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

Optimal. Leaf size=155 \[ -\frac {5 b^3 (b B-8 A c) \tanh ^{-1}\left (\frac {\sqrt {c} x}{\sqrt {b x+c x^2}}\right )}{64 c^{3/2}}+\frac {\left (b x+c x^2\right )^{5/2} (b B-8 A c)}{4 b x}+\frac {5}{24} \left (b x+c x^2\right )^{3/2} (b B-8 A c)+\frac {5 b (b+2 c x) \sqrt {b x+c x^2} (b B-8 A c)}{64 c}+\frac {2 A \left (b x+c x^2\right )^{7/2}}{b x^3} \]

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Rubi [A]  time = 0.17, antiderivative size = 155, normalized size of antiderivative = 1.00, number of steps used = 6, 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 {5 b^3 (b B-8 A c) \tanh ^{-1}\left (\frac {\sqrt {c} x}{\sqrt {b x+c x^2}}\right )}{64 c^{3/2}}+\frac {\left (b x+c x^2\right )^{5/2} (b B-8 A c)}{4 b x}+\frac {5}{24} \left (b x+c x^2\right )^{3/2} (b B-8 A c)+\frac {5 b (b+2 c x) \sqrt {b x+c x^2} (b B-8 A c)}{64 c}+\frac {2 A \left (b x+c x^2\right )^{7/2}}{b x^3} \end {gather*}

Antiderivative was successfully verified.

[In]

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

[Out]

(5*b*(b*B - 8*A*c)*(b + 2*c*x)*Sqrt[b*x + c*x^2])/(64*c) + (5*(b*B - 8*A*c)*(b*x + c*x^2)^(3/2))/24 + ((b*B -
8*A*c)*(b*x + c*x^2)^(5/2))/(4*b*x) + (2*A*(b*x + c*x^2)^(7/2))/(b*x^3) - (5*b^3*(b*B - 8*A*c)*ArcTanh[(Sqrt[c
]*x)/Sqrt[b*x + c*x^2]])/(64*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 )^{5/2}}{x^3} \, dx &=\frac {2 A \left (b x+c x^2\right )^{7/2}}{b x^3}-\frac {\left (2 \left (-3 (-b B+A c)+\frac {7}{2} (-b B+2 A c)\right )\right ) \int \frac {\left (b x+c x^2\right )^{5/2}}{x^2} \, dx}{b}\\ &=\frac {(b B-8 A c) \left (b x+c x^2\right )^{5/2}}{4 b x}+\frac {2 A \left (b x+c x^2\right )^{7/2}}{b x^3}+\frac {1}{8} (5 (b B-8 A c)) \int \frac {\left (b x+c x^2\right )^{3/2}}{x} \, dx\\ &=\frac {5}{24} (b B-8 A c) \left (b x+c x^2\right )^{3/2}+\frac {(b B-8 A c) \left (b x+c x^2\right )^{5/2}}{4 b x}+\frac {2 A \left (b x+c x^2\right )^{7/2}}{b x^3}+\frac {1}{16} (5 b (b B-8 A c)) \int \sqrt {b x+c x^2} \, dx\\ &=\frac {5 b (b B-8 A c) (b+2 c x) \sqrt {b x+c x^2}}{64 c}+\frac {5}{24} (b B-8 A c) \left (b x+c x^2\right )^{3/2}+\frac {(b B-8 A c) \left (b x+c x^2\right )^{5/2}}{4 b x}+\frac {2 A \left (b x+c x^2\right )^{7/2}}{b x^3}-\frac {\left (5 b^3 (b B-8 A c)\right ) \int \frac {1}{\sqrt {b x+c x^2}} \, dx}{128 c}\\ &=\frac {5 b (b B-8 A c) (b+2 c x) \sqrt {b x+c x^2}}{64 c}+\frac {5}{24} (b B-8 A c) \left (b x+c x^2\right )^{3/2}+\frac {(b B-8 A c) \left (b x+c x^2\right )^{5/2}}{4 b x}+\frac {2 A \left (b x+c x^2\right )^{7/2}}{b x^3}-\frac {\left (5 b^3 (b B-8 A c)\right ) \operatorname {Subst}\left (\int \frac {1}{1-c x^2} \, dx,x,\frac {x}{\sqrt {b x+c x^2}}\right )}{64 c}\\ &=\frac {5 b (b B-8 A c) (b+2 c x) \sqrt {b x+c x^2}}{64 c}+\frac {5}{24} (b B-8 A c) \left (b x+c x^2\right )^{3/2}+\frac {(b B-8 A c) \left (b x+c x^2\right )^{5/2}}{4 b x}+\frac {2 A \left (b x+c x^2\right )^{7/2}}{b x^3}-\frac {5 b^3 (b B-8 A c) \tanh ^{-1}\left (\frac {\sqrt {c} x}{\sqrt {b x+c x^2}}\right )}{64 c^{3/2}}\\ \end {align*}

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Mathematica [A]  time = 0.22, size = 128, normalized size = 0.83 \begin {gather*} \frac {\sqrt {x (b+c x)} \left (\sqrt {c} \left (2 b^2 c (132 A+59 B x)+8 b c^2 x (26 A+17 B x)+16 c^3 x^2 (4 A+3 B x)+15 b^3 B\right )-\frac {15 b^{5/2} (b B-8 A c) \sinh ^{-1}\left (\frac {\sqrt {c} \sqrt {x}}{\sqrt {b}}\right )}{\sqrt {x} \sqrt {\frac {c x}{b}+1}}\right )}{192 c^{3/2}} \end {gather*}

Antiderivative was successfully verified.

[In]

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

[Out]

(Sqrt[x*(b + c*x)]*(Sqrt[c]*(15*b^3*B + 16*c^3*x^2*(4*A + 3*B*x) + 8*b*c^2*x*(26*A + 17*B*x) + 2*b^2*c*(132*A
+ 59*B*x)) - (15*b^(5/2)*(b*B - 8*A*c)*ArcSinh[(Sqrt[c]*Sqrt[x])/Sqrt[b]])/(Sqrt[x]*Sqrt[1 + (c*x)/b])))/(192*
c^(3/2))

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IntegrateAlgebraic [A]  time = 0.58, size = 132, normalized size = 0.85 \begin {gather*} \frac {5 \left (b^4 B-8 A b^3 c\right ) \log \left (-2 c^{3/2} \sqrt {b x+c x^2}+b c+2 c^2 x\right )}{128 c^{3/2}}+\frac {\sqrt {b x+c x^2} \left (264 A b^2 c+208 A b c^2 x+64 A c^3 x^2+15 b^3 B+118 b^2 B c x+136 b B c^2 x^2+48 B c^3 x^3\right )}{192 c} \end {gather*}

Antiderivative was successfully verified.

[In]

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

[Out]

(Sqrt[b*x + c*x^2]*(15*b^3*B + 264*A*b^2*c + 118*b^2*B*c*x + 208*A*b*c^2*x + 136*b*B*c^2*x^2 + 64*A*c^3*x^2 +
48*B*c^3*x^3))/(192*c) + (5*(b^4*B - 8*A*b^3*c)*Log[b*c + 2*c^2*x - 2*c^(3/2)*Sqrt[b*x + c*x^2]])/(128*c^(3/2)
)

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fricas [A]  time = 0.43, size = 253, normalized size = 1.63 \begin {gather*} \left [-\frac {15 \, {\left (B b^{4} - 8 \, A b^{3} c\right )} \sqrt {c} \log \left (2 \, c x + b + 2 \, \sqrt {c x^{2} + b x} \sqrt {c}\right ) - 2 \, {\left (48 \, B c^{4} x^{3} + 15 \, B b^{3} c + 264 \, A b^{2} c^{2} + 8 \, {\left (17 \, B b c^{3} + 8 \, A c^{4}\right )} x^{2} + 2 \, {\left (59 \, B b^{2} c^{2} + 104 \, A b c^{3}\right )} x\right )} \sqrt {c x^{2} + b x}}{384 \, c^{2}}, \frac {15 \, {\left (B b^{4} - 8 \, A b^{3} c\right )} \sqrt {-c} \arctan \left (\frac {\sqrt {c x^{2} + b x} \sqrt {-c}}{c x}\right ) + {\left (48 \, B c^{4} x^{3} + 15 \, B b^{3} c + 264 \, A b^{2} c^{2} + 8 \, {\left (17 \, B b c^{3} + 8 \, A c^{4}\right )} x^{2} + 2 \, {\left (59 \, B b^{2} c^{2} + 104 \, A b c^{3}\right )} x\right )} \sqrt {c x^{2} + b x}}{192 \, 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)^(5/2)/x^3,x, algorithm="fricas")

[Out]

[-1/384*(15*(B*b^4 - 8*A*b^3*c)*sqrt(c)*log(2*c*x + b + 2*sqrt(c*x^2 + b*x)*sqrt(c)) - 2*(48*B*c^4*x^3 + 15*B*
b^3*c + 264*A*b^2*c^2 + 8*(17*B*b*c^3 + 8*A*c^4)*x^2 + 2*(59*B*b^2*c^2 + 104*A*b*c^3)*x)*sqrt(c*x^2 + b*x))/c^
2, 1/192*(15*(B*b^4 - 8*A*b^3*c)*sqrt(-c)*arctan(sqrt(c*x^2 + b*x)*sqrt(-c)/(c*x)) + (48*B*c^4*x^3 + 15*B*b^3*
c + 264*A*b^2*c^2 + 8*(17*B*b*c^3 + 8*A*c^4)*x^2 + 2*(59*B*b^2*c^2 + 104*A*b*c^3)*x)*sqrt(c*x^2 + b*x))/c^2]

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giac [A]  time = 0.22, size = 141, normalized size = 0.91 \begin {gather*} \frac {1}{192} \, \sqrt {c x^{2} + b x} {\left (2 \, {\left (4 \, {\left (6 \, B c^{2} x + \frac {17 \, B b c^{4} + 8 \, A c^{5}}{c^{3}}\right )} x + \frac {59 \, B b^{2} c^{3} + 104 \, A b c^{4}}{c^{3}}\right )} x + \frac {3 \, {\left (5 \, B b^{3} c^{2} + 88 \, A b^{2} c^{3}\right )}}{c^{3}}\right )} + \frac {5 \, {\left (B b^{4} - 8 \, A b^{3} c\right )} \log \left ({\left | -2 \, {\left (\sqrt {c} x - \sqrt {c x^{2} + b x}\right )} \sqrt {c} - b \right |}\right )}{128 \, 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)^(5/2)/x^3,x, algorithm="giac")

[Out]

1/192*sqrt(c*x^2 + b*x)*(2*(4*(6*B*c^2*x + (17*B*b*c^4 + 8*A*c^5)/c^3)*x + (59*B*b^2*c^3 + 104*A*b*c^4)/c^3)*x
+ 3*(5*B*b^3*c^2 + 88*A*b^2*c^3)/c^3) + 5/128*(B*b^4 - 8*A*b^3*c)*log(abs(-2*(sqrt(c)*x - sqrt(c*x^2 + b*x))*
sqrt(c) - b))/c^(3/2)

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maple [B]  time = 0.06, size = 306, normalized size = 1.97 \begin {gather*} \frac {5 A \,b^{3} \ln \left (\frac {c x +\frac {b}{2}}{\sqrt {c}}+\sqrt {c \,x^{2}+b x}\right )}{16 \sqrt {c}}-\frac {5 B \,b^{4} \ln \left (\frac {c x +\frac {b}{2}}{\sqrt {c}}+\sqrt {c \,x^{2}+b x}\right )}{128 c^{\frac {3}{2}}}-\frac {5 \sqrt {c \,x^{2}+b x}\, A b c x}{4}+\frac {5 \sqrt {c \,x^{2}+b x}\, B \,b^{2} x}{32}-\frac {5 \sqrt {c \,x^{2}+b x}\, A \,b^{2}}{8}+\frac {10 \left (c \,x^{2}+b x \right )^{\frac {3}{2}} A \,c^{2} x}{3 b}+\frac {5 \sqrt {c \,x^{2}+b x}\, B \,b^{3}}{64 c}-\frac {5 \left (c \,x^{2}+b x \right )^{\frac {3}{2}} B c x}{12}+\frac {5 \left (c \,x^{2}+b x \right )^{\frac {3}{2}} A c}{3}-\frac {5 \left (c \,x^{2}+b x \right )^{\frac {3}{2}} B b}{24}+\frac {16 \left (c \,x^{2}+b x \right )^{\frac {5}{2}} A \,c^{2}}{3 b^{2}}-\frac {2 \left (c \,x^{2}+b x \right )^{\frac {5}{2}} B c}{3 b}-\frac {16 \left (c \,x^{2}+b x \right )^{\frac {7}{2}} A c}{3 b^{2} x^{2}}+\frac {2 \left (c \,x^{2}+b x \right )^{\frac {7}{2}} B}{3 b \,x^{2}}+\frac {2 \left (c \,x^{2}+b x \right )^{\frac {7}{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)^(5/2)/x^3,x)

[Out]

2*A*(c*x^2+b*x)^(7/2)/b/x^3-16/3*A/b^2*c/x^2*(c*x^2+b*x)^(7/2)+16/3*A/b^2*c^2*(c*x^2+b*x)^(5/2)+10/3*A/b*c^2*(
c*x^2+b*x)^(3/2)*x+5/3*A*c*(c*x^2+b*x)^(3/2)-5/4*A*b*c*(c*x^2+b*x)^(1/2)*x-5/8*A*b^2*(c*x^2+b*x)^(1/2)+5/16*A*
b^3/c^(1/2)*ln((c*x+1/2*b)/c^(1/2)+(c*x^2+b*x)^(1/2))+2/3*B/b/x^2*(c*x^2+b*x)^(7/2)-2/3*B/b*c*(c*x^2+b*x)^(5/2
)-5/12*B*c*(c*x^2+b*x)^(3/2)*x-5/24*B*b*(c*x^2+b*x)^(3/2)+5/32*B*b^2*(c*x^2+b*x)^(1/2)*x+5/64*B*b^3/c*(c*x^2+b
*x)^(1/2)-5/128*B*b^4/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 = 0.91, size = 187, normalized size = 1.21 \begin {gather*} \frac {5}{32} \, \sqrt {c x^{2} + b x} B b^{2} x - \frac {5 \, B b^{4} \log \left (2 \, c x + b + 2 \, \sqrt {c x^{2} + b x} \sqrt {c}\right )}{128 \, c^{\frac {3}{2}}} + \frac {5 \, A b^{3} \log \left (2 \, c x + b + 2 \, \sqrt {c x^{2} + b x} \sqrt {c}\right )}{16 \, \sqrt {c}} + \frac {5}{24} \, {\left (c x^{2} + b x\right )}^{\frac {3}{2}} B b + \frac {5}{8} \, \sqrt {c x^{2} + b x} A b^{2} + \frac {5 \, \sqrt {c x^{2} + b x} B b^{3}}{64 \, c} + \frac {{\left (c x^{2} + b x\right )}^{\frac {5}{2}} B}{4 \, x} + \frac {5 \, {\left (c x^{2} + b x\right )}^{\frac {3}{2}} A b}{12 \, x} + \frac {{\left (c x^{2} + b x\right )}^{\frac {5}{2}} A}{3 \, x^{2}} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate((B*x+A)*(c*x^2+b*x)^(5/2)/x^3,x, algorithm="maxima")

[Out]

5/32*sqrt(c*x^2 + b*x)*B*b^2*x - 5/128*B*b^4*log(2*c*x + b + 2*sqrt(c*x^2 + b*x)*sqrt(c))/c^(3/2) + 5/16*A*b^3
*log(2*c*x + b + 2*sqrt(c*x^2 + b*x)*sqrt(c))/sqrt(c) + 5/24*(c*x^2 + b*x)^(3/2)*B*b + 5/8*sqrt(c*x^2 + b*x)*A
*b^2 + 5/64*sqrt(c*x^2 + b*x)*B*b^3/c + 1/4*(c*x^2 + b*x)^(5/2)*B/x + 5/12*(c*x^2 + b*x)^(3/2)*A*b/x + 1/3*(c*
x^2 + b*x)^(5/2)*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 )}^{5/2}\,\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)^(5/2)*(A + B*x))/x^3,x)

[Out]

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

[Out]

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

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