3.201 \(\int \frac{(A+B x) \sqrt{b x+c x^2}}{x^{7/2}} \, dx\)

Optimal. Leaf size=105 \[ -\frac{c (4 b B-A c) \tanh ^{-1}\left (\frac{\sqrt{b x+c x^2}}{\sqrt{b} \sqrt{x}}\right )}{4 b^{3/2}}-\frac{\sqrt{b x+c x^2} (4 b B-A c)}{4 b x^{3/2}}-\frac{A \left (b x+c x^2\right )^{3/2}}{2 b x^{7/2}} \]

[Out]

-((4*b*B - A*c)*Sqrt[b*x + c*x^2])/(4*b*x^(3/2)) - (A*(b*x + c*x^2)^(3/2))/(2*b*x^(7/2)) - (c*(4*b*B - A*c)*Ar
cTanh[Sqrt[b*x + c*x^2]/(Sqrt[b]*Sqrt[x])])/(4*b^(3/2))

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Rubi [A]  time = 0.0924073, antiderivative size = 105, normalized size of antiderivative = 1., number of steps used = 4, number of rules used = 4, integrand size = 24, \(\frac{\text{number of rules}}{\text{integrand size}}\) = 0.167, Rules used = {792, 662, 660, 207} \[ -\frac{c (4 b B-A c) \tanh ^{-1}\left (\frac{\sqrt{b x+c x^2}}{\sqrt{b} \sqrt{x}}\right )}{4 b^{3/2}}-\frac{\sqrt{b x+c x^2} (4 b B-A c)}{4 b x^{3/2}}-\frac{A \left (b x+c x^2\right )^{3/2}}{2 b x^{7/2}} \]

Antiderivative was successfully verified.

[In]

Int[((A + B*x)*Sqrt[b*x + c*x^2])/x^(7/2),x]

[Out]

-((4*b*B - A*c)*Sqrt[b*x + c*x^2])/(4*b*x^(3/2)) - (A*(b*x + c*x^2)^(3/2))/(2*b*x^(7/2)) - (c*(4*b*B - A*c)*Ar
cTanh[Sqrt[b*x + c*x^2]/(Sqrt[b]*Sqrt[x])])/(4*b^(3/2))

Rule 792

Int[((d_.) + (e_.)*(x_))^(m_)*((f_.) + (g_.)*(x_))*((a_.) + (b_.)*(x_) + (c_.)*(x_)^2)^(p_), x_Symbol] :> Simp
[((d*g - e*f)*(d + e*x)^m*(a + b*x + c*x^2)^(p + 1))/((2*c*d - b*e)*(m + p + 1)), x] + Dist[(m*(g*(c*d - b*e)
+ c*e*f) + e*(p + 1)*(2*c*f - b*g))/(e*(2*c*d - b*e)*(m + p + 1)), Int[(d + e*x)^(m + 1)*(a + b*x + c*x^2)^p,
x], x] /; FreeQ[{a, b, c, d, e, f, g, m, p}, x] && NeQ[b^2 - 4*a*c, 0] && EqQ[c*d^2 - b*d*e + a*e^2, 0] && ((L
tQ[m, -1] &&  !IGtQ[m + p + 1, 0]) || (LtQ[m, 0] && LtQ[p, -1]) || EqQ[m + 2*p + 2, 0]) && NeQ[m + p + 1, 0]

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 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{(A+B x) \sqrt{b x+c x^2}}{x^{7/2}} \, dx &=-\frac{A \left (b x+c x^2\right )^{3/2}}{2 b x^{7/2}}+\frac{\left (-\frac{7}{2} (-b B+A c)+\frac{3}{2} (-b B+2 A c)\right ) \int \frac{\sqrt{b x+c x^2}}{x^{5/2}} \, dx}{2 b}\\ &=-\frac{(4 b B-A c) \sqrt{b x+c x^2}}{4 b x^{3/2}}-\frac{A \left (b x+c x^2\right )^{3/2}}{2 b x^{7/2}}+\frac{(c (4 b B-A c)) \int \frac{1}{\sqrt{x} \sqrt{b x+c x^2}} \, dx}{8 b}\\ &=-\frac{(4 b B-A c) \sqrt{b x+c x^2}}{4 b x^{3/2}}-\frac{A \left (b x+c x^2\right )^{3/2}}{2 b x^{7/2}}+\frac{(c (4 b B-A c)) \operatorname{Subst}\left (\int \frac{1}{-b+x^2} \, dx,x,\frac{\sqrt{b x+c x^2}}{\sqrt{x}}\right )}{4 b}\\ &=-\frac{(4 b B-A c) \sqrt{b x+c x^2}}{4 b x^{3/2}}-\frac{A \left (b x+c x^2\right )^{3/2}}{2 b x^{7/2}}-\frac{c (4 b B-A c) \tanh ^{-1}\left (\frac{\sqrt{b x+c x^2}}{\sqrt{b} \sqrt{x}}\right )}{4 b^{3/2}}\\ \end{align*}

Mathematica [A]  time = 0.103393, size = 83, normalized size = 0.79 \[ -\frac{c x^2 \sqrt{\frac{c x}{b}+1} (4 b B-A c) \tanh ^{-1}\left (\sqrt{\frac{c x}{b}+1}\right )+(b+c x) (2 A b+A c x+4 b B x)}{4 b x^{3/2} \sqrt{x (b+c x)}} \]

Antiderivative was successfully verified.

[In]

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

[Out]

-((b + c*x)*(2*A*b + 4*b*B*x + A*c*x) + c*(4*b*B - A*c)*x^2*Sqrt[1 + (c*x)/b]*ArcTanh[Sqrt[1 + (c*x)/b]])/(4*b
*x^(3/2)*Sqrt[x*(b + c*x)])

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Maple [A]  time = 0.017, size = 108, normalized size = 1. \begin{align*}{\frac{1}{4}\sqrt{x \left ( cx+b \right ) } \left ( A{\it Artanh} \left ({\sqrt{cx+b}{\frac{1}{\sqrt{b}}}} \right ){x}^{2}{c}^{2}-4\,B{\it Artanh} \left ({\frac{\sqrt{cx+b}}{\sqrt{b}}} \right ){x}^{2}bc-Axc\sqrt{cx+b}\sqrt{b}-4\,Bx{b}^{3/2}\sqrt{cx+b}-2\,A{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((B*x+A)*(c*x^2+b*x)^(1/2)/x^(7/2),x)

[Out]

1/4*(x*(c*x+b))^(1/2)/b^(3/2)*(A*arctanh((c*x+b)^(1/2)/b^(1/2))*x^2*c^2-4*B*arctanh((c*x+b)^(1/2)/b^(1/2))*x^2
*b*c-A*x*c*(c*x+b)^(1/2)*b^(1/2)-4*B*x*b^(3/2)*(c*x+b)^(1/2)-2*A*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}{\left (B x + A\right )}}{x^{\frac{7}{2}}}\,{d x} \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

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

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Fricas [A]  time = 1.58545, size = 444, normalized size = 4.23 \begin{align*} \left [-\frac{{\left (4 \, B b c - A c^{2}\right )} \sqrt{b} 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 (2 \, A b^{2} +{\left (4 \, B b^{2} + A b c\right )} x\right )} \sqrt{c x^{2} + b x} \sqrt{x}}{8 \, b^{2} x^{3}}, \frac{{\left (4 \, B b c - A c^{2}\right )} \sqrt{-b} x^{3} \arctan \left (\frac{\sqrt{-b} \sqrt{x}}{\sqrt{c x^{2} + b x}}\right ) -{\left (2 \, A b^{2} +{\left (4 \, B b^{2} + A b c\right )} x\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((B*x+A)*(c*x^2+b*x)^(1/2)/x^(7/2),x, algorithm="fricas")

[Out]

[-1/8*((4*B*b*c - A*c^2)*sqrt(b)*x^3*log(-(c*x^2 + 2*b*x + 2*sqrt(c*x^2 + b*x)*sqrt(b)*sqrt(x))/x^2) + 2*(2*A*
b^2 + (4*B*b^2 + A*b*c)*x)*sqrt(c*x^2 + b*x)*sqrt(x))/(b^2*x^3), 1/4*((4*B*b*c - A*c^2)*sqrt(-b)*x^3*arctan(sq
rt(-b)*sqrt(x)/sqrt(c*x^2 + b*x)) - (2*A*b^2 + (4*B*b^2 + A*b*c)*x)*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 )} \left (A + B x\right )}{x^{\frac{7}{2}}}\, dx \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

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

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Giac [A]  time = 1.27428, size = 149, normalized size = 1.42 \begin{align*} \frac{\frac{{\left (4 \, B b c^{2} - A c^{3}\right )} \arctan \left (\frac{\sqrt{c x + b}}{\sqrt{-b}}\right )}{\sqrt{-b} b} - \frac{4 \,{\left (c x + b\right )}^{\frac{3}{2}} B b c^{2} - 4 \, \sqrt{c x + b} B b^{2} c^{2} +{\left (c x + b\right )}^{\frac{3}{2}} A c^{3} + \sqrt{c x + b} A b c^{3}}{b c^{2} x^{2}}}{4 \, c} \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

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