3.46 \(\int \frac{(a x^2+b x^3+c x^4)^{3/2}}{x^7} \, dx\)

Optimal. Leaf size=257 \[ \frac{b x \left (b^2-12 a c\right ) \sqrt{a+b x+c x^2} \tanh ^{-1}\left (\frac{2 a+b x}{2 \sqrt{a} \sqrt{a+b x+c x^2}}\right )}{16 a^{3/2} \sqrt{a x^2+b x^3+c x^4}}+\frac{\left (-8 a c+b^2+2 b c x\right ) \sqrt{a x^2+b x^3+c x^4}}{8 a x^2}+\frac{c^{3/2} x \sqrt{a+b x+c x^2} \tanh ^{-1}\left (\frac{b+2 c x}{2 \sqrt{c} \sqrt{a+b x+c x^2}}\right )}{\sqrt{a x^2+b x^3+c x^4}}-\frac{b \left (a x^2+b x^3+c x^4\right )^{3/2}}{4 a x^5}-\frac{\left (a x^2+b x^3+c x^4\right )^{3/2}}{3 x^6} \]

[Out]

((b^2 - 8*a*c + 2*b*c*x)*Sqrt[a*x^2 + b*x^3 + c*x^4])/(8*a*x^2) - (a*x^2 + b*x^3 + c*x^4)^(3/2)/(3*x^6) - (b*(
a*x^2 + b*x^3 + c*x^4)^(3/2))/(4*a*x^5) + (b*(b^2 - 12*a*c)*x*Sqrt[a + b*x + c*x^2]*ArcTanh[(2*a + b*x)/(2*Sqr
t[a]*Sqrt[a + b*x + c*x^2])])/(16*a^(3/2)*Sqrt[a*x^2 + b*x^3 + c*x^4]) + (c^(3/2)*x*Sqrt[a + b*x + c*x^2]*ArcT
anh[(b + 2*c*x)/(2*Sqrt[c]*Sqrt[a + b*x + c*x^2])])/Sqrt[a*x^2 + b*x^3 + c*x^4]

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Rubi [A]  time = 0.350564, antiderivative size = 257, normalized size of antiderivative = 1., number of steps used = 9, number of rules used = 8, integrand size = 24, \(\frac{\text{number of rules}}{\text{integrand size}}\) = 0.333, Rules used = {1920, 1951, 1941, 1933, 843, 621, 206, 724} \[ \frac{b x \left (b^2-12 a c\right ) \sqrt{a+b x+c x^2} \tanh ^{-1}\left (\frac{2 a+b x}{2 \sqrt{a} \sqrt{a+b x+c x^2}}\right )}{16 a^{3/2} \sqrt{a x^2+b x^3+c x^4}}+\frac{\left (-8 a c+b^2+2 b c x\right ) \sqrt{a x^2+b x^3+c x^4}}{8 a x^2}+\frac{c^{3/2} x \sqrt{a+b x+c x^2} \tanh ^{-1}\left (\frac{b+2 c x}{2 \sqrt{c} \sqrt{a+b x+c x^2}}\right )}{\sqrt{a x^2+b x^3+c x^4}}-\frac{b \left (a x^2+b x^3+c x^4\right )^{3/2}}{4 a x^5}-\frac{\left (a x^2+b x^3+c x^4\right )^{3/2}}{3 x^6} \]

Antiderivative was successfully verified.

[In]

Int[(a*x^2 + b*x^3 + c*x^4)^(3/2)/x^7,x]

[Out]

((b^2 - 8*a*c + 2*b*c*x)*Sqrt[a*x^2 + b*x^3 + c*x^4])/(8*a*x^2) - (a*x^2 + b*x^3 + c*x^4)^(3/2)/(3*x^6) - (b*(
a*x^2 + b*x^3 + c*x^4)^(3/2))/(4*a*x^5) + (b*(b^2 - 12*a*c)*x*Sqrt[a + b*x + c*x^2]*ArcTanh[(2*a + b*x)/(2*Sqr
t[a]*Sqrt[a + b*x + c*x^2])])/(16*a^(3/2)*Sqrt[a*x^2 + b*x^3 + c*x^4]) + (c^(3/2)*x*Sqrt[a + b*x + c*x^2]*ArcT
anh[(b + 2*c*x)/(2*Sqrt[c]*Sqrt[a + b*x + c*x^2])])/Sqrt[a*x^2 + b*x^3 + c*x^4]

Rule 1920

Int[(x_)^(m_.)*((b_.)*(x_)^(n_.) + (a_.)*(x_)^(q_.) + (c_.)*(x_)^(r_.))^(p_), x_Symbol] :> Simp[(x^(m + 1)*(a*
x^q + b*x^n + c*x^(2*n - q))^p)/(m + p*q + 1), x] - Dist[((n - q)*p)/(m + p*q + 1), Int[x^(m + n)*(b + 2*c*x^(
n - q))*(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] && LeQ[m + p*q + 1, -
(n - q) + 1] && NeQ[m + p*q + 1, 0]

Rule 1951

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

Rule 1941

Int[(x_)^(m_.)*((c_.)*(x_)^(j_.) + (b_.)*(x_)^(n_.) + (a_.)*(x_)^(q_.))^(p_.)*((A_) + (B_.)*(x_)^(r_.)), x_Sym
bol] :> Simp[(x^(m + 1)*(A*(m + p*q + (n - q)*(2*p + 1) + 1) + B*(m + p*q + 1)*x^(n - q))*(a*x^q + b*x^n + c*x
^(2*n - q))^p)/((m + p*q + 1)*(m + p*q + (n - q)*(2*p + 1) + 1)), x] + Dist[((n - q)*p)/((m + p*q + 1)*(m + p*
q + (n - q)*(2*p + 1) + 1)), Int[x^(n + m)*Simp[2*a*B*(m + p*q + 1) - A*b*(m + p*q + (n - q)*(2*p + 1) + 1) +
(b*B*(m + p*q + 1) - 2*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
- 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] && LeQ[m + p*q, -(n - q)] && NeQ[m + p*q + 1, 0] && NeQ
[m + p*q + (n - q)*(2*p + 1) + 1, 0]

Rule 1933

Int[((A_) + (B_.)*(x_)^(j_.))/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[(A + B*x^(n - q))/(
x^(q/2)*Sqrt[a + b*x^(n - q) + c*x^(2*(n - q))]), x], x] /; FreeQ[{a, b, c, A, B, n, q}, x] && EqQ[j, n - q] &
& EqQ[r, 2*n - q] && PosQ[n - q] && EqQ[n, 3] && EqQ[q, 2]

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]

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])

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]

Rubi steps

\begin{align*} \int \frac{\left (a x^2+b x^3+c x^4\right )^{3/2}}{x^7} \, dx &=-\frac{\left (a x^2+b x^3+c x^4\right )^{3/2}}{3 x^6}+\frac{1}{2} \int \frac{(b+2 c x) \sqrt{a x^2+b x^3+c x^4}}{x^4} \, dx\\ &=-\frac{\left (a x^2+b x^3+c x^4\right )^{3/2}}{3 x^6}-\frac{b \left (a x^2+b x^3+c x^4\right )^{3/2}}{4 a x^5}-\frac{\int \frac{\left (\frac{1}{2} \left (b^2-8 a c\right )-b c x\right ) \sqrt{a x^2+b x^3+c x^4}}{x^3} \, dx}{4 a}\\ &=\frac{\left (b^2-8 a c+2 b c x\right ) \sqrt{a x^2+b x^3+c x^4}}{8 a x^2}-\frac{\left (a x^2+b x^3+c x^4\right )^{3/2}}{3 x^6}-\frac{b \left (a x^2+b x^3+c x^4\right )^{3/2}}{4 a x^5}+\frac{\int \frac{-\frac{1}{2} b \left (b^2-12 a c\right )+8 a c^2 x}{\sqrt{a x^2+b x^3+c x^4}} \, dx}{8 a}\\ &=\frac{\left (b^2-8 a c+2 b c x\right ) \sqrt{a x^2+b x^3+c x^4}}{8 a x^2}-\frac{\left (a x^2+b x^3+c x^4\right )^{3/2}}{3 x^6}-\frac{b \left (a x^2+b x^3+c x^4\right )^{3/2}}{4 a x^5}+\frac{\left (x \sqrt{a+b x+c x^2}\right ) \int \frac{-\frac{1}{2} b \left (b^2-12 a c\right )+8 a c^2 x}{x \sqrt{a+b x+c x^2}} \, dx}{8 a \sqrt{a x^2+b x^3+c x^4}}\\ &=\frac{\left (b^2-8 a c+2 b c x\right ) \sqrt{a x^2+b x^3+c x^4}}{8 a x^2}-\frac{\left (a x^2+b x^3+c x^4\right )^{3/2}}{3 x^6}-\frac{b \left (a x^2+b x^3+c x^4\right )^{3/2}}{4 a x^5}+\frac{\left (c^2 x \sqrt{a+b x+c x^2}\right ) \int \frac{1}{\sqrt{a+b x+c x^2}} \, dx}{\sqrt{a x^2+b x^3+c x^4}}-\frac{\left (b \left (b^2-12 a c\right ) x \sqrt{a+b x+c x^2}\right ) \int \frac{1}{x \sqrt{a+b x+c x^2}} \, dx}{16 a \sqrt{a x^2+b x^3+c x^4}}\\ &=\frac{\left (b^2-8 a c+2 b c x\right ) \sqrt{a x^2+b x^3+c x^4}}{8 a x^2}-\frac{\left (a x^2+b x^3+c x^4\right )^{3/2}}{3 x^6}-\frac{b \left (a x^2+b x^3+c x^4\right )^{3/2}}{4 a x^5}+\frac{\left (2 c^2 x \sqrt{a+b x+c x^2}\right ) \operatorname{Subst}\left (\int \frac{1}{4 c-x^2} \, dx,x,\frac{b+2 c x}{\sqrt{a+b x+c x^2}}\right )}{\sqrt{a x^2+b x^3+c x^4}}+\frac{\left (b \left (b^2-12 a c\right ) x \sqrt{a+b x+c x^2}\right ) \operatorname{Subst}\left (\int \frac{1}{4 a-x^2} \, dx,x,\frac{2 a+b x}{\sqrt{a+b x+c x^2}}\right )}{8 a \sqrt{a x^2+b x^3+c x^4}}\\ &=\frac{\left (b^2-8 a c+2 b c x\right ) \sqrt{a x^2+b x^3+c x^4}}{8 a x^2}-\frac{\left (a x^2+b x^3+c x^4\right )^{3/2}}{3 x^6}-\frac{b \left (a x^2+b x^3+c x^4\right )^{3/2}}{4 a x^5}+\frac{b \left (b^2-12 a c\right ) x \sqrt{a+b x+c x^2} \tanh ^{-1}\left (\frac{2 a+b x}{2 \sqrt{a} \sqrt{a+b x+c x^2}}\right )}{16 a^{3/2} \sqrt{a x^2+b x^3+c x^4}}+\frac{c^{3/2} x \sqrt{a+b x+c x^2} \tanh ^{-1}\left (\frac{b+2 c x}{2 \sqrt{c} \sqrt{a+b x+c x^2}}\right )}{\sqrt{a x^2+b x^3+c x^4}}\\ \end{align*}

Mathematica [A]  time = 0.306707, size = 175, normalized size = 0.68 \[ \frac{\sqrt{x^2 (a+x (b+c x))} \left (3 b x^3 \left (b^2-12 a c\right ) \tanh ^{-1}\left (\frac{2 a+b x}{2 \sqrt{a} \sqrt{a+x (b+c x)}}\right )-2 \sqrt{a} \left (\sqrt{a+x (b+c x)} \left (8 a^2+2 a x (7 b+16 c x)+3 b^2 x^2\right )-24 a c^{3/2} x^3 \tanh ^{-1}\left (\frac{b+2 c x}{2 \sqrt{c} \sqrt{a+x (b+c x)}}\right )\right )\right )}{48 a^{3/2} x^4 \sqrt{a+x (b+c x)}} \]

Antiderivative was successfully verified.

[In]

Integrate[(a*x^2 + b*x^3 + c*x^4)^(3/2)/x^7,x]

[Out]

(Sqrt[x^2*(a + x*(b + c*x))]*(3*b*(b^2 - 12*a*c)*x^3*ArcTanh[(2*a + b*x)/(2*Sqrt[a]*Sqrt[a + x*(b + c*x)])] -
2*Sqrt[a]*(Sqrt[a + x*(b + c*x)]*(8*a^2 + 3*b^2*x^2 + 2*a*x*(7*b + 16*c*x)) - 24*a*c^(3/2)*x^3*ArcTanh[(b + 2*
c*x)/(2*Sqrt[c]*Sqrt[a + x*(b + c*x)])])))/(48*a^(3/2)*x^4*Sqrt[a + x*(b + c*x)])

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Maple [A]  time = 0.007, size = 435, normalized size = 1.7 \begin{align*}{\frac{1}{48\,{x}^{6}{a}^{3}} \left ( c{x}^{4}+b{x}^{3}+a{x}^{2} \right ) ^{{\frac{3}{2}}} \left ( 32\,{c}^{7/2} \left ( c{x}^{2}+bx+a \right ) ^{3/2}{x}^{4}a-36\,{c}^{5/2}{a}^{5/2}\ln \left ({\frac{2\,a+bx+2\,\sqrt{a}\sqrt{c{x}^{2}+bx+a}}{x}} \right ){x}^{3}b+48\,{c}^{7/2}\sqrt{c{x}^{2}+bx+a}{x}^{4}{a}^{2}-2\,{c}^{5/2} \left ( c{x}^{2}+bx+a \right ) ^{3/2}{x}^{4}{b}^{2}-32\,{c}^{5/2} \left ( c{x}^{2}+bx+a \right ) ^{5/2}{x}^{2}a+28\,{c}^{5/2} \left ( c{x}^{2}+bx+a \right ) ^{3/2}{x}^{3}ab-6\,{c}^{5/2}\sqrt{c{x}^{2}+bx+a}{x}^{4}a{b}^{2}+3\,{c}^{3/2}{a}^{3/2}\ln \left ({\frac{2\,a+bx+2\,\sqrt{a}\sqrt{c{x}^{2}+bx+a}}{x}} \right ){x}^{3}{b}^{3}+60\,{c}^{5/2}\sqrt{c{x}^{2}+bx+a}{x}^{3}{a}^{2}b+2\,{c}^{3/2} \left ( c{x}^{2}+bx+a \right ) ^{5/2}{x}^{2}{b}^{2}-2\,{c}^{3/2} \left ( c{x}^{2}+bx+a \right ) ^{3/2}{x}^{3}{b}^{3}+4\,{c}^{3/2} \left ( c{x}^{2}+bx+a \right ) ^{5/2}xab-6\,{c}^{3/2}\sqrt{c{x}^{2}+bx+a}{x}^{3}a{b}^{3}-16\, \left ( c{x}^{2}+bx+a \right ) ^{5/2}{a}^{2}{c}^{3/2}+48\,\ln \left ( 1/2\,{\frac{2\,\sqrt{c{x}^{2}+bx+a}\sqrt{c}+2\,cx+b}{\sqrt{c}}} \right ){x}^{3}{a}^{3}{c}^{3} \right ) \left ( c{x}^{2}+bx+a \right ) ^{-{\frac{3}{2}}}{c}^{-{\frac{3}{2}}}} \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

int((c*x^4+b*x^3+a*x^2)^(3/2)/x^7,x)

[Out]

1/48*(c*x^4+b*x^3+a*x^2)^(3/2)*(32*c^(7/2)*(c*x^2+b*x+a)^(3/2)*x^4*a-36*c^(5/2)*a^(5/2)*ln((2*a+b*x+2*a^(1/2)*
(c*x^2+b*x+a)^(1/2))/x)*x^3*b+48*c^(7/2)*(c*x^2+b*x+a)^(1/2)*x^4*a^2-2*c^(5/2)*(c*x^2+b*x+a)^(3/2)*x^4*b^2-32*
c^(5/2)*(c*x^2+b*x+a)^(5/2)*x^2*a+28*c^(5/2)*(c*x^2+b*x+a)^(3/2)*x^3*a*b-6*c^(5/2)*(c*x^2+b*x+a)^(1/2)*x^4*a*b
^2+3*c^(3/2)*a^(3/2)*ln((2*a+b*x+2*a^(1/2)*(c*x^2+b*x+a)^(1/2))/x)*x^3*b^3+60*c^(5/2)*(c*x^2+b*x+a)^(1/2)*x^3*
a^2*b+2*c^(3/2)*(c*x^2+b*x+a)^(5/2)*x^2*b^2-2*c^(3/2)*(c*x^2+b*x+a)^(3/2)*x^3*b^3+4*c^(3/2)*(c*x^2+b*x+a)^(5/2
)*x*a*b-6*c^(3/2)*(c*x^2+b*x+a)^(1/2)*x^3*a*b^3-16*(c*x^2+b*x+a)^(5/2)*a^2*c^(3/2)+48*ln(1/2*(2*(c*x^2+b*x+a)^
(1/2)*c^(1/2)+2*c*x+b)/c^(1/2))*x^3*a^3*c^3)/x^6/(c*x^2+b*x+a)^(3/2)/a^3/c^(3/2)

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Maxima [F]  time = 0., size = 0, normalized size = 0. \begin{align*} \int \frac{{\left (c x^{4} + b x^{3} + a x^{2}\right )}^{\frac{3}{2}}}{x^{7}}\,{d x} \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate((c*x^4+b*x^3+a*x^2)^(3/2)/x^7,x, algorithm="maxima")

[Out]

integrate((c*x^4 + b*x^3 + a*x^2)^(3/2)/x^7, x)

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Fricas [A]  time = 2.49978, size = 1871, normalized size = 7.28 \begin{align*} \text{result too large to display} \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate((c*x^4+b*x^3+a*x^2)^(3/2)/x^7,x, algorithm="fricas")

[Out]

[1/96*(48*a^2*c^(3/2)*x^4*log(-(8*c^2*x^3 + 8*b*c*x^2 + 4*sqrt(c*x^4 + b*x^3 + a*x^2)*(2*c*x + b)*sqrt(c) + (b
^2 + 4*a*c)*x)/x) - 3*(b^3 - 12*a*b*c)*sqrt(a)*x^4*log(-(8*a*b*x^2 + (b^2 + 4*a*c)*x^3 + 8*a^2*x - 4*sqrt(c*x^
4 + b*x^3 + a*x^2)*(b*x + 2*a)*sqrt(a))/x^3) - 4*sqrt(c*x^4 + b*x^3 + a*x^2)*(14*a^2*b*x + 8*a^3 + (3*a*b^2 +
32*a^2*c)*x^2))/(a^2*x^4), -1/96*(96*a^2*sqrt(-c)*c*x^4*arctan(1/2*sqrt(c*x^4 + b*x^3 + a*x^2)*(2*c*x + b)*sqr
t(-c)/(c^2*x^3 + b*c*x^2 + a*c*x)) + 3*(b^3 - 12*a*b*c)*sqrt(a)*x^4*log(-(8*a*b*x^2 + (b^2 + 4*a*c)*x^3 + 8*a^
2*x - 4*sqrt(c*x^4 + b*x^3 + a*x^2)*(b*x + 2*a)*sqrt(a))/x^3) + 4*sqrt(c*x^4 + b*x^3 + a*x^2)*(14*a^2*b*x + 8*
a^3 + (3*a*b^2 + 32*a^2*c)*x^2))/(a^2*x^4), 1/48*(24*a^2*c^(3/2)*x^4*log(-(8*c^2*x^3 + 8*b*c*x^2 + 4*sqrt(c*x^
4 + b*x^3 + a*x^2)*(2*c*x + b)*sqrt(c) + (b^2 + 4*a*c)*x)/x) - 3*(b^3 - 12*a*b*c)*sqrt(-a)*x^4*arctan(1/2*sqrt
(c*x^4 + b*x^3 + a*x^2)*(b*x + 2*a)*sqrt(-a)/(a*c*x^3 + a*b*x^2 + a^2*x)) - 2*sqrt(c*x^4 + b*x^3 + a*x^2)*(14*
a^2*b*x + 8*a^3 + (3*a*b^2 + 32*a^2*c)*x^2))/(a^2*x^4), -1/48*(48*a^2*sqrt(-c)*c*x^4*arctan(1/2*sqrt(c*x^4 + b
*x^3 + a*x^2)*(2*c*x + b)*sqrt(-c)/(c^2*x^3 + b*c*x^2 + a*c*x)) + 3*(b^3 - 12*a*b*c)*sqrt(-a)*x^4*arctan(1/2*s
qrt(c*x^4 + b*x^3 + a*x^2)*(b*x + 2*a)*sqrt(-a)/(a*c*x^3 + a*b*x^2 + a^2*x)) + 2*sqrt(c*x^4 + b*x^3 + a*x^2)*(
14*a^2*b*x + 8*a^3 + (3*a*b^2 + 32*a^2*c)*x^2))/(a^2*x^4)]

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Sympy [F]  time = 0., size = 0, normalized size = 0. \begin{align*} \int \frac{\left (x^{2} \left (a + b x + c x^{2}\right )\right )^{\frac{3}{2}}}{x^{7}}\, dx \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate((c*x**4+b*x**3+a*x**2)**(3/2)/x**7,x)

[Out]

Integral((x**2*(a + b*x + c*x**2))**(3/2)/x**7, x)

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

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate((c*x^4+b*x^3+a*x^2)^(3/2)/x^7,x, algorithm="giac")

[Out]

Exception raised: TypeError