3.89 $$\int \frac{(b x+c x^2)^{3/2}}{x^{7/2}} \, dx$$

Optimal. Leaf size=75 $-\frac{\left (b x+c x^2\right )^{3/2}}{x^{5/2}}+\frac{3 c \sqrt{b x+c x^2}}{\sqrt{x}}-3 \sqrt{b} c \tanh ^{-1}\left (\frac{\sqrt{b x+c x^2}}{\sqrt{b} \sqrt{x}}\right )$

[Out]

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

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Rubi [A]  time = 0.0315328, antiderivative size = 75, normalized size of antiderivative = 1., number of steps used = 4, number of rules used = 4, integrand size = 19, $$\frac{\text{number of rules}}{\text{integrand size}}$$ = 0.21, Rules used = {662, 664, 660, 207} $-\frac{\left (b x+c x^2\right )^{3/2}}{x^{5/2}}+\frac{3 c \sqrt{b x+c x^2}}{\sqrt{x}}-3 \sqrt{b} c \tanh ^{-1}\left (\frac{\sqrt{b x+c x^2}}{\sqrt{b} \sqrt{x}}\right )$

Antiderivative was successfully veriﬁed.

[In]

Int[(b*x + c*x^2)^(3/2)/x^(7/2),x]

[Out]

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

Rule 662

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 + p + 1)), x] - Dist[(c*p)/(e^2*(m + p + 1)), Int[(d + e*x)^(m + 2)*(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] && (LtQ[m, -2] || EqQ[m + 2*p + 1, 0]) && NeQ[m + p + 1, 0] && IntegerQ[2*p]

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^{7/2}} \, dx &=-\frac{\left (b x+c x^2\right )^{3/2}}{x^{5/2}}+\frac{1}{2} (3 c) \int \frac{\sqrt{b x+c x^2}}{x^{3/2}} \, dx\\ &=\frac{3 c \sqrt{b x+c x^2}}{\sqrt{x}}-\frac{\left (b x+c x^2\right )^{3/2}}{x^{5/2}}+\frac{1}{2} (3 b c) \int \frac{1}{\sqrt{x} \sqrt{b x+c x^2}} \, dx\\ &=\frac{3 c \sqrt{b x+c x^2}}{\sqrt{x}}-\frac{\left (b x+c x^2\right )^{3/2}}{x^{5/2}}+(3 b c) \operatorname{Subst}\left (\int \frac{1}{-b+x^2} \, dx,x,\frac{\sqrt{b x+c x^2}}{\sqrt{x}}\right )\\ &=\frac{3 c \sqrt{b x+c x^2}}{\sqrt{x}}-\frac{\left (b x+c x^2\right )^{3/2}}{x^{5/2}}-3 \sqrt{b} c \tanh ^{-1}\left (\frac{\sqrt{b x+c x^2}}{\sqrt{b} \sqrt{x}}\right )\\ \end{align*}

Mathematica [C]  time = 0.0162617, size = 40, normalized size = 0.53 $\frac{2 c (x (b+c x))^{5/2} \, _2F_1\left (2,\frac{5}{2};\frac{7}{2};\frac{c x}{b}+1\right )}{5 b^2 x^{5/2}}$

Antiderivative was successfully veriﬁed.

[In]

Integrate[(b*x + c*x^2)^(3/2)/x^(7/2),x]

[Out]

(2*c*(x*(b + c*x))^(5/2)*Hypergeometric2F1[2, 5/2, 7/2, 1 + (c*x)/b])/(5*b^2*x^(5/2))

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Maple [A]  time = 0.195, size = 68, normalized size = 0.9 \begin{align*}{ \left ( 2\,xc\sqrt{cx+b}\sqrt{b}-3\,{\it Artanh} \left ({\frac{\sqrt{cx+b}}{\sqrt{b}}} \right ) xbc-{b}^{{\frac{3}{2}}}\sqrt{cx+b} \right ) \sqrt{x \left ( cx+b \right ) }{x}^{-{\frac{3}{2}}}{\frac{1}{\sqrt{cx+b}}}{\frac{1}{\sqrt{b}}}} \end{align*}

Veriﬁcation of antiderivative is not currently implemented for this CAS.

[In]

int((c*x^2+b*x)^(3/2)/x^(7/2),x)

[Out]

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

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Maxima [F]  time = 0., size = 0, normalized size = 0. \begin{align*} \int \frac{{\left (c x^{2} + b x\right )}^{\frac{3}{2}}}{x^{\frac{7}{2}}}\,{d x} \end{align*}

Veriﬁcation of antiderivative is not currently implemented for this CAS.

[In]

integrate((c*x^2+b*x)^(3/2)/x^(7/2),x, algorithm="maxima")

[Out]

integrate((c*x^2 + b*x)^(3/2)/x^(7/2), x)

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Fricas [A]  time = 2.09823, size = 329, normalized size = 4.39 \begin{align*} \left [\frac{3 \, \sqrt{b} c x^{2} \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 (2 \, c x - b\right )} \sqrt{x}}{2 \, x^{2}}, \frac{3 \, \sqrt{-b} c x^{2} \arctan \left (\frac{\sqrt{-b} \sqrt{x}}{\sqrt{c x^{2} + b x}}\right ) + \sqrt{c x^{2} + b x}{\left (2 \, c x - b\right )} \sqrt{x}}{x^{2}}\right ] \end{align*}

Veriﬁcation of antiderivative is not currently implemented for this CAS.

[In]

integrate((c*x^2+b*x)^(3/2)/x^(7/2),x, algorithm="fricas")

[Out]

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

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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{7}{2}}}\, dx \end{align*}

Veriﬁcation of antiderivative is not currently implemented for this CAS.

[In]

integrate((c*x**2+b*x)**(3/2)/x**(7/2),x)

[Out]

Integral((x*(b + c*x))**(3/2)/x**(7/2), x)

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Giac [A]  time = 1.3444, size = 68, normalized size = 0.91 \begin{align*}{\left (\frac{3 \, b \arctan \left (\frac{\sqrt{c x + b}}{\sqrt{-b}}\right )}{\sqrt{-b}} + 2 \, \sqrt{c x + b} - \frac{\sqrt{c x + b} b}{c x}\right )} c \end{align*}

Veriﬁcation of antiderivative is not currently implemented for this CAS.

[In]

integrate((c*x^2+b*x)^(3/2)/x^(7/2),x, algorithm="giac")

[Out]

(3*b*arctan(sqrt(c*x + b)/sqrt(-b))/sqrt(-b) + 2*sqrt(c*x + b) - sqrt(c*x + b)*b/(c*x))*c