3.1.84 \(\int \frac {(A+B x) (b x+c x^2)^{3/2}}{x^2} \, dx\)

Optimal. Leaf size=126 \[ -\frac {b^2 (b B-6 A c) \tanh ^{-1}\left (\frac {\sqrt {c} x}{\sqrt {b x+c x^2}}\right )}{8 c^{3/2}}+\frac {\left (b x+c x^2\right )^{3/2} (b B-6 A c)}{3 b}+\frac {(b+2 c x) \sqrt {b x+c x^2} (b B-6 A c)}{8 c}+\frac {2 A \left (b x+c x^2\right )^{5/2}}{b x^2} \]

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Rubi [A]  time = 0.14, antiderivative size = 126, normalized size of antiderivative = 1.00, number of steps used = 5, number of rules used = 5, integrand size = 22, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.227, Rules used = {792, 664, 612, 620, 206} \begin {gather*} -\frac {b^2 (b B-6 A c) \tanh ^{-1}\left (\frac {\sqrt {c} x}{\sqrt {b x+c x^2}}\right )}{8 c^{3/2}}+\frac {\left (b x+c x^2\right )^{3/2} (b B-6 A c)}{3 b}+\frac {(b+2 c x) \sqrt {b x+c x^2} (b B-6 A c)}{8 c}+\frac {2 A \left (b x+c x^2\right )^{5/2}}{b x^2} \end {gather*}

Antiderivative was successfully verified.

[In]

Int[((A + B*x)*(b*x + c*x^2)^(3/2))/x^2,x]

[Out]

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

Rule 206

Int[((a_) + (b_.)*(x_)^2)^(-1), x_Symbol] :> Simp[(1*ArcTanh[(Rt[-b, 2]*x)/Rt[a, 2]])/(Rt[a, 2]*Rt[-b, 2]), x]
/; FreeQ[{a, b}, x] && NegQ[a/b] && (GtQ[a, 0] || LtQ[b, 0])

Rule 612

Int[((a_.) + (b_.)*(x_) + (c_.)*(x_)^2)^(p_), x_Symbol] :> Simp[((b + 2*c*x)*(a + b*x + c*x^2)^p)/(2*c*(2*p +
1)), x] - Dist[(p*(b^2 - 4*a*c))/(2*c*(2*p + 1)), Int[(a + b*x + c*x^2)^(p - 1), x], x] /; FreeQ[{a, b, c}, x]
&& NeQ[b^2 - 4*a*c, 0] && GtQ[p, 0] && IntegerQ[4*p]

Rule 620

Int[1/Sqrt[(b_.)*(x_) + (c_.)*(x_)^2], x_Symbol] :> Dist[2, Subst[Int[1/(1 - c*x^2), x], x, x/Sqrt[b*x + c*x^2
]], x] /; FreeQ[{b, c}, x]

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 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]

Rubi steps

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

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Mathematica [A]  time = 0.19, size = 109, normalized size = 0.87 \begin {gather*} \frac {\sqrt {x (b+c x)} \left (\sqrt {c} \left (2 b c (15 A+7 B x)+4 c^2 x (3 A+2 B x)+3 b^2 B\right )-\frac {3 b^{3/2} (b B-6 A c) \sinh ^{-1}\left (\frac {\sqrt {c} \sqrt {x}}{\sqrt {b}}\right )}{\sqrt {x} \sqrt {\frac {c x}{b}+1}}\right )}{24 c^{3/2}} \end {gather*}

Antiderivative was successfully verified.

[In]

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

[Out]

(Sqrt[x*(b + c*x)]*(Sqrt[c]*(3*b^2*B + 4*c^2*x*(3*A + 2*B*x) + 2*b*c*(15*A + 7*B*x)) - (3*b^(3/2)*(b*B - 6*A*c
)*ArcSinh[(Sqrt[c]*Sqrt[x])/Sqrt[b]])/(Sqrt[x]*Sqrt[1 + (c*x)/b])))/(24*c^(3/2))

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IntegrateAlgebraic [A]  time = 0.51, size = 108, normalized size = 0.86 \begin {gather*} \frac {\sqrt {b x+c x^2} \left (30 A b c+12 A c^2 x+3 b^2 B+14 b B c x+8 B c^2 x^2\right )}{24 c}+\frac {\left (b^3 B-6 A b^2 c\right ) \log \left (-2 c^{3/2} \sqrt {b x+c x^2}+b c+2 c^2 x\right )}{16 c^{3/2}} \end {gather*}

Antiderivative was successfully verified.

[In]

IntegrateAlgebraic[((A + B*x)*(b*x + c*x^2)^(3/2))/x^2,x]

[Out]

(Sqrt[b*x + c*x^2]*(3*b^2*B + 30*A*b*c + 14*b*B*c*x + 12*A*c^2*x + 8*B*c^2*x^2))/(24*c) + ((b^3*B - 6*A*b^2*c)
*Log[b*c + 2*c^2*x - 2*c^(3/2)*Sqrt[b*x + c*x^2]])/(16*c^(3/2))

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fricas [A]  time = 0.43, size = 205, normalized size = 1.63 \begin {gather*} \left [-\frac {3 \, {\left (B b^{3} - 6 \, A b^{2} c\right )} \sqrt {c} \log \left (2 \, c x + b + 2 \, \sqrt {c x^{2} + b x} \sqrt {c}\right ) - 2 \, {\left (8 \, B c^{3} x^{2} + 3 \, B b^{2} c + 30 \, A b c^{2} + 2 \, {\left (7 \, B b c^{2} + 6 \, A c^{3}\right )} x\right )} \sqrt {c x^{2} + b x}}{48 \, c^{2}}, \frac {3 \, {\left (B b^{3} - 6 \, A b^{2} c\right )} \sqrt {-c} \arctan \left (\frac {\sqrt {c x^{2} + b x} \sqrt {-c}}{c x}\right ) + {\left (8 \, B c^{3} x^{2} + 3 \, B b^{2} c + 30 \, A b c^{2} + 2 \, {\left (7 \, B b c^{2} + 6 \, A c^{3}\right )} x\right )} \sqrt {c x^{2} + b x}}{24 \, c^{2}}\right ] \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

[-1/48*(3*(B*b^3 - 6*A*b^2*c)*sqrt(c)*log(2*c*x + b + 2*sqrt(c*x^2 + b*x)*sqrt(c)) - 2*(8*B*c^3*x^2 + 3*B*b^2*
c + 30*A*b*c^2 + 2*(7*B*b*c^2 + 6*A*c^3)*x)*sqrt(c*x^2 + b*x))/c^2, 1/24*(3*(B*b^3 - 6*A*b^2*c)*sqrt(-c)*arcta
n(sqrt(c*x^2 + b*x)*sqrt(-c)/(c*x)) + (8*B*c^3*x^2 + 3*B*b^2*c + 30*A*b*c^2 + 2*(7*B*b*c^2 + 6*A*c^3)*x)*sqrt(
c*x^2 + b*x))/c^2]

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giac [A]  time = 0.24, size = 109, normalized size = 0.87 \begin {gather*} \frac {1}{24} \, \sqrt {c x^{2} + b x} {\left (2 \, {\left (4 \, B c x + \frac {7 \, B b c^{2} + 6 \, A c^{3}}{c^{2}}\right )} x + \frac {3 \, {\left (B b^{2} c + 10 \, A b c^{2}\right )}}{c^{2}}\right )} + \frac {{\left (B b^{3} - 6 \, A b^{2} c\right )} \log \left ({\left | -2 \, {\left (\sqrt {c} x - \sqrt {c x^{2} + b x}\right )} \sqrt {c} - b \right |}\right )}{16 \, c^{\frac {3}{2}}} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

1/24*sqrt(c*x^2 + b*x)*(2*(4*B*c*x + (7*B*b*c^2 + 6*A*c^3)/c^2)*x + 3*(B*b^2*c + 10*A*b*c^2)/c^2) + 1/16*(B*b^
3 - 6*A*b^2*c)*log(abs(-2*(sqrt(c)*x - sqrt(c*x^2 + b*x))*sqrt(c) - b))/c^(3/2)

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maple [A]  time = 0.06, size = 187, normalized size = 1.48 \begin {gather*} \frac {3 A \,b^{2} \ln \left (\frac {c x +\frac {b}{2}}{\sqrt {c}}+\sqrt {c \,x^{2}+b x}\right )}{8 \sqrt {c}}-\frac {B \,b^{3} \ln \left (\frac {c x +\frac {b}{2}}{\sqrt {c}}+\sqrt {c \,x^{2}+b x}\right )}{16 c^{\frac {3}{2}}}-\frac {3 \sqrt {c \,x^{2}+b x}\, A c x}{2}+\frac {\sqrt {c \,x^{2}+b x}\, B b x}{4}-\frac {3 \sqrt {c \,x^{2}+b x}\, A b}{4}+\frac {\sqrt {c \,x^{2}+b x}\, B \,b^{2}}{8 c}-\frac {2 \left (c \,x^{2}+b x \right )^{\frac {3}{2}} A c}{b}+\frac {\left (c \,x^{2}+b x \right )^{\frac {3}{2}} B}{3}+\frac {2 \left (c \,x^{2}+b x \right )^{\frac {5}{2}} A}{b \,x^{2}} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

2*A*(c*x^2+b*x)^(5/2)/b/x^2-2*A/b*c*(c*x^2+b*x)^(3/2)-3/2*A*c*(c*x^2+b*x)^(1/2)*x-3/4*A*b*(c*x^2+b*x)^(1/2)+3/
8*A*b^2/c^(1/2)*ln((c*x+1/2*b)/c^(1/2)+(c*x^2+b*x)^(1/2))+1/3*B*(c*x^2+b*x)^(3/2)+1/4*B*b*(c*x^2+b*x)^(1/2)*x+
1/8*B/c*(c*x^2+b*x)^(1/2)*b^2-1/16*B*b^3/c^(3/2)*ln((c*x+1/2*b)/c^(1/2)+(c*x^2+b*x)^(1/2))

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maxima [A]  time = 0.90, size = 147, normalized size = 1.17 \begin {gather*} \frac {1}{4} \, \sqrt {c x^{2} + b x} B b x - \frac {B b^{3} \log \left (2 \, c x + b + 2 \, \sqrt {c x^{2} + b x} \sqrt {c}\right )}{16 \, c^{\frac {3}{2}}} + \frac {3 \, A b^{2} \log \left (2 \, c x + b + 2 \, \sqrt {c x^{2} + b x} \sqrt {c}\right )}{8 \, \sqrt {c}} + \frac {1}{3} \, {\left (c x^{2} + b x\right )}^{\frac {3}{2}} B + \frac {3}{4} \, \sqrt {c x^{2} + b x} A b + \frac {\sqrt {c x^{2} + b x} B b^{2}}{8 \, c} + \frac {{\left (c x^{2} + b x\right )}^{\frac {3}{2}} A}{2 \, x} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

1/4*sqrt(c*x^2 + b*x)*B*b*x - 1/16*B*b^3*log(2*c*x + b + 2*sqrt(c*x^2 + b*x)*sqrt(c))/c^(3/2) + 3/8*A*b^2*log(
2*c*x + b + 2*sqrt(c*x^2 + b*x)*sqrt(c))/sqrt(c) + 1/3*(c*x^2 + b*x)^(3/2)*B + 3/4*sqrt(c*x^2 + b*x)*A*b + 1/8
*sqrt(c*x^2 + b*x)*B*b^2/c + 1/2*(c*x^2 + b*x)^(3/2)*A/x

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mupad [F]  time = 0.00, size = -1, normalized size = -0.01 \begin {gather*} \int \frac {{\left (c\,x^2+b\,x\right )}^{3/2}\,\left (A+B\,x\right )}{x^2} \,d x \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

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

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sympy [F]  time = 0.00, size = 0, normalized size = 0.00 \begin {gather*} \int \frac {\left (x \left (b + c x\right )\right )^{\frac {3}{2}} \left (A + B x\right )}{x^{2}}\, dx \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

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

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