3.1.87 \(\int \frac {(A+B x) (b x+c x^2)^{3/2}}{x^5} \, dx\)

Optimal. Leaf size=95 \[ -\frac {2 A \left (b x+c x^2\right )^{5/2}}{5 b x^5}+2 B c^{3/2} \tanh ^{-1}\left (\frac {\sqrt {c} x}{\sqrt {b x+c x^2}}\right )-\frac {2 B c \sqrt {b x+c x^2}}{x}-\frac {2 B \left (b x+c x^2\right )^{3/2}}{3 x^3} \]

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Rubi [A]  time = 0.10, antiderivative size = 95, 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, 662, 620, 206} \begin {gather*} -\frac {2 A \left (b x+c x^2\right )^{5/2}}{5 b x^5}+2 B c^{3/2} \tanh ^{-1}\left (\frac {\sqrt {c} x}{\sqrt {b x+c x^2}}\right )-\frac {2 B \left (b x+c x^2\right )^{3/2}}{3 x^3}-\frac {2 B c \sqrt {b x+c x^2}}{x} \end {gather*}

Antiderivative was successfully verified.

[In]

Int[((A + B*x)*(b*x + c*x^2)^(3/2))/x^5,x]

[Out]

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

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Mathematica [C]  time = 0.08, size = 88, normalized size = 0.93 \begin {gather*} \frac {2 \sqrt {x (b+c x)} \left ((b+c x)^2 \sqrt {\frac {c x}{b}+1} (b B-A c)-b^3 B \, _2F_1\left (-\frac {5}{2},-\frac {5}{2};-\frac {3}{2};-\frac {c x}{b}\right )\right )}{5 b c x^3 \sqrt {\frac {c x}{b}+1}} \end {gather*}

Antiderivative was successfully verified.

[In]

Integrate[((A + B*x)*(b*x + c*x^2)^(3/2))/x^5,x]

[Out]

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

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IntegrateAlgebraic [A]  time = 0.39, size = 96, normalized size = 1.01 \begin {gather*} -\frac {2 \sqrt {b x+c x^2} \left (3 A b^2+6 A b c x+3 A c^2 x^2+5 b^2 B x+20 b B c x^2\right )}{15 b x^3}-B c^{3/2} \log \left (-2 \sqrt {c} \sqrt {b x+c x^2}+b+2 c x\right ) \end {gather*}

Antiderivative was successfully verified.

[In]

IntegrateAlgebraic[((A + B*x)*(b*x + c*x^2)^(3/2))/x^5,x]

[Out]

(-2*Sqrt[b*x + c*x^2]*(3*A*b^2 + 5*b^2*B*x + 6*A*b*c*x + 20*b*B*c*x^2 + 3*A*c^2*x^2))/(15*b*x^3) - B*c^(3/2)*L
og[b + 2*c*x - 2*Sqrt[c]*Sqrt[b*x + c*x^2]]

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fricas [A]  time = 0.43, size = 188, normalized size = 1.98 \begin {gather*} \left [\frac {15 \, B b c^{\frac {3}{2}} x^{3} \log \left (2 \, c x + b + 2 \, \sqrt {c x^{2} + b x} \sqrt {c}\right ) - 2 \, {\left (3 \, A b^{2} + {\left (20 \, B b c + 3 \, A c^{2}\right )} x^{2} + {\left (5 \, B b^{2} + 6 \, A b c\right )} x\right )} \sqrt {c x^{2} + b x}}{15 \, b x^{3}}, -\frac {2 \, {\left (15 \, B b \sqrt {-c} c x^{3} \arctan \left (\frac {\sqrt {c x^{2} + b x} \sqrt {-c}}{c x}\right ) + {\left (3 \, A b^{2} + {\left (20 \, B b c + 3 \, A c^{2}\right )} x^{2} + {\left (5 \, B b^{2} + 6 \, A b c\right )} x\right )} \sqrt {c x^{2} + b x}\right )}}{15 \, b x^{3}}\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^5,x, algorithm="fricas")

[Out]

[1/15*(15*B*b*c^(3/2)*x^3*log(2*c*x + b + 2*sqrt(c*x^2 + b*x)*sqrt(c)) - 2*(3*A*b^2 + (20*B*b*c + 3*A*c^2)*x^2
+ (5*B*b^2 + 6*A*b*c)*x)*sqrt(c*x^2 + b*x))/(b*x^3), -2/15*(15*B*b*sqrt(-c)*c*x^3*arctan(sqrt(c*x^2 + b*x)*sq
rt(-c)/(c*x)) + (3*A*b^2 + (20*B*b*c + 3*A*c^2)*x^2 + (5*B*b^2 + 6*A*b*c)*x)*sqrt(c*x^2 + b*x))/(b*x^3)]

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giac [B]  time = 0.25, size = 270, normalized size = 2.84 \begin {gather*} -B c^{\frac {3}{2}} \log \left ({\left | -2 \, {\left (\sqrt {c} x - \sqrt {c x^{2} + b x}\right )} \sqrt {c} - b \right |}\right ) + \frac {2 \, {\left (30 \, {\left (\sqrt {c} x - \sqrt {c x^{2} + b x}\right )}^{4} B b c^{\frac {3}{2}} + 15 \, {\left (\sqrt {c} x - \sqrt {c x^{2} + b x}\right )}^{4} A c^{\frac {5}{2}} + 15 \, {\left (\sqrt {c} x - \sqrt {c x^{2} + b x}\right )}^{3} B b^{2} c + 30 \, {\left (\sqrt {c} x - \sqrt {c x^{2} + b x}\right )}^{3} A b c^{2} + 5 \, {\left (\sqrt {c} x - \sqrt {c x^{2} + b x}\right )}^{2} B b^{3} \sqrt {c} + 30 \, {\left (\sqrt {c} x - \sqrt {c x^{2} + b x}\right )}^{2} A b^{2} c^{\frac {3}{2}} + 15 \, {\left (\sqrt {c} x - \sqrt {c x^{2} + b x}\right )} A b^{3} c + 3 \, A b^{4} \sqrt {c}\right )}}{15 \, {\left (\sqrt {c} x - \sqrt {c x^{2} + b x}\right )}^{5} \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^5,x, algorithm="giac")

[Out]

-B*c^(3/2)*log(abs(-2*(sqrt(c)*x - sqrt(c*x^2 + b*x))*sqrt(c) - b)) + 2/15*(30*(sqrt(c)*x - sqrt(c*x^2 + b*x))
^4*B*b*c^(3/2) + 15*(sqrt(c)*x - sqrt(c*x^2 + b*x))^4*A*c^(5/2) + 15*(sqrt(c)*x - sqrt(c*x^2 + b*x))^3*B*b^2*c
+ 30*(sqrt(c)*x - sqrt(c*x^2 + b*x))^3*A*b*c^2 + 5*(sqrt(c)*x - sqrt(c*x^2 + b*x))^2*B*b^3*sqrt(c) + 30*(sqrt
(c)*x - sqrt(c*x^2 + b*x))^2*A*b^2*c^(3/2) + 15*(sqrt(c)*x - sqrt(c*x^2 + b*x))*A*b^3*c + 3*A*b^4*sqrt(c))/((s
qrt(c)*x - sqrt(c*x^2 + b*x))^5*sqrt(c))

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maple [B]  time = 0.05, size = 176, normalized size = 1.85 \begin {gather*} B \,c^{\frac {3}{2}} \ln \left (\frac {c x +\frac {b}{2}}{\sqrt {c}}+\sqrt {c \,x^{2}+b x}\right )-\frac {4 \sqrt {c \,x^{2}+b x}\, B \,c^{3} x}{b^{2}}-\frac {2 \sqrt {c \,x^{2}+b x}\, B \,c^{2}}{b}-\frac {16 \left (c \,x^{2}+b x \right )^{\frac {3}{2}} B \,c^{3}}{3 b^{3}}+\frac {16 \left (c \,x^{2}+b x \right )^{\frac {5}{2}} B \,c^{2}}{3 b^{3} x^{2}}-\frac {4 \left (c \,x^{2}+b x \right )^{\frac {5}{2}} B c}{3 b^{2} x^{3}}-\frac {2 \left (c \,x^{2}+b x \right )^{\frac {5}{2}} B}{3 b \,x^{4}}-\frac {2 \left (c \,x^{2}+b x \right )^{\frac {5}{2}} A}{5 b \,x^{5}} \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^5,x)

[Out]

-2/5*A*(c*x^2+b*x)^(5/2)/b/x^5-2/3*B/b/x^4*(c*x^2+b*x)^(5/2)-4/3*B/b^2*c/x^3*(c*x^2+b*x)^(5/2)+16/3*B/b^3*c^2/
x^2*(c*x^2+b*x)^(5/2)-16/3*B/b^3*c^3*(c*x^2+b*x)^(3/2)-4*B/b^2*c^3*(c*x^2+b*x)^(1/2)*x-2*B/b*c^2*(c*x^2+b*x)^(
1/2)+B*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 = 0.95, size = 158, normalized size = 1.66 \begin {gather*} B c^{\frac {3}{2}} \log \left (2 \, c x + b + 2 \, \sqrt {c x^{2} + b x} \sqrt {c}\right ) - \frac {7 \, \sqrt {c x^{2} + b x} B c}{3 \, x} - \frac {2 \, \sqrt {c x^{2} + b x} A c^{2}}{5 \, b x} - \frac {\sqrt {c x^{2} + b x} B b}{3 \, x^{2}} + \frac {\sqrt {c x^{2} + b x} A c}{5 \, x^{2}} - \frac {{\left (c x^{2} + b x\right )}^{\frac {3}{2}} B}{3 \, x^{3}} + \frac {3 \, \sqrt {c x^{2} + b x} A b}{5 \, x^{3}} - \frac {{\left (c x^{2} + b x\right )}^{\frac {3}{2}} A}{x^{4}} \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^5,x, algorithm="maxima")

[Out]

B*c^(3/2)*log(2*c*x + b + 2*sqrt(c*x^2 + b*x)*sqrt(c)) - 7/3*sqrt(c*x^2 + b*x)*B*c/x - 2/5*sqrt(c*x^2 + b*x)*A
*c^2/(b*x) - 1/3*sqrt(c*x^2 + b*x)*B*b/x^2 + 1/5*sqrt(c*x^2 + b*x)*A*c/x^2 - 1/3*(c*x^2 + b*x)^(3/2)*B/x^3 + 3
/5*sqrt(c*x^2 + b*x)*A*b/x^3 - (c*x^2 + b*x)^(3/2)*A/x^4

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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^5} \,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^5,x)

[Out]

int(((b*x + c*x^2)^(3/2)*(A + B*x))/x^5, 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^{5}}\, 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**5,x)

[Out]

Integral((x*(b + c*x))**(3/2)*(A + B*x)/x**5, x)

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