3.2.29 \(\int \frac {x^5 (A+B x)}{(b x+c x^2)^{5/2}} \, dx\) [129]

Optimal. Leaf size=172 \[ -\frac {2 (b B-A c) x^5}{3 b c \left (b x+c x^2\right )^{3/2}}-\frac {2 (7 b B-4 A c) x^3}{3 b c^2 \sqrt {b x+c x^2}}-\frac {5 (7 b B-4 A c) \sqrt {b x+c x^2}}{4 c^4}+\frac {5 (7 b B-4 A c) x \sqrt {b x+c x^2}}{6 b c^3}+\frac {5 b (7 b B-4 A c) \tanh ^{-1}\left (\frac {\sqrt {c} x}{\sqrt {b x+c x^2}}\right )}{4 c^{9/2}} \]

[Out]

-2/3*(-A*c+B*b)*x^5/b/c/(c*x^2+b*x)^(3/2)+5/4*b*(-4*A*c+7*B*b)*arctanh(x*c^(1/2)/(c*x^2+b*x)^(1/2))/c^(9/2)-2/
3*(-4*A*c+7*B*b)*x^3/b/c^2/(c*x^2+b*x)^(1/2)-5/4*(-4*A*c+7*B*b)*(c*x^2+b*x)^(1/2)/c^4+5/6*(-4*A*c+7*B*b)*x*(c*
x^2+b*x)^(1/2)/b/c^3

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Rubi [A]
time = 0.11, antiderivative size = 172, normalized size of antiderivative = 1.00, number of steps used = 6, number of rules used = 6, integrand size = 22, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.273, Rules used = {802, 682, 684, 654, 634, 212} \begin {gather*} \frac {5 b (7 b B-4 A c) \tanh ^{-1}\left (\frac {\sqrt {c} x}{\sqrt {b x+c x^2}}\right )}{4 c^{9/2}}-\frac {5 \sqrt {b x+c x^2} (7 b B-4 A c)}{4 c^4}+\frac {5 x \sqrt {b x+c x^2} (7 b B-4 A c)}{6 b c^3}-\frac {2 x^3 (7 b B-4 A c)}{3 b c^2 \sqrt {b x+c x^2}}-\frac {2 x^5 (b B-A c)}{3 b c \left (b x+c x^2\right )^{3/2}} \end {gather*}

Antiderivative was successfully verified.

[In]

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

[Out]

(-2*(b*B - A*c)*x^5)/(3*b*c*(b*x + c*x^2)^(3/2)) - (2*(7*b*B - 4*A*c)*x^3)/(3*b*c^2*Sqrt[b*x + c*x^2]) - (5*(7
*b*B - 4*A*c)*Sqrt[b*x + c*x^2])/(4*c^4) + (5*(7*b*B - 4*A*c)*x*Sqrt[b*x + c*x^2])/(6*b*c^3) + (5*b*(7*b*B - 4
*A*c)*ArcTanh[(Sqrt[c]*x)/Sqrt[b*x + c*x^2]])/(4*c^(9/2))

Rule 212

Int[((a_) + (b_.)*(x_)^2)^(-1), x_Symbol] :> Simp[(1/(Rt[a, 2]*Rt[-b, 2]))*ArcTanh[Rt[-b, 2]*(x/Rt[a, 2])], x]
/; FreeQ[{a, b}, x] && NegQ[a/b] && (GtQ[a, 0] || LtQ[b, 0])

Rule 634

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 654

Int[((d_.) + (e_.)*(x_))*((a_.) + (b_.)*(x_) + (c_.)*(x_)^2)^(p_), x_Symbol] :> Simp[e*((a + b*x + c*x^2)^(p +
1)/(2*c*(p + 1))), x] + Dist[(2*c*d - b*e)/(2*c), Int[(a + b*x + c*x^2)^p, x], x] /; FreeQ[{a, b, c, d, e, p}
, x] && NeQ[2*c*d - b*e, 0] && NeQ[p, -1]

Rule 682

Int[((d_.) + (e_.)*(x_))^(m_)*((a_.) + (b_.)*(x_) + (c_.)*(x_)^2)^(p_), x_Symbol] :> Simp[e*(d + e*x)^(m - 1)*
((a + b*x + c*x^2)^(p + 1)/(c*(p + 1))), x] - Dist[e^2*((m + p)/(c*(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] &
& LtQ[p, -1] && GtQ[m, 1] && IntegerQ[2*p]

Rule 684

Int[((d_.) + (e_.)*(x_))^(m_)*((a_.) + (b_.)*(x_) + (c_.)*(x_)^2)^(p_), x_Symbol] :> Simp[e*(d + e*x)^(m - 1)*
((a + b*x + c*x^2)^(p + 1)/(c*(m + 2*p + 1))), x] + Dist[(m + p)*((2*c*d - b*e)/(c*(m + 2*p + 1))), Int[(d + e
*x)^(m - 1)*(a + b*x + c*x^2)^p, x], x] /; FreeQ[{a, b, c, d, e, p}, x] && NeQ[b^2 - 4*a*c, 0] && EqQ[c*d^2 -
b*d*e + a*e^2, 0] && GtQ[m, 1] && NeQ[m + 2*p + 1, 0] && IntegerQ[2*p]

Rule 802

Int[((d_.) + (e_.)*(x_))^(m_)*((f_.) + (g_.)*(x_))*((a_.) + (b_.)*(x_) + (c_.)*(x_)^2)^(p_), x_Symbol] :> Simp
[(g*(c*d - b*e) + c*e*f)*(d + e*x)^m*((a + b*x + c*x^2)^(p + 1)/(c*(p + 1)*(2*c*d - b*e))), x] - Dist[e*((m*(g
*(c*d - b*e) + c*e*f) + e*(p + 1)*(2*c*f - b*g))/(c*(p + 1)*(2*c*d - b*e))), Int[(d + e*x)^(m - 1)*(a + b*x +
c*x^2)^(p + 1), x], x] /; FreeQ[{a, b, c, d, e, f, g}, x] && NeQ[b^2 - 4*a*c, 0] && EqQ[c*d^2 - b*d*e + a*e^2,
0] && LtQ[p, -1] && GtQ[m, 0]

Rubi steps

\begin {align*} \int \frac {x^5 (A+B x)}{\left (b x+c x^2\right )^{5/2}} \, dx &=-\frac {2 (b B-A c) x^5}{3 b c \left (b x+c x^2\right )^{3/2}}-\frac {1}{3} \left (\frac {4 A}{b}-\frac {7 B}{c}\right ) \int \frac {x^4}{\left (b x+c x^2\right )^{3/2}} \, dx\\ &=-\frac {2 (b B-A c) x^5}{3 b c \left (b x+c x^2\right )^{3/2}}-\frac {2 (7 b B-4 A c) x^3}{3 b c^2 \sqrt {b x+c x^2}}+\frac {(5 (7 b B-4 A c)) \int \frac {x^2}{\sqrt {b x+c x^2}} \, dx}{3 b c^2}\\ &=-\frac {2 (b B-A c) x^5}{3 b c \left (b x+c x^2\right )^{3/2}}-\frac {2 (7 b B-4 A c) x^3}{3 b c^2 \sqrt {b x+c x^2}}+\frac {5 (7 b B-4 A c) x \sqrt {b x+c x^2}}{6 b c^3}-\frac {(5 (7 b B-4 A c)) \int \frac {x}{\sqrt {b x+c x^2}} \, dx}{4 c^3}\\ &=-\frac {2 (b B-A c) x^5}{3 b c \left (b x+c x^2\right )^{3/2}}-\frac {2 (7 b B-4 A c) x^3}{3 b c^2 \sqrt {b x+c x^2}}-\frac {5 (7 b B-4 A c) \sqrt {b x+c x^2}}{4 c^4}+\frac {5 (7 b B-4 A c) x \sqrt {b x+c x^2}}{6 b c^3}+\frac {(5 b (7 b B-4 A c)) \int \frac {1}{\sqrt {b x+c x^2}} \, dx}{8 c^4}\\ &=-\frac {2 (b B-A c) x^5}{3 b c \left (b x+c x^2\right )^{3/2}}-\frac {2 (7 b B-4 A c) x^3}{3 b c^2 \sqrt {b x+c x^2}}-\frac {5 (7 b B-4 A c) \sqrt {b x+c x^2}}{4 c^4}+\frac {5 (7 b B-4 A c) x \sqrt {b x+c x^2}}{6 b c^3}+\frac {(5 b (7 b B-4 A c)) \text {Subst}\left (\int \frac {1}{1-c x^2} \, dx,x,\frac {x}{\sqrt {b x+c x^2}}\right )}{4 c^4}\\ &=-\frac {2 (b B-A c) x^5}{3 b c \left (b x+c x^2\right )^{3/2}}-\frac {2 (7 b B-4 A c) x^3}{3 b c^2 \sqrt {b x+c x^2}}-\frac {5 (7 b B-4 A c) \sqrt {b x+c x^2}}{4 c^4}+\frac {5 (7 b B-4 A c) x \sqrt {b x+c x^2}}{6 b c^3}+\frac {5 b (7 b B-4 A c) \tanh ^{-1}\left (\frac {\sqrt {c} x}{\sqrt {b x+c x^2}}\right )}{4 c^{9/2}}\\ \end {align*}

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Mathematica [A]
time = 0.27, size = 128, normalized size = 0.74 \begin {gather*} \frac {x \left (\sqrt {c} x \left (-105 b^3 B+b c^2 x (80 A-21 B x)+20 b^2 c (3 A-7 B x)+6 c^3 x^2 (2 A+B x)\right )-15 b (7 b B-4 A c) \sqrt {x} (b+c x)^{3/2} \log \left (-\sqrt {c} \sqrt {x}+\sqrt {b+c x}\right )\right )}{12 c^{9/2} (x (b+c x))^{3/2}} \end {gather*}

Antiderivative was successfully verified.

[In]

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

[Out]

(x*(Sqrt[c]*x*(-105*b^3*B + b*c^2*x*(80*A - 21*B*x) + 20*b^2*c*(3*A - 7*B*x) + 6*c^3*x^2*(2*A + B*x)) - 15*b*(
7*b*B - 4*A*c)*Sqrt[x]*(b + c*x)^(3/2)*Log[-(Sqrt[c]*Sqrt[x]) + Sqrt[b + c*x]]))/(12*c^(9/2)*(x*(b + c*x))^(3/
2))

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Maple [B] Leaf count of result is larger than twice the leaf count of optimal. \(565\) vs. \(2(148)=296\).
time = 0.56, size = 566, normalized size = 3.29

method	result	size
risch	\(\frac {\left (2 B c x +4 A c -11 B b \right ) x \left (c x +b \right )}{4 c^{4} \sqrt {x \left (c x +b \right )}}-\frac {5 b \ln \left (\frac {\frac {b}{2}+c x}{\sqrt {c}}+\sqrt {c \,x^{2}+b x}\right ) A}{2 c^{\frac {7}{2}}}+\frac {35 b^{2} \ln \left (\frac {\frac {b}{2}+c x}{\sqrt {c}}+\sqrt {c \,x^{2}+b x}\right ) B}{8 c^{\frac {9}{2}}}-\frac {2 b^{2} \sqrt {c \left (\frac {b}{c}+x \right )^{2}-\left (\frac {b}{c}+x \right ) b}\, A}{3 c^{5} \left (\frac {b}{c}+x \right )^{2}}+\frac {2 b^{3} \sqrt {c \left (\frac {b}{c}+x \right )^{2}-\left (\frac {b}{c}+x \right ) b}\, B}{3 c^{6} \left (\frac {b}{c}+x \right )^{2}}+\frac {14 b \sqrt {c \left (\frac {b}{c}+x \right )^{2}-\left (\frac {b}{c}+x \right ) b}\, A}{3 c^{4} \left (\frac {b}{c}+x \right )}-\frac {20 b^{2} \sqrt {c \left (\frac {b}{c}+x \right )^{2}-\left (\frac {b}{c}+x \right ) b}\, B}{3 c^{5} \left (\frac {b}{c}+x \right )}\)	\(266\)
default	\(B \left (\frac {x^{5}}{2 c \left (c \,x^{2}+b x \right )^{\frac {3}{2}}}-\frac {7 b \left (\frac {x^{4}}{c \left (c \,x^{2}+b x \right )^{\frac {3}{2}}}-\frac {5 b \left (-\frac {x^{3}}{3 c \left (c \,x^{2}+b x \right )^{\frac {3}{2}}}-\frac {b \left (-\frac {x^{2}}{c \left (c \,x^{2}+b x \right )^{\frac {3}{2}}}+\frac {b \left (-\frac {x}{2 c \left (c \,x^{2}+b x \right )^{\frac {3}{2}}}-\frac {b \left (-\frac {1}{3 c \left (c \,x^{2}+b x \right )^{\frac {3}{2}}}-\frac {b \left (-\frac {2 \left (2 c x +b \right )}{3 b^{2} \left (c \,x^{2}+b x \right )^{\frac {3}{2}}}+\frac {16 c \left (2 c x +b \right )}{3 b^{4} \sqrt {c \,x^{2}+b x}}\right )}{2 c}\right )}{4 c}\right )}{2 c}\right )}{2 c}+\frac {-\frac {x}{c \sqrt {c \,x^{2}+b x}}-\frac {b \left (-\frac {1}{c \sqrt {c \,x^{2}+b x}}+\frac {2 c x +b}{b c \sqrt {c \,x^{2}+b x}}\right )}{2 c}+\frac {\ln \left (\frac {\frac {b}{2}+c x}{\sqrt {c}}+\sqrt {c \,x^{2}+b x}\right )}{c^{\frac {3}{2}}}}{c}\right )}{2 c}\right )}{4 c}\right )+A \left (\frac {x^{4}}{c \left (c \,x^{2}+b x \right )^{\frac {3}{2}}}-\frac {5 b \left (-\frac {x^{3}}{3 c \left (c \,x^{2}+b x \right )^{\frac {3}{2}}}-\frac {b \left (-\frac {x^{2}}{c \left (c \,x^{2}+b x \right )^{\frac {3}{2}}}+\frac {b \left (-\frac {x}{2 c \left (c \,x^{2}+b x \right )^{\frac {3}{2}}}-\frac {b \left (-\frac {1}{3 c \left (c \,x^{2}+b x \right )^{\frac {3}{2}}}-\frac {b \left (-\frac {2 \left (2 c x +b \right )}{3 b^{2} \left (c \,x^{2}+b x \right )^{\frac {3}{2}}}+\frac {16 c \left (2 c x +b \right )}{3 b^{4} \sqrt {c \,x^{2}+b x}}\right )}{2 c}\right )}{4 c}\right )}{2 c}\right )}{2 c}+\frac {-\frac {x}{c \sqrt {c \,x^{2}+b x}}-\frac {b \left (-\frac {1}{c \sqrt {c \,x^{2}+b x}}+\frac {2 c x +b}{b c \sqrt {c \,x^{2}+b x}}\right )}{2 c}+\frac {\ln \left (\frac {\frac {b}{2}+c x}{\sqrt {c}}+\sqrt {c \,x^{2}+b x}\right )}{c^{\frac {3}{2}}}}{c}\right )}{2 c}\right )\)	\(566\)

Verification of antiderivative is not currently implemented for this CAS.

[In]

int(x^5*(B*x+A)/(c*x^2+b*x)^(5/2),x,method=_RETURNVERBOSE)

[Out]

B*(1/2*x^5/c/(c*x^2+b*x)^(3/2)-7/4*b/c*(x^4/c/(c*x^2+b*x)^(3/2)-5/2*b/c*(-1/3*x^3/c/(c*x^2+b*x)^(3/2)-1/2*b/c*
(-x^2/c/(c*x^2+b*x)^(3/2)+1/2*b/c*(-1/2*x/c/(c*x^2+b*x)^(3/2)-1/4*b/c*(-1/3/c/(c*x^2+b*x)^(3/2)-1/2*b/c*(-2/3*
(2*c*x+b)/b^2/(c*x^2+b*x)^(3/2)+16/3*c*(2*c*x+b)/b^4/(c*x^2+b*x)^(1/2)))))+1/c*(-x/c/(c*x^2+b*x)^(1/2)-1/2*b/c
*(-1/c/(c*x^2+b*x)^(1/2)+1/b/c*(2*c*x+b)/(c*x^2+b*x)^(1/2))+1/c^(3/2)*ln((1/2*b+c*x)/c^(1/2)+(c*x^2+b*x)^(1/2)
)))))+A*(x^4/c/(c*x^2+b*x)^(3/2)-5/2*b/c*(-1/3*x^3/c/(c*x^2+b*x)^(3/2)-1/2*b/c*(-x^2/c/(c*x^2+b*x)^(3/2)+1/2*b
/c*(-1/2*x/c/(c*x^2+b*x)^(3/2)-1/4*b/c*(-1/3/c/(c*x^2+b*x)^(3/2)-1/2*b/c*(-2/3*(2*c*x+b)/b^2/(c*x^2+b*x)^(3/2)
+16/3*c*(2*c*x+b)/b^4/(c*x^2+b*x)^(1/2)))))+1/c*(-x/c/(c*x^2+b*x)^(1/2)-1/2*b/c*(-1/c/(c*x^2+b*x)^(1/2)+1/b/c*
(2*c*x+b)/(c*x^2+b*x)^(1/2))+1/c^(3/2)*ln((1/2*b+c*x)/c^(1/2)+(c*x^2+b*x)^(1/2)))))

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Maxima [B] Leaf count of result is larger than twice the leaf count of optimal. 362 vs. \(2 (148) = 296\).
time = 0.28, size = 362, normalized size = 2.10 \begin {gather*} \frac {B x^{5}}{2 \, {\left (c x^{2} + b x\right )}^{\frac {3}{2}} c} - \frac {35 \, B b^{2} x {\left (\frac {3 \, x^{2}}{{\left (c x^{2} + b x\right )}^{\frac {3}{2}} c} + \frac {b x}{{\left (c x^{2} + b x\right )}^{\frac {3}{2}} c^{2}} - \frac {2 \, x}{\sqrt {c x^{2} + b x} b c} - \frac {1}{\sqrt {c x^{2} + b x} c^{2}}\right )}}{24 \, c^{2}} + \frac {5 \, A b x {\left (\frac {3 \, x^{2}}{{\left (c x^{2} + b x\right )}^{\frac {3}{2}} c} + \frac {b x}{{\left (c x^{2} + b x\right )}^{\frac {3}{2}} c^{2}} - \frac {2 \, x}{\sqrt {c x^{2} + b x} b c} - \frac {1}{\sqrt {c x^{2} + b x} c^{2}}\right )}}{6 \, c} - \frac {7 \, B b x^{4}}{4 \, {\left (c x^{2} + b x\right )}^{\frac {3}{2}} c^{2}} + \frac {A x^{4}}{{\left (c x^{2} + b x\right )}^{\frac {3}{2}} c} - \frac {35 \, B b^{2} x}{6 \, \sqrt {c x^{2} + b x} c^{4}} + \frac {10 \, A b x}{3 \, \sqrt {c x^{2} + b x} c^{3}} + \frac {35 \, B b^{2} \log \left (2 \, c x + b + 2 \, \sqrt {c x^{2} + b x} \sqrt {c}\right )}{8 \, c^{\frac {9}{2}}} - \frac {5 \, A b \log \left (2 \, c x + b + 2 \, \sqrt {c x^{2} + b x} \sqrt {c}\right )}{2 \, c^{\frac {7}{2}}} - \frac {35 \, \sqrt {c x^{2} + b x} B b}{12 \, c^{4}} + \frac {5 \, \sqrt {c x^{2} + b x} A}{3 \, c^{3}} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

1/2*B*x^5/((c*x^2 + b*x)^(3/2)*c) - 35/24*B*b^2*x*(3*x^2/((c*x^2 + b*x)^(3/2)*c) + b*x/((c*x^2 + b*x)^(3/2)*c^
2) - 2*x/(sqrt(c*x^2 + b*x)*b*c) - 1/(sqrt(c*x^2 + b*x)*c^2))/c^2 + 5/6*A*b*x*(3*x^2/((c*x^2 + b*x)^(3/2)*c) +
b*x/((c*x^2 + b*x)^(3/2)*c^2) - 2*x/(sqrt(c*x^2 + b*x)*b*c) - 1/(sqrt(c*x^2 + b*x)*c^2))/c - 7/4*B*b*x^4/((c*
x^2 + b*x)^(3/2)*c^2) + A*x^4/((c*x^2 + b*x)^(3/2)*c) - 35/6*B*b^2*x/(sqrt(c*x^2 + b*x)*c^4) + 10/3*A*b*x/(sqr
t(c*x^2 + b*x)*c^3) + 35/8*B*b^2*log(2*c*x + b + 2*sqrt(c*x^2 + b*x)*sqrt(c))/c^(9/2) - 5/2*A*b*log(2*c*x + b
+ 2*sqrt(c*x^2 + b*x)*sqrt(c))/c^(7/2) - 35/12*sqrt(c*x^2 + b*x)*B*b/c^4 + 5/3*sqrt(c*x^2 + b*x)*A/c^3

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Fricas [A]
time = 3.53, size = 380, normalized size = 2.21 \begin {gather*} \left [-\frac {15 \, {\left (7 \, B b^{4} - 4 \, A b^{3} c + {\left (7 \, B b^{2} c^{2} - 4 \, A b c^{3}\right )} x^{2} + 2 \, {\left (7 \, B b^{3} c - 4 \, A b^{2} c^{2}\right )} x\right )} \sqrt {c} \log \left (2 \, c x + b - 2 \, \sqrt {c x^{2} + b x} \sqrt {c}\right ) - 2 \, {\left (6 \, B c^{4} x^{3} - 105 \, B b^{3} c + 60 \, A b^{2} c^{2} - 3 \, {\left (7 \, B b c^{3} - 4 \, A c^{4}\right )} x^{2} - 20 \, {\left (7 \, B b^{2} c^{2} - 4 \, A b c^{3}\right )} x\right )} \sqrt {c x^{2} + b x}}{24 \, {\left (c^{7} x^{2} + 2 \, b c^{6} x + b^{2} c^{5}\right )}}, -\frac {15 \, {\left (7 \, B b^{4} - 4 \, A b^{3} c + {\left (7 \, B b^{2} c^{2} - 4 \, A b c^{3}\right )} x^{2} + 2 \, {\left (7 \, B b^{3} c - 4 \, A b^{2} c^{2}\right )} x\right )} \sqrt {-c} \arctan \left (\frac {\sqrt {c x^{2} + b x} \sqrt {-c}}{c x}\right ) - {\left (6 \, B c^{4} x^{3} - 105 \, B b^{3} c + 60 \, A b^{2} c^{2} - 3 \, {\left (7 \, B b c^{3} - 4 \, A c^{4}\right )} x^{2} - 20 \, {\left (7 \, B b^{2} c^{2} - 4 \, A b c^{3}\right )} x\right )} \sqrt {c x^{2} + b x}}{12 \, {\left (c^{7} x^{2} + 2 \, b c^{6} x + b^{2} c^{5}\right )}}\right ] \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

[-1/24*(15*(7*B*b^4 - 4*A*b^3*c + (7*B*b^2*c^2 - 4*A*b*c^3)*x^2 + 2*(7*B*b^3*c - 4*A*b^2*c^2)*x)*sqrt(c)*log(2
*c*x + b - 2*sqrt(c*x^2 + b*x)*sqrt(c)) - 2*(6*B*c^4*x^3 - 105*B*b^3*c + 60*A*b^2*c^2 - 3*(7*B*b*c^3 - 4*A*c^4
)*x^2 - 20*(7*B*b^2*c^2 - 4*A*b*c^3)*x)*sqrt(c*x^2 + b*x))/(c^7*x^2 + 2*b*c^6*x + b^2*c^5), -1/12*(15*(7*B*b^4
- 4*A*b^3*c + (7*B*b^2*c^2 - 4*A*b*c^3)*x^2 + 2*(7*B*b^3*c - 4*A*b^2*c^2)*x)*sqrt(-c)*arctan(sqrt(c*x^2 + b*x
)*sqrt(-c)/(c*x)) - (6*B*c^4*x^3 - 105*B*b^3*c + 60*A*b^2*c^2 - 3*(7*B*b*c^3 - 4*A*c^4)*x^2 - 20*(7*B*b^2*c^2
- 4*A*b*c^3)*x)*sqrt(c*x^2 + b*x))/(c^7*x^2 + 2*b*c^6*x + b^2*c^5)]

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Sympy [F]
time = 0.00, size = 0, normalized size = 0.00 \begin {gather*} \int \frac {x^{5} \left (A + B x\right )}{\left (x \left (b + c x\right )\right )^{\frac {5}{2}}}\, dx \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

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

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Giac [A]
time = 0.93, size = 253, normalized size = 1.47 \begin {gather*} \frac {1}{4} \, \sqrt {c x^{2} + b x} {\left (\frac {2 \, B x}{c^{3}} - \frac {11 \, B b c^{7} - 4 \, A c^{8}}{c^{11}}\right )} - \frac {5 \, {\left (7 \, B b^{2} - 4 \, A b c\right )} \log \left ({\left | -2 \, {\left (\sqrt {c} x - \sqrt {c x^{2} + b x}\right )} \sqrt {c} - b \right |}\right )}{8 \, c^{\frac {9}{2}}} - \frac {2 \, {\left (12 \, {\left (\sqrt {c} x - \sqrt {c x^{2} + b x}\right )}^{2} B b^{3} c^{\frac {3}{2}} - 9 \, {\left (\sqrt {c} x - \sqrt {c x^{2} + b x}\right )}^{2} A b^{2} c^{\frac {5}{2}} + 21 \, {\left (\sqrt {c} x - \sqrt {c x^{2} + b x}\right )} B b^{4} c - 15 \, {\left (\sqrt {c} x - \sqrt {c x^{2} + b x}\right )} A b^{3} c^{2} + 10 \, B b^{5} \sqrt {c} - 7 \, A b^{4} c^{\frac {3}{2}}\right )}}{3 \, {\left ({\left (\sqrt {c} x - \sqrt {c x^{2} + b x}\right )} \sqrt {c} + b\right )}^{3} c^{5}} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

1/4*sqrt(c*x^2 + b*x)*(2*B*x/c^3 - (11*B*b*c^7 - 4*A*c^8)/c^11) - 5/8*(7*B*b^2 - 4*A*b*c)*log(abs(-2*(sqrt(c)*
x - sqrt(c*x^2 + b*x))*sqrt(c) - b))/c^(9/2) - 2/3*(12*(sqrt(c)*x - sqrt(c*x^2 + b*x))^2*B*b^3*c^(3/2) - 9*(sq
rt(c)*x - sqrt(c*x^2 + b*x))^2*A*b^2*c^(5/2) + 21*(sqrt(c)*x - sqrt(c*x^2 + b*x))*B*b^4*c - 15*(sqrt(c)*x - sq
rt(c*x^2 + b*x))*A*b^3*c^2 + 10*B*b^5*sqrt(c) - 7*A*b^4*c^(3/2))/(((sqrt(c)*x - sqrt(c*x^2 + b*x))*sqrt(c) + b
)^3*c^5)

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Mupad [F]
time = 0.00, size = -1, normalized size = -0.01 \begin {gather*} \int \frac {x^5\,\left (A+B\,x\right )}{{\left (c\,x^2+b\,x\right )}^{5/2}} \,d x \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

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

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