3.3.78 \(\int \frac {(b x+c x^2)^{3/2}}{(d+e x)^2} \, dx\)

Optimal. Leaf size=198 \[ \frac {3 \left (b^2 e^2-8 b c d e+8 c^2 d^2\right ) \tanh ^{-1}\left (\frac {\sqrt {c} x}{\sqrt {b x+c x^2}}\right )}{4 \sqrt {c} e^4}-\frac {3 \sqrt {d} \sqrt {c d-b e} (2 c d-b e) \tanh ^{-1}\left (\frac {x (2 c d-b e)+b d}{2 \sqrt {d} \sqrt {b x+c x^2} \sqrt {c d-b e}}\right )}{2 e^4}-\frac {3 \sqrt {b x+c x^2} (-3 b e+4 c d-2 c e x)}{4 e^3}-\frac {\left (b x+c x^2\right )^{3/2}}{e (d+e x)} \]

________________________________________________________________________________________

Rubi [A]  time = 0.23, antiderivative size = 198, normalized size of antiderivative = 1.00, number of steps used = 7, number of rules used = 6, integrand size = 21, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.286, Rules used = {732, 814, 843, 620, 206, 724} \begin {gather*} \frac {3 \left (b^2 e^2-8 b c d e+8 c^2 d^2\right ) \tanh ^{-1}\left (\frac {\sqrt {c} x}{\sqrt {b x+c x^2}}\right )}{4 \sqrt {c} e^4}-\frac {3 \sqrt {b x+c x^2} (-3 b e+4 c d-2 c e x)}{4 e^3}-\frac {3 \sqrt {d} \sqrt {c d-b e} (2 c d-b e) \tanh ^{-1}\left (\frac {x (2 c d-b e)+b d}{2 \sqrt {d} \sqrt {b x+c x^2} \sqrt {c d-b e}}\right )}{2 e^4}-\frac {\left (b x+c x^2\right )^{3/2}}{e (d+e x)} \end {gather*}

Antiderivative was successfully verified.

[In]

Int[(b*x + c*x^2)^(3/2)/(d + e*x)^2,x]

[Out]

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

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 724

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

Rule 732

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 + 1)), x] - Dist[p/(e*(m + 1)), Int[(d + e*x)^(m + 1)*(b + 2*c*x)*(a + b*x + c*x^2)^
(p - 1), x], x] /; FreeQ[{a, b, c, d, e, m}, 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] && GtQ[p, 0] && (IntegerQ[p] || LtQ[m, -1]) && NeQ[m, -1] &&  !ILtQ[m + 2*p + 1, 0] && IntQua
draticQ[a, b, c, d, e, m, p, x]

Rule 814

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

Rule 843

Int[((d_.) + (e_.)*(x_))^(m_)*((f_.) + (g_.)*(x_))*((a_.) + (b_.)*(x_) + (c_.)*(x_)^2)^(p_.), x_Symbol] :> Dis
t[g/e, Int[(d + e*x)^(m + 1)*(a + b*x + c*x^2)^p, x], x] + Dist[(e*f - d*g)/e, Int[(d + e*x)^m*(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] && NeQ[c*d^2 - b*d*e + a*e^2, 0]
&&  !IGtQ[m, 0]

Rubi steps

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

________________________________________________________________________________________

Mathematica [A]  time = 0.74, size = 205, normalized size = 1.04 \begin {gather*} \frac {\sqrt {x (b+c x)} \left (\frac {3 \left (b^2 e^2-8 b c d e+8 c^2 d^2\right ) \sinh ^{-1}\left (\frac {\sqrt {c} \sqrt {x}}{\sqrt {b}}\right )}{\sqrt {b} \sqrt {c} \sqrt {\frac {c x}{b}+1}}+\frac {e \sqrt {x} \left (b e (9 d+5 e x)-2 c \left (6 d^2+3 d e x-e^2 x^2\right )\right )}{d+e x}-\frac {12 \sqrt {d} \sqrt {c d-b e} (2 c d-b e) \tanh ^{-1}\left (\frac {\sqrt {x} \sqrt {c d-b e}}{\sqrt {d} \sqrt {b+c x}}\right )}{\sqrt {b+c x}}\right )}{4 e^4 \sqrt {x}} \end {gather*}

Antiderivative was successfully verified.

[In]

Integrate[(b*x + c*x^2)^(3/2)/(d + e*x)^2,x]

[Out]

(Sqrt[x*(b + c*x)]*((e*Sqrt[x]*(b*e*(9*d + 5*e*x) - 2*c*(6*d^2 + 3*d*e*x - e^2*x^2)))/(d + e*x) + (3*(8*c^2*d^
2 - 8*b*c*d*e + b^2*e^2)*ArcSinh[(Sqrt[c]*Sqrt[x])/Sqrt[b]])/(Sqrt[b]*Sqrt[c]*Sqrt[1 + (c*x)/b]) - (12*Sqrt[d]
*Sqrt[c*d - b*e]*(2*c*d - b*e)*ArcTanh[(Sqrt[c*d - b*e]*Sqrt[x])/(Sqrt[d]*Sqrt[b + c*x])])/Sqrt[b + c*x]))/(4*
e^4*Sqrt[x])

________________________________________________________________________________________

IntegrateAlgebraic [A]  time = 1.36, size = 209, normalized size = 1.06 \begin {gather*} -\frac {3 \left (b^2 e^2-8 b c d e+8 c^2 d^2\right ) \log \left (-2 \sqrt {c} \sqrt {b x+c x^2}+b+2 c x\right )}{8 \sqrt {c} e^4}-\frac {3 \sqrt {c d-b e} \left (2 c d^{3/2}-b \sqrt {d} e\right ) \tanh ^{-1}\left (\frac {-e \sqrt {b x+c x^2}+\sqrt {c} d+\sqrt {c} e x}{\sqrt {d} \sqrt {c d-b e}}\right )}{e^4}+\frac {\sqrt {b x+c x^2} \left (9 b d e+5 b e^2 x-12 c d^2-6 c d e x+2 c e^2 x^2\right )}{4 e^3 (d+e x)} \end {gather*}

Antiderivative was successfully verified.

[In]

IntegrateAlgebraic[(b*x + c*x^2)^(3/2)/(d + e*x)^2,x]

[Out]

(Sqrt[b*x + c*x^2]*(-12*c*d^2 + 9*b*d*e - 6*c*d*e*x + 5*b*e^2*x + 2*c*e^2*x^2))/(4*e^3*(d + e*x)) - (3*Sqrt[c*
d - b*e]*(2*c*d^(3/2) - b*Sqrt[d]*e)*ArcTanh[(Sqrt[c]*d + Sqrt[c]*e*x - e*Sqrt[b*x + c*x^2])/(Sqrt[d]*Sqrt[c*d
- b*e])])/e^4 - (3*(8*c^2*d^2 - 8*b*c*d*e + b^2*e^2)*Log[b + 2*c*x - 2*Sqrt[c]*Sqrt[b*x + c*x^2]])/(8*Sqrt[c]
*e^4)

________________________________________________________________________________________

fricas [A]  time = 0.51, size = 1015, normalized size = 5.13 \begin {gather*} \left [\frac {3 \, {\left (8 \, c^{2} d^{3} - 8 \, b c d^{2} e + b^{2} d e^{2} + {\left (8 \, c^{2} d^{2} e - 8 \, b c d e^{2} + b^{2} e^{3}\right )} x\right )} \sqrt {c} \log \left (2 \, c x + b + 2 \, \sqrt {c x^{2} + b x} \sqrt {c}\right ) - 12 \, {\left (2 \, c^{2} d^{2} - b c d e + {\left (2 \, c^{2} d e - b c e^{2}\right )} x\right )} \sqrt {c d^{2} - b d e} \log \left (\frac {b d + {\left (2 \, c d - b e\right )} x + 2 \, \sqrt {c d^{2} - b d e} \sqrt {c x^{2} + b x}}{e x + d}\right ) + 2 \, {\left (2 \, c^{2} e^{3} x^{2} - 12 \, c^{2} d^{2} e + 9 \, b c d e^{2} - {\left (6 \, c^{2} d e^{2} - 5 \, b c e^{3}\right )} x\right )} \sqrt {c x^{2} + b x}}{8 \, {\left (c e^{5} x + c d e^{4}\right )}}, -\frac {24 \, {\left (2 \, c^{2} d^{2} - b c d e + {\left (2 \, c^{2} d e - b c e^{2}\right )} x\right )} \sqrt {-c d^{2} + b d e} \arctan \left (-\frac {\sqrt {-c d^{2} + b d e} \sqrt {c x^{2} + b x}}{{\left (c d - b e\right )} x}\right ) - 3 \, {\left (8 \, c^{2} d^{3} - 8 \, b c d^{2} e + b^{2} d e^{2} + {\left (8 \, c^{2} d^{2} e - 8 \, b c d e^{2} + b^{2} e^{3}\right )} x\right )} \sqrt {c} \log \left (2 \, c x + b + 2 \, \sqrt {c x^{2} + b x} \sqrt {c}\right ) - 2 \, {\left (2 \, c^{2} e^{3} x^{2} - 12 \, c^{2} d^{2} e + 9 \, b c d e^{2} - {\left (6 \, c^{2} d e^{2} - 5 \, b c e^{3}\right )} x\right )} \sqrt {c x^{2} + b x}}{8 \, {\left (c e^{5} x + c d e^{4}\right )}}, -\frac {3 \, {\left (8 \, c^{2} d^{3} - 8 \, b c d^{2} e + b^{2} d e^{2} + {\left (8 \, c^{2} d^{2} e - 8 \, b c d e^{2} + b^{2} e^{3}\right )} x\right )} \sqrt {-c} \arctan \left (\frac {\sqrt {c x^{2} + b x} \sqrt {-c}}{c x}\right ) + 6 \, {\left (2 \, c^{2} d^{2} - b c d e + {\left (2 \, c^{2} d e - b c e^{2}\right )} x\right )} \sqrt {c d^{2} - b d e} \log \left (\frac {b d + {\left (2 \, c d - b e\right )} x + 2 \, \sqrt {c d^{2} - b d e} \sqrt {c x^{2} + b x}}{e x + d}\right ) - {\left (2 \, c^{2} e^{3} x^{2} - 12 \, c^{2} d^{2} e + 9 \, b c d e^{2} - {\left (6 \, c^{2} d e^{2} - 5 \, b c e^{3}\right )} x\right )} \sqrt {c x^{2} + b x}}{4 \, {\left (c e^{5} x + c d e^{4}\right )}}, -\frac {12 \, {\left (2 \, c^{2} d^{2} - b c d e + {\left (2 \, c^{2} d e - b c e^{2}\right )} x\right )} \sqrt {-c d^{2} + b d e} \arctan \left (-\frac {\sqrt {-c d^{2} + b d e} \sqrt {c x^{2} + b x}}{{\left (c d - b e\right )} x}\right ) + 3 \, {\left (8 \, c^{2} d^{3} - 8 \, b c d^{2} e + b^{2} d e^{2} + {\left (8 \, c^{2} d^{2} e - 8 \, b c d e^{2} + b^{2} e^{3}\right )} x\right )} \sqrt {-c} \arctan \left (\frac {\sqrt {c x^{2} + b x} \sqrt {-c}}{c x}\right ) - {\left (2 \, c^{2} e^{3} x^{2} - 12 \, c^{2} d^{2} e + 9 \, b c d e^{2} - {\left (6 \, c^{2} d e^{2} - 5 \, b c e^{3}\right )} x\right )} \sqrt {c x^{2} + b x}}{4 \, {\left (c e^{5} x + c d e^{4}\right )}}\right ] \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate((c*x^2+b*x)^(3/2)/(e*x+d)^2,x, algorithm="fricas")

[Out]

[1/8*(3*(8*c^2*d^3 - 8*b*c*d^2*e + b^2*d*e^2 + (8*c^2*d^2*e - 8*b*c*d*e^2 + b^2*e^3)*x)*sqrt(c)*log(2*c*x + b
+ 2*sqrt(c*x^2 + b*x)*sqrt(c)) - 12*(2*c^2*d^2 - b*c*d*e + (2*c^2*d*e - b*c*e^2)*x)*sqrt(c*d^2 - b*d*e)*log((b
*d + (2*c*d - b*e)*x + 2*sqrt(c*d^2 - b*d*e)*sqrt(c*x^2 + b*x))/(e*x + d)) + 2*(2*c^2*e^3*x^2 - 12*c^2*d^2*e +
9*b*c*d*e^2 - (6*c^2*d*e^2 - 5*b*c*e^3)*x)*sqrt(c*x^2 + b*x))/(c*e^5*x + c*d*e^4), -1/8*(24*(2*c^2*d^2 - b*c*
d*e + (2*c^2*d*e - b*c*e^2)*x)*sqrt(-c*d^2 + b*d*e)*arctan(-sqrt(-c*d^2 + b*d*e)*sqrt(c*x^2 + b*x)/((c*d - b*e
)*x)) - 3*(8*c^2*d^3 - 8*b*c*d^2*e + b^2*d*e^2 + (8*c^2*d^2*e - 8*b*c*d*e^2 + b^2*e^3)*x)*sqrt(c)*log(2*c*x +
b + 2*sqrt(c*x^2 + b*x)*sqrt(c)) - 2*(2*c^2*e^3*x^2 - 12*c^2*d^2*e + 9*b*c*d*e^2 - (6*c^2*d*e^2 - 5*b*c*e^3)*x
)*sqrt(c*x^2 + b*x))/(c*e^5*x + c*d*e^4), -1/4*(3*(8*c^2*d^3 - 8*b*c*d^2*e + b^2*d*e^2 + (8*c^2*d^2*e - 8*b*c*
d*e^2 + b^2*e^3)*x)*sqrt(-c)*arctan(sqrt(c*x^2 + b*x)*sqrt(-c)/(c*x)) + 6*(2*c^2*d^2 - b*c*d*e + (2*c^2*d*e -
b*c*e^2)*x)*sqrt(c*d^2 - b*d*e)*log((b*d + (2*c*d - b*e)*x + 2*sqrt(c*d^2 - b*d*e)*sqrt(c*x^2 + b*x))/(e*x + d
)) - (2*c^2*e^3*x^2 - 12*c^2*d^2*e + 9*b*c*d*e^2 - (6*c^2*d*e^2 - 5*b*c*e^3)*x)*sqrt(c*x^2 + b*x))/(c*e^5*x +
c*d*e^4), -1/4*(12*(2*c^2*d^2 - b*c*d*e + (2*c^2*d*e - b*c*e^2)*x)*sqrt(-c*d^2 + b*d*e)*arctan(-sqrt(-c*d^2 +
b*d*e)*sqrt(c*x^2 + b*x)/((c*d - b*e)*x)) + 3*(8*c^2*d^3 - 8*b*c*d^2*e + b^2*d*e^2 + (8*c^2*d^2*e - 8*b*c*d*e^
2 + b^2*e^3)*x)*sqrt(-c)*arctan(sqrt(c*x^2 + b*x)*sqrt(-c)/(c*x)) - (2*c^2*e^3*x^2 - 12*c^2*d^2*e + 9*b*c*d*e^
2 - (6*c^2*d*e^2 - 5*b*c*e^3)*x)*sqrt(c*x^2 + b*x))/(c*e^5*x + c*d*e^4)]

________________________________________________________________________________________

giac [F(-1)]  time = 0.00, size = 0, normalized size = 0.00 \begin {gather*} \text {Timed out} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate((c*x^2+b*x)^(3/2)/(e*x+d)^2,x, algorithm="giac")

[Out]

Timed out

________________________________________________________________________________________

maple [B]  time = 0.05, size = 1569, normalized size = 7.92

result too large to display

Verification of antiderivative is not currently implemented for this CAS.

[In]

int((c*x^2+b*x)^(3/2)/(e*x+d)^2,x)

[Out]

1/(b*e-c*d)/d/(x+d/e)*((x+d/e)^2*c-(b*e-c*d)*d/e^2+(b*e-2*c*d)*(x+d/e)/e)^(5/2)+6/e^3/(b*e-c*d)*d^2*ln(((x+d/e
)*c+1/2*(b*e-2*c*d)/e)/c^(1/2)+((x+d/e)^2*c-(b*e-c*d)*d/e^2+(b*e-2*c*d)*(x+d/e)/e)^(1/2))*c^(3/2)*b-3/e^4/(b*e
-c*d)*d^3*ln(((x+d/e)*c+1/2*(b*e-2*c*d)/e)/c^(1/2)+((x+d/e)^2*c-(b*e-c*d)*d/e^2+(b*e-2*c*d)*(x+d/e)/e)^(1/2))*
c^(5/2)+3/2/e^2/(b*e-c*d)*d/(-(b*e-c*d)*d/e^2)^(1/2)*ln((-2*(b*e-c*d)*d/e^2+(b*e-2*c*d)*(x+d/e)/e+2*(-(b*e-c*d
)*d/e^2)^(1/2)*((x+d/e)^2*c-(b*e-c*d)*d/e^2+(b*e-2*c*d)*(x+d/e)/e)^(1/2))/(x+d/e))*b^3-3/e^5/(b*e-c*d)*d^4/(-(
b*e-c*d)*d/e^2)^(1/2)*ln((-2*(b*e-c*d)*d/e^2+(b*e-2*c*d)*(x+d/e)/e+2*(-(b*e-c*d)*d/e^2)^(1/2)*((x+d/e)^2*c-(b*
e-c*d)*d/e^2+(b*e-2*c*d)*(x+d/e)/e)^(1/2))/(x+d/e))*c^3+3/e^3/(b*e-c*d)*d^2*((x+d/e)^2*c-(b*e-c*d)*d/e^2+(b*e-
2*c*d)*(x+d/e)/e)^(1/2)*c^2+9/4/e/(b*e-c*d)*((x+d/e)^2*c-(b*e-c*d)*d/e^2+(b*e-2*c*d)*(x+d/e)/e)^(1/2)*b^2-6/e^
3/(b*e-c*d)*d^2/(-(b*e-c*d)*d/e^2)^(1/2)*ln((-2*(b*e-c*d)*d/e^2+(b*e-2*c*d)*(x+d/e)/e+2*(-(b*e-c*d)*d/e^2)^(1/
2)*((x+d/e)^2*c-(b*e-c*d)*d/e^2+(b*e-2*c*d)*(x+d/e)/e)^(1/2))/(x+d/e))*b^2*c+15/2/e^4/(b*e-c*d)*d^3/(-(b*e-c*d
)*d/e^2)^(1/2)*ln((-2*(b*e-c*d)*d/e^2+(b*e-2*c*d)*(x+d/e)/e+2*(-(b*e-c*d)*d/e^2)^(1/2)*((x+d/e)^2*c-(b*e-c*d)*
d/e^2+(b*e-2*c*d)*(x+d/e)/e)^(1/2))/(x+d/e))*b*c^2-27/8/e^2/(b*e-c*d)*d*ln(((x+d/e)*c+1/2*(b*e-2*c*d)/e)/c^(1/
2)+((x+d/e)^2*c-(b*e-c*d)*d/e^2+(b*e-2*c*d)*(x+d/e)/e)^(1/2))*c^(1/2)*b^2+1/e/(b*e-c*d)*((x+d/e)^2*c-(b*e-c*d)
*d/e^2+(b*e-2*c*d)*(x+d/e)/e)^(3/2)*c-1/(b*e-c*d)/d*((x+d/e)^2*c-(b*e-c*d)*d/e^2+(b*e-2*c*d)*(x+d/e)/e)^(3/2)*
b+3/2/e/(b*e-c*d)*((x+d/e)^2*c-(b*e-c*d)*d/e^2+(b*e-2*c*d)*(x+d/e)/e)^(1/2)*x*b*c-3/2/e^2/(b*e-c*d)*d*((x+d/e)
^2*c-(b*e-c*d)*d/e^2+(b*e-2*c*d)*(x+d/e)/e)^(1/2)*x*c^2-21/4/e^2/(b*e-c*d)*d*((x+d/e)^2*c-(b*e-c*d)*d/e^2+(b*e
-2*c*d)*(x+d/e)/e)^(1/2)*b*c+3/8/e/(b*e-c*d)*ln(((x+d/e)*c+1/2*(b*e-2*c*d)/e)/c^(1/2)+((x+d/e)^2*c-(b*e-c*d)*d
/e^2+(b*e-2*c*d)*(x+d/e)/e)^(1/2))/c^(1/2)*b^3-c/(b*e-c*d)/d*((x+d/e)^2*c-(b*e-c*d)*d/e^2+(b*e-2*c*d)*(x+d/e)/
e)^(3/2)*x

________________________________________________________________________________________

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

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate((c*x^2+b*x)^(3/2)/(e*x+d)^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 zero or nonzero?

________________________________________________________________________________________

mupad [F]  time = 0.00, size = -1, normalized size = -0.01 \begin {gather*} \int \frac {{\left (c\,x^2+b\,x\right )}^{3/2}}{{\left (d+e\,x\right )}^2} \,d x \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

int((b*x + c*x^2)^(3/2)/(d + e*x)^2,x)

[Out]

int((b*x + c*x^2)^(3/2)/(d + e*x)^2, x)

________________________________________________________________________________________

sympy [F]  time = 0.00, size = 0, normalized size = 0.00 \begin {gather*} \int \frac {\left (x \left (b + c x\right )\right )^{\frac {3}{2}}}{\left (d + e x\right )^{2}}\, dx \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate((c*x**2+b*x)**(3/2)/(e*x+d)**2,x)

[Out]

Integral((x*(b + c*x))**(3/2)/(d + e*x)**2, x)

________________________________________________________________________________________