3.1.88 \(\int \frac {(b x+c x^2)^{3/2}}{x^{5/2}} \, dx\)

Optimal. Leaf size=76 \[ -2 b^{3/2} \tanh ^{-1}\left (\frac {\sqrt {b x+c x^2}}{\sqrt {b} \sqrt {x}}\right )+\frac {2 b \sqrt {b x+c x^2}}{\sqrt {x}}+\frac {2 \left (b x+c x^2\right )^{3/2}}{3 x^{3/2}} \]

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Rubi [A]  time = 0.03, antiderivative size = 76, normalized size of antiderivative = 1.00, number of steps used = 4, number of rules used = 3, integrand size = 19, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.158, Rules used = {664, 660, 207} \begin {gather*} -2 b^{3/2} \tanh ^{-1}\left (\frac {\sqrt {b x+c x^2}}{\sqrt {b} \sqrt {x}}\right )+\frac {2 b \sqrt {b x+c x^2}}{\sqrt {x}}+\frac {2 \left (b x+c x^2\right )^{3/2}}{3 x^{3/2}} \end {gather*}

Antiderivative was successfully verified.

[In]

Int[(b*x + c*x^2)^(3/2)/x^(5/2),x]

[Out]

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

Rule 207

Int[((a_) + (b_.)*(x_)^2)^(-1), x_Symbol] :> -Simp[ArcTanh[(Rt[b, 2]*x)/Rt[-a, 2]]/(Rt[-a, 2]*Rt[b, 2]), x] /;
FreeQ[{a, b}, x] && NegQ[a/b] && (LtQ[a, 0] || GtQ[b, 0])

Rule 660

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

Rule 664

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

Rubi steps

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

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Mathematica [A]  time = 0.04, size = 70, normalized size = 0.92 \begin {gather*} \frac {2 \sqrt {x} \sqrt {b+c x} \left (\sqrt {b+c x} (4 b+c x)-3 b^{3/2} \tanh ^{-1}\left (\frac {\sqrt {b+c x}}{\sqrt {b}}\right )\right )}{3 \sqrt {x (b+c x)}} \end {gather*}

Antiderivative was successfully verified.

[In]

Integrate[(b*x + c*x^2)^(3/2)/x^(5/2),x]

[Out]

(2*Sqrt[x]*Sqrt[b + c*x]*(Sqrt[b + c*x]*(4*b + c*x) - 3*b^(3/2)*ArcTanh[Sqrt[b + c*x]/Sqrt[b]]))/(3*Sqrt[x*(b
+ c*x)])

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IntegrateAlgebraic [A]  time = 0.44, size = 62, normalized size = 0.82 \begin {gather*} \frac {2 (4 b+c x) \sqrt {b x+c x^2}}{3 \sqrt {x}}-2 b^{3/2} \tanh ^{-1}\left (\frac {\sqrt {b} \sqrt {x}}{\sqrt {b x+c x^2}}\right ) \end {gather*}

Antiderivative was successfully verified.

[In]

IntegrateAlgebraic[(b*x + c*x^2)^(3/2)/x^(5/2),x]

[Out]

(2*(4*b + c*x)*Sqrt[b*x + c*x^2])/(3*Sqrt[x]) - 2*b^(3/2)*ArcTanh[(Sqrt[b]*Sqrt[x])/Sqrt[b*x + c*x^2]]

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fricas [A]  time = 0.41, size = 129, normalized size = 1.70 \begin {gather*} \left [\frac {3 \, b^{\frac {3}{2}} x \log \left (-\frac {c x^{2} + 2 \, b x - 2 \, \sqrt {c x^{2} + b x} \sqrt {b} \sqrt {x}}{x^{2}}\right ) + 2 \, \sqrt {c x^{2} + b x} {\left (c x + 4 \, b\right )} \sqrt {x}}{3 \, x}, \frac {2 \, {\left (3 \, \sqrt {-b} b x \arctan \left (\frac {\sqrt {-b} \sqrt {x}}{\sqrt {c x^{2} + b x}}\right ) + \sqrt {c x^{2} + b x} {\left (c x + 4 \, b\right )} \sqrt {x}\right )}}{3 \, x}\right ] \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate((c*x^2+b*x)^(3/2)/x^(5/2),x, algorithm="fricas")

[Out]

[1/3*(3*b^(3/2)*x*log(-(c*x^2 + 2*b*x - 2*sqrt(c*x^2 + b*x)*sqrt(b)*sqrt(x))/x^2) + 2*sqrt(c*x^2 + b*x)*(c*x +
4*b)*sqrt(x))/x, 2/3*(3*sqrt(-b)*b*x*arctan(sqrt(-b)*sqrt(x)/sqrt(c*x^2 + b*x)) + sqrt(c*x^2 + b*x)*(c*x + 4*
b)*sqrt(x))/x]

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giac [A]  time = 0.20, size = 77, normalized size = 1.01 \begin {gather*} \frac {2 \, b^{2} \arctan \left (\frac {\sqrt {c x + b}}{\sqrt {-b}}\right )}{\sqrt {-b}} + \frac {2}{3} \, {\left (c x + b\right )}^{\frac {3}{2}} + 2 \, \sqrt {c x + b} b - \frac {2 \, {\left (3 \, b^{2} \arctan \left (\frac {\sqrt {b}}{\sqrt {-b}}\right ) + 4 \, \sqrt {-b} b^{\frac {3}{2}}\right )}}{3 \, \sqrt {-b}} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate((c*x^2+b*x)^(3/2)/x^(5/2),x, algorithm="giac")

[Out]

2*b^2*arctan(sqrt(c*x + b)/sqrt(-b))/sqrt(-b) + 2/3*(c*x + b)^(3/2) + 2*sqrt(c*x + b)*b - 2/3*(3*b^2*arctan(sq
rt(b)/sqrt(-b)) + 4*sqrt(-b)*b^(3/2))/sqrt(-b)

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maple [A]  time = 0.06, size = 61, normalized size = 0.80 \begin {gather*} -\frac {2 \sqrt {\left (c x +b \right ) x}\, \left (3 b^{\frac {3}{2}} \arctanh \left (\frac {\sqrt {c x +b}}{\sqrt {b}}\right )-\sqrt {c x +b}\, c x -4 \sqrt {c x +b}\, b \right )}{3 \sqrt {c x +b}\, \sqrt {x}} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

int((c*x^2+b*x)^(3/2)/x^(5/2),x)

[Out]

-2/3*((c*x+b)*x)^(1/2)*(3*b^(3/2)*arctanh((c*x+b)^(1/2)/b^(1/2))-x*c*(c*x+b)^(1/2)-4*(c*x+b)^(1/2)*b)/x^(1/2)/
(c*x+b)^(1/2)

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

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate((c*x^2+b*x)^(3/2)/x^(5/2),x, algorithm="maxima")

[Out]

b*integrate(sqrt(c*x + b)/x, x) + 2/3*(c*x + b)^(3/2)

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mupad [F]  time = 0.00, size = -1, normalized size = -0.01 \begin {gather*} \int \frac {{\left (c\,x^2+b\,x\right )}^{3/2}}{x^{5/2}} \,d x \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

int((b*x + c*x^2)^(3/2)/x^(5/2),x)

[Out]

int((b*x + c*x^2)^(3/2)/x^(5/2), x)

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sympy [F]  time = 0.00, size = 0, normalized size = 0.00 \begin {gather*} \int \frac {\left (x \left (b + c x\right )\right )^{\frac {3}{2}}}{x^{\frac {5}{2}}}\, dx \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate((c*x**2+b*x)**(3/2)/x**(5/2),x)

[Out]

Integral((x*(b + c*x))**(3/2)/x**(5/2), x)

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