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

Optimal. Leaf size=118 \[ \frac {\left (b x+c x^2\right )^{3/2} (4 A c+b B)}{2 b x}+\frac {3}{4} \sqrt {b x+c x^2} (4 A c+b B)+\frac {3 b (4 A c+b B) \tanh ^{-1}\left (\frac {\sqrt {c} x}{\sqrt {b x+c x^2}}\right )}{4 \sqrt {c}}-\frac {2 A \left (b x+c x^2\right )^{5/2}}{b x^3} \]

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Rubi [A]  time = 0.12, antiderivative size = 118, normalized size of antiderivative = 1.00, number of steps used = 5, number of rules used = 4, integrand size = 22, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.182, Rules used = {792, 664, 620, 206} \begin {gather*} \frac {\left (b x+c x^2\right )^{3/2} (4 A c+b B)}{2 b x}+\frac {3}{4} \sqrt {b x+c x^2} (4 A c+b B)+\frac {3 b (4 A c+b B) \tanh ^{-1}\left (\frac {\sqrt {c} x}{\sqrt {b x+c x^2}}\right )}{4 \sqrt {c}}-\frac {2 A \left (b x+c x^2\right )^{5/2}}{b x^3} \end {gather*}

Antiderivative was successfully verified.

[In]

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

[Out]

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

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

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Mathematica [A]  time = 0.17, size = 94, normalized size = 0.80 \begin {gather*} \frac {\sqrt {x (b+c x)} \left (\frac {3 \sqrt {b} \sqrt {x} (4 A c+b B) \sinh ^{-1}\left (\frac {\sqrt {c} \sqrt {x}}{\sqrt {b}}\right )}{\sqrt {c} \sqrt {\frac {c x}{b}+1}}+A (4 c x-8 b)+B x (5 b+2 c x)\right )}{4 x} \end {gather*}

Antiderivative was successfully verified.

[In]

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

[Out]

(Sqrt[x*(b + c*x)]*(B*x*(5*b + 2*c*x) + A*(-8*b + 4*c*x) + (3*Sqrt[b]*(b*B + 4*A*c)*Sqrt[x]*ArcSinh[(Sqrt[c]*S
qrt[x])/Sqrt[b]])/(Sqrt[c]*Sqrt[1 + (c*x)/b])))/(4*x)

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IntegrateAlgebraic [A]  time = 0.50, size = 90, normalized size = 0.76 \begin {gather*} \frac {\sqrt {b x+c x^2} \left (-8 A b+4 A c x+5 b B x+2 B c x^2\right )}{4 x}-\frac {3 \left (4 A b c+b^2 B\right ) \log \left (-2 \sqrt {c} \sqrt {b x+c x^2}+b+2 c x\right )}{8 \sqrt {c}} \end {gather*}

Antiderivative was successfully verified.

[In]

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

[Out]

(Sqrt[b*x + c*x^2]*(-8*A*b + 5*b*B*x + 4*A*c*x + 2*B*c*x^2))/(4*x) - (3*(b^2*B + 4*A*b*c)*Log[b + 2*c*x - 2*Sq
rt[c]*Sqrt[b*x + c*x^2]])/(8*Sqrt[c])

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fricas [A]  time = 0.43, size = 186, normalized size = 1.58 \begin {gather*} \left [\frac {3 \, {\left (B b^{2} + 4 \, A b c\right )} \sqrt {c} x \log \left (2 \, c x + b + 2 \, \sqrt {c x^{2} + b x} \sqrt {c}\right ) + 2 \, {\left (2 \, B c^{2} x^{2} - 8 \, A b c + {\left (5 \, B b c + 4 \, A c^{2}\right )} x\right )} \sqrt {c x^{2} + b x}}{8 \, c x}, -\frac {3 \, {\left (B b^{2} + 4 \, A b c\right )} \sqrt {-c} x \arctan \left (\frac {\sqrt {c x^{2} + b x} \sqrt {-c}}{c x}\right ) - {\left (2 \, B c^{2} x^{2} - 8 \, A b c + {\left (5 \, B b c + 4 \, A c^{2}\right )} x\right )} \sqrt {c x^{2} + b x}}{4 \, c x}\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^3,x, algorithm="fricas")

[Out]

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

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giac [A]  time = 0.26, size = 109, normalized size = 0.92 \begin {gather*} \frac {2 \, A b^{2}}{\sqrt {c} x - \sqrt {c x^{2} + b x}} + \frac {1}{4} \, {\left (2 \, B c x + \frac {5 \, B b c + 4 \, A c^{2}}{c}\right )} \sqrt {c x^{2} + b x} - \frac {3 \, {\left (B b^{2} + 4 \, A b c\right )} \log \left ({\left | -2 \, {\left (\sqrt {c} x - \sqrt {c x^{2} + b x}\right )} \sqrt {c} - b \right |}\right )}{8 \, \sqrt {c}} \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^3,x, algorithm="giac")

[Out]

2*A*b^2/(sqrt(c)*x - sqrt(c*x^2 + b*x)) + 1/4*(2*B*c*x + (5*B*b*c + 4*A*c^2)/c)*sqrt(c*x^2 + b*x) - 3/8*(B*b^2
+ 4*A*b*c)*log(abs(-2*(sqrt(c)*x - sqrt(c*x^2 + b*x))*sqrt(c) - b))/sqrt(c)

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maple [B]  time = 0.06, size = 232, normalized size = 1.97 \begin {gather*} \frac {3 A b \sqrt {c}\, \ln \left (\frac {c x +\frac {b}{2}}{\sqrt {c}}+\sqrt {c \,x^{2}+b x}\right )}{2}+\frac {3 B \,b^{2} \ln \left (\frac {c x +\frac {b}{2}}{\sqrt {c}}+\sqrt {c \,x^{2}+b x}\right )}{8 \sqrt {c}}-\frac {6 \sqrt {c \,x^{2}+b x}\, A \,c^{2} x}{b}-\frac {3 \sqrt {c \,x^{2}+b x}\, B c x}{2}-3 \sqrt {c \,x^{2}+b x}\, A c -\frac {3 \sqrt {c \,x^{2}+b x}\, B b}{4}-\frac {8 \left (c \,x^{2}+b x \right )^{\frac {3}{2}} A \,c^{2}}{b^{2}}-\frac {2 \left (c \,x^{2}+b x \right )^{\frac {3}{2}} B c}{b}+\frac {8 \left (c \,x^{2}+b x \right )^{\frac {5}{2}} A c}{b^{2} x^{2}}+\frac {2 \left (c \,x^{2}+b x \right )^{\frac {5}{2}} B}{b \,x^{2}}-\frac {2 \left (c \,x^{2}+b x \right )^{\frac {5}{2}} A}{b \,x^{3}} \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^3,x)

[Out]

-2*A*(c*x^2+b*x)^(5/2)/b/x^3+8*A/b^2*c/x^2*(c*x^2+b*x)^(5/2)-8*A/b^2*c^2*(c*x^2+b*x)^(3/2)-6*A/b*c^2*(c*x^2+b*
x)^(1/2)*x-3*A*c*(c*x^2+b*x)^(1/2)+3/2*A*b*c^(1/2)*ln((c*x+1/2*b)/c^(1/2)+(c*x^2+b*x)^(1/2))+2*B/b/x^2*(c*x^2+
b*x)^(5/2)-2*B/b*c*(c*x^2+b*x)^(3/2)-3/2*B*c*(c*x^2+b*x)^(1/2)*x-3/4*B*b*(c*x^2+b*x)^(1/2)+3/8*B*b^2/c^(1/2)*l
n((c*x+1/2*b)/c^(1/2)+(c*x^2+b*x)^(1/2))

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maxima [A]  time = 0.87, size = 129, normalized size = 1.09 \begin {gather*} \frac {3 \, B b^{2} \log \left (2 \, c x + b + 2 \, \sqrt {c x^{2} + b x} \sqrt {c}\right )}{8 \, \sqrt {c}} + \frac {3}{2} \, A b \sqrt {c} \log \left (2 \, c x + b + 2 \, \sqrt {c x^{2} + b x} \sqrt {c}\right ) + \frac {3}{4} \, \sqrt {c x^{2} + b x} B b + \frac {{\left (c x^{2} + b x\right )}^{\frac {3}{2}} B}{2 \, x} - \frac {3 \, \sqrt {c x^{2} + b x} A b}{x} + \frac {{\left (c x^{2} + b x\right )}^{\frac {3}{2}} A}{x^{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^3,x, algorithm="maxima")

[Out]

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

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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^3} \,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^3,x)

[Out]

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

[Out]

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

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