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

Optimal. Leaf size=223 \[ -\frac {3 b \left (b^2-4 a c\right )^2 \left (3 b^2-4 a c\right ) \tanh ^{-1}\left (\frac {b+2 c x^2}{2 \sqrt {c} \sqrt {a+b x^2+c x^4}}\right )}{4096 c^{11/2}}+\frac {3 b \left (b^2-4 a c\right ) \left (3 b^2-4 a c\right ) \left (b+2 c x^2\right ) \sqrt {a+b x^2+c x^4}}{2048 c^5}-\frac {b \left (3 b^2-4 a c\right ) \left (b+2 c x^2\right ) \left (a+b x^2+c x^4\right )^{3/2}}{256 c^4}+\frac {\left (-16 a c+21 b^2-30 b c x^2\right ) \left (a+b x^2+c x^4\right )^{5/2}}{560 c^3}+\frac {x^4 \left (a+b x^2+c x^4\right )^{5/2}}{14 c} \]

[Out]

-1/256*b*(-4*a*c+3*b^2)*(2*c*x^2+b)*(c*x^4+b*x^2+a)^(3/2)/c^4+1/14*x^4*(c*x^4+b*x^2+a)^(5/2)/c+1/560*(-30*b*c*
x^2-16*a*c+21*b^2)*(c*x^4+b*x^2+a)^(5/2)/c^3-3/4096*b*(-4*a*c+b^2)^2*(-4*a*c+3*b^2)*arctanh(1/2*(2*c*x^2+b)/c^
(1/2)/(c*x^4+b*x^2+a)^(1/2))/c^(11/2)+3/2048*b*(-4*a*c+b^2)*(-4*a*c+3*b^2)*(2*c*x^2+b)*(c*x^4+b*x^2+a)^(1/2)/c
^5

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Rubi [A]  time = 0.21, antiderivative size = 223, normalized size of antiderivative = 1.00, number of steps used = 7, number of rules used = 6, integrand size = 20, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.300, Rules used = {1114, 742, 779, 612, 621, 206} \[ \frac {\left (-16 a c+21 b^2-30 b c x^2\right ) \left (a+b x^2+c x^4\right )^{5/2}}{560 c^3}-\frac {b \left (3 b^2-4 a c\right ) \left (b+2 c x^2\right ) \left (a+b x^2+c x^4\right )^{3/2}}{256 c^4}+\frac {3 b \left (b^2-4 a c\right ) \left (3 b^2-4 a c\right ) \left (b+2 c x^2\right ) \sqrt {a+b x^2+c x^4}}{2048 c^5}-\frac {3 b \left (b^2-4 a c\right )^2 \left (3 b^2-4 a c\right ) \tanh ^{-1}\left (\frac {b+2 c x^2}{2 \sqrt {c} \sqrt {a+b x^2+c x^4}}\right )}{4096 c^{11/2}}+\frac {x^4 \left (a+b x^2+c x^4\right )^{5/2}}{14 c} \]

Antiderivative was successfully verified.

[In]

Int[x^7*(a + b*x^2 + c*x^4)^(3/2),x]

[Out]

(3*b*(b^2 - 4*a*c)*(3*b^2 - 4*a*c)*(b + 2*c*x^2)*Sqrt[a + b*x^2 + c*x^4])/(2048*c^5) - (b*(3*b^2 - 4*a*c)*(b +
2*c*x^2)*(a + b*x^2 + c*x^4)^(3/2))/(256*c^4) + (x^4*(a + b*x^2 + c*x^4)^(5/2))/(14*c) + ((21*b^2 - 16*a*c -
30*b*c*x^2)*(a + b*x^2 + c*x^4)^(5/2))/(560*c^3) - (3*b*(b^2 - 4*a*c)^2*(3*b^2 - 4*a*c)*ArcTanh[(b + 2*c*x^2)/
(2*Sqrt[c]*Sqrt[a + b*x^2 + c*x^4])])/(4096*c^(11/2))

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 612

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

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 742

Int[((d_.) + (e_.)*(x_))^(m_)*((a_.) + (b_.)*(x_) + (c_.)*(x_)^2)^(p_), x_Symbol] :> Simp[(e*(d + e*x)^(m - 1)
*(a + b*x + c*x^2)^(p + 1))/(c*(m + 2*p + 1)), x] + Dist[1/(c*(m + 2*p + 1)), Int[(d + e*x)^(m - 2)*Simp[c*d^2
*(m + 2*p + 1) - e*(a*e*(m - 1) + b*d*(p + 1)) + e*(2*c*d - b*e)*(m + p)*x, x]*(a + b*x + c*x^2)^p, x], x] /;
FreeQ[{a, b, c, d, e, m, p}, 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]
&& If[RationalQ[m], GtQ[m, 1], SumSimplerQ[m, -2]] && NeQ[m + 2*p + 1, 0] && IntQuadraticQ[a, b, c, d, e, m,
p, x]

Rule 779

Int[((d_.) + (e_.)*(x_))*((f_.) + (g_.)*(x_))*((a_.) + (b_.)*(x_) + (c_.)*(x_)^2)^(p_), x_Symbol] :> -Simp[((b
*e*g*(p + 2) - c*(e*f + d*g)*(2*p + 3) - 2*c*e*g*(p + 1)*x)*(a + b*x + c*x^2)^(p + 1))/(2*c^2*(p + 1)*(2*p + 3
)), x] + Dist[(b^2*e*g*(p + 2) - 2*a*c*e*g + c*(2*c*d*f - b*(e*f + d*g))*(2*p + 3))/(2*c^2*(2*p + 3)), Int[(a
+ b*x + c*x^2)^p, x], x] /; FreeQ[{a, b, c, d, e, f, g, p}, x] && NeQ[b^2 - 4*a*c, 0] &&  !LeQ[p, -1]

Rule 1114

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

Rubi steps

\begin {align*} \int x^7 \left (a+b x^2+c x^4\right )^{3/2} \, dx &=\frac {1}{2} \operatorname {Subst}\left (\int x^3 \left (a+b x+c x^2\right )^{3/2} \, dx,x,x^2\right )\\ &=\frac {x^4 \left (a+b x^2+c x^4\right )^{5/2}}{14 c}+\frac {\operatorname {Subst}\left (\int x \left (-2 a-\frac {9 b x}{2}\right ) \left (a+b x+c x^2\right )^{3/2} \, dx,x,x^2\right )}{14 c}\\ &=\frac {x^4 \left (a+b x^2+c x^4\right )^{5/2}}{14 c}+\frac {\left (21 b^2-16 a c-30 b c x^2\right ) \left (a+b x^2+c x^4\right )^{5/2}}{560 c^3}-\frac {\left (b \left (3 b^2-4 a c\right )\right ) \operatorname {Subst}\left (\int \left (a+b x+c x^2\right )^{3/2} \, dx,x,x^2\right )}{32 c^3}\\ &=-\frac {b \left (3 b^2-4 a c\right ) \left (b+2 c x^2\right ) \left (a+b x^2+c x^4\right )^{3/2}}{256 c^4}+\frac {x^4 \left (a+b x^2+c x^4\right )^{5/2}}{14 c}+\frac {\left (21 b^2-16 a c-30 b c x^2\right ) \left (a+b x^2+c x^4\right )^{5/2}}{560 c^3}+\frac {\left (3 b \left (b^2-4 a c\right ) \left (3 b^2-4 a c\right )\right ) \operatorname {Subst}\left (\int \sqrt {a+b x+c x^2} \, dx,x,x^2\right )}{512 c^4}\\ &=\frac {3 b \left (b^2-4 a c\right ) \left (3 b^2-4 a c\right ) \left (b+2 c x^2\right ) \sqrt {a+b x^2+c x^4}}{2048 c^5}-\frac {b \left (3 b^2-4 a c\right ) \left (b+2 c x^2\right ) \left (a+b x^2+c x^4\right )^{3/2}}{256 c^4}+\frac {x^4 \left (a+b x^2+c x^4\right )^{5/2}}{14 c}+\frac {\left (21 b^2-16 a c-30 b c x^2\right ) \left (a+b x^2+c x^4\right )^{5/2}}{560 c^3}-\frac {\left (3 b \left (b^2-4 a c\right )^2 \left (3 b^2-4 a c\right )\right ) \operatorname {Subst}\left (\int \frac {1}{\sqrt {a+b x+c x^2}} \, dx,x,x^2\right )}{4096 c^5}\\ &=\frac {3 b \left (b^2-4 a c\right ) \left (3 b^2-4 a c\right ) \left (b+2 c x^2\right ) \sqrt {a+b x^2+c x^4}}{2048 c^5}-\frac {b \left (3 b^2-4 a c\right ) \left (b+2 c x^2\right ) \left (a+b x^2+c x^4\right )^{3/2}}{256 c^4}+\frac {x^4 \left (a+b x^2+c x^4\right )^{5/2}}{14 c}+\frac {\left (21 b^2-16 a c-30 b c x^2\right ) \left (a+b x^2+c x^4\right )^{5/2}}{560 c^3}-\frac {\left (3 b \left (b^2-4 a c\right )^2 \left (3 b^2-4 a c\right )\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 )}{2048 c^5}\\ &=\frac {3 b \left (b^2-4 a c\right ) \left (3 b^2-4 a c\right ) \left (b+2 c x^2\right ) \sqrt {a+b x^2+c x^4}}{2048 c^5}-\frac {b \left (3 b^2-4 a c\right ) \left (b+2 c x^2\right ) \left (a+b x^2+c x^4\right )^{3/2}}{256 c^4}+\frac {x^4 \left (a+b x^2+c x^4\right )^{5/2}}{14 c}+\frac {\left (21 b^2-16 a c-30 b c x^2\right ) \left (a+b x^2+c x^4\right )^{5/2}}{560 c^3}-\frac {3 b \left (b^2-4 a c\right )^2 \left (3 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 )}{4096 c^{11/2}}\\ \end {align*}

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Mathematica [A]  time = 0.25, size = 192, normalized size = 0.86 \[ \frac {-\frac {\left (16 a c-21 b^2+30 b c x^2\right ) \left (a+b x^2+c x^4\right )^{5/2}}{40 c^2}+\frac {7 \left (4 a b c-3 b^3\right ) \left (2 \sqrt {c} \left (b+2 c x^2\right ) \sqrt {a+b x^2+c x^4} \left (4 c \left (5 a+2 c x^4\right )-3 b^2+8 b c x^2\right )+3 \left (b^2-4 a c\right )^2 \tanh ^{-1}\left (\frac {b+2 c x^2}{2 \sqrt {c} \sqrt {a+b x^2+c x^4}}\right )\right )}{2048 c^{9/2}}+x^4 \left (a+b x^2+c x^4\right )^{5/2}}{14 c} \]

Antiderivative was successfully verified.

[In]

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

[Out]

(x^4*(a + b*x^2 + c*x^4)^(5/2) - ((-21*b^2 + 16*a*c + 30*b*c*x^2)*(a + b*x^2 + c*x^4)^(5/2))/(40*c^2) + (7*(-3
*b^3 + 4*a*b*c)*(2*Sqrt[c]*(b + 2*c*x^2)*Sqrt[a + b*x^2 + c*x^4]*(-3*b^2 + 8*b*c*x^2 + 4*c*(5*a + 2*c*x^4)) +
3*(b^2 - 4*a*c)^2*ArcTanh[(b + 2*c*x^2)/(2*Sqrt[c]*Sqrt[a + b*x^2 + c*x^4])]))/(2048*c^(9/2)))/(14*c)

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fricas [A]  time = 0.93, size = 535, normalized size = 2.40 \[ \left [-\frac {105 \, {\left (3 \, b^{7} - 28 \, a b^{5} c + 80 \, a^{2} b^{3} c^{2} - 64 \, a^{3} b c^{3}\right )} \sqrt {c} \log \left (-8 \, c^{2} x^{4} - 8 \, b c x^{2} - b^{2} - 4 \, \sqrt {c x^{4} + b x^{2} + a} {\left (2 \, c x^{2} + b\right )} \sqrt {c} - 4 \, a c\right ) - 4 \, {\left (5120 \, c^{7} x^{12} + 6400 \, b c^{6} x^{10} + 128 \, {\left (b^{2} c^{5} + 64 \, a c^{6}\right )} x^{8} + 315 \, b^{6} c - 2520 \, a b^{4} c^{2} + 5488 \, a^{2} b^{2} c^{3} - 2048 \, a^{3} c^{4} - 16 \, {\left (9 \, b^{3} c^{4} - 44 \, a b c^{5}\right )} x^{6} + 8 \, {\left (21 \, b^{4} c^{3} - 124 \, a b^{2} c^{4} + 128 \, a^{2} c^{5}\right )} x^{4} - 2 \, {\left (105 \, b^{5} c^{2} - 728 \, a b^{3} c^{3} + 1168 \, a^{2} b c^{4}\right )} x^{2}\right )} \sqrt {c x^{4} + b x^{2} + a}}{286720 \, c^{6}}, \frac {105 \, {\left (3 \, b^{7} - 28 \, a b^{5} c + 80 \, a^{2} b^{3} c^{2} - 64 \, a^{3} b c^{3}\right )} \sqrt {-c} \arctan \left (\frac {\sqrt {c x^{4} + b x^{2} + a} {\left (2 \, c x^{2} + b\right )} \sqrt {-c}}{2 \, {\left (c^{2} x^{4} + b c x^{2} + a c\right )}}\right ) + 2 \, {\left (5120 \, c^{7} x^{12} + 6400 \, b c^{6} x^{10} + 128 \, {\left (b^{2} c^{5} + 64 \, a c^{6}\right )} x^{8} + 315 \, b^{6} c - 2520 \, a b^{4} c^{2} + 5488 \, a^{2} b^{2} c^{3} - 2048 \, a^{3} c^{4} - 16 \, {\left (9 \, b^{3} c^{4} - 44 \, a b c^{5}\right )} x^{6} + 8 \, {\left (21 \, b^{4} c^{3} - 124 \, a b^{2} c^{4} + 128 \, a^{2} c^{5}\right )} x^{4} - 2 \, {\left (105 \, b^{5} c^{2} - 728 \, a b^{3} c^{3} + 1168 \, a^{2} b c^{4}\right )} x^{2}\right )} \sqrt {c x^{4} + b x^{2} + a}}{143360 \, c^{6}}\right ] \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

[-1/286720*(105*(3*b^7 - 28*a*b^5*c + 80*a^2*b^3*c^2 - 64*a^3*b*c^3)*sqrt(c)*log(-8*c^2*x^4 - 8*b*c*x^2 - b^2
- 4*sqrt(c*x^4 + b*x^2 + a)*(2*c*x^2 + b)*sqrt(c) - 4*a*c) - 4*(5120*c^7*x^12 + 6400*b*c^6*x^10 + 128*(b^2*c^5
+ 64*a*c^6)*x^8 + 315*b^6*c - 2520*a*b^4*c^2 + 5488*a^2*b^2*c^3 - 2048*a^3*c^4 - 16*(9*b^3*c^4 - 44*a*b*c^5)*
x^6 + 8*(21*b^4*c^3 - 124*a*b^2*c^4 + 128*a^2*c^5)*x^4 - 2*(105*b^5*c^2 - 728*a*b^3*c^3 + 1168*a^2*b*c^4)*x^2)
*sqrt(c*x^4 + b*x^2 + a))/c^6, 1/143360*(105*(3*b^7 - 28*a*b^5*c + 80*a^2*b^3*c^2 - 64*a^3*b*c^3)*sqrt(-c)*arc
tan(1/2*sqrt(c*x^4 + b*x^2 + a)*(2*c*x^2 + b)*sqrt(-c)/(c^2*x^4 + b*c*x^2 + a*c)) + 2*(5120*c^7*x^12 + 6400*b*
c^6*x^10 + 128*(b^2*c^5 + 64*a*c^6)*x^8 + 315*b^6*c - 2520*a*b^4*c^2 + 5488*a^2*b^2*c^3 - 2048*a^3*c^4 - 16*(9
*b^3*c^4 - 44*a*b*c^5)*x^6 + 8*(21*b^4*c^3 - 124*a*b^2*c^4 + 128*a^2*c^5)*x^4 - 2*(105*b^5*c^2 - 728*a*b^3*c^3
+ 1168*a^2*b*c^4)*x^2)*sqrt(c*x^4 + b*x^2 + a))/c^6]

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giac [B]  time = 0.40, size = 669, normalized size = 3.00 \[ \frac {1}{7680} \, {\left (2 \, \sqrt {c x^{4} + b x^{2} + a} {\left (2 \, {\left (4 \, {\left (6 \, {\left (8 \, x^{2} + \frac {b}{c}\right )} x^{2} - \frac {7 \, b^{2} c^{2} - 16 \, a c^{3}}{c^{4}}\right )} x^{2} + \frac {35 \, b^{3} c - 116 \, a b c^{2}}{c^{4}}\right )} x^{2} - \frac {105 \, b^{4} - 460 \, a b^{2} c + 256 \, a^{2} c^{2}}{c^{4}}\right )} - \frac {15 \, {\left (7 \, b^{5} - 40 \, a b^{3} c + 48 \, a^{2} b c^{2}\right )} \log \left ({\left | -2 \, {\left (\sqrt {c} x^{2} - \sqrt {c x^{4} + b x^{2} + a}\right )} \sqrt {c} - b \right |}\right )}{c^{\frac {9}{2}}}\right )} a + \frac {1}{30720} \, {\left (2 \, \sqrt {c x^{4} + b x^{2} + a} {\left (2 \, {\left (4 \, {\left (2 \, {\left (8 \, {\left (10 \, x^{2} + \frac {b}{c}\right )} x^{2} - \frac {9 \, b^{2} c^{3} - 20 \, a c^{4}}{c^{5}}\right )} x^{2} + \frac {21 \, b^{3} c^{2} - 68 \, a b c^{3}}{c^{5}}\right )} x^{2} - \frac {105 \, b^{4} c - 448 \, a b^{2} c^{2} + 240 \, a^{2} c^{3}}{c^{5}}\right )} x^{2} + \frac {315 \, b^{5} - 1680 \, a b^{3} c + 1808 \, a^{2} b c^{2}}{c^{5}}\right )} + \frac {15 \, {\left (21 \, b^{6} - 140 \, a b^{4} c + 240 \, a^{2} b^{2} c^{2} - 64 \, a^{3} c^{3}\right )} \log \left ({\left | -2 \, {\left (\sqrt {c} x^{2} - \sqrt {c x^{4} + b x^{2} + a}\right )} \sqrt {c} - b \right |}\right )}{c^{\frac {11}{2}}}\right )} b + \frac {1}{430080} \, {\left (2 \, \sqrt {c x^{4} + b x^{2} + a} {\left (2 \, {\left (4 \, {\left (2 \, {\left (8 \, {\left (10 \, {\left (12 \, x^{2} + \frac {b}{c}\right )} x^{2} - \frac {11 \, b^{2} c^{4} - 24 \, a c^{5}}{c^{6}}\right )} x^{2} + \frac {99 \, b^{3} c^{3} - 316 \, a b c^{4}}{c^{6}}\right )} x^{2} - \frac {231 \, b^{4} c^{2} - 972 \, a b^{2} c^{3} + 512 \, a^{2} c^{4}}{c^{6}}\right )} x^{2} + \frac {1155 \, b^{5} c - 6048 \, a b^{3} c^{2} + 6352 \, a^{2} b c^{3}}{c^{6}}\right )} x^{2} - \frac {3465 \, b^{6} - 21840 \, a b^{4} c + 34608 \, a^{2} b^{2} c^{2} - 8192 \, a^{3} c^{3}}{c^{6}}\right )} - \frac {105 \, {\left (33 \, b^{7} - 252 \, a b^{5} c + 560 \, a^{2} b^{3} c^{2} - 320 \, a^{3} b c^{3}\right )} \log \left ({\left | -2 \, {\left (\sqrt {c} x^{2} - \sqrt {c x^{4} + b x^{2} + a}\right )} \sqrt {c} - b \right |}\right )}{c^{\frac {13}{2}}}\right )} c \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

1/7680*(2*sqrt(c*x^4 + b*x^2 + a)*(2*(4*(6*(8*x^2 + b/c)*x^2 - (7*b^2*c^2 - 16*a*c^3)/c^4)*x^2 + (35*b^3*c - 1
16*a*b*c^2)/c^4)*x^2 - (105*b^4 - 460*a*b^2*c + 256*a^2*c^2)/c^4) - 15*(7*b^5 - 40*a*b^3*c + 48*a^2*b*c^2)*log
(abs(-2*(sqrt(c)*x^2 - sqrt(c*x^4 + b*x^2 + a))*sqrt(c) - b))/c^(9/2))*a + 1/30720*(2*sqrt(c*x^4 + b*x^2 + a)*
(2*(4*(2*(8*(10*x^2 + b/c)*x^2 - (9*b^2*c^3 - 20*a*c^4)/c^5)*x^2 + (21*b^3*c^2 - 68*a*b*c^3)/c^5)*x^2 - (105*b
^4*c - 448*a*b^2*c^2 + 240*a^2*c^3)/c^5)*x^2 + (315*b^5 - 1680*a*b^3*c + 1808*a^2*b*c^2)/c^5) + 15*(21*b^6 - 1
40*a*b^4*c + 240*a^2*b^2*c^2 - 64*a^3*c^3)*log(abs(-2*(sqrt(c)*x^2 - sqrt(c*x^4 + b*x^2 + a))*sqrt(c) - b))/c^
(11/2))*b + 1/430080*(2*sqrt(c*x^4 + b*x^2 + a)*(2*(4*(2*(8*(10*(12*x^2 + b/c)*x^2 - (11*b^2*c^4 - 24*a*c^5)/c
^6)*x^2 + (99*b^3*c^3 - 316*a*b*c^4)/c^6)*x^2 - (231*b^4*c^2 - 972*a*b^2*c^3 + 512*a^2*c^4)/c^6)*x^2 + (1155*b
^5*c - 6048*a*b^3*c^2 + 6352*a^2*b*c^3)/c^6)*x^2 - (3465*b^6 - 21840*a*b^4*c + 34608*a^2*b^2*c^2 - 8192*a^3*c^
3)/c^6) - 105*(33*b^7 - 252*a*b^5*c + 560*a^2*b^3*c^2 - 320*a^3*b*c^3)*log(abs(-2*(sqrt(c)*x^2 - sqrt(c*x^4 +
b*x^2 + a))*sqrt(c) - b))/c^(13/2))*c

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maple [B]  time = 0.04, size = 534, normalized size = 2.39 \[ \frac {\sqrt {c \,x^{4}+b \,x^{2}+a}\, c \,x^{12}}{14}+\frac {5 \sqrt {c \,x^{4}+b \,x^{2}+a}\, b \,x^{10}}{56}+\frac {4 \sqrt {c \,x^{4}+b \,x^{2}+a}\, a \,x^{8}}{35}+\frac {\sqrt {c \,x^{4}+b \,x^{2}+a}\, b^{2} x^{8}}{560 c}+\frac {11 \sqrt {c \,x^{4}+b \,x^{2}+a}\, a b \,x^{6}}{1120 c}-\frac {9 \sqrt {c \,x^{4}+b \,x^{2}+a}\, b^{3} x^{6}}{4480 c^{2}}+\frac {\sqrt {c \,x^{4}+b \,x^{2}+a}\, a^{2} x^{4}}{70 c}-\frac {31 \sqrt {c \,x^{4}+b \,x^{2}+a}\, a \,b^{2} x^{4}}{2240 c^{2}}+\frac {3 \sqrt {c \,x^{4}+b \,x^{2}+a}\, b^{4} x^{4}}{1280 c^{3}}-\frac {73 \sqrt {c \,x^{4}+b \,x^{2}+a}\, a^{2} b \,x^{2}}{2240 c^{2}}+\frac {13 \sqrt {c \,x^{4}+b \,x^{2}+a}\, a \,b^{3} x^{2}}{640 c^{3}}-\frac {3 \sqrt {c \,x^{4}+b \,x^{2}+a}\, b^{5} x^{2}}{1024 c^{4}}+\frac {3 a^{3} b \ln \left (\frac {c \,x^{2}+\frac {b}{2}}{\sqrt {c}}+\sqrt {c \,x^{4}+b \,x^{2}+a}\right )}{64 c^{\frac {5}{2}}}-\frac {15 a^{2} b^{3} \ln \left (\frac {c \,x^{2}+\frac {b}{2}}{\sqrt {c}}+\sqrt {c \,x^{4}+b \,x^{2}+a}\right )}{256 c^{\frac {7}{2}}}+\frac {21 a \,b^{5} \ln \left (\frac {c \,x^{2}+\frac {b}{2}}{\sqrt {c}}+\sqrt {c \,x^{4}+b \,x^{2}+a}\right )}{1024 c^{\frac {9}{2}}}-\frac {9 b^{7} \ln \left (\frac {c \,x^{2}+\frac {b}{2}}{\sqrt {c}}+\sqrt {c \,x^{4}+b \,x^{2}+a}\right )}{4096 c^{\frac {11}{2}}}-\frac {\sqrt {c \,x^{4}+b \,x^{2}+a}\, a^{3}}{35 c^{2}}+\frac {49 \sqrt {c \,x^{4}+b \,x^{2}+a}\, a^{2} b^{2}}{640 c^{3}}-\frac {9 \sqrt {c \,x^{4}+b \,x^{2}+a}\, a \,b^{4}}{256 c^{4}}+\frac {9 \sqrt {c \,x^{4}+b \,x^{2}+a}\, b^{6}}{2048 c^{5}} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

1/560*b^2*x^8/c*(c*x^4+b*x^2+a)^(1/2)-9/4480*b^3/c^2*x^6*(c*x^4+b*x^2+a)^(1/2)+3/1280*b^4/c^3*x^4*(c*x^4+b*x^2
+a)^(1/2)-3/1024*b^5/c^4*x^2*(c*x^4+b*x^2+a)^(1/2)+1/70*a^2*x^4/c*(c*x^4+b*x^2+a)^(1/2)+49/640*a^2*b^2/c^3*(c*
x^4+b*x^2+a)^(1/2)+3/64*a^3*b/c^(5/2)*ln((c*x^2+1/2*b)/c^(1/2)+(c*x^4+b*x^2+a)^(1/2))-9/256*a*b^4/c^4*(c*x^4+b
*x^2+a)^(1/2)+21/1024*a*b^5/c^(9/2)*ln((c*x^2+1/2*b)/c^(1/2)+(c*x^4+b*x^2+a)^(1/2))-15/256*a^2*b^3/c^(7/2)*ln(
(c*x^2+1/2*b)/c^(1/2)+(c*x^4+b*x^2+a)^(1/2))+4/35*a*x^8*(c*x^4+b*x^2+a)^(1/2)+9/2048*b^6/c^5*(c*x^4+b*x^2+a)^(
1/2)+1/14*c*x^12*(c*x^4+b*x^2+a)^(1/2)+11/1120*a*b*x^6/c*(c*x^4+b*x^2+a)^(1/2)-31/2240*a*b^2/c^2*x^4*(c*x^4+b*
x^2+a)^(1/2)+13/640*a*b^3/c^3*x^2*(c*x^4+b*x^2+a)^(1/2)-73/2240*a^2*b/c^2*x^2*(c*x^4+b*x^2+a)^(1/2)+5/56*b*x^1
0*(c*x^4+b*x^2+a)^(1/2)-9/4096*b^7/c^(11/2)*ln((c*x^2+1/2*b)/c^(1/2)+(c*x^4+b*x^2+a)^(1/2))-1/35*a^3/c^2*(c*x^
4+b*x^2+a)^(1/2)

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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(x^7*(c*x^4+b*x^2+a)^(3/2),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(4*a*c-b^2>0)', see `assume?` f
or more details)Is 4*a*c-b^2 positive, negative or zero?

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mupad [F]  time = 0.00, size = -1, normalized size = -0.00 \[ \int x^7\,{\left (c\,x^4+b\,x^2+a\right )}^{3/2} \,d x \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

int(x^7*(a + b*x^2 + c*x^4)^(3/2),x)

[Out]

int(x^7*(a + b*x^2 + c*x^4)^(3/2), x)

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sympy [F]  time = 0.00, size = 0, normalized size = 0.00 \[ \int x^{7} \left (a + b x^{2} + c x^{4}\right )^{\frac {3}{2}}\, dx \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

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

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