### 3.79 $$\int \frac{\sqrt{b x+c x^2}}{x^{7/2}} \, dx$$

Optimal. Leaf size=86 $\frac{c^2 \tanh ^{-1}\left (\frac{\sqrt{b x+c x^2}}{\sqrt{b} \sqrt{x}}\right )}{4 b^{3/2}}-\frac{c \sqrt{b x+c x^2}}{4 b x^{3/2}}-\frac{\sqrt{b x+c x^2}}{2 x^{5/2}}$

[Out]

-Sqrt[b*x + c*x^2]/(2*x^(5/2)) - (c*Sqrt[b*x + c*x^2])/(4*b*x^(3/2)) + (c^2*ArcTanh[Sqrt[b*x + c*x^2]/(Sqrt[b]
*Sqrt[x])])/(4*b^(3/2))

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Rubi [A]  time = 0.0331965, antiderivative size = 86, 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, 672, 660, 207} $\frac{c^2 \tanh ^{-1}\left (\frac{\sqrt{b x+c x^2}}{\sqrt{b} \sqrt{x}}\right )}{4 b^{3/2}}-\frac{c \sqrt{b x+c x^2}}{4 b x^{3/2}}-\frac{\sqrt{b x+c x^2}}{2 x^{5/2}}$

Antiderivative was successfully veriﬁed.

[In]

Int[Sqrt[b*x + c*x^2]/x^(7/2),x]

[Out]

-Sqrt[b*x + c*x^2]/(2*x^(5/2)) - (c*Sqrt[b*x + c*x^2])/(4*b*x^(3/2)) + (c^2*ArcTanh[Sqrt[b*x + c*x^2]/(Sqrt[b]
*Sqrt[x])])/(4*b^(3/2))

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 672

Int[((d_.) + (e_.)*(x_))^(m_)*((a_.) + (b_.)*(x_) + (c_.)*(x_)^2)^(p_), x_Symbol] :> -Simp[(e*(d + e*x)^m*(a +
b*x + c*x^2)^(p + 1))/((m + p + 1)*(2*c*d - b*e)), x] + Dist[(c*(m + 2*p + 2))/((m + p + 1)*(2*c*d - b*e)), I
nt[(d + e*x)^(m + 1)*(a + b*x + c*x^2)^p, x], x] /; FreeQ[{a, b, c, d, e, p}, x] && NeQ[b^2 - 4*a*c, 0] && EqQ
[c*d^2 - b*d*e + a*e^2, 0] && LtQ[m, 0] && NeQ[m + 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{\sqrt{b x+c x^2}}{x^{7/2}} \, dx &=-\frac{\sqrt{b x+c x^2}}{2 x^{5/2}}+\frac{1}{4} c \int \frac{1}{x^{3/2} \sqrt{b x+c x^2}} \, dx\\ &=-\frac{\sqrt{b x+c x^2}}{2 x^{5/2}}-\frac{c \sqrt{b x+c x^2}}{4 b x^{3/2}}-\frac{c^2 \int \frac{1}{\sqrt{x} \sqrt{b x+c x^2}} \, dx}{8 b}\\ &=-\frac{\sqrt{b x+c x^2}}{2 x^{5/2}}-\frac{c \sqrt{b x+c x^2}}{4 b x^{3/2}}-\frac{c^2 \operatorname{Subst}\left (\int \frac{1}{-b+x^2} \, dx,x,\frac{\sqrt{b x+c x^2}}{\sqrt{x}}\right )}{4 b}\\ &=-\frac{\sqrt{b x+c x^2}}{2 x^{5/2}}-\frac{c \sqrt{b x+c x^2}}{4 b x^{3/2}}+\frac{c^2 \tanh ^{-1}\left (\frac{\sqrt{b x+c x^2}}{\sqrt{b} \sqrt{x}}\right )}{4 b^{3/2}}\\ \end{align*}

Mathematica [C]  time = 0.0139201, size = 42, normalized size = 0.49 $-\frac{2 c^2 (x (b+c x))^{3/2} \, _2F_1\left (\frac{3}{2},3;\frac{5}{2};\frac{c x}{b}+1\right )}{3 b^3 x^{3/2}}$

Antiderivative was successfully veriﬁed.

[In]

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

[Out]

(-2*c^2*(x*(b + c*x))^(3/2)*Hypergeometric2F1[3/2, 3, 5/2, 1 + (c*x)/b])/(3*b^3*x^(3/2))

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Maple [A]  time = 0.182, size = 71, normalized size = 0.8 \begin{align*}{\frac{1}{4}\sqrt{x \left ( cx+b \right ) } \left ({\it Artanh} \left ({\sqrt{cx+b}{\frac{1}{\sqrt{b}}}} \right ){x}^{2}{c}^{2}-xc\sqrt{cx+b}\sqrt{b}-2\,{b}^{3/2}\sqrt{cx+b} \right ){b}^{-{\frac{3}{2}}}{x}^{-{\frac{5}{2}}}{\frac{1}{\sqrt{cx+b}}}} \end{align*}

Veriﬁcation of antiderivative is not currently implemented for this CAS.

[In]

int((c*x^2+b*x)^(1/2)/x^(7/2),x)

[Out]

1/4*(x*(c*x+b))^(1/2)/b^(3/2)*(arctanh((c*x+b)^(1/2)/b^(1/2))*x^2*c^2-x*c*(c*x+b)^(1/2)*b^(1/2)-2*b^(3/2)*(c*x
+b)^(1/2))/x^(5/2)/(c*x+b)^(1/2)

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Maxima [F]  time = 0., size = 0, normalized size = 0. \begin{align*} \int \frac{\sqrt{c x^{2} + b x}}{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)^(1/2)/x^(7/2),x, algorithm="maxima")

[Out]

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

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Fricas [A]  time = 2.13271, size = 363, normalized size = 4.22 \begin{align*} \left [\frac{\sqrt{b} c^{2} x^{3} \log \left (-\frac{c x^{2} + 2 \, b x + 2 \, \sqrt{c x^{2} + b x} \sqrt{b} \sqrt{x}}{x^{2}}\right ) - 2 \,{\left (b c x + 2 \, b^{2}\right )} \sqrt{c x^{2} + b x} \sqrt{x}}{8 \, b^{2} x^{3}}, -\frac{\sqrt{-b} c^{2} x^{3} \arctan \left (\frac{\sqrt{-b} \sqrt{x}}{\sqrt{c x^{2} + b x}}\right ) +{\left (b c x + 2 \, b^{2}\right )} \sqrt{c x^{2} + b x} \sqrt{x}}{4 \, b^{2} x^{3}}\right ] \end{align*}

Veriﬁcation of antiderivative is not currently implemented for this CAS.

[In]

integrate((c*x^2+b*x)^(1/2)/x^(7/2),x, algorithm="fricas")

[Out]

[1/8*(sqrt(b)*c^2*x^3*log(-(c*x^2 + 2*b*x + 2*sqrt(c*x^2 + b*x)*sqrt(b)*sqrt(x))/x^2) - 2*(b*c*x + 2*b^2)*sqrt
(c*x^2 + b*x)*sqrt(x))/(b^2*x^3), -1/4*(sqrt(-b)*c^2*x^3*arctan(sqrt(-b)*sqrt(x)/sqrt(c*x^2 + b*x)) + (b*c*x +
2*b^2)*sqrt(c*x^2 + b*x)*sqrt(x))/(b^2*x^3)]

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

Veriﬁcation of antiderivative is not currently implemented for this CAS.

[In]

integrate((c*x**2+b*x)**(1/2)/x**(7/2),x)

[Out]

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

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Giac [A]  time = 1.18345, size = 76, normalized size = 0.88 \begin{align*} -\frac{1}{4} \, c^{2}{\left (\frac{\arctan \left (\frac{\sqrt{c x + b}}{\sqrt{-b}}\right )}{\sqrt{-b} b} + \frac{{\left (c x + b\right )}^{\frac{3}{2}} + \sqrt{c x + b} b}{b c^{2} x^{2}}\right )} \end{align*}

Veriﬁcation of antiderivative is not currently implemented for this CAS.

[In]

integrate((c*x^2+b*x)^(1/2)/x^(7/2),x, algorithm="giac")

[Out]

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