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

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

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Rubi [A]  time = 0.09, 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 = {738, 779, 620, 206} \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 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 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]

Rule 779

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 ) \operatorname {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 = 0.10, size = 129, normalized size = 0.93 \begin {gather*} \frac {\sqrt {c} \left (3 b^3 e^3 x+b^2 c e^2 x (e x-6 d)-2 b c^2 d^2 (d-3 e x)-4 c^3 d^3 x\right )-3 b^{5/2} e^2 \sqrt {x} \sqrt {\frac {c x}{b}+1} (b e-2 c d) \sinh ^{-1}\left (\frac {\sqrt {c} \sqrt {x}}{\sqrt {b}}\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^(5/2)*e^2*(-2
*c*d + b*e)*Sqrt[x]*Sqrt[1 + (c*x)/b]*ArcSinh[(Sqrt[c]*Sqrt[x])/Sqrt[b]])/(b^2*c^(5/2)*Sqrt[x*(b + c*x)])

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IntegrateAlgebraic [A]  time = 0.61, size = 148, normalized size = 1.06 \begin {gather*} \frac {\sqrt {b x+c x^2} \left (3 b^3 e^3 x-6 b^2 c d e^2 x+b^2 c e^3 x^2-2 b c^2 d^3+6 b c^2 d^2 e x-4 c^3 d^3 x\right )}{b^2 c^2 x (b+c x)}-\frac {3 \left (2 c d e^2-b e^3\right ) \log \left (-2 c^{5/2} \sqrt {b x+c x^2}+b c^2+2 c^3 x\right )}{2 c^{5/2}} \end {gather*}

Antiderivative was successfully verified.

[In]

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

[Out]

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

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fricas [A]  time = 0.43, size = 362, normalized size = 2.60 \begin {gather*} \left [-\frac {3 \, {\left ({\left (2 \, b^{2} c^{2} d e^{2} - b^{3} c e^{3}\right )} x^{2} + {\left (2 \, b^{3} c d e^{2} - b^{4} e^{3}\right )} x\right )} \sqrt {c} \log \left (2 \, c x + b - 2 \, \sqrt {c x^{2} + b x} \sqrt {c}\right ) - 2 \, {\left (b^{2} c^{2} e^{3} x^{2} - 2 \, b c^{3} d^{3} - {\left (4 \, c^{4} d^{3} - 6 \, b c^{3} d^{2} e + 6 \, b^{2} c^{2} d e^{2} - 3 \, b^{3} c e^{3}\right )} x\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 (2 \, b^{2} c^{2} d e^{2} - b^{3} c e^{3}\right )} x^{2} + {\left (2 \, b^{3} c d e^{2} - b^{4} e^{3}\right )} x\right )} \sqrt {-c} \arctan \left (\frac {\sqrt {c x^{2} + b x} \sqrt {-c}}{c x}\right ) - {\left (b^{2} c^{2} e^{3} x^{2} - 2 \, b c^{3} d^{3} - {\left (4 \, c^{4} d^{3} - 6 \, b c^{3} d^{2} e + 6 \, b^{2} c^{2} d e^{2} - 3 \, b^{3} c e^{3}\right )} x\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*((2*b^2*c^2*d*e^2 - b^3*c*e^3)*x^2 + (2*b^3*c*d*e^2 - b^4*e^3)*x)*sqrt(c)*log(2*c*x + b - 2*sqrt(c*x^
2 + b*x)*sqrt(c)) - 2*(b^2*c^2*e^3*x^2 - 2*b*c^3*d^3 - (4*c^4*d^3 - 6*b*c^3*d^2*e + 6*b^2*c^2*d*e^2 - 3*b^3*c*
e^3)*x)*sqrt(c*x^2 + b*x))/(b^2*c^4*x^2 + b^3*c^3*x), -(3*((2*b^2*c^2*d*e^2 - b^3*c*e^3)*x^2 + (2*b^3*c*d*e^2
- b^4*e^3)*x)*sqrt(-c)*arctan(sqrt(c*x^2 + b*x)*sqrt(-c)/(c*x)) - (b^2*c^2*e^3*x^2 - 2*b*c^3*d^3 - (4*c^4*d^3
- 6*b*c^3*d^2*e + 6*b^2*c^2*d*e^2 - 3*b^3*c*e^3)*x)*sqrt(c*x^2 + b*x))/(b^2*c^4*x^2 + b^3*c^3*x)]

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giac [A]  time = 0.27, 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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maple [A]  time = 0.05, size = 177, normalized size = 1.27 \begin {gather*} \frac {e^{3} x^{2}}{\sqrt {c \,x^{2}+b x}\, c}+\frac {3 b \,e^{3} x}{\sqrt {c \,x^{2}+b x}\, c^{2}}+\frac {6 d^{2} e x}{\sqrt {c \,x^{2}+b x}\, b}-\frac {6 d \,e^{2} x}{\sqrt {c \,x^{2}+b x}\, c}-\frac {3 b \,e^{3} \ln \left (\frac {c x +\frac {b}{2}}{\sqrt {c}}+\sqrt {c \,x^{2}+b x}\right )}{2 c^{\frac {5}{2}}}+\frac {3 d \,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^{3}}{\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)^3/(c*x^2+b*x)^(3/2),x)

[Out]

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

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maxima [A]  time = 1.36, size = 189, normalized size = 1.36 \begin {gather*} \frac {e^{3} x^{2}}{\sqrt {c x^{2} + b x} c} - \frac {4 \, c d^{3} x}{\sqrt {c x^{2} + b x} b^{2}} + \frac {6 \, d^{2} e x}{\sqrt {c x^{2} + b x} b} - \frac {6 \, d e^{2} x}{\sqrt {c x^{2} + b x} c} + \frac {3 \, b e^{3} x}{\sqrt {c x^{2} + b x} c^{2}} + \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 e^{3} \log \left (2 \, c x + b + 2 \, \sqrt {c x^{2} + b x} \sqrt {c}\right )}{2 \, c^{\frac {5}{2}}} - \frac {2 \, d^{3}}{\sqrt {c x^{2} + b x} b} \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]

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

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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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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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