3.1.86 \(\int \frac {(A+B x) (b x+c x^2)^{3/2}}{x^4} \, dx\)

Optimal. Leaf size=120 \[ -\frac {2 \left (b x+c x^2\right )^{3/2} (2 A c+3 b B)}{3 b x^2}+\frac {c \sqrt {b x+c x^2} (2 A c+3 b B)}{b}+\sqrt {c} (2 A c+3 b B) \tanh ^{-1}\left (\frac {\sqrt {c} x}{\sqrt {b x+c x^2}}\right )-\frac {2 A \left (b x+c x^2\right )^{5/2}}{3 b x^4} \]

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Rubi [A]  time = 0.12, antiderivative size = 120, 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, 662, 664, 620, 206} \begin {gather*} -\frac {2 \left (b x+c x^2\right )^{3/2} (2 A c+3 b B)}{3 b x^2}+\frac {c \sqrt {b x+c x^2} (2 A c+3 b B)}{b}+\sqrt {c} (2 A c+3 b B) \tanh ^{-1}\left (\frac {\sqrt {c} x}{\sqrt {b x+c x^2}}\right )-\frac {2 A \left (b x+c x^2\right )^{5/2}}{3 b x^4} \end {gather*}

Antiderivative was successfully verified.

[In]

Int[((A + B*x)*(b*x + c*x^2)^(3/2))/x^4,x]

[Out]

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

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Mathematica [C]  time = 0.04, size = 84, normalized size = 0.70 \begin {gather*} -\frac {2 \sqrt {x (b+c x)} \left (b x (2 A c+3 b B) \, _2F_1\left (-\frac {3}{2},-\frac {1}{2};\frac {1}{2};-\frac {c x}{b}\right )+A \sqrt {\frac {c x}{b}+1} (b+c x)^2\right )}{3 b x^2 \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^4,x]

[Out]

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

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IntegrateAlgebraic [A]  time = 0.47, size = 92, normalized size = 0.77 \begin {gather*} \frac {1}{2} \left (-2 A c^{3/2}-3 b B \sqrt {c}\right ) \log \left (-2 \sqrt {c} \sqrt {b x+c x^2}+b+2 c x\right )+\frac {\sqrt {b x+c x^2} \left (-2 A b-8 A c x-6 b B x+3 B c x^2\right )}{3 x^2} \end {gather*}

Antiderivative was successfully verified.

[In]

IntegrateAlgebraic[((A + B*x)*(b*x + c*x^2)^(3/2))/x^4,x]

[Out]

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

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fricas [A]  time = 0.43, size = 170, normalized size = 1.42 \begin {gather*} \left [\frac {3 \, {\left (3 \, B b + 2 \, A c\right )} \sqrt {c} x^{2} \log \left (2 \, c x + b + 2 \, \sqrt {c x^{2} + b x} \sqrt {c}\right ) + 2 \, {\left (3 \, B c x^{2} - 2 \, A b - 2 \, {\left (3 \, B b + 4 \, A c\right )} x\right )} \sqrt {c x^{2} + b x}}{6 \, x^{2}}, -\frac {3 \, {\left (3 \, B b + 2 \, A c\right )} \sqrt {-c} x^{2} \arctan \left (\frac {\sqrt {c x^{2} + b x} \sqrt {-c}}{c x}\right ) - {\left (3 \, B c x^{2} - 2 \, A b - 2 \, {\left (3 \, B b + 4 \, A c\right )} x\right )} \sqrt {c x^{2} + b x}}{3 \, x^{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^4,x, algorithm="fricas")

[Out]

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

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giac [A]  time = 0.27, size = 181, normalized size = 1.51 \begin {gather*} \sqrt {c x^{2} + b x} B c - \frac {{\left (3 \, B b c + 2 \, A c^{2}\right )} \log \left ({\left | -2 \, {\left (\sqrt {c} x - \sqrt {c x^{2} + b x}\right )} \sqrt {c} - b \right |}\right )}{2 \, \sqrt {c}} + \frac {2 \, {\left (3 \, {\left (\sqrt {c} x - \sqrt {c x^{2} + b x}\right )}^{2} B b^{2} \sqrt {c} + 6 \, {\left (\sqrt {c} x - \sqrt {c x^{2} + b x}\right )}^{2} A b c^{\frac {3}{2}} + 3 \, {\left (\sqrt {c} x - \sqrt {c x^{2} + b x}\right )} A b^{2} c + A b^{3} \sqrt {c}\right )}}{3 \, {\left (\sqrt {c} x - \sqrt {c x^{2} + b x}\right )}^{3} \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^4,x, algorithm="giac")

[Out]

sqrt(c*x^2 + b*x)*B*c - 1/2*(3*B*b*c + 2*A*c^2)*log(abs(-2*(sqrt(c)*x - sqrt(c*x^2 + b*x))*sqrt(c) - b))/sqrt(
c) + 2/3*(3*(sqrt(c)*x - sqrt(c*x^2 + b*x))^2*B*b^2*sqrt(c) + 6*(sqrt(c)*x - sqrt(c*x^2 + b*x))^2*A*b*c^(3/2)
+ 3*(sqrt(c)*x - sqrt(c*x^2 + b*x))*A*b^2*c + A*b^3*sqrt(c))/((sqrt(c)*x - sqrt(c*x^2 + b*x))^3*sqrt(c))

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maple [B]  time = 0.06, size = 284, normalized size = 2.37 \begin {gather*} A \,c^{\frac {3}{2}} \ln \left (\frac {c x +\frac {b}{2}}{\sqrt {c}}+\sqrt {c \,x^{2}+b x}\right )+\frac {3 B b \sqrt {c}\, \ln \left (\frac {c x +\frac {b}{2}}{\sqrt {c}}+\sqrt {c \,x^{2}+b x}\right )}{2}-\frac {4 \sqrt {c \,x^{2}+b x}\, A \,c^{3} x}{b^{2}}-\frac {6 \sqrt {c \,x^{2}+b x}\, B \,c^{2} x}{b}-\frac {2 \sqrt {c \,x^{2}+b x}\, A \,c^{2}}{b}-3 \sqrt {c \,x^{2}+b x}\, B c -\frac {16 \left (c \,x^{2}+b x \right )^{\frac {3}{2}} A \,c^{3}}{3 b^{3}}-\frac {8 \left (c \,x^{2}+b x \right )^{\frac {3}{2}} B \,c^{2}}{b^{2}}+\frac {16 \left (c \,x^{2}+b x \right )^{\frac {5}{2}} A \,c^{2}}{3 b^{3} x^{2}}+\frac {8 \left (c \,x^{2}+b x \right )^{\frac {5}{2}} B c}{b^{2} x^{2}}-\frac {4 \left (c \,x^{2}+b x \right )^{\frac {5}{2}} A c}{3 b^{2} x^{3}}-\frac {2 \left (c \,x^{2}+b x \right )^{\frac {5}{2}} B}{b \,x^{3}}-\frac {2 \left (c \,x^{2}+b x \right )^{\frac {5}{2}} A}{3 b \,x^{4}} \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^4,x)

[Out]

-2/3*A*(c*x^2+b*x)^(5/2)/b/x^4-4/3*A/b^2*c/x^3*(c*x^2+b*x)^(5/2)+16/3*A/b^3*c^2/x^2*(c*x^2+b*x)^(5/2)-16/3*A/b
^3*c^3*(c*x^2+b*x)^(3/2)-4*A/b^2*c^3*(c*x^2+b*x)^(1/2)*x-2*A/b*c^2*(c*x^2+b*x)^(1/2)+A*c^(3/2)*ln((c*x+1/2*b)/
c^(1/2)+(c*x^2+b*x)^(1/2))-2*B/b/x^3*(c*x^2+b*x)^(5/2)+8*B/b^2*c/x^2*(c*x^2+b*x)^(5/2)-8*B/b^2*c^2*(c*x^2+b*x)
^(3/2)-6*B/b*c^2*(c*x^2+b*x)^(1/2)*x-3*B*c*(c*x^2+b*x)^(1/2)+3/2*B*b*c^(1/2)*ln((c*x+1/2*b)/c^(1/2)+(c*x^2+b*x
)^(1/2))

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maxima [A]  time = 1.00, size = 146, normalized size = 1.22 \begin {gather*} \frac {3}{2} \, B b \sqrt {c} \log \left (2 \, c x + b + 2 \, \sqrt {c x^{2} + b x} \sqrt {c}\right ) + A c^{\frac {3}{2}} \log \left (2 \, c x + b + 2 \, \sqrt {c x^{2} + b x} \sqrt {c}\right ) - \frac {3 \, \sqrt {c x^{2} + b x} B b}{x} - \frac {7 \, \sqrt {c x^{2} + b x} A c}{3 \, x} + \frac {{\left (c x^{2} + b x\right )}^{\frac {3}{2}} B}{x^{2}} - \frac {\sqrt {c x^{2} + b x} A b}{3 \, x^{2}} - \frac {{\left (c x^{2} + b x\right )}^{\frac {3}{2}} A}{3 \, x^{3}} \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^4,x, algorithm="maxima")

[Out]

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

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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^4} \,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^4,x)

[Out]

int(((b*x + c*x^2)^(3/2)*(A + B*x))/x^4, 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^{4}}\, 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**4,x)

[Out]

Integral((x*(b + c*x))**(3/2)*(A + B*x)/x**4, x)

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