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

Optimal. Leaf size=142 \[ -\frac{c^2 (6 b B-5 A c) \tanh ^{-1}\left (\frac{\sqrt{b x+c x^2}}{\sqrt{b} \sqrt{x}}\right )}{8 b^{7/2}}+\frac{c \sqrt{b x+c x^2} (6 b B-5 A c)}{8 b^3 x^{3/2}}-\frac{\sqrt{b x+c x^2} (6 b B-5 A c)}{12 b^2 x^{5/2}}-\frac{A \sqrt{b x+c x^2}}{3 b x^{7/2}} \]

[Out]

-(A*Sqrt[b*x + c*x^2])/(3*b*x^(7/2)) - ((6*b*B - 5*A*c)*Sqrt[b*x + c*x^2])/(12*b^2*x^(5/2)) + (c*(6*b*B - 5*A*
c)*Sqrt[b*x + c*x^2])/(8*b^3*x^(3/2)) - (c^2*(6*b*B - 5*A*c)*ArcTanh[Sqrt[b*x + c*x^2]/(Sqrt[b]*Sqrt[x])])/(8*
b^(7/2))

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Rubi [A]  time = 0.120299, antiderivative size = 142, normalized size of antiderivative = 1., number of steps used = 5, number of rules used = 4, integrand size = 24, \(\frac{\text{number of rules}}{\text{integrand size}}\) = 0.167, Rules used = {792, 672, 660, 207} \[ -\frac{c^2 (6 b B-5 A c) \tanh ^{-1}\left (\frac{\sqrt{b x+c x^2}}{\sqrt{b} \sqrt{x}}\right )}{8 b^{7/2}}+\frac{c \sqrt{b x+c x^2} (6 b B-5 A c)}{8 b^3 x^{3/2}}-\frac{\sqrt{b x+c x^2} (6 b B-5 A c)}{12 b^2 x^{5/2}}-\frac{A \sqrt{b x+c x^2}}{3 b x^{7/2}} \]

Antiderivative was successfully verified.

[In]

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

[Out]

-(A*Sqrt[b*x + c*x^2])/(3*b*x^(7/2)) - ((6*b*B - 5*A*c)*Sqrt[b*x + c*x^2])/(12*b^2*x^(5/2)) + (c*(6*b*B - 5*A*
c)*Sqrt[b*x + c*x^2])/(8*b^3*x^(3/2)) - (c^2*(6*b*B - 5*A*c)*ArcTanh[Sqrt[b*x + c*x^2]/(Sqrt[b]*Sqrt[x])])/(8*
b^(7/2))

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 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{A+B x}{x^{7/2} \sqrt{b x+c x^2}} \, dx &=-\frac{A \sqrt{b x+c x^2}}{3 b x^{7/2}}+\frac{\left (-\frac{7}{2} (-b B+A c)+\frac{1}{2} (-b B+2 A c)\right ) \int \frac{1}{x^{5/2} \sqrt{b x+c x^2}} \, dx}{3 b}\\ &=-\frac{A \sqrt{b x+c x^2}}{3 b x^{7/2}}-\frac{(6 b B-5 A c) \sqrt{b x+c x^2}}{12 b^2 x^{5/2}}-\frac{(c (6 b B-5 A c)) \int \frac{1}{x^{3/2} \sqrt{b x+c x^2}} \, dx}{8 b^2}\\ &=-\frac{A \sqrt{b x+c x^2}}{3 b x^{7/2}}-\frac{(6 b B-5 A c) \sqrt{b x+c x^2}}{12 b^2 x^{5/2}}+\frac{c (6 b B-5 A c) \sqrt{b x+c x^2}}{8 b^3 x^{3/2}}+\frac{\left (c^2 (6 b B-5 A c)\right ) \int \frac{1}{\sqrt{x} \sqrt{b x+c x^2}} \, dx}{16 b^3}\\ &=-\frac{A \sqrt{b x+c x^2}}{3 b x^{7/2}}-\frac{(6 b B-5 A c) \sqrt{b x+c x^2}}{12 b^2 x^{5/2}}+\frac{c (6 b B-5 A c) \sqrt{b x+c x^2}}{8 b^3 x^{3/2}}+\frac{\left (c^2 (6 b B-5 A c)\right ) \operatorname{Subst}\left (\int \frac{1}{-b+x^2} \, dx,x,\frac{\sqrt{b x+c x^2}}{\sqrt{x}}\right )}{8 b^3}\\ &=-\frac{A \sqrt{b x+c x^2}}{3 b x^{7/2}}-\frac{(6 b B-5 A c) \sqrt{b x+c x^2}}{12 b^2 x^{5/2}}+\frac{c (6 b B-5 A c) \sqrt{b x+c x^2}}{8 b^3 x^{3/2}}-\frac{c^2 (6 b B-5 A c) \tanh ^{-1}\left (\frac{\sqrt{b x+c x^2}}{\sqrt{b} \sqrt{x}}\right )}{8 b^{7/2}}\\ \end{align*}

Mathematica [C]  time = 0.0276484, size = 61, normalized size = 0.43 \[ -\frac{\sqrt{x (b+c x)} \left (A b^3+c^2 x^3 (6 b B-5 A c) \, _2F_1\left (\frac{1}{2},3;\frac{3}{2};\frac{c x}{b}+1\right )\right )}{3 b^4 x^{7/2}} \]

Antiderivative was successfully verified.

[In]

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

[Out]

-(Sqrt[x*(b + c*x)]*(A*b^3 + c^2*(6*b*B - 5*A*c)*x^3*Hypergeometric2F1[1/2, 3, 3/2, 1 + (c*x)/b]))/(3*b^4*x^(7
/2))

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Maple [A]  time = 0.017, size = 147, normalized size = 1. \begin{align*}{\frac{1}{24}\sqrt{x \left ( cx+b \right ) } \left ( 15\,A{\it Artanh} \left ({\frac{\sqrt{cx+b}}{\sqrt{b}}} \right ){x}^{3}{c}^{3}-18\,B{\it Artanh} \left ({\frac{\sqrt{cx+b}}{\sqrt{b}}} \right ){x}^{3}b{c}^{2}-15\,A{x}^{2}{c}^{2}\sqrt{b}\sqrt{cx+b}+18\,B{x}^{2}{b}^{3/2}c\sqrt{cx+b}+10\,Ax{b}^{3/2}c\sqrt{cx+b}-12\,Bx{b}^{5/2}\sqrt{cx+b}-8\,A{b}^{5/2}\sqrt{cx+b} \right ){b}^{-{\frac{7}{2}}}{x}^{-{\frac{7}{2}}}{\frac{1}{\sqrt{cx+b}}}} \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

1/24*(x*(c*x+b))^(1/2)/b^(7/2)*(15*A*arctanh((c*x+b)^(1/2)/b^(1/2))*x^3*c^3-18*B*arctanh((c*x+b)^(1/2)/b^(1/2)
)*x^3*b*c^2-15*A*x^2*c^2*b^(1/2)*(c*x+b)^(1/2)+18*B*x^2*b^(3/2)*c*(c*x+b)^(1/2)+10*A*x*b^(3/2)*c*(c*x+b)^(1/2)
-12*B*x*b^(5/2)*(c*x+b)^(1/2)-8*A*b^(5/2)*(c*x+b)^(1/2))/x^(7/2)/(c*x+b)^(1/2)

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

[Out]

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

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Fricas [A]  time = 1.98262, size = 566, normalized size = 3.99 \begin{align*} \left [-\frac{3 \,{\left (6 \, B b c^{2} - 5 \, A c^{3}\right )} \sqrt{b} x^{4} \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 (8 \, A b^{3} - 3 \,{\left (6 \, B b^{2} c - 5 \, A b c^{2}\right )} x^{2} + 2 \,{\left (6 \, B b^{3} - 5 \, A b^{2} c\right )} x\right )} \sqrt{c x^{2} + b x} \sqrt{x}}{48 \, b^{4} x^{4}}, \frac{3 \,{\left (6 \, B b c^{2} - 5 \, A c^{3}\right )} \sqrt{-b} x^{4} \arctan \left (\frac{\sqrt{-b} \sqrt{x}}{\sqrt{c x^{2} + b x}}\right ) -{\left (8 \, A b^{3} - 3 \,{\left (6 \, B b^{2} c - 5 \, A b c^{2}\right )} x^{2} + 2 \,{\left (6 \, B b^{3} - 5 \, A b^{2} c\right )} x\right )} \sqrt{c x^{2} + b x} \sqrt{x}}{24 \, b^{4} x^{4}}\right ] \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

[-1/48*(3*(6*B*b*c^2 - 5*A*c^3)*sqrt(b)*x^4*log(-(c*x^2 + 2*b*x + 2*sqrt(c*x^2 + b*x)*sqrt(b)*sqrt(x))/x^2) +
2*(8*A*b^3 - 3*(6*B*b^2*c - 5*A*b*c^2)*x^2 + 2*(6*B*b^3 - 5*A*b^2*c)*x)*sqrt(c*x^2 + b*x)*sqrt(x))/(b^4*x^4),
1/24*(3*(6*B*b*c^2 - 5*A*c^3)*sqrt(-b)*x^4*arctan(sqrt(-b)*sqrt(x)/sqrt(c*x^2 + b*x)) - (8*A*b^3 - 3*(6*B*b^2*
c - 5*A*b*c^2)*x^2 + 2*(6*B*b^3 - 5*A*b^2*c)*x)*sqrt(c*x^2 + b*x)*sqrt(x))/(b^4*x^4)]

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

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

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

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Giac [A]  time = 1.23999, size = 194, normalized size = 1.37 \begin{align*} \frac{\frac{3 \,{\left (6 \, B b c^{3} - 5 \, A c^{4}\right )} \arctan \left (\frac{\sqrt{c x + b}}{\sqrt{-b}}\right )}{\sqrt{-b} b^{3}} + \frac{18 \,{\left (c x + b\right )}^{\frac{5}{2}} B b c^{3} - 48 \,{\left (c x + b\right )}^{\frac{3}{2}} B b^{2} c^{3} + 30 \, \sqrt{c x + b} B b^{3} c^{3} - 15 \,{\left (c x + b\right )}^{\frac{5}{2}} A c^{4} + 40 \,{\left (c x + b\right )}^{\frac{3}{2}} A b c^{4} - 33 \, \sqrt{c x + b} A b^{2} c^{4}}{b^{3} c^{3} x^{3}}}{24 \, c} \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

1/24*(3*(6*B*b*c^3 - 5*A*c^4)*arctan(sqrt(c*x + b)/sqrt(-b))/(sqrt(-b)*b^3) + (18*(c*x + b)^(5/2)*B*b*c^3 - 48
*(c*x + b)^(3/2)*B*b^2*c^3 + 30*sqrt(c*x + b)*B*b^3*c^3 - 15*(c*x + b)^(5/2)*A*c^4 + 40*(c*x + b)^(3/2)*A*b*c^
4 - 33*sqrt(c*x + b)*A*b^2*c^4)/(b^3*c^3*x^3))/c