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

Optimal. Leaf size=216 \[ \frac {\sqrt {b x+c x^2} \left (b^2 e^2-2 c e x (2 c d-b e)-10 b c d e+8 c^2 d^2\right )}{8 c e^3}-\frac {(2 c d-b e) \left (-b^2 e^2-8 b c d e+8 c^2 d^2\right ) \tanh ^{-1}\left (\frac {\sqrt {c} x}{\sqrt {b x+c x^2}}\right )}{8 c^{3/2} e^4}+\frac {d^{3/2} (c d-b e)^{3/2} \tanh ^{-1}\left (\frac {x (2 c d-b e)+b d}{2 \sqrt {d} \sqrt {b x+c x^2} \sqrt {c d-b e}}\right )}{e^4}+\frac {\left (b x+c x^2\right )^{3/2}}{3 e} \]

________________________________________________________________________________________

Rubi [A]  time = 0.24, antiderivative size = 216, 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 = {734, 814, 843, 620, 206, 724} \begin {gather*} \frac {\sqrt {b x+c x^2} \left (b^2 e^2-2 c e x (2 c d-b e)-10 b c d e+8 c^2 d^2\right )}{8 c e^3}-\frac {(2 c d-b e) \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 )}{8 c^{3/2} e^4}+\frac {d^{3/2} (c d-b e)^{3/2} \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 )}{e^4}+\frac {\left (b x+c x^2\right )^{3/2}}{3 e} \end {gather*}

Antiderivative was successfully verified.

[In]

Int[(b*x + c*x^2)^(3/2)/(d + e*x),x]

[Out]

((8*c^2*d^2 - 10*b*c*d*e + b^2*e^2 - 2*c*e*(2*c*d - b*e)*x)*Sqrt[b*x + c*x^2])/(8*c*e^3) + (b*x + c*x^2)^(3/2)
/(3*e) - ((2*c*d - b*e)*(8*c^2*d^2 - 8*b*c*d*e - b^2*e^2)*ArcTanh[(Sqrt[c]*x)/Sqrt[b*x + c*x^2]])/(8*c^(3/2)*e
^4) + (d^(3/2)*(c*d - b*e)^(3/2)*ArcTanh[(b*d + (2*c*d - b*e)*x)/(2*Sqrt[d]*Sqrt[c*d - b*e]*Sqrt[b*x + c*x^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 734

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 + 2*p + 1)), x] - Dist[p/(e*(m + 2*p + 1)), Int[(d + e*x)^m*Simp[b*d - 2*a*e + (2*c*
d - b*e)*x, 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] && NeQ[m + 2*p + 1, 0] && ( !RationalQ[m] || Lt
Q[m, 1]) &&  !ILtQ[m + 2*p, 0] && IntQuadraticQ[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} \, dx &=\frac {\left (b x+c x^2\right )^{3/2}}{3 e}-\frac {\int \frac {(b d+(2 c d-b e) x) \sqrt {b x+c x^2}}{d+e x} \, dx}{2 e}\\ &=\frac {\left (8 c^2 d^2-10 b c d e+b^2 e^2-2 c e (2 c d-b e) x\right ) \sqrt {b x+c x^2}}{8 c e^3}+\frac {\left (b x+c x^2\right )^{3/2}}{3 e}+\frac {\int \frac {-\frac {1}{2} b d \left (8 c^2 d^2-10 b c d e+b^2 e^2\right )-\frac {1}{2} (2 c d-b e) \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 {\left (8 c^2 d^2-10 b c d e+b^2 e^2-2 c e (2 c d-b e) x\right ) \sqrt {b x+c x^2}}{8 c e^3}+\frac {\left (b x+c x^2\right )^{3/2}}{3 e}+\frac {\left (d^2 (c d-b e)^2\right ) \int \frac {1}{(d+e x) \sqrt {b x+c x^2}} \, dx}{e^4}-\frac {\left ((2 c d-b e) \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}{16 c e^4}\\ &=\frac {\left (8 c^2 d^2-10 b c d e+b^2 e^2-2 c e (2 c d-b e) x\right ) \sqrt {b x+c x^2}}{8 c e^3}+\frac {\left (b x+c x^2\right )^{3/2}}{3 e}-\frac {\left (2 d^2 (c d-b e)^2\right ) \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 ((2 c d-b e) \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 )}{8 c e^4}\\ &=\frac {\left (8 c^2 d^2-10 b c d e+b^2 e^2-2 c e (2 c d-b e) x\right ) \sqrt {b x+c x^2}}{8 c e^3}+\frac {\left (b x+c x^2\right )^{3/2}}{3 e}-\frac {(2 c d-b e) \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 )}{8 c^{3/2} e^4}+\frac {d^{3/2} (c d-b e)^{3/2} \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 )}{e^4}\\ \end {align*}

________________________________________________________________________________________

Mathematica [A]  time = 0.55, size = 225, normalized size = 1.04 \begin {gather*} \frac {\sqrt {x (b+c x)} \left (\sqrt {c} e \sqrt {x} \left (3 b^2 e^2+2 b c e (7 e x-15 d)+4 c^2 \left (6 d^2-3 d e x+2 e^2 x^2\right )\right )-\frac {3 \left (b^3 e^3+6 b^2 c d e^2-24 b c^2 d^2 e+16 c^3 d^3\right ) \sinh ^{-1}\left (\frac {\sqrt {c} \sqrt {x}}{\sqrt {b}}\right )}{\sqrt {b} \sqrt {\frac {c x}{b}+1}}+\frac {48 c^{3/2} d^{3/2} (c d-b e)^{3/2} \tanh ^{-1}\left (\frac {\sqrt {x} \sqrt {c d-b e}}{\sqrt {d} \sqrt {b+c x}}\right )}{\sqrt {b+c x}}\right )}{24 c^{3/2} e^4 \sqrt {x}} \end {gather*}

Antiderivative was successfully verified.

[In]

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

[Out]

(Sqrt[x*(b + c*x)]*(Sqrt[c]*e*Sqrt[x]*(3*b^2*e^2 + 2*b*c*e*(-15*d + 7*e*x) + 4*c^2*(6*d^2 - 3*d*e*x + 2*e^2*x^
2)) - (3*(16*c^3*d^3 - 24*b*c^2*d^2*e + 6*b^2*c*d*e^2 + b^3*e^3)*ArcSinh[(Sqrt[c]*Sqrt[x])/Sqrt[b]])/(Sqrt[b]*
Sqrt[1 + (c*x)/b]) + (48*c^(3/2)*d^(3/2)*(c*d - b*e)^(3/2)*ArcTanh[(Sqrt[c*d - b*e]*Sqrt[x])/(Sqrt[d]*Sqrt[b +
c*x])])/Sqrt[b + c*x]))/(24*c^(3/2)*e^4*Sqrt[x])

________________________________________________________________________________________

IntegrateAlgebraic [A]  time = 1.27, size = 238, normalized size = 1.10 \begin {gather*} \frac {\sqrt {b x+c x^2} \left (3 b^2 e^2-30 b c d e+14 b c e^2 x+24 c^2 d^2-12 c^2 d e x+8 c^2 e^2 x^2\right )}{24 c e^3}+\frac {\left (b^3 e^3+6 b^2 c d e^2-24 b c^2 d^2 e+16 c^3 d^3\right ) \log \left (-2 c^{3/2} \sqrt {b x+c x^2}+b c+2 c^2 x\right )}{16 c^{3/2} e^4}+\frac {2 \sqrt {c d-b e} \left (c d^{5/2}-b d^{3/2} 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} \end {gather*}

Antiderivative was successfully verified.

[In]

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

[Out]

(Sqrt[b*x + c*x^2]*(24*c^2*d^2 - 30*b*c*d*e + 3*b^2*e^2 - 12*c^2*d*e*x + 14*b*c*e^2*x + 8*c^2*e^2*x^2))/(24*c*
e^3) + (2*Sqrt[c*d - b*e]*(c*d^(5/2) - b*d^(3/2)*e)*ArcTanh[(Sqrt[c]*d + Sqrt[c]*e*x - e*Sqrt[b*x + c*x^2])/(S
qrt[d]*Sqrt[c*d - b*e])])/e^4 + ((16*c^3*d^3 - 24*b*c^2*d^2*e + 6*b^2*c*d*e^2 + b^3*e^3)*Log[b*c + 2*c^2*x - 2
*c^(3/2)*Sqrt[b*x + c*x^2]])/(16*c^(3/2)*e^4)

________________________________________________________________________________________

fricas [A]  time = 0.79, size = 901, normalized size = 4.17 \begin {gather*} \left [\frac {3 \, {\left (16 \, c^{3} d^{3} - 24 \, b c^{2} d^{2} e + 6 \, b^{2} c d e^{2} + b^{3} e^{3}\right )} \sqrt {c} \log \left (2 \, c x + b - 2 \, \sqrt {c x^{2} + b x} \sqrt {c}\right ) - 48 \, {\left (c^{3} d^{2} - b c^{2} d e\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 (8 \, c^{3} e^{3} x^{2} + 24 \, c^{3} d^{2} e - 30 \, b c^{2} d e^{2} + 3 \, b^{2} c e^{3} - 2 \, {\left (6 \, c^{3} d e^{2} - 7 \, b c^{2} e^{3}\right )} x\right )} \sqrt {c x^{2} + b x}}{48 \, c^{2} e^{4}}, \frac {96 \, {\left (c^{3} d^{2} - b c^{2} d e\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 (16 \, c^{3} d^{3} - 24 \, b c^{2} d^{2} e + 6 \, b^{2} c d e^{2} + b^{3} e^{3}\right )} \sqrt {c} \log \left (2 \, c x + b - 2 \, \sqrt {c x^{2} + b x} \sqrt {c}\right ) + 2 \, {\left (8 \, c^{3} e^{3} x^{2} + 24 \, c^{3} d^{2} e - 30 \, b c^{2} d e^{2} + 3 \, b^{2} c e^{3} - 2 \, {\left (6 \, c^{3} d e^{2} - 7 \, b c^{2} e^{3}\right )} x\right )} \sqrt {c x^{2} + b x}}{48 \, c^{2} e^{4}}, \frac {3 \, {\left (16 \, c^{3} d^{3} - 24 \, b c^{2} d^{2} e + 6 \, b^{2} c d e^{2} + b^{3} e^{3}\right )} \sqrt {-c} \arctan \left (\frac {\sqrt {c x^{2} + b x} \sqrt {-c}}{c x}\right ) - 24 \, {\left (c^{3} d^{2} - b c^{2} d e\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 (8 \, c^{3} e^{3} x^{2} + 24 \, c^{3} d^{2} e - 30 \, b c^{2} d e^{2} + 3 \, b^{2} c e^{3} - 2 \, {\left (6 \, c^{3} d e^{2} - 7 \, b c^{2} e^{3}\right )} x\right )} \sqrt {c x^{2} + b x}}{24 \, c^{2} e^{4}}, \frac {48 \, {\left (c^{3} d^{2} - b c^{2} d e\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 (16 \, c^{3} d^{3} - 24 \, b c^{2} d^{2} e + 6 \, b^{2} c d e^{2} + b^{3} e^{3}\right )} \sqrt {-c} \arctan \left (\frac {\sqrt {c x^{2} + b x} \sqrt {-c}}{c x}\right ) + {\left (8 \, c^{3} e^{3} x^{2} + 24 \, c^{3} d^{2} e - 30 \, b c^{2} d e^{2} + 3 \, b^{2} c e^{3} - 2 \, {\left (6 \, c^{3} d e^{2} - 7 \, b c^{2} e^{3}\right )} x\right )} \sqrt {c x^{2} + b x}}{24 \, c^{2} e^{4}}\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),x, algorithm="fricas")

[Out]

[1/48*(3*(16*c^3*d^3 - 24*b*c^2*d^2*e + 6*b^2*c*d*e^2 + b^3*e^3)*sqrt(c)*log(2*c*x + b - 2*sqrt(c*x^2 + b*x)*s
qrt(c)) - 48*(c^3*d^2 - b*c^2*d*e)*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*(8*c^3*e^3*x^2 + 24*c^3*d^2*e - 30*b*c^2*d*e^2 + 3*b^2*c*e^3 - 2*(6*c^3*d*e^2 -
7*b*c^2*e^3)*x)*sqrt(c*x^2 + b*x))/(c^2*e^4), 1/48*(96*(c^3*d^2 - b*c^2*d*e)*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*(16*c^3*d^3 - 24*b*c^2*d^2*e + 6*b^2*c*d*e^2 + b^3*e^3
)*sqrt(c)*log(2*c*x + b - 2*sqrt(c*x^2 + b*x)*sqrt(c)) + 2*(8*c^3*e^3*x^2 + 24*c^3*d^2*e - 30*b*c^2*d*e^2 + 3*
b^2*c*e^3 - 2*(6*c^3*d*e^2 - 7*b*c^2*e^3)*x)*sqrt(c*x^2 + b*x))/(c^2*e^4), 1/24*(3*(16*c^3*d^3 - 24*b*c^2*d^2*
e + 6*b^2*c*d*e^2 + b^3*e^3)*sqrt(-c)*arctan(sqrt(c*x^2 + b*x)*sqrt(-c)/(c*x)) - 24*(c^3*d^2 - b*c^2*d*e)*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)) + (8*c^3*e^3*
x^2 + 24*c^3*d^2*e - 30*b*c^2*d*e^2 + 3*b^2*c*e^3 - 2*(6*c^3*d*e^2 - 7*b*c^2*e^3)*x)*sqrt(c*x^2 + b*x))/(c^2*e
^4), 1/24*(48*(c^3*d^2 - b*c^2*d*e)*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*(16*c^3*d^3 - 24*b*c^2*d^2*e + 6*b^2*c*d*e^2 + b^3*e^3)*sqrt(-c)*arctan(sqrt(c*x^2 + b*x)*sqrt(
-c)/(c*x)) + (8*c^3*e^3*x^2 + 24*c^3*d^2*e - 30*b*c^2*d*e^2 + 3*b^2*c*e^3 - 2*(6*c^3*d*e^2 - 7*b*c^2*e^3)*x)*s
qrt(c*x^2 + b*x))/(c^2*e^4)]

________________________________________________________________________________________

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

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

Exception raised: TypeError >> An error occurred running a Giac command:INPUT:sage2:=int(sage0,x):;OUTPUT:Erro
r: Bad Argument Type

________________________________________________________________________________________

maple [B]  time = 0.06, size = 1090, normalized size = 5.05

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),x)

[Out]

1/3/e*((x+d/e)^2*c-(b*e-c*d)*d/e^2+(b*e-2*c*d)*(x+d/e)/e)^(3/2)+1/4/e*((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-1/2/e^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*c*d+1/8/e/c*((x+d/e)
^2*c-(b*e-c*d)*d/e^2+(b*e-2*c*d)*(x+d/e)/e)^(1/2)*b^2-5/4/e^2*((x+d/e)^2*c-(b*e-c*d)*d/e^2+(b*e-2*c*d)*(x+d/e)
/e)^(1/2)*b*d-1/16/e/c^(3/2)*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))*b^3-3/8/e^2*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^3*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+3/2/e
^3*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^(
1/2)*b-1/e^4*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^(3/2)-1/e^3*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+2/e^4*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-1/e^5*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^
2

________________________________________________________________________________________

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),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 = -0.00 \begin {gather*} \int \frac {{\left (c\,x^2+b\,x\right )}^{3/2}}{d+e\,x} \,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),x)

[Out]

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

[Out]

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

________________________________________________________________________________________