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

Optimal. Leaf size=105 \[ -2 A b^{3/2} \tanh ^{-1}\left (\frac {\sqrt {b x+c x^2}}{\sqrt {b} \sqrt {x}}\right )+\frac {2 A b \sqrt {b x+c x^2}}{\sqrt {x}}+\frac {2 A \left (b x+c x^2\right )^{3/2}}{3 x^{3/2}}+\frac {2 B \left (b x+c x^2\right )^{5/2}}{5 c x^{5/2}} \]

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Rubi [A]  time = 0.09, antiderivative size = 105, normalized size of antiderivative = 1.00, number of steps used = 5, number of rules used = 4, integrand size = 24, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.167, Rules used = {794, 664, 660, 207} \begin {gather*} -2 A b^{3/2} \tanh ^{-1}\left (\frac {\sqrt {b x+c x^2}}{\sqrt {b} \sqrt {x}}\right )+\frac {2 A \left (b x+c x^2\right )^{3/2}}{3 x^{3/2}}+\frac {2 A b \sqrt {b x+c x^2}}{\sqrt {x}}+\frac {2 B \left (b x+c x^2\right )^{5/2}}{5 c x^{5/2}} \end {gather*}

Antiderivative was successfully verified.

[In]

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

[Out]

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

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])

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 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 794

Int[((d_.) + (e_.)*(x_))^(m_)*((f_.) + (g_.)*(x_))*((a_.) + (b_.)*(x_) + (c_.)*(x_)^2)^(p_), x_Symbol] :> Simp
[(g*(d + e*x)^m*(a + b*x + c*x^2)^(p + 1))/(c*(m + 2*p + 2)), x] + Dist[(m*(g*(c*d - b*e) + c*e*f) + e*(p + 1)
*(2*c*f - b*g))/(c*e*(m + 2*p + 2)), Int[(d + e*x)^m*(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] && NeQ[m + 2*p + 2, 0] && (NeQ[m, 2] || Eq
Q[d, 0])

Rubi steps

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

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Mathematica [A]  time = 0.08, size = 100, normalized size = 0.95 \begin {gather*} \frac {2 \sqrt {x} \sqrt {b+c x} \left (\sqrt {b+c x} \left (b (20 A c+6 B c x)+c^2 x (5 A+3 B x)+3 b^2 B\right )-15 A b^{3/2} c \tanh ^{-1}\left (\frac {\sqrt {b+c x}}{\sqrt {b}}\right )\right )}{15 c \sqrt {x (b+c x)}} \end {gather*}

Antiderivative was successfully verified.

[In]

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

[Out]

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

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IntegrateAlgebraic [A]  time = 0.68, size = 93, normalized size = 0.89 \begin {gather*} \frac {2 \sqrt {b x+c x^2} \left (20 A b c+5 A c^2 x+3 b^2 B+6 b B c x+3 B c^2 x^2\right )}{15 c \sqrt {x}}-2 A b^{3/2} \tanh ^{-1}\left (\frac {\sqrt {b} \sqrt {x}}{\sqrt {b x+c x^2}}\right ) \end {gather*}

Antiderivative was successfully verified.

[In]

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

[Out]

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

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fricas [A]  time = 0.42, size = 195, normalized size = 1.86 \begin {gather*} \left [\frac {15 \, A b^{\frac {3}{2}} c x \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 \, B c^{2} x^{2} + 3 \, B b^{2} + 20 \, A b c + {\left (6 \, B b c + 5 \, A c^{2}\right )} x\right )} \sqrt {c x^{2} + b x} \sqrt {x}}{15 \, c x}, \frac {2 \, {\left (15 \, A \sqrt {-b} b c x \arctan \left (\frac {\sqrt {-b} \sqrt {x}}{\sqrt {c x^{2} + b x}}\right ) + {\left (3 \, B c^{2} x^{2} + 3 \, B b^{2} + 20 \, A b c + {\left (6 \, B b c + 5 \, A c^{2}\right )} x\right )} \sqrt {c x^{2} + b x} \sqrt {x}\right )}}{15 \, 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^(5/2),x, algorithm="fricas")

[Out]

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

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giac [A]  time = 0.20, size = 123, normalized size = 1.17 \begin {gather*} \frac {2 \, A b^{2} \arctan \left (\frac {\sqrt {c x + b}}{\sqrt {-b}}\right )}{\sqrt {-b}} - \frac {2 \, {\left (15 \, A b^{2} c \arctan \left (\frac {\sqrt {b}}{\sqrt {-b}}\right ) + 3 \, B \sqrt {-b} b^{\frac {5}{2}} + 20 \, A \sqrt {-b} b^{\frac {3}{2}} c\right )}}{15 \, \sqrt {-b} c} + \frac {2 \, {\left (3 \, {\left (c x + b\right )}^{\frac {5}{2}} B c^{4} + 5 \, {\left (c x + b\right )}^{\frac {3}{2}} A c^{5} + 15 \, \sqrt {c x + b} A b c^{5}\right )}}{15 \, c^{5}} \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^(5/2),x, algorithm="giac")

[Out]

2*A*b^2*arctan(sqrt(c*x + b)/sqrt(-b))/sqrt(-b) - 2/15*(15*A*b^2*c*arctan(sqrt(b)/sqrt(-b)) + 3*B*sqrt(-b)*b^(
5/2) + 20*A*sqrt(-b)*b^(3/2)*c)/(sqrt(-b)*c) + 2/15*(3*(c*x + b)^(5/2)*B*c^4 + 5*(c*x + b)^(3/2)*A*c^5 + 15*sq
rt(c*x + b)*A*b*c^5)/c^5

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maple [A]  time = 0.06, size = 113, normalized size = 1.08 \begin {gather*} -\frac {2 \sqrt {\left (c x +b \right ) x}\, \left (-3 \sqrt {c x +b}\, B \,c^{2} x^{2}+15 A \,b^{\frac {3}{2}} c \arctanh \left (\frac {\sqrt {c x +b}}{\sqrt {b}}\right )-5 \sqrt {c x +b}\, A \,c^{2} x -6 \sqrt {c x +b}\, B b c x -20 \sqrt {c x +b}\, A b c -3 \sqrt {c x +b}\, B \,b^{2}\right )}{15 \sqrt {c x +b}\, c \sqrt {x}} \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^(5/2),x)

[Out]

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

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maxima [F]  time = 0.00, size = 0, normalized size = 0.00 \begin {gather*} A b \int \frac {\sqrt {c x + b}}{x}\,{d x} + \frac {2 \, {\left (5 \, {\left (B b c + A c^{2}\right )} x^{2} + {\left (3 \, B c^{2} x^{2} + B b c x - 2 \, B b^{2}\right )} x + 5 \, {\left (B b^{2} + A b c\right )} x\right )} \sqrt {c x + b}}{15 \, c 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^(5/2),x, algorithm="maxima")

[Out]

A*b*integrate(sqrt(c*x + b)/x, x) + 2/15*(5*(B*b*c + A*c^2)*x^2 + (3*B*c^2*x^2 + B*b*c*x - 2*B*b^2)*x + 5*(B*b
^2 + A*b*c)*x)*sqrt(c*x + b)/(c*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^{5/2}} \,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^(5/2),x)

[Out]

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

[Out]

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

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