3.109 \(\int x^{3/2} (a x+b x^3+c x^5)^{3/2} \, dx\)

Optimal. Leaf size=244 \[ \frac{\left (128 a^2 c^2-100 a b^2 c+15 b^4\right ) \sqrt{a x+b x^3+c x^5}}{1280 c^3 \sqrt{x}}-\frac{x^{3/2} \left (4 c x^2 \left (5 b^2-16 a c\right )+b \left (5 b^2-4 a c\right )\right ) \sqrt{a x+b x^3+c x^5}}{640 c^2}-\frac{3 b \sqrt{x} \left (b^2-4 a c\right )^2 \sqrt{a+b x^2+c x^4} \tanh ^{-1}\left (\frac{b+2 c x^2}{2 \sqrt{c} \sqrt{a+b x^2+c x^4}}\right )}{512 c^{7/2} \sqrt{a x+b x^3+c x^5}}+\frac{\sqrt{x} \left (3 b+8 c x^2\right ) \left (a x+b x^3+c x^5\right )^{3/2}}{80 c} \]

[Out]

((15*b^4 - 100*a*b^2*c + 128*a^2*c^2)*Sqrt[a*x + b*x^3 + c*x^5])/(1280*c^3*Sqrt[x]) - (x^(3/2)*(b*(5*b^2 - 4*a
*c) + 4*c*(5*b^2 - 16*a*c)*x^2)*Sqrt[a*x + b*x^3 + c*x^5])/(640*c^2) + (Sqrt[x]*(3*b + 8*c*x^2)*(a*x + b*x^3 +
c*x^5)^(3/2))/(80*c) - (3*b*(b^2 - 4*a*c)^2*Sqrt[x]*Sqrt[a + b*x^2 + c*x^4]*ArcTanh[(b + 2*c*x^2)/(2*Sqrt[c]*
Sqrt[a + b*x^2 + c*x^4])])/(512*c^(7/2)*Sqrt[a*x + b*x^3 + c*x^5])

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Rubi [A]  time = 0.357206, antiderivative size = 244, normalized size of antiderivative = 1., number of steps used = 8, number of rules used = 8, integrand size = 24, \(\frac{\text{number of rules}}{\text{integrand size}}\) = 0.333, Rules used = {1919, 1945, 1949, 12, 1914, 1107, 621, 206} \[ \frac{\left (128 a^2 c^2-100 a b^2 c+15 b^4\right ) \sqrt{a x+b x^3+c x^5}}{1280 c^3 \sqrt{x}}-\frac{x^{3/2} \left (4 c x^2 \left (5 b^2-16 a c\right )+b \left (5 b^2-4 a c\right )\right ) \sqrt{a x+b x^3+c x^5}}{640 c^2}-\frac{3 b \sqrt{x} \left (b^2-4 a c\right )^2 \sqrt{a+b x^2+c x^4} \tanh ^{-1}\left (\frac{b+2 c x^2}{2 \sqrt{c} \sqrt{a+b x^2+c x^4}}\right )}{512 c^{7/2} \sqrt{a x+b x^3+c x^5}}+\frac{\sqrt{x} \left (3 b+8 c x^2\right ) \left (a x+b x^3+c x^5\right )^{3/2}}{80 c} \]

Antiderivative was successfully verified.

[In]

Int[x^(3/2)*(a*x + b*x^3 + c*x^5)^(3/2),x]

[Out]

((15*b^4 - 100*a*b^2*c + 128*a^2*c^2)*Sqrt[a*x + b*x^3 + c*x^5])/(1280*c^3*Sqrt[x]) - (x^(3/2)*(b*(5*b^2 - 4*a
*c) + 4*c*(5*b^2 - 16*a*c)*x^2)*Sqrt[a*x + b*x^3 + c*x^5])/(640*c^2) + (Sqrt[x]*(3*b + 8*c*x^2)*(a*x + b*x^3 +
c*x^5)^(3/2))/(80*c) - (3*b*(b^2 - 4*a*c)^2*Sqrt[x]*Sqrt[a + b*x^2 + c*x^4]*ArcTanh[(b + 2*c*x^2)/(2*Sqrt[c]*
Sqrt[a + b*x^2 + c*x^4])])/(512*c^(7/2)*Sqrt[a*x + b*x^3 + c*x^5])

Rule 1919

Int[(x_)^(m_.)*((b_.)*(x_)^(n_.) + (a_.)*(x_)^(q_.) + (c_.)*(x_)^(r_.))^(p_), x_Symbol] :> Simp[(x^(m - n + q
+ 1)*(b*(n - q)*p + c*(m + p*q + (n - q)*(2*p - 1) + 1)*x^(n - q))*(a*x^q + b*x^n + c*x^(2*n - q))^p)/(c*(m +
p*(2*n - q) + 1)*(m + p*q + (n - q)*(2*p - 1) + 1)), x] + Dist[((n - q)*p)/(c*(m + p*(2*n - q) + 1)*(m + p*q +
(n - q)*(2*p - 1) + 1)), Int[x^(m - (n - 2*q))*Simp[-(a*b*(m + p*q - n + q + 1)) + (2*a*c*(m + p*q + (n - q)*
(2*p - 1) + 1) - b^2*(m + p*q + (n - q)*(p - 1) + 1))*x^(n - q), x]*(a*x^q + b*x^n + c*x^(2*n - q))^(p - 1), x
], x] /; FreeQ[{a, b, c}, x] && EqQ[r, 2*n - q] && PosQ[n - q] &&  !IntegerQ[p] && NeQ[b^2 - 4*a*c, 0] && IGtQ
[n, 0] && GtQ[p, 0] && RationalQ[m, q] && GtQ[m + p*q + 1, n - q] && NeQ[m + p*(2*n - q) + 1, 0] && NeQ[m + p*
q + (n - q)*(2*p - 1) + 1, 0]

Rule 1945

Int[(x_)^(m_.)*((c_.)*(x_)^(j_.) + (b_.)*(x_)^(n_.) + (a_.)*(x_)^(q_.))^(p_.)*((A_) + (B_.)*(x_)^(r_.)), x_Sym
bol] :> Simp[(x^(m + 1)*(b*B*(n - q)*p + A*c*(m + p*q + (n - q)*(2*p + 1) + 1) + B*c*(m + p*q + 2*(n - q)*p +
1)*x^(n - q))*(a*x^q + b*x^n + c*x^(2*n - q))^p)/(c*(m + p*(2*n - q) + 1)*(m + p*q + (n - q)*(2*p + 1) + 1)),
x] + Dist[((n - q)*p)/(c*(m + p*(2*n - q) + 1)*(m + p*q + (n - q)*(2*p + 1) + 1)), Int[x^(m + q)*Simp[2*a*A*c*
(m + p*q + (n - q)*(2*p + 1) + 1) - a*b*B*(m + p*q + 1) + (2*a*B*c*(m + p*q + 2*(n - q)*p + 1) + A*b*c*(m + p*
q + (n - q)*(2*p + 1) + 1) - b^2*B*(m + p*q + (n - q)*p + 1))*x^(n - q), x]*(a*x^q + b*x^n + c*x^(2*n - q))^(p
- 1), x], x] /; FreeQ[{a, b, c, A, B}, x] && EqQ[r, n - q] && EqQ[j, 2*n - q] &&  !IntegerQ[p] && NeQ[b^2 - 4
*a*c, 0] && IGtQ[n, 0] && GtQ[p, 0] && RationalQ[m, q] && GtQ[m + p*q, -(n - q) - 1] && NeQ[m + p*(2*n - q) +
1, 0] && NeQ[m + p*q + (n - q)*(2*p + 1) + 1, 0]

Rule 1949

Int[(x_)^(m_.)*((c_.)*(x_)^(j_.) + (b_.)*(x_)^(n_.) + (a_.)*(x_)^(q_.))^(p_.)*((A_) + (B_.)*(x_)^(r_.)), x_Sym
bol] :> Simp[(B*x^(m - n + 1)*(a*x^q + b*x^n + c*x^(2*n - q))^(p + 1))/(c*(m + p*q + (n - q)*(2*p + 1) + 1)),
x] - Dist[1/(c*(m + p*q + (n - q)*(2*p + 1) + 1)), Int[x^(m - n + q)*Simp[a*B*(m + p*q - n + q + 1) + (b*B*(m
+ p*q + (n - q)*p + 1) - A*c*(m + p*q + (n - q)*(2*p + 1) + 1))*x^(n - q), x]*(a*x^q + b*x^n + c*x^(2*n - q))^
p, x], x] /; FreeQ[{a, b, c, A, B}, x] && EqQ[r, n - q] && EqQ[j, 2*n - q] &&  !IntegerQ[p] && NeQ[b^2 - 4*a*c
, 0] && IGtQ[n, 0] && GeQ[p, -1] && LtQ[p, 0] && RationalQ[m, q] && GeQ[m + p*q, n - q - 1] && NeQ[m + p*q + (
n - q)*(2*p + 1) + 1, 0]

Rule 12

Int[(a_)*(u_), x_Symbol] :> Dist[a, Int[u, x], x] /; FreeQ[a, x] &&  !MatchQ[u, (b_)*(v_) /; FreeQ[b, x]]

Rule 1914

Int[(x_)^(m_.)/Sqrt[(b_.)*(x_)^(n_.) + (a_.)*(x_)^(q_.) + (c_.)*(x_)^(r_.)], x_Symbol] :> Dist[(x^(q/2)*Sqrt[a
+ b*x^(n - q) + c*x^(2*(n - q))])/Sqrt[a*x^q + b*x^n + c*x^(2*n - q)], Int[x^(m - q/2)/Sqrt[a + b*x^(n - q) +
c*x^(2*(n - q))], x], x] /; FreeQ[{a, b, c, m, n, q}, x] && EqQ[r, 2*n - q] && PosQ[n - q] && ((EqQ[m, 1] &&
EqQ[n, 3] && EqQ[q, 2]) || ((EqQ[m + 1/2] || EqQ[m, 3/2] || EqQ[m, 1/2] || EqQ[m, 5/2]) && EqQ[n, 3] && EqQ[q,
1]))

Rule 1107

Int[(x_)*((a_) + (b_.)*(x_)^2 + (c_.)*(x_)^4)^(p_.), x_Symbol] :> Dist[1/2, Subst[Int[(a + b*x + c*x^2)^p, x],
x, x^2], x] /; FreeQ[{a, b, c, p}, x]

Rule 621

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

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])

Rubi steps

\begin{align*} \int x^{3/2} \left (a x+b x^3+c x^5\right )^{3/2} \, dx &=\frac{\sqrt{x} \left (3 b+8 c x^2\right ) \left (a x+b x^3+c x^5\right )^{3/2}}{80 c}+\frac{3 \int \sqrt{x} \left (-2 a b-\left (5 b^2-16 a c\right ) x^2\right ) \sqrt{a x+b x^3+c x^5} \, dx}{80 c}\\ &=-\frac{x^{3/2} \left (b \left (5 b^2-4 a c\right )+4 c \left (5 b^2-16 a c\right ) x^2\right ) \sqrt{a x+b x^3+c x^5}}{640 c^2}+\frac{\sqrt{x} \left (3 b+8 c x^2\right ) \left (a x+b x^3+c x^5\right )^{3/2}}{80 c}+\frac{\int \frac{x^{3/2} \left (2 a b \left (5 b^2-28 a c\right )+\left (15 b^4-100 a b^2 c+128 a^2 c^2\right ) x^2\right )}{\sqrt{a x+b x^3+c x^5}} \, dx}{640 c^2}\\ &=\frac{\left (15 b^4-100 a b^2 c+128 a^2 c^2\right ) \sqrt{a x+b x^3+c x^5}}{1280 c^3 \sqrt{x}}-\frac{x^{3/2} \left (b \left (5 b^2-4 a c\right )+4 c \left (5 b^2-16 a c\right ) x^2\right ) \sqrt{a x+b x^3+c x^5}}{640 c^2}+\frac{\sqrt{x} \left (3 b+8 c x^2\right ) \left (a x+b x^3+c x^5\right )^{3/2}}{80 c}-\frac{\int \frac{15 b \left (b^2-4 a c\right )^2 x^{3/2}}{\sqrt{a x+b x^3+c x^5}} \, dx}{1280 c^3}\\ &=\frac{\left (15 b^4-100 a b^2 c+128 a^2 c^2\right ) \sqrt{a x+b x^3+c x^5}}{1280 c^3 \sqrt{x}}-\frac{x^{3/2} \left (b \left (5 b^2-4 a c\right )+4 c \left (5 b^2-16 a c\right ) x^2\right ) \sqrt{a x+b x^3+c x^5}}{640 c^2}+\frac{\sqrt{x} \left (3 b+8 c x^2\right ) \left (a x+b x^3+c x^5\right )^{3/2}}{80 c}-\frac{\left (3 b \left (b^2-4 a c\right )^2\right ) \int \frac{x^{3/2}}{\sqrt{a x+b x^3+c x^5}} \, dx}{256 c^3}\\ &=\frac{\left (15 b^4-100 a b^2 c+128 a^2 c^2\right ) \sqrt{a x+b x^3+c x^5}}{1280 c^3 \sqrt{x}}-\frac{x^{3/2} \left (b \left (5 b^2-4 a c\right )+4 c \left (5 b^2-16 a c\right ) x^2\right ) \sqrt{a x+b x^3+c x^5}}{640 c^2}+\frac{\sqrt{x} \left (3 b+8 c x^2\right ) \left (a x+b x^3+c x^5\right )^{3/2}}{80 c}-\frac{\left (3 b \left (b^2-4 a c\right )^2 \sqrt{x} \sqrt{a+b x^2+c x^4}\right ) \int \frac{x}{\sqrt{a+b x^2+c x^4}} \, dx}{256 c^3 \sqrt{a x+b x^3+c x^5}}\\ &=\frac{\left (15 b^4-100 a b^2 c+128 a^2 c^2\right ) \sqrt{a x+b x^3+c x^5}}{1280 c^3 \sqrt{x}}-\frac{x^{3/2} \left (b \left (5 b^2-4 a c\right )+4 c \left (5 b^2-16 a c\right ) x^2\right ) \sqrt{a x+b x^3+c x^5}}{640 c^2}+\frac{\sqrt{x} \left (3 b+8 c x^2\right ) \left (a x+b x^3+c x^5\right )^{3/2}}{80 c}-\frac{\left (3 b \left (b^2-4 a c\right )^2 \sqrt{x} \sqrt{a+b x^2+c x^4}\right ) \operatorname{Subst}\left (\int \frac{1}{\sqrt{a+b x+c x^2}} \, dx,x,x^2\right )}{512 c^3 \sqrt{a x+b x^3+c x^5}}\\ &=\frac{\left (15 b^4-100 a b^2 c+128 a^2 c^2\right ) \sqrt{a x+b x^3+c x^5}}{1280 c^3 \sqrt{x}}-\frac{x^{3/2} \left (b \left (5 b^2-4 a c\right )+4 c \left (5 b^2-16 a c\right ) x^2\right ) \sqrt{a x+b x^3+c x^5}}{640 c^2}+\frac{\sqrt{x} \left (3 b+8 c x^2\right ) \left (a x+b x^3+c x^5\right )^{3/2}}{80 c}-\frac{\left (3 b \left (b^2-4 a c\right )^2 \sqrt{x} \sqrt{a+b x^2+c x^4}\right ) \operatorname{Subst}\left (\int \frac{1}{4 c-x^2} \, dx,x,\frac{b+2 c x^2}{\sqrt{a+b x^2+c x^4}}\right )}{256 c^3 \sqrt{a x+b x^3+c x^5}}\\ &=\frac{\left (15 b^4-100 a b^2 c+128 a^2 c^2\right ) \sqrt{a x+b x^3+c x^5}}{1280 c^3 \sqrt{x}}-\frac{x^{3/2} \left (b \left (5 b^2-4 a c\right )+4 c \left (5 b^2-16 a c\right ) x^2\right ) \sqrt{a x+b x^3+c x^5}}{640 c^2}+\frac{\sqrt{x} \left (3 b+8 c x^2\right ) \left (a x+b x^3+c x^5\right )^{3/2}}{80 c}-\frac{3 b \left (b^2-4 a c\right )^2 \sqrt{x} \sqrt{a+b x^2+c x^4} \tanh ^{-1}\left (\frac{b+2 c x^2}{2 \sqrt{c} \sqrt{a+b x^2+c x^4}}\right )}{512 c^{7/2} \sqrt{a x+b x^3+c x^5}}\\ \end{align*}

Mathematica [A]  time = 0.212016, size = 192, normalized size = 0.79 \[ \frac{\left (x \left (a+b x^2+c x^4\right )\right )^{3/2} \left (-\frac{3 b \left (b^2-4 a c\right ) \left (\left (b^2-4 a c\right ) \tanh ^{-1}\left (\frac{b+2 c x^2}{2 \sqrt{c} \sqrt{a+b x^2+c x^4}}\right )-2 \sqrt{c} \left (b+2 c x^2\right ) \sqrt{a+b x^2+c x^4}\right )}{256 c^{7/2}}-\frac{b \left (b+2 c x^2\right ) \left (a+b x^2+c x^4\right )^{3/2}}{16 c^2}+\frac{\left (a+b x^2+c x^4\right )^{5/2}}{5 c}\right )}{2 x^{3/2} \left (a+b x^2+c x^4\right )^{3/2}} \]

Antiderivative was successfully verified.

[In]

Integrate[x^(3/2)*(a*x + b*x^3 + c*x^5)^(3/2),x]

[Out]

((x*(a + b*x^2 + c*x^4))^(3/2)*(-(b*(b + 2*c*x^2)*(a + b*x^2 + c*x^4)^(3/2))/(16*c^2) + (a + b*x^2 + c*x^4)^(5
/2)/(5*c) - (3*b*(b^2 - 4*a*c)*(-2*Sqrt[c]*(b + 2*c*x^2)*Sqrt[a + b*x^2 + c*x^4] + (b^2 - 4*a*c)*ArcTanh[(b +
2*c*x^2)/(2*Sqrt[c]*Sqrt[a + b*x^2 + c*x^4])]))/(256*c^(7/2))))/(2*x^(3/2)*(a + b*x^2 + c*x^4)^(3/2))

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Maple [A]  time = 0.026, size = 369, normalized size = 1.5 \begin{align*} -{\frac{1}{2560}\sqrt{x \left ( c{x}^{4}+b{x}^{2}+a \right ) } \left ( -256\,{x}^{8}{c}^{9/2}\sqrt{c{x}^{4}+b{x}^{2}+a}-352\,{x}^{6}b{c}^{7/2}\sqrt{c{x}^{4}+b{x}^{2}+a}-512\,{x}^{4}a{c}^{7/2}\sqrt{c{x}^{4}+b{x}^{2}+a}-16\,{x}^{4}{b}^{2}{c}^{5/2}\sqrt{c{x}^{4}+b{x}^{2}+a}-112\,{x}^{2}ab{c}^{5/2}\sqrt{c{x}^{4}+b{x}^{2}+a}+20\,{x}^{2}{b}^{3}{c}^{3/2}\sqrt{c{x}^{4}+b{x}^{2}+a}+240\,\ln \left ( 1/2\,{\frac{2\,c{x}^{2}+2\,\sqrt{c{x}^{4}+b{x}^{2}+a}\sqrt{c}+b}{\sqrt{c}}} \right ){a}^{2}b{c}^{2}-120\,\ln \left ( 1/2\,{\frac{2\,c{x}^{2}+2\,\sqrt{c{x}^{4}+b{x}^{2}+a}\sqrt{c}+b}{\sqrt{c}}} \right ) a{b}^{3}c+15\,\ln \left ( 1/2\,{\frac{2\,c{x}^{2}+2\,\sqrt{c{x}^{4}+b{x}^{2}+a}\sqrt{c}+b}{\sqrt{c}}} \right ){b}^{5}-256\,{a}^{2}{c}^{5/2}\sqrt{c{x}^{4}+b{x}^{2}+a}+200\,a{b}^{2}{c}^{3/2}\sqrt{c{x}^{4}+b{x}^{2}+a}-30\,{b}^{4}\sqrt{c}\sqrt{c{x}^{4}+b{x}^{2}+a} \right ){c}^{-{\frac{7}{2}}}{\frac{1}{\sqrt{x}}}{\frac{1}{\sqrt{c{x}^{4}+b{x}^{2}+a}}}} \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

int(x^(3/2)*(c*x^5+b*x^3+a*x)^(3/2),x)

[Out]

-1/2560*(x*(c*x^4+b*x^2+a))^(1/2)/c^(7/2)*(-256*x^8*c^(9/2)*(c*x^4+b*x^2+a)^(1/2)-352*x^6*b*c^(7/2)*(c*x^4+b*x
^2+a)^(1/2)-512*x^4*a*c^(7/2)*(c*x^4+b*x^2+a)^(1/2)-16*x^4*b^2*c^(5/2)*(c*x^4+b*x^2+a)^(1/2)-112*x^2*a*b*c^(5/
2)*(c*x^4+b*x^2+a)^(1/2)+20*x^2*b^3*c^(3/2)*(c*x^4+b*x^2+a)^(1/2)+240*ln(1/2*(2*c*x^2+2*(c*x^4+b*x^2+a)^(1/2)*
c^(1/2)+b)/c^(1/2))*a^2*b*c^2-120*ln(1/2*(2*c*x^2+2*(c*x^4+b*x^2+a)^(1/2)*c^(1/2)+b)/c^(1/2))*a*b^3*c+15*ln(1/
2*(2*c*x^2+2*(c*x^4+b*x^2+a)^(1/2)*c^(1/2)+b)/c^(1/2))*b^5-256*a^2*c^(5/2)*(c*x^4+b*x^2+a)^(1/2)+200*a*b^2*c^(
3/2)*(c*x^4+b*x^2+a)^(1/2)-30*b^4*c^(1/2)*(c*x^4+b*x^2+a)^(1/2))/x^(1/2)/(c*x^4+b*x^2+a)^(1/2)

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Maxima [F]  time = 0., size = 0, normalized size = 0. \begin{align*} \int{\left (c x^{5} + b x^{3} + a x\right )}^{\frac{3}{2}} x^{\frac{3}{2}}\,{d x} \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate(x^(3/2)*(c*x^5+b*x^3+a*x)^(3/2),x, algorithm="maxima")

[Out]

integrate((c*x^5 + b*x^3 + a*x)^(3/2)*x^(3/2), x)

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Fricas [A]  time = 1.44359, size = 921, normalized size = 3.77 \begin{align*} \left [\frac{15 \,{\left (b^{5} - 8 \, a b^{3} c + 16 \, a^{2} b c^{2}\right )} \sqrt{c} x \log \left (-\frac{8 \, c^{2} x^{5} + 8 \, b c x^{3} - 4 \, \sqrt{c x^{5} + b x^{3} + a x}{\left (2 \, c x^{2} + b\right )} \sqrt{c} \sqrt{x} +{\left (b^{2} + 4 \, a c\right )} x}{x}\right ) + 4 \,{\left (128 \, c^{5} x^{8} + 176 \, b c^{4} x^{6} + 15 \, b^{4} c - 100 \, a b^{2} c^{2} + 128 \, a^{2} c^{3} + 8 \,{\left (b^{2} c^{3} + 32 \, a c^{4}\right )} x^{4} - 2 \,{\left (5 \, b^{3} c^{2} - 28 \, a b c^{3}\right )} x^{2}\right )} \sqrt{c x^{5} + b x^{3} + a x} \sqrt{x}}{5120 \, c^{4} x}, \frac{15 \,{\left (b^{5} - 8 \, a b^{3} c + 16 \, a^{2} b c^{2}\right )} \sqrt{-c} x \arctan \left (\frac{\sqrt{c x^{5} + b x^{3} + a x}{\left (2 \, c x^{2} + b\right )} \sqrt{-c} \sqrt{x}}{2 \,{\left (c^{2} x^{5} + b c x^{3} + a c x\right )}}\right ) + 2 \,{\left (128 \, c^{5} x^{8} + 176 \, b c^{4} x^{6} + 15 \, b^{4} c - 100 \, a b^{2} c^{2} + 128 \, a^{2} c^{3} + 8 \,{\left (b^{2} c^{3} + 32 \, a c^{4}\right )} x^{4} - 2 \,{\left (5 \, b^{3} c^{2} - 28 \, a b c^{3}\right )} x^{2}\right )} \sqrt{c x^{5} + b x^{3} + a x} \sqrt{x}}{2560 \, c^{4} x}\right ] \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate(x^(3/2)*(c*x^5+b*x^3+a*x)^(3/2),x, algorithm="fricas")

[Out]

[1/5120*(15*(b^5 - 8*a*b^3*c + 16*a^2*b*c^2)*sqrt(c)*x*log(-(8*c^2*x^5 + 8*b*c*x^3 - 4*sqrt(c*x^5 + b*x^3 + a*
x)*(2*c*x^2 + b)*sqrt(c)*sqrt(x) + (b^2 + 4*a*c)*x)/x) + 4*(128*c^5*x^8 + 176*b*c^4*x^6 + 15*b^4*c - 100*a*b^2
*c^2 + 128*a^2*c^3 + 8*(b^2*c^3 + 32*a*c^4)*x^4 - 2*(5*b^3*c^2 - 28*a*b*c^3)*x^2)*sqrt(c*x^5 + b*x^3 + a*x)*sq
rt(x))/(c^4*x), 1/2560*(15*(b^5 - 8*a*b^3*c + 16*a^2*b*c^2)*sqrt(-c)*x*arctan(1/2*sqrt(c*x^5 + b*x^3 + a*x)*(2
*c*x^2 + b)*sqrt(-c)*sqrt(x)/(c^2*x^5 + b*c*x^3 + a*c*x)) + 2*(128*c^5*x^8 + 176*b*c^4*x^6 + 15*b^4*c - 100*a*
b^2*c^2 + 128*a^2*c^3 + 8*(b^2*c^3 + 32*a*c^4)*x^4 - 2*(5*b^3*c^2 - 28*a*b*c^3)*x^2)*sqrt(c*x^5 + b*x^3 + a*x)
*sqrt(x))/(c^4*x)]

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Sympy [F(-1)]  time = 0., size = 0, normalized size = 0. \begin{align*} \text{Timed out} \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate(x**(3/2)*(c*x**5+b*x**3+a*x)**(3/2),x)

[Out]

Timed out

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Giac [F]  time = 0., size = 0, normalized size = 0. \begin{align*} \int{\left (c x^{5} + b x^{3} + a x\right )}^{\frac{3}{2}} x^{\frac{3}{2}}\,{d x} \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate(x^(3/2)*(c*x^5+b*x^3+a*x)^(3/2),x, algorithm="giac")

[Out]

integrate((c*x^5 + b*x^3 + a*x)^(3/2)*x^(3/2), x)