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

Optimal. Leaf size=169 \[ \frac {b^4 (3 b B-10 A c) \tanh ^{-1}\left (\frac {\sqrt {c} x}{\sqrt {b x+c x^2}}\right )}{128 c^{5/2}}-\frac {b^2 (b+2 c x) \sqrt {b x+c x^2} (3 b B-10 A c)}{128 c^2}+\frac {(b+2 c x) \left (b x+c x^2\right )^{3/2} (3 b B-10 A c)}{48 c}+\frac {\left (b x+c x^2\right )^{5/2} (3 b B-10 A c)}{15 b}+\frac {2 A \left (b x+c x^2\right )^{7/2}}{3 b x^2} \]

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Rubi [A]  time = 0.17, antiderivative size = 169, 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 {b^2 (b+2 c x) \sqrt {b x+c x^2} (3 b B-10 A c)}{128 c^2}+\frac {b^4 (3 b B-10 A c) \tanh ^{-1}\left (\frac {\sqrt {c} x}{\sqrt {b x+c x^2}}\right )}{128 c^{5/2}}+\frac {(b+2 c x) \left (b x+c x^2\right )^{3/2} (3 b B-10 A c)}{48 c}+\frac {\left (b x+c x^2\right )^{5/2} (3 b B-10 A c)}{15 b}+\frac {2 A \left (b x+c x^2\right )^{7/2}}{3 b x^2} \end {gather*}

Antiderivative was successfully verified.

[In]

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

[Out]

-(b^2*(3*b*B - 10*A*c)*(b + 2*c*x)*Sqrt[b*x + c*x^2])/(128*c^2) + ((3*b*B - 10*A*c)*(b + 2*c*x)*(b*x + c*x^2)^
(3/2))/(48*c) + ((3*b*B - 10*A*c)*(b*x + c*x^2)^(5/2))/(15*b) + (2*A*(b*x + c*x^2)^(7/2))/(3*b*x^2) + (b^4*(3*
b*B - 10*A*c)*ArcTanh[(Sqrt[c]*x)/Sqrt[b*x + c*x^2]])/(128*c^(5/2))

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^2} \, dx &=\frac {2 A \left (b x+c x^2\right )^{7/2}}{3 b x^2}-\frac {\left (2 \left (-2 (-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} \, dx}{3 b}\\ &=\frac {(3 b B-10 A c) \left (b x+c x^2\right )^{5/2}}{15 b}+\frac {2 A \left (b x+c x^2\right )^{7/2}}{3 b x^2}-\frac {1}{6} (-3 b B+10 A c) \int \left (b x+c x^2\right )^{3/2} \, dx\\ &=\frac {(3 b B-10 A c) (b+2 c x) \left (b x+c x^2\right )^{3/2}}{48 c}+\frac {(3 b B-10 A c) \left (b x+c x^2\right )^{5/2}}{15 b}+\frac {2 A \left (b x+c x^2\right )^{7/2}}{3 b x^2}-\frac {\left (b^2 (3 b B-10 A c)\right ) \int \sqrt {b x+c x^2} \, dx}{32 c}\\ &=-\frac {b^2 (3 b B-10 A c) (b+2 c x) \sqrt {b x+c x^2}}{128 c^2}+\frac {(3 b B-10 A c) (b+2 c x) \left (b x+c x^2\right )^{3/2}}{48 c}+\frac {(3 b B-10 A c) \left (b x+c x^2\right )^{5/2}}{15 b}+\frac {2 A \left (b x+c x^2\right )^{7/2}}{3 b x^2}+\frac {\left (b^4 (3 b B-10 A c)\right ) \int \frac {1}{\sqrt {b x+c x^2}} \, dx}{256 c^2}\\ &=-\frac {b^2 (3 b B-10 A c) (b+2 c x) \sqrt {b x+c x^2}}{128 c^2}+\frac {(3 b B-10 A c) (b+2 c x) \left (b x+c x^2\right )^{3/2}}{48 c}+\frac {(3 b B-10 A c) \left (b x+c x^2\right )^{5/2}}{15 b}+\frac {2 A \left (b x+c x^2\right )^{7/2}}{3 b x^2}+\frac {\left (b^4 (3 b B-10 A c)\right ) \operatorname {Subst}\left (\int \frac {1}{1-c x^2} \, dx,x,\frac {x}{\sqrt {b x+c x^2}}\right )}{128 c^2}\\ &=-\frac {b^2 (3 b B-10 A c) (b+2 c x) \sqrt {b x+c x^2}}{128 c^2}+\frac {(3 b B-10 A c) (b+2 c x) \left (b x+c x^2\right )^{3/2}}{48 c}+\frac {(3 b B-10 A c) \left (b x+c x^2\right )^{5/2}}{15 b}+\frac {2 A \left (b x+c x^2\right )^{7/2}}{3 b x^2}+\frac {b^4 (3 b B-10 A c) \tanh ^{-1}\left (\frac {\sqrt {c} x}{\sqrt {b x+c x^2}}\right )}{128 c^{5/2}}\\ \end {align*}

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Mathematica [A]  time = 0.26, size = 147, normalized size = 0.87 \begin {gather*} \frac {\sqrt {x (b+c x)} \left (\frac {15 b^{7/2} (3 b B-10 A c) \sinh ^{-1}\left (\frac {\sqrt {c} \sqrt {x}}{\sqrt {b}}\right )}{\sqrt {x} \sqrt {\frac {c x}{b}+1}}+\sqrt {c} \left (30 b^3 c (5 A+B x)+4 b^2 c^2 x (295 A+186 B x)+16 b c^3 x^2 (85 A+63 B x)+96 c^4 x^3 (5 A+4 B x)-45 b^4 B\right )\right )}{1920 c^{5/2}} \end {gather*}

Antiderivative was successfully verified.

[In]

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

[Out]

(Sqrt[x*(b + c*x)]*(Sqrt[c]*(-45*b^4*B + 30*b^3*c*(5*A + B*x) + 96*c^4*x^3*(5*A + 4*B*x) + 16*b*c^3*x^2*(85*A
+ 63*B*x) + 4*b^2*c^2*x*(295*A + 186*B*x)) + (15*b^(7/2)*(3*b*B - 10*A*c)*ArcSinh[(Sqrt[c]*Sqrt[x])/Sqrt[b]])/
(Sqrt[x]*Sqrt[1 + (c*x)/b])))/(1920*c^(5/2))

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IntegrateAlgebraic [A]  time = 0.70, size = 153, normalized size = 0.91 \begin {gather*} \frac {\left (10 A b^4 c-3 b^5 B\right ) \log \left (-2 \sqrt {c} \sqrt {b x+c x^2}+b+2 c x\right )}{256 c^{5/2}}+\frac {\sqrt {b x+c x^2} \left (150 A b^3 c+1180 A b^2 c^2 x+1360 A b c^3 x^2+480 A c^4 x^3-45 b^4 B+30 b^3 B c x+744 b^2 B c^2 x^2+1008 b B c^3 x^3+384 B c^4 x^4\right )}{1920 c^2} \end {gather*}

Antiderivative was successfully verified.

[In]

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

[Out]

(Sqrt[b*x + c*x^2]*(-45*b^4*B + 150*A*b^3*c + 30*b^3*B*c*x + 1180*A*b^2*c^2*x + 744*b^2*B*c^2*x^2 + 1360*A*b*c
^3*x^2 + 1008*b*B*c^3*x^3 + 480*A*c^4*x^3 + 384*B*c^4*x^4))/(1920*c^2) + ((-3*b^5*B + 10*A*b^4*c)*Log[b + 2*c*
x - 2*Sqrt[c]*Sqrt[b*x + c*x^2]])/(256*c^(5/2))

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fricas [A]  time = 0.42, size = 304, normalized size = 1.80 \begin {gather*} \left [-\frac {15 \, {\left (3 \, B b^{5} - 10 \, A b^{4} c\right )} \sqrt {c} \log \left (2 \, c x + b - 2 \, \sqrt {c x^{2} + b x} \sqrt {c}\right ) - 2 \, {\left (384 \, B c^{5} x^{4} - 45 \, B b^{4} c + 150 \, A b^{3} c^{2} + 48 \, {\left (21 \, B b c^{4} + 10 \, A c^{5}\right )} x^{3} + 8 \, {\left (93 \, B b^{2} c^{3} + 170 \, A b c^{4}\right )} x^{2} + 10 \, {\left (3 \, B b^{3} c^{2} + 118 \, A b^{2} c^{3}\right )} x\right )} \sqrt {c x^{2} + b x}}{3840 \, c^{3}}, -\frac {15 \, {\left (3 \, B b^{5} - 10 \, A b^{4} c\right )} \sqrt {-c} \arctan \left (\frac {\sqrt {c x^{2} + b x} \sqrt {-c}}{c x}\right ) - {\left (384 \, B c^{5} x^{4} - 45 \, B b^{4} c + 150 \, A b^{3} c^{2} + 48 \, {\left (21 \, B b c^{4} + 10 \, A c^{5}\right )} x^{3} + 8 \, {\left (93 \, B b^{2} c^{3} + 170 \, A b c^{4}\right )} x^{2} + 10 \, {\left (3 \, B b^{3} c^{2} + 118 \, A b^{2} c^{3}\right )} x\right )} \sqrt {c x^{2} + b x}}{1920 \, c^{3}}\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^2,x, algorithm="fricas")

[Out]

[-1/3840*(15*(3*B*b^5 - 10*A*b^4*c)*sqrt(c)*log(2*c*x + b - 2*sqrt(c*x^2 + b*x)*sqrt(c)) - 2*(384*B*c^5*x^4 -
45*B*b^4*c + 150*A*b^3*c^2 + 48*(21*B*b*c^4 + 10*A*c^5)*x^3 + 8*(93*B*b^2*c^3 + 170*A*b*c^4)*x^2 + 10*(3*B*b^3
*c^2 + 118*A*b^2*c^3)*x)*sqrt(c*x^2 + b*x))/c^3, -1/1920*(15*(3*B*b^5 - 10*A*b^4*c)*sqrt(-c)*arctan(sqrt(c*x^2
+ b*x)*sqrt(-c)/(c*x)) - (384*B*c^5*x^4 - 45*B*b^4*c + 150*A*b^3*c^2 + 48*(21*B*b*c^4 + 10*A*c^5)*x^3 + 8*(93
*B*b^2*c^3 + 170*A*b*c^4)*x^2 + 10*(3*B*b^3*c^2 + 118*A*b^2*c^3)*x)*sqrt(c*x^2 + b*x))/c^3]

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giac [A]  time = 0.24, size = 170, normalized size = 1.01 \begin {gather*} \frac {1}{1920} \, \sqrt {c x^{2} + b x} {\left (2 \, {\left (4 \, {\left (6 \, {\left (8 \, B c^{2} x + \frac {21 \, B b c^{5} + 10 \, A c^{6}}{c^{4}}\right )} x + \frac {93 \, B b^{2} c^{4} + 170 \, A b c^{5}}{c^{4}}\right )} x + \frac {5 \, {\left (3 \, B b^{3} c^{3} + 118 \, A b^{2} c^{4}\right )}}{c^{4}}\right )} x - \frac {15 \, {\left (3 \, B b^{4} c^{2} - 10 \, A b^{3} c^{3}\right )}}{c^{4}}\right )} - \frac {{\left (3 \, B b^{5} - 10 \, A b^{4} c\right )} \log \left ({\left | -2 \, {\left (\sqrt {c} x - \sqrt {c x^{2} + b x}\right )} \sqrt {c} - b \right |}\right )}{256 \, c^{\frac {5}{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^2,x, algorithm="giac")

[Out]

1/1920*sqrt(c*x^2 + b*x)*(2*(4*(6*(8*B*c^2*x + (21*B*b*c^5 + 10*A*c^6)/c^4)*x + (93*B*b^2*c^4 + 170*A*b*c^5)/c
^4)*x + 5*(3*B*b^3*c^3 + 118*A*b^2*c^4)/c^4)*x - 15*(3*B*b^4*c^2 - 10*A*b^3*c^3)/c^4) - 1/256*(3*B*b^5 - 10*A*
b^4*c)*log(abs(-2*(sqrt(c)*x - sqrt(c*x^2 + b*x))*sqrt(c) - b))/c^(5/2)

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maple [A]  time = 0.05, size = 266, normalized size = 1.57 \begin {gather*} -\frac {5 A \,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 {3 B \,b^{5} \ln \left (\frac {c x +\frac {b}{2}}{\sqrt {c}}+\sqrt {c \,x^{2}+b x}\right )}{256 c^{\frac {5}{2}}}+\frac {5 \sqrt {c \,x^{2}+b x}\, A \,b^{2} x}{32}-\frac {3 \sqrt {c \,x^{2}+b x}\, B \,b^{3} x}{64 c}+\frac {5 \sqrt {c \,x^{2}+b x}\, A \,b^{3}}{64 c}-\frac {5 \left (c \,x^{2}+b x \right )^{\frac {3}{2}} A c x}{12}-\frac {3 \sqrt {c \,x^{2}+b x}\, B \,b^{4}}{128 c^{2}}+\frac {\left (c \,x^{2}+b x \right )^{\frac {3}{2}} B b x}{8}-\frac {5 \left (c \,x^{2}+b x \right )^{\frac {3}{2}} A b}{24}+\frac {\left (c \,x^{2}+b x \right )^{\frac {3}{2}} B \,b^{2}}{16 c}-\frac {2 \left (c \,x^{2}+b x \right )^{\frac {5}{2}} A c}{3 b}+\frac {\left (c \,x^{2}+b x \right )^{\frac {5}{2}} B}{5}+\frac {2 \left (c \,x^{2}+b x \right )^{\frac {7}{2}} A}{3 b \,x^{2}} \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^2,x)

[Out]

2/3*A*(c*x^2+b*x)^(7/2)/b/x^2-2/3*A/b*c*(c*x^2+b*x)^(5/2)-5/12*A*c*(c*x^2+b*x)^(3/2)*x-5/24*A*b*(c*x^2+b*x)^(3
/2)+5/32*A*b^2*(c*x^2+b*x)^(1/2)*x+5/64*A*b^3/c*(c*x^2+b*x)^(1/2)-5/128*A*b^4/c^(3/2)*ln((c*x+1/2*b)/c^(1/2)+(
c*x^2+b*x)^(1/2))+1/5*B*(c*x^2+b*x)^(5/2)+1/8*B*b*(c*x^2+b*x)^(3/2)*x+1/16*B/c*(c*x^2+b*x)^(3/2)*b^2-3/64*B*b^
3/c*(c*x^2+b*x)^(1/2)*x-3/128*B*b^4/c^2*(c*x^2+b*x)^(1/2)+3/256*B*b^5/c^(5/2)*ln((c*x+1/2*b)/c^(1/2)+(c*x^2+b*
x)^(1/2))

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maxima [A]  time = 0.94, size = 226, normalized size = 1.34 \begin {gather*} \frac {1}{8} \, {\left (c x^{2} + b x\right )}^{\frac {3}{2}} B b x + \frac {5}{32} \, \sqrt {c x^{2} + b x} A b^{2} x - \frac {3 \, \sqrt {c x^{2} + b x} B b^{3} x}{64 \, c} + \frac {3 \, B b^{5} \log \left (2 \, c x + b + 2 \, \sqrt {c x^{2} + b x} \sqrt {c}\right )}{256 \, c^{\frac {5}{2}}} - \frac {5 \, A b^{4} \log \left (2 \, c x + b + 2 \, \sqrt {c x^{2} + b x} \sqrt {c}\right )}{128 \, c^{\frac {3}{2}}} + \frac {1}{5} \, {\left (c x^{2} + b x\right )}^{\frac {5}{2}} B + \frac {5}{24} \, {\left (c x^{2} + b x\right )}^{\frac {3}{2}} A b - \frac {3 \, \sqrt {c x^{2} + b x} B b^{4}}{128 \, c^{2}} + \frac {{\left (c x^{2} + b x\right )}^{\frac {3}{2}} B b^{2}}{16 \, c} + \frac {5 \, \sqrt {c x^{2} + b x} A b^{3}}{64 \, c} + \frac {{\left (c x^{2} + b x\right )}^{\frac {5}{2}} A}{4 \, x} \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^2,x, algorithm="maxima")

[Out]

1/8*(c*x^2 + b*x)^(3/2)*B*b*x + 5/32*sqrt(c*x^2 + b*x)*A*b^2*x - 3/64*sqrt(c*x^2 + b*x)*B*b^3*x/c + 3/256*B*b^
5*log(2*c*x + b + 2*sqrt(c*x^2 + b*x)*sqrt(c))/c^(5/2) - 5/128*A*b^4*log(2*c*x + b + 2*sqrt(c*x^2 + b*x)*sqrt(
c))/c^(3/2) + 1/5*(c*x^2 + b*x)^(5/2)*B + 5/24*(c*x^2 + b*x)^(3/2)*A*b - 3/128*sqrt(c*x^2 + b*x)*B*b^4/c^2 + 1
/16*(c*x^2 + b*x)^(3/2)*B*b^2/c + 5/64*sqrt(c*x^2 + b*x)*A*b^3/c + 1/4*(c*x^2 + b*x)^(5/2)*A/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 )}^{5/2}\,\left (A+B\,x\right )}{x^2} \,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^2,x)

[Out]

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

[Out]

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

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