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

Optimal. Leaf size=159 \[ -\frac {(c d-b e)^{3/2} (b e+4 c d) \tanh ^{-1}\left (\frac {\sqrt {c} \sqrt {d+e x}}{\sqrt {c d-b e}}\right )}{b^3 c^{3/2}}+\frac {d^{3/2} (4 c d-5 b e) \tanh ^{-1}\left (\frac {\sqrt {d+e x}}{\sqrt {d}}\right )}{b^3}-\frac {(d+e x)^{3/2} (x (2 c d-b e)+b d)}{b^2 \left (b x+c x^2\right )}+\frac {e \sqrt {d+e x} (2 c d-b e)}{b^2 c} \]

[Out]

-(e*x+d)^(3/2)*(b*d+(-b*e+2*c*d)*x)/b^2/(c*x^2+b*x)+d^(3/2)*(-5*b*e+4*c*d)*arctanh((e*x+d)^(1/2)/d^(1/2))/b^3-
(-b*e+c*d)^(3/2)*(b*e+4*c*d)*arctanh(c^(1/2)*(e*x+d)^(1/2)/(-b*e+c*d)^(1/2))/b^3/c^(3/2)+e*(-b*e+2*c*d)*(e*x+d
)^(1/2)/b^2/c

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Rubi [A]  time = 0.30, antiderivative size = 159, normalized size of antiderivative = 1.00, number of steps used = 6, number of rules used = 5, integrand size = 21, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.238, Rules used = {738, 824, 826, 1166, 208} \[ -\frac {(c d-b e)^{3/2} (b e+4 c d) \tanh ^{-1}\left (\frac {\sqrt {c} \sqrt {d+e x}}{\sqrt {c d-b e}}\right )}{b^3 c^{3/2}}+\frac {d^{3/2} (4 c d-5 b e) \tanh ^{-1}\left (\frac {\sqrt {d+e x}}{\sqrt {d}}\right )}{b^3}-\frac {(d+e x)^{3/2} (x (2 c d-b e)+b d)}{b^2 \left (b x+c x^2\right )}+\frac {e \sqrt {d+e x} (2 c d-b e)}{b^2 c} \]

Antiderivative was successfully verified.

[In]

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

[Out]

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

Rule 208

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

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 824

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

Rule 826

Int[((f_.) + (g_.)*(x_))/(Sqrt[(d_.) + (e_.)*(x_)]*((a_.) + (b_.)*(x_) + (c_.)*(x_)^2)), x_Symbol] :> Dist[2,
Subst[Int[(e*f - d*g + g*x^2)/(c*d^2 - b*d*e + a*e^2 - (2*c*d - b*e)*x^2 + c*x^4), x], x, Sqrt[d + e*x]], x] /
; FreeQ[{a, b, c, d, e, f, g}, x] && NeQ[b^2 - 4*a*c, 0] && NeQ[c*d^2 - b*d*e + a*e^2, 0]

Rule 1166

Int[((d_) + (e_.)*(x_)^2)/((a_) + (b_.)*(x_)^2 + (c_.)*(x_)^4), x_Symbol] :> With[{q = Rt[b^2 - 4*a*c, 2]}, Di
st[e/2 + (2*c*d - b*e)/(2*q), Int[1/(b/2 - q/2 + c*x^2), x], x] + Dist[e/2 - (2*c*d - b*e)/(2*q), Int[1/(b/2 +
q/2 + c*x^2), x], x]] /; FreeQ[{a, b, c, d, e}, x] && NeQ[b^2 - 4*a*c, 0] && NeQ[c*d^2 - a*e^2, 0] && PosQ[b^
2 - 4*a*c]

Rubi steps

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

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Mathematica [A]  time = 0.26, size = 159, normalized size = 1.00 \[ \frac {-\frac {b \sqrt {d+e x} \left (b^2 e^2 x+b c d (d-2 e x)+2 c^2 d^2 x\right )}{c x (b+c x)}-\frac {\sqrt {c d-b e} \left (-b^2 e^2-3 b c d e+4 c^2 d^2\right ) \tanh ^{-1}\left (\frac {\sqrt {c} \sqrt {d+e x}}{\sqrt {c d-b e}}\right )}{c^{3/2}}+d^{3/2} (4 c d-5 b e) \tanh ^{-1}\left (\frac {\sqrt {d+e x}}{\sqrt {d}}\right )}{b^3} \]

Antiderivative was successfully verified.

[In]

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

[Out]

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

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fricas [A]  time = 1.18, size = 1002, normalized size = 6.30 \[ \left [-\frac {{\left ({\left (4 \, c^{3} d^{2} - 3 \, b c^{2} d e - b^{2} c e^{2}\right )} x^{2} + {\left (4 \, b c^{2} d^{2} - 3 \, b^{2} c d e - b^{3} e^{2}\right )} x\right )} \sqrt {\frac {c d - b e}{c}} \log \left (\frac {c e x + 2 \, c d - b e + 2 \, \sqrt {e x + d} c \sqrt {\frac {c d - b e}{c}}}{c x + b}\right ) + {\left ({\left (4 \, c^{3} d^{2} - 5 \, b c^{2} d e\right )} x^{2} + {\left (4 \, b c^{2} d^{2} - 5 \, b^{2} c d e\right )} x\right )} \sqrt {d} \log \left (\frac {e x - 2 \, \sqrt {e x + d} \sqrt {d} + 2 \, d}{x}\right ) + 2 \, {\left (b^{2} c d^{2} + {\left (2 \, b c^{2} d^{2} - 2 \, b^{2} c d e + b^{3} e^{2}\right )} x\right )} \sqrt {e x + d}}{2 \, {\left (b^{3} c^{2} x^{2} + b^{4} c x\right )}}, -\frac {2 \, {\left ({\left (4 \, c^{3} d^{2} - 3 \, b c^{2} d e - b^{2} c e^{2}\right )} x^{2} + {\left (4 \, b c^{2} d^{2} - 3 \, b^{2} c d e - b^{3} e^{2}\right )} x\right )} \sqrt {-\frac {c d - b e}{c}} \arctan \left (-\frac {\sqrt {e x + d} c \sqrt {-\frac {c d - b e}{c}}}{c d - b e}\right ) + {\left ({\left (4 \, c^{3} d^{2} - 5 \, b c^{2} d e\right )} x^{2} + {\left (4 \, b c^{2} d^{2} - 5 \, b^{2} c d e\right )} x\right )} \sqrt {d} \log \left (\frac {e x - 2 \, \sqrt {e x + d} \sqrt {d} + 2 \, d}{x}\right ) + 2 \, {\left (b^{2} c d^{2} + {\left (2 \, b c^{2} d^{2} - 2 \, b^{2} c d e + b^{3} e^{2}\right )} x\right )} \sqrt {e x + d}}{2 \, {\left (b^{3} c^{2} x^{2} + b^{4} c x\right )}}, -\frac {2 \, {\left ({\left (4 \, c^{3} d^{2} - 5 \, b c^{2} d e\right )} x^{2} + {\left (4 \, b c^{2} d^{2} - 5 \, b^{2} c d e\right )} x\right )} \sqrt {-d} \arctan \left (\frac {\sqrt {e x + d} \sqrt {-d}}{d}\right ) + {\left ({\left (4 \, c^{3} d^{2} - 3 \, b c^{2} d e - b^{2} c e^{2}\right )} x^{2} + {\left (4 \, b c^{2} d^{2} - 3 \, b^{2} c d e - b^{3} e^{2}\right )} x\right )} \sqrt {\frac {c d - b e}{c}} \log \left (\frac {c e x + 2 \, c d - b e + 2 \, \sqrt {e x + d} c \sqrt {\frac {c d - b e}{c}}}{c x + b}\right ) + 2 \, {\left (b^{2} c d^{2} + {\left (2 \, b c^{2} d^{2} - 2 \, b^{2} c d e + b^{3} e^{2}\right )} x\right )} \sqrt {e x + d}}{2 \, {\left (b^{3} c^{2} x^{2} + b^{4} c x\right )}}, -\frac {{\left ({\left (4 \, c^{3} d^{2} - 3 \, b c^{2} d e - b^{2} c e^{2}\right )} x^{2} + {\left (4 \, b c^{2} d^{2} - 3 \, b^{2} c d e - b^{3} e^{2}\right )} x\right )} \sqrt {-\frac {c d - b e}{c}} \arctan \left (-\frac {\sqrt {e x + d} c \sqrt {-\frac {c d - b e}{c}}}{c d - b e}\right ) + {\left ({\left (4 \, c^{3} d^{2} - 5 \, b c^{2} d e\right )} x^{2} + {\left (4 \, b c^{2} d^{2} - 5 \, b^{2} c d e\right )} x\right )} \sqrt {-d} \arctan \left (\frac {\sqrt {e x + d} \sqrt {-d}}{d}\right ) + {\left (b^{2} c d^{2} + {\left (2 \, b c^{2} d^{2} - 2 \, b^{2} c d e + b^{3} e^{2}\right )} x\right )} \sqrt {e x + d}}{b^{3} c^{2} x^{2} + b^{4} c x}\right ] \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

[-1/2*(((4*c^3*d^2 - 3*b*c^2*d*e - b^2*c*e^2)*x^2 + (4*b*c^2*d^2 - 3*b^2*c*d*e - b^3*e^2)*x)*sqrt((c*d - b*e)/
c)*log((c*e*x + 2*c*d - b*e + 2*sqrt(e*x + d)*c*sqrt((c*d - b*e)/c))/(c*x + b)) + ((4*c^3*d^2 - 5*b*c^2*d*e)*x
^2 + (4*b*c^2*d^2 - 5*b^2*c*d*e)*x)*sqrt(d)*log((e*x - 2*sqrt(e*x + d)*sqrt(d) + 2*d)/x) + 2*(b^2*c*d^2 + (2*b
*c^2*d^2 - 2*b^2*c*d*e + b^3*e^2)*x)*sqrt(e*x + d))/(b^3*c^2*x^2 + b^4*c*x), -1/2*(2*((4*c^3*d^2 - 3*b*c^2*d*e
- b^2*c*e^2)*x^2 + (4*b*c^2*d^2 - 3*b^2*c*d*e - b^3*e^2)*x)*sqrt(-(c*d - b*e)/c)*arctan(-sqrt(e*x + d)*c*sqrt
(-(c*d - b*e)/c)/(c*d - b*e)) + ((4*c^3*d^2 - 5*b*c^2*d*e)*x^2 + (4*b*c^2*d^2 - 5*b^2*c*d*e)*x)*sqrt(d)*log((e
*x - 2*sqrt(e*x + d)*sqrt(d) + 2*d)/x) + 2*(b^2*c*d^2 + (2*b*c^2*d^2 - 2*b^2*c*d*e + b^3*e^2)*x)*sqrt(e*x + d)
)/(b^3*c^2*x^2 + b^4*c*x), -1/2*(2*((4*c^3*d^2 - 5*b*c^2*d*e)*x^2 + (4*b*c^2*d^2 - 5*b^2*c*d*e)*x)*sqrt(-d)*ar
ctan(sqrt(e*x + d)*sqrt(-d)/d) + ((4*c^3*d^2 - 3*b*c^2*d*e - b^2*c*e^2)*x^2 + (4*b*c^2*d^2 - 3*b^2*c*d*e - b^3
*e^2)*x)*sqrt((c*d - b*e)/c)*log((c*e*x + 2*c*d - b*e + 2*sqrt(e*x + d)*c*sqrt((c*d - b*e)/c))/(c*x + b)) + 2*
(b^2*c*d^2 + (2*b*c^2*d^2 - 2*b^2*c*d*e + b^3*e^2)*x)*sqrt(e*x + d))/(b^3*c^2*x^2 + b^4*c*x), -(((4*c^3*d^2 -
3*b*c^2*d*e - b^2*c*e^2)*x^2 + (4*b*c^2*d^2 - 3*b^2*c*d*e - b^3*e^2)*x)*sqrt(-(c*d - b*e)/c)*arctan(-sqrt(e*x
+ d)*c*sqrt(-(c*d - b*e)/c)/(c*d - b*e)) + ((4*c^3*d^2 - 5*b*c^2*d*e)*x^2 + (4*b*c^2*d^2 - 5*b^2*c*d*e)*x)*sqr
t(-d)*arctan(sqrt(e*x + d)*sqrt(-d)/d) + (b^2*c*d^2 + (2*b*c^2*d^2 - 2*b^2*c*d*e + b^3*e^2)*x)*sqrt(e*x + d))/
(b^3*c^2*x^2 + b^4*c*x)]

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giac [A]  time = 0.20, size = 273, normalized size = 1.72 \[ -\frac {{\left (4 \, c d^{3} - 5 \, b d^{2} e\right )} \arctan \left (\frac {\sqrt {x e + d}}{\sqrt {-d}}\right )}{b^{3} \sqrt {-d}} + \frac {{\left (4 \, c^{3} d^{3} - 7 \, b c^{2} d^{2} e + 2 \, b^{2} c d e^{2} + b^{3} e^{3}\right )} \arctan \left (\frac {\sqrt {x e + d} c}{\sqrt {-c^{2} d + b c e}}\right )}{\sqrt {-c^{2} d + b c e} b^{3} c} - \frac {2 \, {\left (x e + d\right )}^{\frac {3}{2}} c^{2} d^{2} e - 2 \, \sqrt {x e + d} c^{2} d^{3} e - 2 \, {\left (x e + d\right )}^{\frac {3}{2}} b c d e^{2} + 3 \, \sqrt {x e + d} b c d^{2} e^{2} + {\left (x e + d\right )}^{\frac {3}{2}} b^{2} e^{3} - \sqrt {x e + d} b^{2} d e^{3}}{{\left ({\left (x e + d\right )}^{2} c - 2 \, {\left (x e + d\right )} c d + c d^{2} + {\left (x e + d\right )} b e - b d e\right )} b^{2} c} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

-(4*c*d^3 - 5*b*d^2*e)*arctan(sqrt(x*e + d)/sqrt(-d))/(b^3*sqrt(-d)) + (4*c^3*d^3 - 7*b*c^2*d^2*e + 2*b^2*c*d*
e^2 + b^3*e^3)*arctan(sqrt(x*e + d)*c/sqrt(-c^2*d + b*c*e))/(sqrt(-c^2*d + b*c*e)*b^3*c) - (2*(x*e + d)^(3/2)*
c^2*d^2*e - 2*sqrt(x*e + d)*c^2*d^3*e - 2*(x*e + d)^(3/2)*b*c*d*e^2 + 3*sqrt(x*e + d)*b*c*d^2*e^2 + (x*e + d)^
(3/2)*b^2*e^3 - sqrt(x*e + d)*b^2*d*e^3)/(((x*e + d)^2*c - 2*(x*e + d)*c*d + c*d^2 + (x*e + d)*b*e - b*d*e)*b^
2*c)

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maple [B]  time = 0.07, size = 313, normalized size = 1.97 \[ \frac {2 d \,e^{2} \arctan \left (\frac {\sqrt {e x +d}\, c}{\sqrt {\left (b e -c d \right ) c}}\right )}{\sqrt {\left (b e -c d \right ) c}\, b}-\frac {7 c \,d^{2} e \arctan \left (\frac {\sqrt {e x +d}\, c}{\sqrt {\left (b e -c d \right ) c}}\right )}{\sqrt {\left (b e -c d \right ) c}\, b^{2}}+\frac {4 c^{2} d^{3} \arctan \left (\frac {\sqrt {e x +d}\, c}{\sqrt {\left (b e -c d \right ) c}}\right )}{\sqrt {\left (b e -c d \right ) c}\, b^{3}}+\frac {e^{3} \arctan \left (\frac {\sqrt {e x +d}\, c}{\sqrt {\left (b e -c d \right ) c}}\right )}{\sqrt {\left (b e -c d \right ) c}\, c}+\frac {2 \sqrt {e x +d}\, d \,e^{2}}{\left (c e x +b e \right ) b}-\frac {\sqrt {e x +d}\, c \,d^{2} e}{\left (c e x +b e \right ) b^{2}}-\frac {\sqrt {e x +d}\, e^{3}}{\left (c e x +b e \right ) c}-\frac {5 d^{\frac {3}{2}} e \arctanh \left (\frac {\sqrt {e x +d}}{\sqrt {d}}\right )}{b^{2}}+\frac {4 c \,d^{\frac {5}{2}} \arctanh \left (\frac {\sqrt {e x +d}}{\sqrt {d}}\right )}{b^{3}}-\frac {\sqrt {e x +d}\, d^{2}}{b^{2} x} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

-e^3/c*(e*x+d)^(1/2)/(c*e*x+b*e)+2*e^2/b*(e*x+d)^(1/2)/(c*e*x+b*e)*d-e/b^2*c*(e*x+d)^(1/2)/(c*e*x+b*e)*d^2+e^3
/c/((b*e-c*d)*c)^(1/2)*arctan((e*x+d)^(1/2)/((b*e-c*d)*c)^(1/2)*c)+2*e^2/b/((b*e-c*d)*c)^(1/2)*arctan((e*x+d)^
(1/2)/((b*e-c*d)*c)^(1/2)*c)*d-7*e/b^2*c/((b*e-c*d)*c)^(1/2)*arctan((e*x+d)^(1/2)/((b*e-c*d)*c)^(1/2)*c)*d^2+4
/b^3/((b*e-c*d)*c)^(1/2)*arctan((e*x+d)^(1/2)/((b*e-c*d)*c)^(1/2)*c)*d^3*c^2-d^2/b^2*(e*x+d)^(1/2)/x-5*e*d^(3/
2)/b^2*arctanh((e*x+d)^(1/2)/d^(1/2))+4*d^(5/2)/b^3*arctanh((e*x+d)^(1/2)/d^(1/2))*c

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maxima [F(-2)]  time = 0.00, size = 0, normalized size = 0.00 \[ \text {Exception raised: ValueError} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

Exception raised: ValueError >> Computation failed since Maxima requested additional constraints; using the 'a
ssume' command before evaluation *may* help (example of legal syntax is 'assume(b*e-c*d>0)', see `assume?` for
more details)Is b*e-c*d positive or negative?

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mupad [B]  time = 0.50, size = 1127, normalized size = 7.09 \[ \frac {\frac {\sqrt {d+e\,x}\,\left (b^2\,d\,e^3-3\,b\,c\,d^2\,e^2+2\,c^2\,d^3\,e\right )}{b^2\,c}-\frac {e\,{\left (d+e\,x\right )}^{3/2}\,\left (b^2\,e^2-2\,b\,c\,d\,e+2\,c^2\,d^2\right )}{b^2\,c}}{\left (b\,e-2\,c\,d\right )\,\left (d+e\,x\right )+c\,{\left (d+e\,x\right )}^2+c\,d^2-b\,d\,e}-\frac {\mathrm {atanh}\left (\frac {10\,e^9\,\sqrt {d^3}\,\sqrt {d+e\,x}}{10\,d^2\,e^9+\frac {32\,c\,d^3\,e^8}{b}-\frac {132\,c^2\,d^4\,e^7}{b^2}+\frac {130\,c^3\,d^5\,e^6}{b^3}-\frac {40\,c^4\,d^6\,e^5}{b^4}}+\frac {32\,d\,e^8\,\sqrt {d^3}\,\sqrt {d+e\,x}}{32\,d^3\,e^8+\frac {10\,b\,d^2\,e^9}{c}-\frac {132\,c\,d^4\,e^7}{b}+\frac {130\,c^2\,d^5\,e^6}{b^2}-\frac {40\,c^3\,d^6\,e^5}{b^3}}-\frac {132\,c\,d^2\,e^7\,\sqrt {d^3}\,\sqrt {d+e\,x}}{32\,b\,d^3\,e^8-132\,c\,d^4\,e^7+\frac {130\,c^2\,d^5\,e^6}{b}+\frac {10\,b^2\,d^2\,e^9}{c}-\frac {40\,c^3\,d^6\,e^5}{b^2}}+\frac {130\,c^2\,d^3\,e^6\,\sqrt {d^3}\,\sqrt {d+e\,x}}{32\,b^2\,d^3\,e^8+130\,c^2\,d^5\,e^6-\frac {40\,c^3\,d^6\,e^5}{b}+\frac {10\,b^3\,d^2\,e^9}{c}-132\,b\,c\,d^4\,e^7}-\frac {40\,c^3\,d^4\,e^5\,\sqrt {d^3}\,\sqrt {d+e\,x}}{32\,b^3\,d^3\,e^8-40\,c^3\,d^6\,e^5+130\,b\,c^2\,d^5\,e^6-132\,b^2\,c\,d^4\,e^7+\frac {10\,b^4\,d^2\,e^9}{c}}\right )\,\left (5\,b\,e-4\,c\,d\right )\,\sqrt {d^3}}{b^3}-\frac {\mathrm {atanh}\left (\frac {30\,d^3\,e^6\,\sqrt {d+e\,x}\,\sqrt {-b^3\,c^3\,e^3+3\,b^2\,c^4\,d\,e^2-3\,b\,c^5\,d^2\,e+c^6\,d^3}}{14\,b^3\,d^2\,e^9+110\,c^3\,d^5\,e^6-82\,b\,c^2\,d^4\,e^7-4\,b^2\,c\,d^3\,e^8+\frac {2\,b^4\,d\,e^{10}}{c}-\frac {40\,c^4\,d^6\,e^5}{b}}-\frac {2\,d\,e^8\,\sqrt {d+e\,x}\,\sqrt {-b^3\,c^3\,e^3+3\,b^2\,c^4\,d\,e^2-3\,b\,c^5\,d^2\,e+c^6\,d^3}}{4\,c^3\,d^3\,e^8-14\,b\,c^2\,d^2\,e^9+\frac {82\,c^4\,d^4\,e^7}{b}-\frac {110\,c^5\,d^5\,e^6}{b^2}+\frac {40\,c^6\,d^6\,e^5}{b^3}-2\,b^2\,c\,d\,e^{10}}+\frac {18\,d^2\,e^7\,\sqrt {d+e\,x}\,\sqrt {-b^3\,c^3\,e^3+3\,b^2\,c^4\,d\,e^2-3\,b\,c^5\,d^2\,e+c^6\,d^3}}{2\,b^3\,d\,e^{10}-82\,c^3\,d^4\,e^7-4\,b\,c^2\,d^3\,e^8+14\,b^2\,c\,d^2\,e^9+\frac {110\,c^4\,d^5\,e^6}{b}-\frac {40\,c^5\,d^6\,e^5}{b^2}}+\frac {40\,d^4\,e^5\,\sqrt {d+e\,x}\,\sqrt {-b^3\,c^3\,e^3+3\,b^2\,c^4\,d\,e^2-3\,b\,c^5\,d^2\,e+c^6\,d^3}}{4\,b^3\,d^3\,e^8+40\,c^3\,d^6\,e^5-110\,b\,c^2\,d^5\,e^6+82\,b^2\,c\,d^4\,e^7-\frac {2\,b^5\,d\,e^{10}}{c^2}-\frac {14\,b^4\,d^2\,e^9}{c}}\right )\,\sqrt {-c^3\,{\left (b\,e-c\,d\right )}^3}\,\left (b\,e+4\,c\,d\right )}{b^3\,c^3} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

(((d + e*x)^(1/2)*(b^2*d*e^3 + 2*c^2*d^3*e - 3*b*c*d^2*e^2))/(b^2*c) - (e*(d + e*x)^(3/2)*(b^2*e^2 + 2*c^2*d^2
- 2*b*c*d*e))/(b^2*c))/((b*e - 2*c*d)*(d + e*x) + c*(d + e*x)^2 + c*d^2 - b*d*e) - (atanh((10*e^9*(d^3)^(1/2)
*(d + e*x)^(1/2))/(10*d^2*e^9 + (32*c*d^3*e^8)/b - (132*c^2*d^4*e^7)/b^2 + (130*c^3*d^5*e^6)/b^3 - (40*c^4*d^6
*e^5)/b^4) + (32*d*e^8*(d^3)^(1/2)*(d + e*x)^(1/2))/(32*d^3*e^8 + (10*b*d^2*e^9)/c - (132*c*d^4*e^7)/b + (130*
c^2*d^5*e^6)/b^2 - (40*c^3*d^6*e^5)/b^3) - (132*c*d^2*e^7*(d^3)^(1/2)*(d + e*x)^(1/2))/(32*b*d^3*e^8 - 132*c*d
^4*e^7 + (130*c^2*d^5*e^6)/b + (10*b^2*d^2*e^9)/c - (40*c^3*d^6*e^5)/b^2) + (130*c^2*d^3*e^6*(d^3)^(1/2)*(d +
e*x)^(1/2))/(32*b^2*d^3*e^8 + 130*c^2*d^5*e^6 - (40*c^3*d^6*e^5)/b + (10*b^3*d^2*e^9)/c - 132*b*c*d^4*e^7) - (
40*c^3*d^4*e^5*(d^3)^(1/2)*(d + e*x)^(1/2))/(32*b^3*d^3*e^8 - 40*c^3*d^6*e^5 + 130*b*c^2*d^5*e^6 - 132*b^2*c*d
^4*e^7 + (10*b^4*d^2*e^9)/c))*(5*b*e - 4*c*d)*(d^3)^(1/2))/b^3 - (atanh((30*d^3*e^6*(d + e*x)^(1/2)*(c^6*d^3 -
b^3*c^3*e^3 + 3*b^2*c^4*d*e^2 - 3*b*c^5*d^2*e)^(1/2))/(14*b^3*d^2*e^9 + 110*c^3*d^5*e^6 - 82*b*c^2*d^4*e^7 -
4*b^2*c*d^3*e^8 + (2*b^4*d*e^10)/c - (40*c^4*d^6*e^5)/b) - (2*d*e^8*(d + e*x)^(1/2)*(c^6*d^3 - b^3*c^3*e^3 + 3
*b^2*c^4*d*e^2 - 3*b*c^5*d^2*e)^(1/2))/(4*c^3*d^3*e^8 - 14*b*c^2*d^2*e^9 + (82*c^4*d^4*e^7)/b - (110*c^5*d^5*e
^6)/b^2 + (40*c^6*d^6*e^5)/b^3 - 2*b^2*c*d*e^10) + (18*d^2*e^7*(d + e*x)^(1/2)*(c^6*d^3 - b^3*c^3*e^3 + 3*b^2*
c^4*d*e^2 - 3*b*c^5*d^2*e)^(1/2))/(2*b^3*d*e^10 - 82*c^3*d^4*e^7 - 4*b*c^2*d^3*e^8 + 14*b^2*c*d^2*e^9 + (110*c
^4*d^5*e^6)/b - (40*c^5*d^6*e^5)/b^2) + (40*d^4*e^5*(d + e*x)^(1/2)*(c^6*d^3 - b^3*c^3*e^3 + 3*b^2*c^4*d*e^2 -
3*b*c^5*d^2*e)^(1/2))/(4*b^3*d^3*e^8 + 40*c^3*d^6*e^5 - 110*b*c^2*d^5*e^6 + 82*b^2*c*d^4*e^7 - (2*b^5*d*e^10)
/c^2 - (14*b^4*d^2*e^9)/c))*(-c^3*(b*e - c*d)^3)^(1/2)*(b*e + 4*c*d))/(b^3*c^3)

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sympy [F(-1)]  time = 0.00, size = 0, normalized size = 0.00 \[ \text {Timed out} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

Timed out

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