3.247 \(\int \frac{\sqrt{x} (A+B x)}{(b x+c x^2)^{5/2}} \, dx\)

Optimal. Leaf size=146 \[ \frac{\sqrt{x} (2 b B-5 A c)}{b^3 \sqrt{b x+c x^2}}+\frac{2 b B-5 A c}{3 b^2 c \sqrt{x} \sqrt{b x+c x^2}}-\frac{(2 b B-5 A c) \tanh ^{-1}\left (\frac{\sqrt{b x+c x^2}}{\sqrt{b} \sqrt{x}}\right )}{b^{7/2}}-\frac{2 \sqrt{x} (b B-A c)}{3 b c \left (b x+c x^2\right )^{3/2}} \]

[Out]

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

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Rubi [A]  time = 0.118561, antiderivative size = 146, normalized size of antiderivative = 1., number of steps used = 5, number of rules used = 5, integrand size = 24, \(\frac{\text{number of rules}}{\text{integrand size}}\) = 0.208, Rules used = {788, 672, 666, 660, 207} \[ \frac{\sqrt{x} (2 b B-5 A c)}{b^3 \sqrt{b x+c x^2}}+\frac{2 b B-5 A c}{3 b^2 c \sqrt{x} \sqrt{b x+c x^2}}-\frac{(2 b B-5 A c) \tanh ^{-1}\left (\frac{\sqrt{b x+c x^2}}{\sqrt{b} \sqrt{x}}\right )}{b^{7/2}}-\frac{2 \sqrt{x} (b B-A c)}{3 b c \left (b x+c x^2\right )^{3/2}} \]

Antiderivative was successfully verified.

[In]

Int[(Sqrt[x]*(A + B*x))/(b*x + c*x^2)^(5/2),x]

[Out]

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

Rule 788

Int[((d_.) + (e_.)*(x_))^(m_)*((f_.) + (g_.)*(x_))*((a_.) + (b_.)*(x_) + (c_.)*(x_)^2)^(p_), x_Symbol] :> Simp
[((g*(c*d - b*e) + c*e*f)*(d + e*x)^m*(a + b*x + c*x^2)^(p + 1))/(c*(p + 1)*(2*c*d - b*e)), x] - Dist[(e*(m*(g
*(c*d - b*e) + c*e*f) + e*(p + 1)*(2*c*f - b*g)))/(c*(p + 1)*(2*c*d - b*e)), Int[(d + e*x)^(m - 1)*(a + b*x +
c*x^2)^(p + 1), x], x] /; FreeQ[{a, b, c, d, e, f, g}, x] && NeQ[b^2 - 4*a*c, 0] && EqQ[c*d^2 - b*d*e + a*e^2,
0] && LtQ[p, -1] && GtQ[m, 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 666

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

Mathematica [C]  time = 0.026761, size = 55, normalized size = 0.38 \[ \frac{\sqrt{x} \left (x (2 b B-5 A c) \, _2F_1\left (-\frac{3}{2},1;-\frac{1}{2};\frac{c x}{b}+1\right )-3 A b\right )}{3 b^2 (x (b+c x))^{3/2}} \]

Antiderivative was successfully verified.

[In]

Integrate[(Sqrt[x]*(A + B*x))/(b*x + c*x^2)^(5/2),x]

[Out]

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

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Maple [A]  time = 0.021, size = 175, normalized size = 1.2 \begin{align*}{\frac{1}{3\, \left ( cx+b \right ) ^{2}}\sqrt{x \left ( cx+b \right ) } \left ( 15\,A{\it Artanh} \left ({\frac{\sqrt{cx+b}}{\sqrt{b}}} \right ) \sqrt{cx+b}{x}^{2}{c}^{2}-6\,B{\it Artanh} \left ({\frac{\sqrt{cx+b}}{\sqrt{b}}} \right ) \sqrt{cx+b}{x}^{2}bc+15\,A{\it Artanh} \left ({\frac{\sqrt{cx+b}}{\sqrt{b}}} \right ) xbc\sqrt{cx+b}-15\,A\sqrt{b}{x}^{2}{c}^{2}-6\,B{\it Artanh} \left ({\frac{\sqrt{cx+b}}{\sqrt{b}}} \right ) x{b}^{2}\sqrt{cx+b}+6\,B{b}^{3/2}{x}^{2}c-20\,A{b}^{3/2}xc+8\,B{b}^{5/2}x-3\,A{b}^{5/2} \right ){x}^{-{\frac{3}{2}}}{b}^{-{\frac{7}{2}}}} \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

int((B*x+A)*x^(1/2)/(c*x^2+b*x)^(5/2),x)

[Out]

1/3*(x*(c*x+b))^(1/2)*(15*A*arctanh((c*x+b)^(1/2)/b^(1/2))*(c*x+b)^(1/2)*x^2*c^2-6*B*arctanh((c*x+b)^(1/2)/b^(
1/2))*(c*x+b)^(1/2)*x^2*b*c+15*A*arctanh((c*x+b)^(1/2)/b^(1/2))*x*b*c*(c*x+b)^(1/2)-15*A*b^(1/2)*x^2*c^2-6*B*a
rctanh((c*x+b)^(1/2)/b^(1/2))*x*b^2*(c*x+b)^(1/2)+6*B*b^(3/2)*x^2*c-20*A*b^(3/2)*x*c+8*B*b^(5/2)*x-3*A*b^(5/2)
)/x^(3/2)/(c*x+b)^2/b^(7/2)

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Maxima [F]  time = 0., size = 0, normalized size = 0. \begin{align*} \int \frac{{\left (B x + A\right )} \sqrt{x}}{{\left (c x^{2} + b x\right )}^{\frac{5}{2}}}\,{d x} \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate((B*x+A)*x^(1/2)/(c*x^2+b*x)^(5/2),x, algorithm="maxima")

[Out]

integrate((B*x + A)*sqrt(x)/(c*x^2 + b*x)^(5/2), x)

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Fricas [A]  time = 1.91668, size = 806, normalized size = 5.52 \begin{align*} \left [-\frac{3 \,{\left ({\left (2 \, B b c^{2} - 5 \, A c^{3}\right )} x^{4} + 2 \,{\left (2 \, B b^{2} c - 5 \, A b c^{2}\right )} x^{3} +{\left (2 \, B b^{3} - 5 \, A b^{2} c\right )} x^{2}\right )} \sqrt{b} \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 (3 \, A b^{3} - 3 \,{\left (2 \, B b^{2} c - 5 \, A b c^{2}\right )} x^{2} - 4 \,{\left (2 \, B b^{3} - 5 \, A b^{2} c\right )} x\right )} \sqrt{c x^{2} + b x} \sqrt{x}}{6 \,{\left (b^{4} c^{2} x^{4} + 2 \, b^{5} c x^{3} + b^{6} x^{2}\right )}}, \frac{3 \,{\left ({\left (2 \, B b c^{2} - 5 \, A c^{3}\right )} x^{4} + 2 \,{\left (2 \, B b^{2} c - 5 \, A b c^{2}\right )} x^{3} +{\left (2 \, B b^{3} - 5 \, A b^{2} c\right )} x^{2}\right )} \sqrt{-b} \arctan \left (\frac{\sqrt{-b} \sqrt{x}}{\sqrt{c x^{2} + b x}}\right ) -{\left (3 \, A b^{3} - 3 \,{\left (2 \, B b^{2} c - 5 \, A b c^{2}\right )} x^{2} - 4 \,{\left (2 \, B b^{3} - 5 \, A b^{2} c\right )} x\right )} \sqrt{c x^{2} + b x} \sqrt{x}}{3 \,{\left (b^{4} c^{2} x^{4} + 2 \, b^{5} c x^{3} + b^{6} x^{2}\right )}}\right ] \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate((B*x+A)*x^(1/2)/(c*x^2+b*x)^(5/2),x, algorithm="fricas")

[Out]

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

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

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate((B*x+A)*x**(1/2)/(c*x**2+b*x)**(5/2),x)

[Out]

Integral(sqrt(x)*(A + B*x)/(x*(b + c*x))**(5/2), x)

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Giac [A]  time = 1.26472, size = 122, normalized size = 0.84 \begin{align*} \frac{{\left (2 \, B b - 5 \, A c\right )} \arctan \left (\frac{\sqrt{c x + b}}{\sqrt{-b}}\right )}{\sqrt{-b} b^{3}} - \frac{\sqrt{c x + b} A}{b^{3} x} + \frac{2 \,{\left (3 \,{\left (c x + b\right )} B b + B b^{2} - 6 \,{\left (c x + b\right )} A c - A b c\right )}}{3 \,{\left (c x + b\right )}^{\frac{3}{2}} b^{3}} \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate((B*x+A)*x^(1/2)/(c*x^2+b*x)^(5/2),x, algorithm="giac")

[Out]

(2*B*b - 5*A*c)*arctan(sqrt(c*x + b)/sqrt(-b))/(sqrt(-b)*b^3) - sqrt(c*x + b)*A/(b^3*x) + 2/3*(3*(c*x + b)*B*b
+ B*b^2 - 6*(c*x + b)*A*c - A*b*c)/((c*x + b)^(3/2)*b^3)