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

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

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Rubi [A]  time = 0.17, 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 = {788, 668, 670, 640, 620, 206} \begin {gather*} -\frac {2 x^3 (7 b B-4 A c)}{3 b c^2 \sqrt {b x+c x^2}}+\frac {5 x \sqrt {b x+c x^2} (7 b B-4 A c)}{6 b c^3}-\frac {5 \sqrt {b x+c x^2} (7 b B-4 A c)}{4 c^4}+\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 {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 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 640

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 668

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 670

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 788

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)) \operatorname {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 [C]  time = 0.06, size = 80, normalized size = 0.47 \begin {gather*} \frac {2 x^5 \left ((b+c x) \sqrt {\frac {c x}{b}+1} (7 b B-4 A c) \, _2F_1\left (\frac {3}{2},\frac {7}{2};\frac {9}{2};-\frac {c x}{b}\right )+7 b (A c-b B)\right )}{21 b^2 c (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]

(2*x^5*(7*b*(-(b*B) + A*c) + (7*b*B - 4*A*c)*(b + c*x)*Sqrt[1 + (c*x)/b]*Hypergeometric2F1[3/2, 7/2, 9/2, -((c
*x)/b)]))/(21*b^2*c*(x*(b + c*x))^(3/2))

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IntegrateAlgebraic [A]  time = 0.61, size = 140, normalized size = 0.81 \begin {gather*} \frac {\sqrt {b x+c x^2} \left (60 A b^2 c+80 A b c^2 x+12 A c^3 x^2-105 b^3 B-140 b^2 B c x-21 b B c^2 x^2+6 B c^3 x^3\right )}{12 c^4 (b+c x)^2}-\frac {5 \left (7 b^2 B-4 A b c\right ) \log \left (-2 c^{9/2} \sqrt {b x+c x^2}+b c^4+2 c^5 x\right )}{8 c^{9/2}} \end {gather*}

Antiderivative was successfully verified.

[In]

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

[Out]

(Sqrt[b*x + c*x^2]*(-105*b^3*B + 60*A*b^2*c - 140*b^2*B*c*x + 80*A*b*c^2*x - 21*b*B*c^2*x^2 + 12*A*c^3*x^2 + 6
*B*c^3*x^3))/(12*c^4*(b + c*x)^2) - (5*(7*b^2*B - 4*A*b*c)*Log[b*c^4 + 2*c^5*x - 2*c^(9/2)*Sqrt[b*x + c*x^2]])
/(8*c^(9/2))

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fricas [A]  time = 0.43, 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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giac [A]  time = 0.49, 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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maple [B]  time = 0.05, size = 338, normalized size = 1.97 \begin {gather*} \frac {B \,x^{5}}{2 \left (c \,x^{2}+b x \right )^{\frac {3}{2}} c}+\frac {A \,x^{4}}{\left (c \,x^{2}+b x \right )^{\frac {3}{2}} c}-\frac {7 B b \,x^{4}}{4 \left (c \,x^{2}+b x \right )^{\frac {3}{2}} c^{2}}+\frac {5 A b \,x^{3}}{6 \left (c \,x^{2}+b x \right )^{\frac {3}{2}} c^{2}}-\frac {35 B \,b^{2} x^{3}}{24 \left (c \,x^{2}+b x \right )^{\frac {3}{2}} c^{3}}-\frac {5 A \,b^{2} x^{2}}{4 \left (c \,x^{2}+b x \right )^{\frac {3}{2}} c^{3}}+\frac {35 B \,b^{3} x^{2}}{16 \left (c \,x^{2}+b x \right )^{\frac {3}{2}} c^{4}}-\frac {5 A \,b^{3} x}{12 \left (c \,x^{2}+b x \right )^{\frac {3}{2}} c^{4}}+\frac {35 B \,b^{4} x}{48 \left (c \,x^{2}+b x \right )^{\frac {3}{2}} c^{5}}+\frac {35 A b x}{6 \sqrt {c \,x^{2}+b x}\, c^{3}}-\frac {245 B \,b^{2} x}{24 \sqrt {c \,x^{2}+b x}\, c^{4}}-\frac {5 A b \ln \left (\frac {c x +\frac {b}{2}}{\sqrt {c}}+\sqrt {c \,x^{2}+b x}\right )}{2 c^{\frac {7}{2}}}+\frac {35 B \,b^{2} \ln \left (\frac {c x +\frac {b}{2}}{\sqrt {c}}+\sqrt {c \,x^{2}+b x}\right )}{8 c^{\frac {9}{2}}}+\frac {5 A \,b^{2}}{12 \sqrt {c \,x^{2}+b x}\, c^{4}}-\frac {35 B \,b^{3}}{48 \sqrt {c \,x^{2}+b x}\, c^{5}} \end {gather*}

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)

[Out]

1/2*B*x^5/c/(c*x^2+b*x)^(3/2)-7/4*B*b/c^2*x^4/(c*x^2+b*x)^(3/2)-35/24*B*b^2/c^3*x^3/(c*x^2+b*x)^(3/2)+35/16*B*
b^3/c^4*x^2/(c*x^2+b*x)^(3/2)+35/48*B*b^4/c^5/(c*x^2+b*x)^(3/2)*x-245/24*B*b^2/c^4/(c*x^2+b*x)^(1/2)*x-35/48*B
*b^3/c^5/(c*x^2+b*x)^(1/2)+35/8*B*b^2/c^(9/2)*ln((c*x+1/2*b)/c^(1/2)+(c*x^2+b*x)^(1/2))+A*x^4/c/(c*x^2+b*x)^(3
/2)+5/6*A*b/c^2*x^3/(c*x^2+b*x)^(3/2)-5/4*A*b^2/c^3*x^2/(c*x^2+b*x)^(3/2)-5/12*A*b^3/c^4/(c*x^2+b*x)^(3/2)*x+3
5/6*A*b/c^3/(c*x^2+b*x)^(1/2)*x+5/12*A*b^2/c^4/(c*x^2+b*x)^(1/2)-5/2*A*b/c^(7/2)*ln((c*x+1/2*b)/c^(1/2)+(c*x^2
+b*x)^(1/2))

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maxima [B]  time = 0.68, 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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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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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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