### 3.88 $$\int \frac{(b x+c x^2)^{3/2}}{x^{5/2}} \, dx$$

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

[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])]

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Rubi [A]  time = 0.0335446, antiderivative size = 76, normalized size of antiderivative = 1., 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} $-2 b^{3/2} \tanh ^{-1}\left (\frac{\sqrt{b x+c x^2}}{\sqrt{b} \sqrt{x}}\right )+\frac{2 b \sqrt{b x+c x^2}}{\sqrt{x}}+\frac{2 \left (b x+c x^2\right )^{3/2}}{3 x^{3/2}}$

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 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]

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 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])

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*}

Mathematica [A]  time = 0.0444276, size = 70, normalized size = 0.92 $\frac{2 \sqrt{x} \sqrt{b+c x} \left (\sqrt{b+c x} (4 b+c x)-3 b^{3/2} \tanh ^{-1}\left (\frac{\sqrt{b+c x}}{\sqrt{b}}\right )\right )}{3 \sqrt{x (b+c x)}}$

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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Maple [A]  time = 0.193, size = 61, normalized size = 0.8 \begin{align*} -{\frac{2}{3}\sqrt{x \left ( cx+b \right ) } \left ( 3\,{b}^{3/2}{\it Artanh} \left ({\frac{\sqrt{cx+b}}{\sqrt{b}}} \right ) -xc\sqrt{cx+b}-4\,b\sqrt{cx+b} \right ){\frac{1}{\sqrt{x}}}{\frac{1}{\sqrt{cx+b}}}} \end{align*}

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

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

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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Fricas [A]  time = 2.08767, size = 321, normalized size = 4.22 \begin{align*} \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{align*}

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

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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Giac [A]  time = 1.26144, size = 104, normalized size = 1.37 \begin{align*} \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{align*}

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)