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

Optimal. Leaf size=356 \[ \frac {\left (b x+c x^2\right )^{3/2} \left (3 b^2 e^2-6 c e x (2 c d-b e)-22 b c d e+16 c^2 d^2\right )}{48 c e^3}+\frac {\sqrt {b x+c x^2} \left (-3 b^4 e^4-10 b^3 c d e^3-2 c e x (2 c d-b e) \left (-3 b^2 e^2-16 b c d e+16 c^2 d^2\right )+176 b^2 c^2 d^2 e^2-288 b c^3 d^3 e+128 c^4 d^4\right )}{128 c^2 e^5}-\frac {(2 c d-b e) \left (3 b^4 e^4+16 b^3 c d e^3+112 b^2 c^2 d^2 e^2-256 b c^3 d^3 e+128 c^4 d^4\right ) \tanh ^{-1}\left (\frac {\sqrt {c} x}{\sqrt {b x+c x^2}}\right )}{128 c^{5/2} e^6}+\frac {d^{5/2} (c d-b e)^{5/2} \tanh ^{-1}\left (\frac {x (2 c d-b e)+b d}{2 \sqrt {d} \sqrt {b x+c x^2} \sqrt {c d-b e}}\right )}{e^6}+\frac {\left (b x+c x^2\right )^{5/2}}{5 e} \]

[Out]

1/48*(16*c^2*d^2-22*b*c*d*e+3*b^2*e^2-6*c*e*(-b*e+2*c*d)*x)*(c*x^2+b*x)^(3/2)/c/e^3+1/5*(c*x^2+b*x)^(5/2)/e-1/
128*(-b*e+2*c*d)*(3*b^4*e^4+16*b^3*c*d*e^3+112*b^2*c^2*d^2*e^2-256*b*c^3*d^3*e+128*c^4*d^4)*arctanh(x*c^(1/2)/
(c*x^2+b*x)^(1/2))/c^(5/2)/e^6+d^(5/2)*(-b*e+c*d)^(5/2)*arctanh(1/2*(b*d+(-b*e+2*c*d)*x)/d^(1/2)/(-b*e+c*d)^(1
/2)/(c*x^2+b*x)^(1/2))/e^6+1/128*(128*c^4*d^4-288*b*c^3*d^3*e+176*b^2*c^2*d^2*e^2-10*b^3*c*d*e^3-3*b^4*e^4-2*c
*e*(-b*e+2*c*d)*(-3*b^2*e^2-16*b*c*d*e+16*c^2*d^2)*x)*(c*x^2+b*x)^(1/2)/c^2/e^5

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Rubi [A]  time = 0.43, antiderivative size = 356, normalized size of antiderivative = 1.00, number of steps used = 8, 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} \[ \frac {\left (b x+c x^2\right )^{3/2} \left (3 b^2 e^2-6 c e x (2 c d-b e)-22 b c d e+16 c^2 d^2\right )}{48 c e^3}+\frac {\sqrt {b x+c x^2} \left (-2 c e x (2 c d-b e) \left (-3 b^2 e^2-16 b c d e+16 c^2 d^2\right )+176 b^2 c^2 d^2 e^2-10 b^3 c d e^3-3 b^4 e^4-288 b c^3 d^3 e+128 c^4 d^4\right )}{128 c^2 e^5}-\frac {(2 c d-b e) \left (112 b^2 c^2 d^2 e^2+16 b^3 c d e^3+3 b^4 e^4-256 b c^3 d^3 e+128 c^4 d^4\right ) \tanh ^{-1}\left (\frac {\sqrt {c} x}{\sqrt {b x+c x^2}}\right )}{128 c^{5/2} e^6}+\frac {d^{5/2} (c d-b e)^{5/2} \tanh ^{-1}\left (\frac {x (2 c d-b e)+b d}{2 \sqrt {d} \sqrt {b x+c x^2} \sqrt {c d-b e}}\right )}{e^6}+\frac {\left (b x+c x^2\right )^{5/2}}{5 e} \]

Antiderivative was successfully verified.

[In]

Int[(b*x + c*x^2)^(5/2)/(d + e*x),x]

[Out]

((128*c^4*d^4 - 288*b*c^3*d^3*e + 176*b^2*c^2*d^2*e^2 - 10*b^3*c*d*e^3 - 3*b^4*e^4 - 2*c*e*(2*c*d - b*e)*(16*c
^2*d^2 - 16*b*c*d*e - 3*b^2*e^2)*x)*Sqrt[b*x + c*x^2])/(128*c^2*e^5) + ((16*c^2*d^2 - 22*b*c*d*e + 3*b^2*e^2 -
6*c*e*(2*c*d - b*e)*x)*(b*x + c*x^2)^(3/2))/(48*c*e^3) + (b*x + c*x^2)^(5/2)/(5*e) - ((2*c*d - b*e)*(128*c^4*
d^4 - 256*b*c^3*d^3*e + 112*b^2*c^2*d^2*e^2 + 16*b^3*c*d*e^3 + 3*b^4*e^4)*ArcTanh[(Sqrt[c]*x)/Sqrt[b*x + c*x^2
]])/(128*c^(5/2)*e^6) + (d^(5/2)*(c*d - b*e)^(5/2)*ArcTanh[(b*d + (2*c*d - b*e)*x)/(2*Sqrt[d]*Sqrt[c*d - b*e]*
Sqrt[b*x + c*x^2])])/e^6

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 )^{5/2}}{d+e x} \, dx &=\frac {\left (b x+c x^2\right )^{5/2}}{5 e}-\frac {\int \frac {(b d+(2 c d-b e) x) \left (b x+c x^2\right )^{3/2}}{d+e x} \, dx}{2 e}\\ &=\frac {\left (16 c^2 d^2-22 b c d e+3 b^2 e^2-6 c e (2 c d-b e) x\right ) \left (b x+c x^2\right )^{3/2}}{48 c e^3}+\frac {\left (b x+c x^2\right )^{5/2}}{5 e}+\frac {\int \frac {\left (-\frac {1}{2} b d \left (16 c^2 d^2-22 b c d e+3 b^2 e^2\right )-\frac {1}{2} (2 c d-b e) \left (16 c^2 d^2-16 b c d e-3 b^2 e^2\right ) x\right ) \sqrt {b x+c x^2}}{d+e x} \, dx}{16 c e^3}\\ &=\frac {\left (128 c^4 d^4-288 b c^3 d^3 e+176 b^2 c^2 d^2 e^2-10 b^3 c d e^3-3 b^4 e^4-2 c e (2 c d-b e) \left (16 c^2 d^2-16 b c d e-3 b^2 e^2\right ) x\right ) \sqrt {b x+c x^2}}{128 c^2 e^5}+\frac {\left (16 c^2 d^2-22 b c d e+3 b^2 e^2-6 c e (2 c d-b e) x\right ) \left (b x+c x^2\right )^{3/2}}{48 c e^3}+\frac {\left (b x+c x^2\right )^{5/2}}{5 e}-\frac {\int \frac {\frac {1}{4} b d \left (128 c^4 d^4-288 b c^3 d^3 e+176 b^2 c^2 d^2 e^2-10 b^3 c d e^3-3 b^4 e^4\right )+\frac {1}{4} (2 c d-b e) \left (128 c^4 d^4-256 b c^3 d^3 e+112 b^2 c^2 d^2 e^2+16 b^3 c d e^3+3 b^4 e^4\right ) x}{(d+e x) \sqrt {b x+c x^2}} \, dx}{64 c^2 e^5}\\ &=\frac {\left (128 c^4 d^4-288 b c^3 d^3 e+176 b^2 c^2 d^2 e^2-10 b^3 c d e^3-3 b^4 e^4-2 c e (2 c d-b e) \left (16 c^2 d^2-16 b c d e-3 b^2 e^2\right ) x\right ) \sqrt {b x+c x^2}}{128 c^2 e^5}+\frac {\left (16 c^2 d^2-22 b c d e+3 b^2 e^2-6 c e (2 c d-b e) x\right ) \left (b x+c x^2\right )^{3/2}}{48 c e^3}+\frac {\left (b x+c x^2\right )^{5/2}}{5 e}+\frac {\left (d^3 (c d-b e)^3\right ) \int \frac {1}{(d+e x) \sqrt {b x+c x^2}} \, dx}{e^6}-\frac {\left ((2 c d-b e) \left (128 c^4 d^4-256 b c^3 d^3 e+112 b^2 c^2 d^2 e^2+16 b^3 c d e^3+3 b^4 e^4\right )\right ) \int \frac {1}{\sqrt {b x+c x^2}} \, dx}{256 c^2 e^6}\\ &=\frac {\left (128 c^4 d^4-288 b c^3 d^3 e+176 b^2 c^2 d^2 e^2-10 b^3 c d e^3-3 b^4 e^4-2 c e (2 c d-b e) \left (16 c^2 d^2-16 b c d e-3 b^2 e^2\right ) x\right ) \sqrt {b x+c x^2}}{128 c^2 e^5}+\frac {\left (16 c^2 d^2-22 b c d e+3 b^2 e^2-6 c e (2 c d-b e) x\right ) \left (b x+c x^2\right )^{3/2}}{48 c e^3}+\frac {\left (b x+c x^2\right )^{5/2}}{5 e}-\frac {\left (2 d^3 (c d-b e)^3\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^6}-\frac {\left ((2 c d-b e) \left (128 c^4 d^4-256 b c^3 d^3 e+112 b^2 c^2 d^2 e^2+16 b^3 c d e^3+3 b^4 e^4\right )\right ) \operatorname {Subst}\left (\int \frac {1}{1-c x^2} \, dx,x,\frac {x}{\sqrt {b x+c x^2}}\right )}{128 c^2 e^6}\\ &=\frac {\left (128 c^4 d^4-288 b c^3 d^3 e+176 b^2 c^2 d^2 e^2-10 b^3 c d e^3-3 b^4 e^4-2 c e (2 c d-b e) \left (16 c^2 d^2-16 b c d e-3 b^2 e^2\right ) x\right ) \sqrt {b x+c x^2}}{128 c^2 e^5}+\frac {\left (16 c^2 d^2-22 b c d e+3 b^2 e^2-6 c e (2 c d-b e) x\right ) \left (b x+c x^2\right )^{3/2}}{48 c e^3}+\frac {\left (b x+c x^2\right )^{5/2}}{5 e}-\frac {(2 c d-b e) \left (128 c^4 d^4-256 b c^3 d^3 e+112 b^2 c^2 d^2 e^2+16 b^3 c d e^3+3 b^4 e^4\right ) \tanh ^{-1}\left (\frac {\sqrt {c} x}{\sqrt {b x+c x^2}}\right )}{128 c^{5/2} e^6}+\frac {d^{5/2} (c d-b e)^{5/2} \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^6}\\ \end {align*}

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Mathematica [A]  time = 0.97, size = 346, normalized size = 0.97 \[ \frac {\sqrt {x (b+c x)} \left (\sqrt {c} e \sqrt {x} \left (-45 b^4 e^4+30 b^3 c e^3 (e x-5 d)+4 b^2 c^2 e^2 \left (660 d^2-295 d e x+186 e^2 x^2\right )+16 b c^3 e \left (-270 d^3+130 d^2 e x-85 d e^2 x^2+63 e^3 x^3\right )+32 c^4 \left (60 d^4-30 d^3 e x+20 d^2 e^2 x^2-15 d e^3 x^3+12 e^4 x^4\right )\right )+\frac {15 \left (3 b^5 e^5+10 b^4 c d e^4+80 b^3 c^2 d^2 e^3-480 b^2 c^3 d^3 e^2+640 b c^4 d^4 e-256 c^5 d^5\right ) \sinh ^{-1}\left (\frac {\sqrt {c} \sqrt {x}}{\sqrt {b}}\right )}{\sqrt {b} \sqrt {\frac {c x}{b}+1}}+\frac {3840 c^{5/2} d^{5/2} (c d-b e)^{5/2} \tanh ^{-1}\left (\frac {\sqrt {x} \sqrt {c d-b e}}{\sqrt {d} \sqrt {b+c x}}\right )}{\sqrt {b+c x}}\right )}{1920 c^{5/2} e^6 \sqrt {x}} \]

Antiderivative was successfully verified.

[In]

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

[Out]

(Sqrt[x*(b + c*x)]*(Sqrt[c]*e*Sqrt[x]*(-45*b^4*e^4 + 30*b^3*c*e^3*(-5*d + e*x) + 4*b^2*c^2*e^2*(660*d^2 - 295*
d*e*x + 186*e^2*x^2) + 16*b*c^3*e*(-270*d^3 + 130*d^2*e*x - 85*d*e^2*x^2 + 63*e^3*x^3) + 32*c^4*(60*d^4 - 30*d
^3*e*x + 20*d^2*e^2*x^2 - 15*d*e^3*x^3 + 12*e^4*x^4)) + (15*(-256*c^5*d^5 + 640*b*c^4*d^4*e - 480*b^2*c^3*d^3*
e^2 + 80*b^3*c^2*d^2*e^3 + 10*b^4*c*d*e^4 + 3*b^5*e^5)*ArcSinh[(Sqrt[c]*Sqrt[x])/Sqrt[b]])/(Sqrt[b]*Sqrt[1 + (
c*x)/b]) + (3840*c^(5/2)*d^(5/2)*(c*d - b*e)^(5/2)*ArcTanh[(Sqrt[c*d - b*e]*Sqrt[x])/(Sqrt[d]*Sqrt[b + c*x])])
/Sqrt[b + c*x]))/(1920*c^(5/2)*e^6*Sqrt[x])

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fricas [A]  time = 9.81, size = 1549, normalized size = 4.35 \[ \text {result too large to display} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

[-1/3840*(15*(256*c^5*d^5 - 640*b*c^4*d^4*e + 480*b^2*c^3*d^3*e^2 - 80*b^3*c^2*d^2*e^3 - 10*b^4*c*d*e^4 - 3*b^
5*e^5)*sqrt(c)*log(2*c*x + b + 2*sqrt(c*x^2 + b*x)*sqrt(c)) - 3840*(c^5*d^4 - 2*b*c^4*d^3*e + b^2*c^3*d^2*e^2)
*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*(384
*c^5*e^5*x^4 + 1920*c^5*d^4*e - 4320*b*c^4*d^3*e^2 + 2640*b^2*c^3*d^2*e^3 - 150*b^3*c^2*d*e^4 - 45*b^4*c*e^5 -
48*(10*c^5*d*e^4 - 21*b*c^4*e^5)*x^3 + 8*(80*c^5*d^2*e^3 - 170*b*c^4*d*e^4 + 93*b^2*c^3*e^5)*x^2 - 10*(96*c^5
*d^3*e^2 - 208*b*c^4*d^2*e^3 + 118*b^2*c^3*d*e^4 - 3*b^3*c^2*e^5)*x)*sqrt(c*x^2 + b*x))/(c^3*e^6), 1/3840*(768
0*(c^5*d^4 - 2*b*c^4*d^3*e + b^2*c^3*d^2*e^2)*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)) - 15*(256*c^5*d^5 - 640*b*c^4*d^4*e + 480*b^2*c^3*d^3*e^2 - 80*b^3*c^2*d^2*e^3 - 10*b^4*c
*d*e^4 - 3*b^5*e^5)*sqrt(c)*log(2*c*x + b + 2*sqrt(c*x^2 + b*x)*sqrt(c)) + 2*(384*c^5*e^5*x^4 + 1920*c^5*d^4*e
- 4320*b*c^4*d^3*e^2 + 2640*b^2*c^3*d^2*e^3 - 150*b^3*c^2*d*e^4 - 45*b^4*c*e^5 - 48*(10*c^5*d*e^4 - 21*b*c^4*
e^5)*x^3 + 8*(80*c^5*d^2*e^3 - 170*b*c^4*d*e^4 + 93*b^2*c^3*e^5)*x^2 - 10*(96*c^5*d^3*e^2 - 208*b*c^4*d^2*e^3
+ 118*b^2*c^3*d*e^4 - 3*b^3*c^2*e^5)*x)*sqrt(c*x^2 + b*x))/(c^3*e^6), 1/1920*(15*(256*c^5*d^5 - 640*b*c^4*d^4*
e + 480*b^2*c^3*d^3*e^2 - 80*b^3*c^2*d^2*e^3 - 10*b^4*c*d*e^4 - 3*b^5*e^5)*sqrt(-c)*arctan(sqrt(c*x^2 + b*x)*s
qrt(-c)/(c*x)) + 1920*(c^5*d^4 - 2*b*c^4*d^3*e + b^2*c^3*d^2*e^2)*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)) + (384*c^5*e^5*x^4 + 1920*c^5*d^4*e - 4320*b*c^4*d^3*
e^2 + 2640*b^2*c^3*d^2*e^3 - 150*b^3*c^2*d*e^4 - 45*b^4*c*e^5 - 48*(10*c^5*d*e^4 - 21*b*c^4*e^5)*x^3 + 8*(80*c
^5*d^2*e^3 - 170*b*c^4*d*e^4 + 93*b^2*c^3*e^5)*x^2 - 10*(96*c^5*d^3*e^2 - 208*b*c^4*d^2*e^3 + 118*b^2*c^3*d*e^
4 - 3*b^3*c^2*e^5)*x)*sqrt(c*x^2 + b*x))/(c^3*e^6), 1/1920*(3840*(c^5*d^4 - 2*b*c^4*d^3*e + b^2*c^3*d^2*e^2)*s
qrt(-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)) + 15*(256*c^5*d^5 - 640*b*
c^4*d^4*e + 480*b^2*c^3*d^3*e^2 - 80*b^3*c^2*d^2*e^3 - 10*b^4*c*d*e^4 - 3*b^5*e^5)*sqrt(-c)*arctan(sqrt(c*x^2
+ b*x)*sqrt(-c)/(c*x)) + (384*c^5*e^5*x^4 + 1920*c^5*d^4*e - 4320*b*c^4*d^3*e^2 + 2640*b^2*c^3*d^2*e^3 - 150*b
^3*c^2*d*e^4 - 45*b^4*c*e^5 - 48*(10*c^5*d*e^4 - 21*b*c^4*e^5)*x^3 + 8*(80*c^5*d^2*e^3 - 170*b*c^4*d*e^4 + 93*
b^2*c^3*e^5)*x^2 - 10*(96*c^5*d^3*e^2 - 208*b*c^4*d^2*e^3 + 118*b^2*c^3*d*e^4 - 3*b^3*c^2*e^5)*x)*sqrt(c*x^2 +
b*x))/(c^3*e^6)]

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

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate((c*x^2+b*x)^(5/2)/(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

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maple [B]  time = 0.05, size = 1932, normalized size = 5.43 \[ \text {result too large to display} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

3/256/e/c^(5/2)*b^5*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))+1/8/e*((x+d/e)^2*c-(b*e-c*d)*d/e^2+(b*e-2*c*d)*(x+d/e)/e)^(3/2)*x*b+1/16/e/c*((x+d/e)^2*c-(b*e-c*d)*
d/e^2+(b*e-2*c*d)*(x+d/e)/e)^(3/2)*b^2-11/24/e^2*((x+d/e)^2*c-(b*e-c*d)*d/e^2+(b*e-2*c*d)*(x+d/e)/e)^(3/2)*b*d
-3/128/e/c^2*b^4*((x+d/e)^2*c-(b*e-c*d)*d/e^2+(b*e-2*c*d)*(x+d/e)/e)^(1/2)+1/e^5*d^4*((x+d/e)^2*c-(b*e-c*d)*d/
e^2+(b*e-2*c*d)*(x+d/e)/e)^(1/2)*c^2-1/e^6*d^5*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)+1/3/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)^(
3/2)*c+11/8/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)*b^2+3/4/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)*x*b*c-3/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
))*b^2*c+3/e^6*d^5/(-(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+1/5/e*((x+d/e)^2*c-(b*e-c*d)*d
/e^2+(b*e-2*c*d)*(x+d/e)/e)^(5/2)+5/128/e^2*d/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^4-1/2/e^4*d^3*((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^2-5/32/e^2*b^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-5/64/e^2/c*b^3*((x+d/
e)^2*c-(b*e-c*d)*d/e^2+(b*e-2*c*d)*(x+d/e)/e)^(1/2)*d-1/e^7*d^6/(-(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+5/2/e^5*d^4*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-1/4/e^2*((x+d/e)^2*c-(b*e-c*d)*d/e^2+(b*e-2*c*d)*(x+d/e)/e)^(3/2)*x*c*d-9/4/e^4*d^3*(
(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+5/16/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^3-15/8/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^(1/2)*b^2+1/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^3-3/64/e/c*b^3*((x+d/e)^2*c-(b*e-c*d)*d/e^2+(b*e-2*c*d)*(x+d/e
)/e)^(1/2)*x

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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((c*x^2+b*x)^(5/2)/(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?

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

Verification of antiderivative is not currently implemented for this CAS.

[In]

int((b*x + c*x^2)^(5/2)/(d + e*x),x)

[Out]

int((b*x + c*x^2)^(5/2)/(d + e*x), x)

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sympy [F]  time = 0.00, size = 0, normalized size = 0.00 \[ \int \frac {\left (x \left (b + c x\right )\right )^{\frac {5}{2}}}{d + e x}\, dx \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

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

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