\left (4 \, c^{4} d^{3} x - 6 \, b c^{3} d^{2} x e + 2 \, b c^{3} d^{3} + 6 \, b^{2} c^{2} d x e^{2} - {\left (b^{2} c^{2} x^{2} + 3 \, b^{3} c x\right )} e^{3}\right )} \sqrt {c x^{2} + b x}}{b^{2} c^{4} x^{2} + b^{3} c^{3} x}\right ] \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

[1/2*(3*((b^3*c*x^2 + b^4*x)*e^3 - 2*(b^2*c^2*d*x^2 + b^3*c*d*x)*e^2)*sqrt(c)*log(2*c*x + b - 2*sqrt(c*x^2 + b
*x)*sqrt(c)) - 2*(4*c^4*d^3*x - 6*b*c^3*d^2*x*e + 2*b*c^3*d^3 + 6*b^2*c^2*d*x*e^2 - (b^2*c^2*x^2 + 3*b^3*c*x)*
e^3)*sqrt(c*x^2 + b*x))/(b^2*c^4*x^2 + b^3*c^3*x), (3*((b^3*c*x^2 + b^4*x)*e^3 - 2*(b^2*c^2*d*x^2 + b^3*c*d*x)
*e^2)*sqrt(-c)*arctan(sqrt(c*x^2 + b*x)*sqrt(-c)/(c*x)) - (4*c^4*d^3*x - 6*b*c^3*d^2*x*e + 2*b*c^3*d^3 + 6*b^2
*c^2*d*x*e^2 - (b^2*c^2*x^2 + 3*b^3*c*x)*e^3)*sqrt(c*x^2 + b*x))/(b^2*c^4*x^2 + b^3*c^3*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 )^{3}}{\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)**3/(c*x**2+b*x)**(3/2),x)

[Out]

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

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Giac [A]
time = 1.58, size = 125, normalized size = 0.90 \begin {gather*} -\frac {\frac {2 \, d^{3}}{b} - x {\left (\frac {x e^{3}}{c} - \frac {4 \, c^{3} d^{3} - 6 \, b c^{2} d^{2} e + 6 \, b^{2} c d e^{2} - 3 \, b^{3} e^{3}}{b^{2} c^{2}}\right )}}{\sqrt {c x^{2} + b x}} - \frac {3 \, {\left (2 \, c d e^{2} - b e^{3}\right )} \log \left ({\left | -2 \, {\left (\sqrt {c} x - \sqrt {c x^{2} + b x}\right )} \sqrt {c} - b \right |}\right )}{2 \, c^{\frac {5}{2}}} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

-(2*d^3/b - x*(x*e^3/c - (4*c^3*d^3 - 6*b*c^2*d^2*e + 6*b^2*c*d*e^2 - 3*b^3*e^3)/(b^2*c^2)))/sqrt(c*x^2 + b*x)
- 3/2*(2*c*d*e^2 - b*e^3)*log(abs(-2*(sqrt(c)*x - sqrt(c*x^2 + b*x))*sqrt(c) - b))/c^(5/2)

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

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

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

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