3.4.7 \(\int \frac {(d+e x)^4}{(b x+c x^2)^{5/2}} \, dx\)

Optimal. Leaf size=208 \[ -\frac {2 (d+e x)^3 (x (2 c d-b e)+b d)}{3 b^2 \left (b x+c x^2\right )^{3/2}}-\frac {2 e \sqrt {b x+c x^2} (2 c d-b e) \left (-3 b^2 e^2-8 b c d e+8 c^2 d^2\right )}{3 b^4 c^2}+\frac {4 (d+e x) \left (x (2 c d-b e) \left (-b^2 e^2-4 b c d e+4 c^2 d^2\right )+b c d^2 (4 c d-5 b e)\right )}{3 b^4 c \sqrt {b x+c x^2}}+\frac {2 e^4 \tanh ^{-1}\left (\frac {\sqrt {c} x}{\sqrt {b x+c x^2}}\right )}{c^{5/2}} \]

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Rubi [A]  time = 0.20, antiderivative size = 208, normalized size of antiderivative = 1.00, number of steps used = 5, number of rules used = 5, integrand size = 21, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.238, Rules used = {738, 818, 640, 620, 206} \begin {gather*} -\frac {2 e \sqrt {b x+c x^2} (2 c d-b e) \left (-3 b^2 e^2-8 b c d e+8 c^2 d^2\right )}{3 b^4 c^2}+\frac {4 (d+e x) \left (x (2 c d-b e) \left (-b^2 e^2-4 b c d e+4 c^2 d^2\right )+b c d^2 (4 c d-5 b e)\right )}{3 b^4 c \sqrt {b x+c x^2}}-\frac {2 (d+e x)^3 (x (2 c d-b e)+b d)}{3 b^2 \left (b x+c x^2\right )^{3/2}}+\frac {2 e^4 \tanh ^{-1}\left (\frac {\sqrt {c} x}{\sqrt {b x+c x^2}}\right )}{c^{5/2}} \end {gather*}

Antiderivative was successfully verified.

[In]

Int[(d + e*x)^4/(b*x + c*x^2)^(5/2),x]

[Out]

(-2*(d + e*x)^3*(b*d + (2*c*d - b*e)*x))/(3*b^2*(b*x + c*x^2)^(3/2)) + (4*(d + e*x)*(b*c*d^2*(4*c*d - 5*b*e) +
(2*c*d - b*e)*(4*c^2*d^2 - 4*b*c*d*e - b^2*e^2)*x))/(3*b^4*c*Sqrt[b*x + c*x^2]) - (2*e*(2*c*d - b*e)*(8*c^2*d
^2 - 8*b*c*d*e - 3*b^2*e^2)*Sqrt[b*x + c*x^2])/(3*b^4*c^2) + (2*e^4*ArcTanh[(Sqrt[c]*x)/Sqrt[b*x + c*x^2]])/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 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 738

Int[((d_.) + (e_.)*(x_))^(m_)*((a_.) + (b_.)*(x_) + (c_.)*(x_)^2)^(p_), x_Symbol] :> Simp[((d + e*x)^(m - 1)*(
d*b - 2*a*e + (2*c*d - b*e)*x)*(a + b*x + c*x^2)^(p + 1))/((p + 1)*(b^2 - 4*a*c)), x] + Dist[1/((p + 1)*(b^2 -
4*a*c)), Int[(d + e*x)^(m - 2)*Simp[e*(2*a*e*(m - 1) + b*d*(2*p - m + 4)) - 2*c*d^2*(2*p + 3) + e*(b*e - 2*d*
c)*(m + 2*p + 2)*x, x]*(a + b*x + c*x^2)^(p + 1), x], x] /; FreeQ[{a, b, c, d, e}, x] && NeQ[b^2 - 4*a*c, 0] &
& NeQ[c*d^2 - b*d*e + a*e^2, 0] && NeQ[2*c*d - b*e, 0] && LtQ[p, -1] && GtQ[m, 1] && IntQuadraticQ[a, b, c, d,
e, m, p, x]

Rule 818

Int[((d_.) + (e_.)*(x_))^(m_)*((f_.) + (g_.)*(x_))*((a_.) + (b_.)*(x_) + (c_.)*(x_)^2)^(p_.), x_Symbol] :> -Si
mp[((d + e*x)^(m - 1)*(a + b*x + c*x^2)^(p + 1)*(2*a*c*(e*f + d*g) - b*(c*d*f + a*e*g) - (2*c^2*d*f + b^2*e*g
- c*(b*e*f + b*d*g + 2*a*e*g))*x))/(c*(p + 1)*(b^2 - 4*a*c)), x] - Dist[1/(c*(p + 1)*(b^2 - 4*a*c)), Int[(d +
e*x)^(m - 2)*(a + b*x + c*x^2)^(p + 1)*Simp[2*c^2*d^2*f*(2*p + 3) + b*e*g*(a*e*(m - 1) + b*d*(p + 2)) - c*(2*a
*e*(e*f*(m - 1) + d*g*m) + b*d*(d*g*(2*p + 3) - e*f*(m - 2*p - 4))) + e*(b^2*e*g*(m + p + 1) + 2*c^2*d*f*(m +
2*p + 2) - c*(2*a*e*g*m + b*(e*f + d*g)*(m + 2*p + 2)))*x, x], x], x] /; FreeQ[{a, b, c, d, e, f, g}, x] && Ne
Q[b^2 - 4*a*c, 0] && NeQ[c*d^2 - b*d*e + a*e^2, 0] && LtQ[p, -1] && GtQ[m, 1] && ((EqQ[m, 2] && EqQ[p, -3] &&
RationalQ[a, b, c, d, e, f, g]) ||  !ILtQ[m + 2*p + 3, 0])

Rubi steps

\begin {align*} \int \frac {(d+e x)^4}{\left (b x+c x^2\right )^{5/2}} \, dx &=-\frac {2 (d+e x)^3 (b d+(2 c d-b e) x)}{3 b^2 \left (b x+c x^2\right )^{3/2}}-\frac {2 \int \frac {(d+e x)^2 (d (4 c d-5 b e)-e (2 c d-b e) x)}{\left (b x+c x^2\right )^{3/2}} \, dx}{3 b^2}\\ &=-\frac {2 (d+e x)^3 (b d+(2 c d-b e) x)}{3 b^2 \left (b x+c x^2\right )^{3/2}}+\frac {4 (d+e x) \left (b c d^2 (4 c d-5 b e)+(2 c d-b e) \left (4 c^2 d^2-4 b c d e-b^2 e^2\right ) x\right )}{3 b^4 c \sqrt {b x+c x^2}}-\frac {4 \int \frac {\frac {1}{2} b d e \left (8 c^2 d^2-12 b c d e+b^2 e^2\right )+\frac {1}{2} e (2 c d-b e) \left (8 c^2 d^2-8 b c d e-3 b^2 e^2\right ) x}{\sqrt {b x+c x^2}} \, dx}{3 b^4 c}\\ &=-\frac {2 (d+e x)^3 (b d+(2 c d-b e) x)}{3 b^2 \left (b x+c x^2\right )^{3/2}}+\frac {4 (d+e x) \left (b c d^2 (4 c d-5 b e)+(2 c d-b e) \left (4 c^2 d^2-4 b c d e-b^2 e^2\right ) x\right )}{3 b^4 c \sqrt {b x+c x^2}}-\frac {2 e (2 c d-b e) \left (8 c^2 d^2-8 b c d e-3 b^2 e^2\right ) \sqrt {b x+c x^2}}{3 b^4 c^2}+\frac {e^4 \int \frac {1}{\sqrt {b x+c x^2}} \, dx}{c^2}\\ &=-\frac {2 (d+e x)^3 (b d+(2 c d-b e) x)}{3 b^2 \left (b x+c x^2\right )^{3/2}}+\frac {4 (d+e x) \left (b c d^2 (4 c d-5 b e)+(2 c d-b e) \left (4 c^2 d^2-4 b c d e-b^2 e^2\right ) x\right )}{3 b^4 c \sqrt {b x+c x^2}}-\frac {2 e (2 c d-b e) \left (8 c^2 d^2-8 b c d e-3 b^2 e^2\right ) \sqrt {b x+c x^2}}{3 b^4 c^2}+\frac {\left (2 e^4\right ) \operatorname {Subst}\left (\int \frac {1}{1-c x^2} \, dx,x,\frac {x}{\sqrt {b x+c x^2}}\right )}{c^2}\\ &=-\frac {2 (d+e x)^3 (b d+(2 c d-b e) x)}{3 b^2 \left (b x+c x^2\right )^{3/2}}+\frac {4 (d+e x) \left (b c d^2 (4 c d-5 b e)+(2 c d-b e) \left (4 c^2 d^2-4 b c d e-b^2 e^2\right ) x\right )}{3 b^4 c \sqrt {b x+c x^2}}-\frac {2 e (2 c d-b e) \left (8 c^2 d^2-8 b c d e-3 b^2 e^2\right ) \sqrt {b x+c x^2}}{3 b^4 c^2}+\frac {2 e^4 \tanh ^{-1}\left (\frac {\sqrt {c} x}{\sqrt {b x+c x^2}}\right )}{c^{5/2}}\\ \end {align*}

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Mathematica [A]  time = 5.17, size = 196, normalized size = 0.94 \begin {gather*} \frac {x^{5/2} (b+c x)^3 \left (-\frac {8 \sqrt {x} (b e-c d)^3 (b e+2 c d)}{3 b^4 c^2 (b+c x)}-\frac {8 d^3 (3 b e-2 c d)}{3 b^4 \sqrt {x}}+\frac {2 \sqrt {x} (b e-c d)^4}{3 b^3 c^2 (b+c x)^2}-\frac {2 d^4}{3 b^3 x^{3/2}}\right )}{(x (b+c x))^{5/2}}+\frac {2 e^4 x^{5/2} (b+c x)^{5/2} \log \left (\sqrt {c} \sqrt {b+c x}+c \sqrt {x}\right )}{c^{5/2} (x (b+c x))^{5/2}} \end {gather*}

Antiderivative was successfully verified.

[In]

Integrate[(d + e*x)^4/(b*x + c*x^2)^(5/2),x]

[Out]

(x^(5/2)*(b + c*x)^3*((-2*d^4)/(3*b^3*x^(3/2)) - (8*d^3*(-2*c*d + 3*b*e))/(3*b^4*Sqrt[x]) + (2*(-(c*d) + b*e)^
4*Sqrt[x])/(3*b^3*c^2*(b + c*x)^2) - (8*(-(c*d) + b*e)^3*(2*c*d + b*e)*Sqrt[x])/(3*b^4*c^2*(b + c*x))))/(x*(b
+ c*x))^(5/2) + (2*e^4*x^(5/2)*(b + c*x)^(5/2)*Log[c*Sqrt[x] + Sqrt[c]*Sqrt[b + c*x]])/(c^(5/2)*(x*(b + c*x))^
(5/2))

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IntegrateAlgebraic [A]  time = 0.65, size = 236, normalized size = 1.13 \begin {gather*} -\frac {2 \sqrt {b x+c x^2} \left (3 b^5 e^4 x^2+4 b^4 c e^4 x^3+b^3 c^2 d^4+12 b^3 c^2 d^3 e x-18 b^3 c^2 d^2 e^2 x^2-4 b^3 c^2 d e^3 x^3-6 b^2 c^3 d^4 x+48 b^2 c^3 d^3 e x^2-12 b^2 c^3 d^2 e^2 x^3-24 b c^4 d^4 x^2+32 b c^4 d^3 e x^3-16 c^5 d^4 x^3\right )}{3 b^4 c^2 x^2 (b+c x)^2}-\frac {e^4 \log \left (-2 c^{5/2} \sqrt {b x+c x^2}+b c^2+2 c^3 x\right )}{c^{5/2}} \end {gather*}

Antiderivative was successfully verified.

[In]

IntegrateAlgebraic[(d + e*x)^4/(b*x + c*x^2)^(5/2),x]

[Out]

(-2*Sqrt[b*x + c*x^2]*(b^3*c^2*d^4 - 6*b^2*c^3*d^4*x + 12*b^3*c^2*d^3*e*x - 24*b*c^4*d^4*x^2 + 48*b^2*c^3*d^3*
e*x^2 - 18*b^3*c^2*d^2*e^2*x^2 + 3*b^5*e^4*x^2 - 16*c^5*d^4*x^3 + 32*b*c^4*d^3*e*x^3 - 12*b^2*c^3*d^2*e^2*x^3
- 4*b^3*c^2*d*e^3*x^3 + 4*b^4*c*e^4*x^3))/(3*b^4*c^2*x^2*(b + c*x)^2) - (e^4*Log[b*c^2 + 2*c^3*x - 2*c^(5/2)*S
qrt[b*x + c*x^2]])/c^(5/2)

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fricas [A]  time = 0.44, size = 525, normalized size = 2.52 \begin {gather*} \left [\frac {3 \, {\left (b^{4} c^{2} e^{4} x^{4} + 2 \, b^{5} c e^{4} x^{3} + b^{6} e^{4} x^{2}\right )} \sqrt {c} \log \left (2 \, c x + b + 2 \, \sqrt {c x^{2} + b x} \sqrt {c}\right ) - 2 \, {\left (b^{3} c^{3} d^{4} - 4 \, {\left (4 \, c^{6} d^{4} - 8 \, b c^{5} d^{3} e + 3 \, b^{2} c^{4} d^{2} e^{2} + b^{3} c^{3} d e^{3} - b^{4} c^{2} e^{4}\right )} x^{3} - 3 \, {\left (8 \, b c^{5} d^{4} - 16 \, b^{2} c^{4} d^{3} e + 6 \, b^{3} c^{3} d^{2} e^{2} - b^{5} c e^{4}\right )} x^{2} - 6 \, {\left (b^{2} c^{4} d^{4} - 2 \, b^{3} c^{3} d^{3} e\right )} x\right )} \sqrt {c x^{2} + b x}}{3 \, {\left (b^{4} c^{5} x^{4} + 2 \, b^{5} c^{4} x^{3} + b^{6} c^{3} x^{2}\right )}}, -\frac {2 \, {\left (3 \, {\left (b^{4} c^{2} e^{4} x^{4} + 2 \, b^{5} c e^{4} x^{3} + b^{6} e^{4} x^{2}\right )} \sqrt {-c} \arctan \left (\frac {\sqrt {c x^{2} + b x} \sqrt {-c}}{c x}\right ) + {\left (b^{3} c^{3} d^{4} - 4 \, {\left (4 \, c^{6} d^{4} - 8 \, b c^{5} d^{3} e + 3 \, b^{2} c^{4} d^{2} e^{2} + b^{3} c^{3} d e^{3} - b^{4} c^{2} e^{4}\right )} x^{3} - 3 \, {\left (8 \, b c^{5} d^{4} - 16 \, b^{2} c^{4} d^{3} e + 6 \, b^{3} c^{3} d^{2} e^{2} - b^{5} c e^{4}\right )} x^{2} - 6 \, {\left (b^{2} c^{4} d^{4} - 2 \, b^{3} c^{3} d^{3} e\right )} x\right )} \sqrt {c x^{2} + b x}\right )}}{3 \, {\left (b^{4} c^{5} x^{4} + 2 \, b^{5} c^{4} x^{3} + b^{6} c^{3} x^{2}\right )}}\right ] \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate((e*x+d)^4/(c*x^2+b*x)^(5/2),x, algorithm="fricas")

[Out]

[1/3*(3*(b^4*c^2*e^4*x^4 + 2*b^5*c*e^4*x^3 + b^6*e^4*x^2)*sqrt(c)*log(2*c*x + b + 2*sqrt(c*x^2 + b*x)*sqrt(c))
- 2*(b^3*c^3*d^4 - 4*(4*c^6*d^4 - 8*b*c^5*d^3*e + 3*b^2*c^4*d^2*e^2 + b^3*c^3*d*e^3 - b^4*c^2*e^4)*x^3 - 3*(8
*b*c^5*d^4 - 16*b^2*c^4*d^3*e + 6*b^3*c^3*d^2*e^2 - b^5*c*e^4)*x^2 - 6*(b^2*c^4*d^4 - 2*b^3*c^3*d^3*e)*x)*sqrt
(c*x^2 + b*x))/(b^4*c^5*x^4 + 2*b^5*c^4*x^3 + b^6*c^3*x^2), -2/3*(3*(b^4*c^2*e^4*x^4 + 2*b^5*c*e^4*x^3 + b^6*e
^4*x^2)*sqrt(-c)*arctan(sqrt(c*x^2 + b*x)*sqrt(-c)/(c*x)) + (b^3*c^3*d^4 - 4*(4*c^6*d^4 - 8*b*c^5*d^3*e + 3*b^
2*c^4*d^2*e^2 + b^3*c^3*d*e^3 - b^4*c^2*e^4)*x^3 - 3*(8*b*c^5*d^4 - 16*b^2*c^4*d^3*e + 6*b^3*c^3*d^2*e^2 - b^5
*c*e^4)*x^2 - 6*(b^2*c^4*d^4 - 2*b^3*c^3*d^3*e)*x)*sqrt(c*x^2 + b*x))/(b^4*c^5*x^4 + 2*b^5*c^4*x^3 + b^6*c^3*x
^2)]

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giac [A]  time = 0.36, size = 209, normalized size = 1.00 \begin {gather*} -\frac {2 \, {\left (\frac {d^{4}}{b} - {\left (x {\left (\frac {4 \, {\left (4 \, c^{5} d^{4} - 8 \, b c^{4} d^{3} e + 3 \, b^{2} c^{3} d^{2} e^{2} + b^{3} c^{2} d e^{3} - b^{4} c e^{4}\right )} x}{b^{4} c^{2}} + \frac {3 \, {\left (8 \, b c^{4} d^{4} - 16 \, b^{2} c^{3} d^{3} e + 6 \, b^{3} c^{2} d^{2} e^{2} - b^{5} e^{4}\right )}}{b^{4} c^{2}}\right )} + \frac {6 \, {\left (b^{2} c^{3} d^{4} - 2 \, b^{3} c^{2} d^{3} e\right )}}{b^{4} c^{2}}\right )} x\right )}}{3 \, {\left (c x^{2} + b x\right )}^{\frac {3}{2}}} - \frac {e^{4} \log \left ({\left | -2 \, {\left (\sqrt {c} x - \sqrt {c x^{2} + b x}\right )} \sqrt {c} - b \right |}\right )}{c^{\frac {5}{2}}} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate((e*x+d)^4/(c*x^2+b*x)^(5/2),x, algorithm="giac")

[Out]

-2/3*(d^4/b - (x*(4*(4*c^5*d^4 - 8*b*c^4*d^3*e + 3*b^2*c^3*d^2*e^2 + b^3*c^2*d*e^3 - b^4*c*e^4)*x/(b^4*c^2) +
3*(8*b*c^4*d^4 - 16*b^2*c^3*d^3*e + 6*b^3*c^2*d^2*e^2 - b^5*e^4)/(b^4*c^2)) + 6*(b^2*c^3*d^4 - 2*b^3*c^2*d^3*e
)/(b^4*c^2))*x)/(c*x^2 + b*x)^(3/2) - e^4*log(abs(-2*(sqrt(c)*x - sqrt(c*x^2 + b*x))*sqrt(c) - b))/c^(5/2)

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maple [B]  time = 0.12, size = 447, normalized size = 2.15 \begin {gather*} -\frac {e^{4} x^{3}}{3 \left (c \,x^{2}+b x \right )^{\frac {3}{2}} c}+\frac {b \,e^{4} x^{2}}{2 \left (c \,x^{2}+b x \right )^{\frac {3}{2}} c^{2}}-\frac {4 d \,e^{3} x^{2}}{\left (c \,x^{2}+b x \right )^{\frac {3}{2}} c}+\frac {b^{2} e^{4} x}{6 \left (c \,x^{2}+b x \right )^{\frac {3}{2}} c^{3}}-\frac {4 b d \,e^{3} x}{3 \left (c \,x^{2}+b x \right )^{\frac {3}{2}} c^{2}}+\frac {8 d^{3} e x}{3 \left (c \,x^{2}+b x \right )^{\frac {3}{2}} b}-\frac {4 c \,d^{4} x}{3 \left (c \,x^{2}+b x \right )^{\frac {3}{2}} b^{2}}-\frac {4 d^{2} e^{2} x}{\left (c \,x^{2}+b x \right )^{\frac {3}{2}} c}+\frac {8 d \,e^{3} x}{3 \sqrt {c \,x^{2}+b x}\, b c}-\frac {2 d^{4}}{3 \left (c \,x^{2}+b x \right )^{\frac {3}{2}} b}+\frac {8 d^{2} e^{2} x}{\sqrt {c \,x^{2}+b x}\, b^{2}}-\frac {64 c \,d^{3} e x}{3 \sqrt {c \,x^{2}+b x}\, b^{3}}+\frac {32 c^{2} d^{4} x}{3 \sqrt {c \,x^{2}+b x}\, b^{4}}-\frac {7 e^{4} x}{3 \sqrt {c \,x^{2}+b x}\, c^{2}}+\frac {e^{4} \ln \left (\frac {c x +\frac {b}{2}}{\sqrt {c}}+\sqrt {c \,x^{2}+b x}\right )}{c^{\frac {5}{2}}}-\frac {b \,e^{4}}{6 \sqrt {c \,x^{2}+b x}\, c^{3}}+\frac {4 d^{2} e^{2}}{\sqrt {c \,x^{2}+b x}\, b c}-\frac {32 d^{3} e}{3 \sqrt {c \,x^{2}+b x}\, b^{2}}+\frac {16 c \,d^{4}}{3 \sqrt {c \,x^{2}+b x}\, b^{3}}+\frac {4 d \,e^{3}}{3 \sqrt {c \,x^{2}+b x}\, c^{2}} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

int((e*x+d)^4/(c*x^2+b*x)^(5/2),x)

[Out]

-1/3*e^4*x^3/c/(c*x^2+b*x)^(3/2)+1/2*e^4*b/c^2*x^2/(c*x^2+b*x)^(3/2)+1/6*e^4*b^2/c^3/(c*x^2+b*x)^(3/2)*x-7/3*e
^4/c^2/(c*x^2+b*x)^(1/2)*x-1/6*e^4*b/c^3/(c*x^2+b*x)^(1/2)+e^4/c^(5/2)*ln((c*x+1/2*b)/c^(1/2)+(c*x^2+b*x)^(1/2
))-4*d*e^3*x^2/c/(c*x^2+b*x)^(3/2)-4/3*d*e^3*b/c^2/(c*x^2+b*x)^(3/2)*x+8/3*d*e^3/b/c/(c*x^2+b*x)^(1/2)*x+4/3*d
*e^3/c^2/(c*x^2+b*x)^(1/2)-4*d^2*e^2/c/(c*x^2+b*x)^(3/2)*x+8*d^2*e^2/b^2/(c*x^2+b*x)^(1/2)*x+4*d^2*e^2/b/c/(c*
x^2+b*x)^(1/2)+8/3*d^3*e/b/(c*x^2+b*x)^(3/2)*x-64/3*d^3*e/b^3*c/(c*x^2+b*x)^(1/2)*x-32/3*d^3*e/b^2/(c*x^2+b*x)
^(1/2)-4/3*d^4/b^2/(c*x^2+b*x)^(3/2)*c*x-2/3*d^4/b/(c*x^2+b*x)^(3/2)+32/3*d^4*c^2/b^4/(c*x^2+b*x)^(1/2)*x+16/3
*d^4*c/b^3/(c*x^2+b*x)^(1/2)

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maxima [B]  time = 1.50, size = 458, normalized size = 2.20 \begin {gather*} -\frac {1}{3} \, e^{4} 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 )} - \frac {4 \, d e^{3} x^{2}}{{\left (c x^{2} + b x\right )}^{\frac {3}{2}} c} - \frac {4 \, c d^{4} x}{3 \, {\left (c x^{2} + b x\right )}^{\frac {3}{2}} b^{2}} + \frac {32 \, c^{2} d^{4} x}{3 \, \sqrt {c x^{2} + b x} b^{4}} + \frac {8 \, d^{3} e x}{3 \, {\left (c x^{2} + b x\right )}^{\frac {3}{2}} b} - \frac {64 \, c d^{3} e x}{3 \, \sqrt {c x^{2} + b x} b^{3}} + \frac {8 \, d^{2} e^{2} x}{\sqrt {c x^{2} + b x} b^{2}} - \frac {4 \, d^{2} e^{2} x}{{\left (c x^{2} + b x\right )}^{\frac {3}{2}} c} - \frac {4 \, b d e^{3} x}{3 \, {\left (c x^{2} + b x\right )}^{\frac {3}{2}} c^{2}} + \frac {8 \, d e^{3} x}{3 \, \sqrt {c x^{2} + b x} b c} - \frac {4 \, e^{4} x}{3 \, \sqrt {c x^{2} + b x} c^{2}} + \frac {e^{4} \log \left (2 \, c x + b + 2 \, \sqrt {c x^{2} + b x} \sqrt {c}\right )}{c^{\frac {5}{2}}} - \frac {2 \, d^{4}}{3 \, {\left (c x^{2} + b x\right )}^{\frac {3}{2}} b} + \frac {16 \, c d^{4}}{3 \, \sqrt {c x^{2} + b x} b^{3}} - \frac {32 \, d^{3} e}{3 \, \sqrt {c x^{2} + b x} b^{2}} + \frac {4 \, d^{2} e^{2}}{\sqrt {c x^{2} + b x} b c} + \frac {4 \, d e^{3}}{3 \, \sqrt {c x^{2} + b x} c^{2}} - \frac {2 \, \sqrt {c x^{2} + b x} e^{4}}{3 \, b c^{2}} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate((e*x+d)^4/(c*x^2+b*x)^(5/2),x, algorithm="maxima")

[Out]

-1/3*e^4*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/(s
qrt(c*x^2 + b*x)*c^2)) - 4*d*e^3*x^2/((c*x^2 + b*x)^(3/2)*c) - 4/3*c*d^4*x/((c*x^2 + b*x)^(3/2)*b^2) + 32/3*c^
2*d^4*x/(sqrt(c*x^2 + b*x)*b^4) + 8/3*d^3*e*x/((c*x^2 + b*x)^(3/2)*b) - 64/3*c*d^3*e*x/(sqrt(c*x^2 + b*x)*b^3)
+ 8*d^2*e^2*x/(sqrt(c*x^2 + b*x)*b^2) - 4*d^2*e^2*x/((c*x^2 + b*x)^(3/2)*c) - 4/3*b*d*e^3*x/((c*x^2 + b*x)^(3
/2)*c^2) + 8/3*d*e^3*x/(sqrt(c*x^2 + b*x)*b*c) - 4/3*e^4*x/(sqrt(c*x^2 + b*x)*c^2) + e^4*log(2*c*x + b + 2*sqr
t(c*x^2 + b*x)*sqrt(c))/c^(5/2) - 2/3*d^4/((c*x^2 + b*x)^(3/2)*b) + 16/3*c*d^4/(sqrt(c*x^2 + b*x)*b^3) - 32/3*
d^3*e/(sqrt(c*x^2 + b*x)*b^2) + 4*d^2*e^2/(sqrt(c*x^2 + b*x)*b*c) + 4/3*d*e^3/(sqrt(c*x^2 + b*x)*c^2) - 2/3*sq
rt(c*x^2 + b*x)*e^4/(b*c^2)

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mupad [F]  time = 0.00, size = -1, normalized size = -0.00 \begin {gather*} \int \frac {{\left (d+e\,x\right )}^4}{{\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((d + e*x)^4/(b*x + c*x^2)^(5/2),x)

[Out]

int((d + e*x)^4/(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 {\left (d + e x\right )^{4}}{\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((e*x+d)**4/(c*x**2+b*x)**(5/2),x)

[Out]

Integral((d + e*x)**4/(x*(b + c*x))**(5/2), x)

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