3.4.38 \(\int \frac {(A+B x) (a+c x^2)^{3/2}}{x^8} \, dx\)

Optimal. Leaf size=147 \[ \frac {B c^3 \tanh ^{-1}\left (\frac {\sqrt {a+c x^2}}{\sqrt {a}}\right )}{16 a^{3/2}}+\frac {2 A c \left (a+c x^2\right )^{5/2}}{35 a^2 x^5}-\frac {A \left (a+c x^2\right )^{5/2}}{7 a x^7}+\frac {B c^2 \sqrt {a+c x^2}}{16 a x^2}-\frac {B \left (a+c x^2\right )^{5/2}}{6 a x^6}+\frac {B c \left (a+c x^2\right )^{3/2}}{24 a x^4} \]

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Rubi [A]  time = 0.11, antiderivative size = 147, normalized size of antiderivative = 1.00, number of steps used = 8, number of rules used = 6, integrand size = 20, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.300, Rules used = {835, 807, 266, 47, 63, 208} \begin {gather*} \frac {2 A c \left (a+c x^2\right )^{5/2}}{35 a^2 x^5}+\frac {B c^3 \tanh ^{-1}\left (\frac {\sqrt {a+c x^2}}{\sqrt {a}}\right )}{16 a^{3/2}}-\frac {A \left (a+c x^2\right )^{5/2}}{7 a x^7}+\frac {B c^2 \sqrt {a+c x^2}}{16 a x^2}+\frac {B c \left (a+c x^2\right )^{3/2}}{24 a x^4}-\frac {B \left (a+c x^2\right )^{5/2}}{6 a x^6} \end {gather*}

Antiderivative was successfully verified.

[In]

Int[((A + B*x)*(a + c*x^2)^(3/2))/x^8,x]

[Out]

(B*c^2*Sqrt[a + c*x^2])/(16*a*x^2) + (B*c*(a + c*x^2)^(3/2))/(24*a*x^4) - (A*(a + c*x^2)^(5/2))/(7*a*x^7) - (B
*(a + c*x^2)^(5/2))/(6*a*x^6) + (2*A*c*(a + c*x^2)^(5/2))/(35*a^2*x^5) + (B*c^3*ArcTanh[Sqrt[a + c*x^2]/Sqrt[a
]])/(16*a^(3/2))

Rule 47

Int[((a_.) + (b_.)*(x_))^(m_)*((c_.) + (d_.)*(x_))^(n_), x_Symbol] :> Simp[((a + b*x)^(m + 1)*(c + d*x)^n)/(b*
(m + 1)), x] - Dist[(d*n)/(b*(m + 1)), Int[(a + b*x)^(m + 1)*(c + d*x)^(n - 1), x], x] /; FreeQ[{a, b, c, d},
x] && NeQ[b*c - a*d, 0] && GtQ[n, 0] && LtQ[m, -1] &&  !(IntegerQ[n] &&  !IntegerQ[m]) &&  !(ILeQ[m + n + 2, 0
] && (FractionQ[m] || GeQ[2*n + m + 1, 0])) && IntLinearQ[a, b, c, d, m, n, x]

Rule 63

Int[((a_.) + (b_.)*(x_))^(m_)*((c_.) + (d_.)*(x_))^(n_), x_Symbol] :> With[{p = Denominator[m]}, Dist[p/b, Sub
st[Int[x^(p*(m + 1) - 1)*(c - (a*d)/b + (d*x^p)/b)^n, x], x, (a + b*x)^(1/p)], x]] /; FreeQ[{a, b, c, d}, x] &
& NeQ[b*c - a*d, 0] && LtQ[-1, m, 0] && LeQ[-1, n, 0] && LeQ[Denominator[n], Denominator[m]] && IntLinearQ[a,
b, c, d, m, n, x]

Rule 208

Int[((a_) + (b_.)*(x_)^2)^(-1), x_Symbol] :> Simp[(Rt[-(a/b), 2]*ArcTanh[x/Rt[-(a/b), 2]])/a, x] /; FreeQ[{a,
b}, x] && NegQ[a/b]

Rule 266

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

Rule 807

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

Rule 835

Int[((d_.) + (e_.)*(x_))^(m_)*((f_.) + (g_.)*(x_))*((a_) + (c_.)*(x_)^2)^(p_.), x_Symbol] :> Simp[((e*f - d*g)
*(d + e*x)^(m + 1)*(a + c*x^2)^(p + 1))/((m + 1)*(c*d^2 + a*e^2)), x] + Dist[1/((m + 1)*(c*d^2 + a*e^2)), Int[
(d + e*x)^(m + 1)*(a + c*x^2)^p*Simp[(c*d*f + a*e*g)*(m + 1) - c*(e*f - d*g)*(m + 2*p + 3)*x, x], x], x] /; Fr
eeQ[{a, c, d, e, f, g, p}, x] && NeQ[c*d^2 + a*e^2, 0] && LtQ[m, -1] && (IntegerQ[m] || IntegerQ[p] || Integer
sQ[2*m, 2*p])

Rubi steps

\begin {align*} \int \frac {(A+B x) \left (a+c x^2\right )^{3/2}}{x^8} \, dx &=-\frac {A \left (a+c x^2\right )^{5/2}}{7 a x^7}-\frac {\int \frac {(-7 a B+2 A c x) \left (a+c x^2\right )^{3/2}}{x^7} \, dx}{7 a}\\ &=-\frac {A \left (a+c x^2\right )^{5/2}}{7 a x^7}-\frac {B \left (a+c x^2\right )^{5/2}}{6 a x^6}+\frac {\int \frac {(-12 a A c-7 a B c x) \left (a+c x^2\right )^{3/2}}{x^6} \, dx}{42 a^2}\\ &=-\frac {A \left (a+c x^2\right )^{5/2}}{7 a x^7}-\frac {B \left (a+c x^2\right )^{5/2}}{6 a x^6}+\frac {2 A c \left (a+c x^2\right )^{5/2}}{35 a^2 x^5}-\frac {(B c) \int \frac {\left (a+c x^2\right )^{3/2}}{x^5} \, dx}{6 a}\\ &=-\frac {A \left (a+c x^2\right )^{5/2}}{7 a x^7}-\frac {B \left (a+c x^2\right )^{5/2}}{6 a x^6}+\frac {2 A c \left (a+c x^2\right )^{5/2}}{35 a^2 x^5}-\frac {(B c) \operatorname {Subst}\left (\int \frac {(a+c x)^{3/2}}{x^3} \, dx,x,x^2\right )}{12 a}\\ &=\frac {B c \left (a+c x^2\right )^{3/2}}{24 a x^4}-\frac {A \left (a+c x^2\right )^{5/2}}{7 a x^7}-\frac {B \left (a+c x^2\right )^{5/2}}{6 a x^6}+\frac {2 A c \left (a+c x^2\right )^{5/2}}{35 a^2 x^5}-\frac {\left (B c^2\right ) \operatorname {Subst}\left (\int \frac {\sqrt {a+c x}}{x^2} \, dx,x,x^2\right )}{16 a}\\ &=\frac {B c^2 \sqrt {a+c x^2}}{16 a x^2}+\frac {B c \left (a+c x^2\right )^{3/2}}{24 a x^4}-\frac {A \left (a+c x^2\right )^{5/2}}{7 a x^7}-\frac {B \left (a+c x^2\right )^{5/2}}{6 a x^6}+\frac {2 A c \left (a+c x^2\right )^{5/2}}{35 a^2 x^5}-\frac {\left (B c^3\right ) \operatorname {Subst}\left (\int \frac {1}{x \sqrt {a+c x}} \, dx,x,x^2\right )}{32 a}\\ &=\frac {B c^2 \sqrt {a+c x^2}}{16 a x^2}+\frac {B c \left (a+c x^2\right )^{3/2}}{24 a x^4}-\frac {A \left (a+c x^2\right )^{5/2}}{7 a x^7}-\frac {B \left (a+c x^2\right )^{5/2}}{6 a x^6}+\frac {2 A c \left (a+c x^2\right )^{5/2}}{35 a^2 x^5}-\frac {\left (B c^2\right ) \operatorname {Subst}\left (\int \frac {1}{-\frac {a}{c}+\frac {x^2}{c}} \, dx,x,\sqrt {a+c x^2}\right )}{16 a}\\ &=\frac {B c^2 \sqrt {a+c x^2}}{16 a x^2}+\frac {B c \left (a+c x^2\right )^{3/2}}{24 a x^4}-\frac {A \left (a+c x^2\right )^{5/2}}{7 a x^7}-\frac {B \left (a+c x^2\right )^{5/2}}{6 a x^6}+\frac {2 A c \left (a+c x^2\right )^{5/2}}{35 a^2 x^5}+\frac {B c^3 \tanh ^{-1}\left (\frac {\sqrt {a+c x^2}}{\sqrt {a}}\right )}{16 a^{3/2}}\\ \end {align*}

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Mathematica [C]  time = 0.03, size = 64, normalized size = 0.44 \begin {gather*} \frac {\left (a+c x^2\right )^{5/2} \left (a^2 A \left (2 c x^2-5 a\right )+7 B c^3 x^7 \, _2F_1\left (\frac {5}{2},4;\frac {7}{2};\frac {c x^2}{a}+1\right )\right )}{35 a^4 x^7} \end {gather*}

Antiderivative was successfully verified.

[In]

Integrate[((A + B*x)*(a + c*x^2)^(3/2))/x^8,x]

[Out]

((a + c*x^2)^(5/2)*(a^2*A*(-5*a + 2*c*x^2) + 7*B*c^3*x^7*Hypergeometric2F1[5/2, 4, 7/2, 1 + (c*x^2)/a]))/(35*a
^4*x^7)

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IntegrateAlgebraic [A]  time = 0.84, size = 130, normalized size = 0.88 \begin {gather*} \frac {\sqrt {a+c x^2} \left (-240 a^3 A-280 a^3 B x-384 a^2 A c x^2-490 a^2 B c x^3-48 a A c^2 x^4-105 a B c^2 x^5+96 A c^3 x^6\right )}{1680 a^2 x^7}-\frac {B c^3 \tanh ^{-1}\left (\frac {\sqrt {c} x}{\sqrt {a}}-\frac {\sqrt {a+c x^2}}{\sqrt {a}}\right )}{8 a^{3/2}} \end {gather*}

Antiderivative was successfully verified.

[In]

IntegrateAlgebraic[((A + B*x)*(a + c*x^2)^(3/2))/x^8,x]

[Out]

(Sqrt[a + c*x^2]*(-240*a^3*A - 280*a^3*B*x - 384*a^2*A*c*x^2 - 490*a^2*B*c*x^3 - 48*a*A*c^2*x^4 - 105*a*B*c^2*
x^5 + 96*A*c^3*x^6))/(1680*a^2*x^7) - (B*c^3*ArcTanh[(Sqrt[c]*x)/Sqrt[a] - Sqrt[a + c*x^2]/Sqrt[a]])/(8*a^(3/2
))

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fricas [A]  time = 0.48, size = 238, normalized size = 1.62 \begin {gather*} \left [\frac {105 \, B \sqrt {a} c^{3} x^{7} \log \left (-\frac {c x^{2} + 2 \, \sqrt {c x^{2} + a} \sqrt {a} + 2 \, a}{x^{2}}\right ) + 2 \, {\left (96 \, A c^{3} x^{6} - 105 \, B a c^{2} x^{5} - 48 \, A a c^{2} x^{4} - 490 \, B a^{2} c x^{3} - 384 \, A a^{2} c x^{2} - 280 \, B a^{3} x - 240 \, A a^{3}\right )} \sqrt {c x^{2} + a}}{3360 \, a^{2} x^{7}}, -\frac {105 \, B \sqrt {-a} c^{3} x^{7} \arctan \left (\frac {\sqrt {-a}}{\sqrt {c x^{2} + a}}\right ) - {\left (96 \, A c^{3} x^{6} - 105 \, B a c^{2} x^{5} - 48 \, A a c^{2} x^{4} - 490 \, B a^{2} c x^{3} - 384 \, A a^{2} c x^{2} - 280 \, B a^{3} x - 240 \, A a^{3}\right )} \sqrt {c x^{2} + a}}{1680 \, a^{2} x^{7}}\right ] \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate((B*x+A)*(c*x^2+a)^(3/2)/x^8,x, algorithm="fricas")

[Out]

[1/3360*(105*B*sqrt(a)*c^3*x^7*log(-(c*x^2 + 2*sqrt(c*x^2 + a)*sqrt(a) + 2*a)/x^2) + 2*(96*A*c^3*x^6 - 105*B*a
*c^2*x^5 - 48*A*a*c^2*x^4 - 490*B*a^2*c*x^3 - 384*A*a^2*c*x^2 - 280*B*a^3*x - 240*A*a^3)*sqrt(c*x^2 + a))/(a^2
*x^7), -1/1680*(105*B*sqrt(-a)*c^3*x^7*arctan(sqrt(-a)/sqrt(c*x^2 + a)) - (96*A*c^3*x^6 - 105*B*a*c^2*x^5 - 48
*A*a*c^2*x^4 - 490*B*a^2*c*x^3 - 384*A*a^2*c*x^2 - 280*B*a^3*x - 240*A*a^3)*sqrt(c*x^2 + a))/(a^2*x^7)]

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giac [B]  time = 0.22, size = 379, normalized size = 2.58 \begin {gather*} -\frac {B c^{3} \arctan \left (-\frac {\sqrt {c} x - \sqrt {c x^{2} + a}}{\sqrt {-a}}\right )}{8 \, \sqrt {-a} a} + \frac {105 \, {\left (\sqrt {c} x - \sqrt {c x^{2} + a}\right )}^{13} B c^{3} + 1540 \, {\left (\sqrt {c} x - \sqrt {c x^{2} + a}\right )}^{11} B a c^{3} + 3360 \, {\left (\sqrt {c} x - \sqrt {c x^{2} + a}\right )}^{10} A a c^{\frac {7}{2}} + 1085 \, {\left (\sqrt {c} x - \sqrt {c x^{2} + a}\right )}^{9} B a^{2} c^{3} + 3360 \, {\left (\sqrt {c} x - \sqrt {c x^{2} + a}\right )}^{8} A a^{2} c^{\frac {7}{2}} + 6720 \, {\left (\sqrt {c} x - \sqrt {c x^{2} + a}\right )}^{6} A a^{3} c^{\frac {7}{2}} - 1085 \, {\left (\sqrt {c} x - \sqrt {c x^{2} + a}\right )}^{5} B a^{4} c^{3} + 1344 \, {\left (\sqrt {c} x - \sqrt {c x^{2} + a}\right )}^{4} A a^{4} c^{\frac {7}{2}} - 1540 \, {\left (\sqrt {c} x - \sqrt {c x^{2} + a}\right )}^{3} B a^{5} c^{3} + 672 \, {\left (\sqrt {c} x - \sqrt {c x^{2} + a}\right )}^{2} A a^{5} c^{\frac {7}{2}} - 105 \, {\left (\sqrt {c} x - \sqrt {c x^{2} + a}\right )} B a^{6} c^{3} - 96 \, A a^{6} c^{\frac {7}{2}}}{840 \, {\left ({\left (\sqrt {c} x - \sqrt {c x^{2} + a}\right )}^{2} - a\right )}^{7} a} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate((B*x+A)*(c*x^2+a)^(3/2)/x^8,x, algorithm="giac")

[Out]

-1/8*B*c^3*arctan(-(sqrt(c)*x - sqrt(c*x^2 + a))/sqrt(-a))/(sqrt(-a)*a) + 1/840*(105*(sqrt(c)*x - sqrt(c*x^2 +
a))^13*B*c^3 + 1540*(sqrt(c)*x - sqrt(c*x^2 + a))^11*B*a*c^3 + 3360*(sqrt(c)*x - sqrt(c*x^2 + a))^10*A*a*c^(7
/2) + 1085*(sqrt(c)*x - sqrt(c*x^2 + a))^9*B*a^2*c^3 + 3360*(sqrt(c)*x - sqrt(c*x^2 + a))^8*A*a^2*c^(7/2) + 67
20*(sqrt(c)*x - sqrt(c*x^2 + a))^6*A*a^3*c^(7/2) - 1085*(sqrt(c)*x - sqrt(c*x^2 + a))^5*B*a^4*c^3 + 1344*(sqrt
(c)*x - sqrt(c*x^2 + a))^4*A*a^4*c^(7/2) - 1540*(sqrt(c)*x - sqrt(c*x^2 + a))^3*B*a^5*c^3 + 672*(sqrt(c)*x - s
qrt(c*x^2 + a))^2*A*a^5*c^(7/2) - 105*(sqrt(c)*x - sqrt(c*x^2 + a))*B*a^6*c^3 - 96*A*a^6*c^(7/2))/(((sqrt(c)*x
- sqrt(c*x^2 + a))^2 - a)^7*a)

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maple [A]  time = 0.07, size = 165, normalized size = 1.12 \begin {gather*} \frac {B \,c^{3} \ln \left (\frac {2 a +2 \sqrt {c \,x^{2}+a}\, \sqrt {a}}{x}\right )}{16 a^{\frac {3}{2}}}-\frac {\sqrt {c \,x^{2}+a}\, B \,c^{3}}{16 a^{2}}-\frac {\left (c \,x^{2}+a \right )^{\frac {3}{2}} B \,c^{3}}{48 a^{3}}+\frac {\left (c \,x^{2}+a \right )^{\frac {5}{2}} B \,c^{2}}{48 a^{3} x^{2}}+\frac {\left (c \,x^{2}+a \right )^{\frac {5}{2}} B c}{24 a^{2} x^{4}}+\frac {2 \left (c \,x^{2}+a \right )^{\frac {5}{2}} A c}{35 a^{2} x^{5}}-\frac {\left (c \,x^{2}+a \right )^{\frac {5}{2}} B}{6 a \,x^{6}}-\frac {\left (c \,x^{2}+a \right )^{\frac {5}{2}} A}{7 a \,x^{7}} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

int((B*x+A)*(c*x^2+a)^(3/2)/x^8,x)

[Out]

-1/7*A*(c*x^2+a)^(5/2)/a/x^7+2/35*A*c*(c*x^2+a)^(5/2)/a^2/x^5-1/6*B*(c*x^2+a)^(5/2)/a/x^6+1/24*B*c/a^2/x^4*(c*
x^2+a)^(5/2)+1/48*B*c^2/a^3/x^2*(c*x^2+a)^(5/2)-1/48*B*c^3/a^3*(c*x^2+a)^(3/2)+1/16*B*c^3/a^(3/2)*ln((2*a+2*(c
*x^2+a)^(1/2)*a^(1/2))/x)-1/16*B*c^3/a^2*(c*x^2+a)^(1/2)

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maxima [A]  time = 0.49, size = 153, normalized size = 1.04 \begin {gather*} \frac {B c^{3} \operatorname {arsinh}\left (\frac {a}{\sqrt {a c} {\left | x \right |}}\right )}{16 \, a^{\frac {3}{2}}} - \frac {{\left (c x^{2} + a\right )}^{\frac {3}{2}} B c^{3}}{48 \, a^{3}} - \frac {\sqrt {c x^{2} + a} B c^{3}}{16 \, a^{2}} + \frac {{\left (c x^{2} + a\right )}^{\frac {5}{2}} B c^{2}}{48 \, a^{3} x^{2}} + \frac {{\left (c x^{2} + a\right )}^{\frac {5}{2}} B c}{24 \, a^{2} x^{4}} + \frac {2 \, {\left (c x^{2} + a\right )}^{\frac {5}{2}} A c}{35 \, a^{2} x^{5}} - \frac {{\left (c x^{2} + a\right )}^{\frac {5}{2}} B}{6 \, a x^{6}} - \frac {{\left (c x^{2} + a\right )}^{\frac {5}{2}} A}{7 \, a x^{7}} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate((B*x+A)*(c*x^2+a)^(3/2)/x^8,x, algorithm="maxima")

[Out]

1/16*B*c^3*arcsinh(a/(sqrt(a*c)*abs(x)))/a^(3/2) - 1/48*(c*x^2 + a)^(3/2)*B*c^3/a^3 - 1/16*sqrt(c*x^2 + a)*B*c
^3/a^2 + 1/48*(c*x^2 + a)^(5/2)*B*c^2/(a^3*x^2) + 1/24*(c*x^2 + a)^(5/2)*B*c/(a^2*x^4) + 2/35*(c*x^2 + a)^(5/2
)*A*c/(a^2*x^5) - 1/6*(c*x^2 + a)^(5/2)*B/(a*x^6) - 1/7*(c*x^2 + a)^(5/2)*A/(a*x^7)

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mupad [B]  time = 3.77, size = 150, normalized size = 1.02 \begin {gather*} \frac {B\,a\,\sqrt {c\,x^2+a}}{16\,x^6}-\frac {A\,a\,\sqrt {c\,x^2+a}}{7\,x^7}-\frac {B\,{\left (c\,x^2+a\right )}^{3/2}}{6\,x^6}-\frac {8\,A\,c\,\sqrt {c\,x^2+a}}{35\,x^5}-\frac {B\,{\left (c\,x^2+a\right )}^{5/2}}{16\,a\,x^6}-\frac {A\,c^2\,\sqrt {c\,x^2+a}}{35\,a\,x^3}+\frac {2\,A\,c^3\,\sqrt {c\,x^2+a}}{35\,a^2\,x}-\frac {B\,c^3\,\mathrm {atan}\left (\frac {\sqrt {c\,x^2+a}\,1{}\mathrm {i}}{\sqrt {a}}\right )\,1{}\mathrm {i}}{16\,a^{3/2}} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

int(((a + c*x^2)^(3/2)*(A + B*x))/x^8,x)

[Out]

(B*a*(a + c*x^2)^(1/2))/(16*x^6) - (B*c^3*atan(((a + c*x^2)^(1/2)*1i)/a^(1/2))*1i)/(16*a^(3/2)) - (A*a*(a + c*
x^2)^(1/2))/(7*x^7) - (B*(a + c*x^2)^(3/2))/(6*x^6) - (8*A*c*(a + c*x^2)^(1/2))/(35*x^5) - (B*(a + c*x^2)^(5/2
))/(16*a*x^6) - (A*c^2*(a + c*x^2)^(1/2))/(35*a*x^3) + (2*A*c^3*(a + c*x^2)^(1/2))/(35*a^2*x)

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sympy [B]  time = 13.99, size = 575, normalized size = 3.91 \begin {gather*} - \frac {15 A a^{6} c^{\frac {9}{2}} \sqrt {\frac {a}{c x^{2}} + 1}}{105 a^{5} c^{4} x^{6} + 210 a^{4} c^{5} x^{8} + 105 a^{3} c^{6} x^{10}} - \frac {33 A a^{5} c^{\frac {11}{2}} x^{2} \sqrt {\frac {a}{c x^{2}} + 1}}{105 a^{5} c^{4} x^{6} + 210 a^{4} c^{5} x^{8} + 105 a^{3} c^{6} x^{10}} - \frac {17 A a^{4} c^{\frac {13}{2}} x^{4} \sqrt {\frac {a}{c x^{2}} + 1}}{105 a^{5} c^{4} x^{6} + 210 a^{4} c^{5} x^{8} + 105 a^{3} c^{6} x^{10}} - \frac {3 A a^{3} c^{\frac {15}{2}} x^{6} \sqrt {\frac {a}{c x^{2}} + 1}}{105 a^{5} c^{4} x^{6} + 210 a^{4} c^{5} x^{8} + 105 a^{3} c^{6} x^{10}} - \frac {12 A a^{2} c^{\frac {17}{2}} x^{8} \sqrt {\frac {a}{c x^{2}} + 1}}{105 a^{5} c^{4} x^{6} + 210 a^{4} c^{5} x^{8} + 105 a^{3} c^{6} x^{10}} - \frac {8 A a c^{\frac {19}{2}} x^{10} \sqrt {\frac {a}{c x^{2}} + 1}}{105 a^{5} c^{4} x^{6} + 210 a^{4} c^{5} x^{8} + 105 a^{3} c^{6} x^{10}} - \frac {A c^{\frac {3}{2}} \sqrt {\frac {a}{c x^{2}} + 1}}{5 x^{4}} - \frac {A c^{\frac {5}{2}} \sqrt {\frac {a}{c x^{2}} + 1}}{15 a x^{2}} + \frac {2 A c^{\frac {7}{2}} \sqrt {\frac {a}{c x^{2}} + 1}}{15 a^{2}} - \frac {B a^{2}}{6 \sqrt {c} x^{7} \sqrt {\frac {a}{c x^{2}} + 1}} - \frac {11 B a \sqrt {c}}{24 x^{5} \sqrt {\frac {a}{c x^{2}} + 1}} - \frac {17 B c^{\frac {3}{2}}}{48 x^{3} \sqrt {\frac {a}{c x^{2}} + 1}} - \frac {B c^{\frac {5}{2}}}{16 a x \sqrt {\frac {a}{c x^{2}} + 1}} + \frac {B c^{3} \operatorname {asinh}{\left (\frac {\sqrt {a}}{\sqrt {c} x} \right )}}{16 a^{\frac {3}{2}}} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate((B*x+A)*(c*x**2+a)**(3/2)/x**8,x)

[Out]

-15*A*a**6*c**(9/2)*sqrt(a/(c*x**2) + 1)/(105*a**5*c**4*x**6 + 210*a**4*c**5*x**8 + 105*a**3*c**6*x**10) - 33*
A*a**5*c**(11/2)*x**2*sqrt(a/(c*x**2) + 1)/(105*a**5*c**4*x**6 + 210*a**4*c**5*x**8 + 105*a**3*c**6*x**10) - 1
7*A*a**4*c**(13/2)*x**4*sqrt(a/(c*x**2) + 1)/(105*a**5*c**4*x**6 + 210*a**4*c**5*x**8 + 105*a**3*c**6*x**10) -
3*A*a**3*c**(15/2)*x**6*sqrt(a/(c*x**2) + 1)/(105*a**5*c**4*x**6 + 210*a**4*c**5*x**8 + 105*a**3*c**6*x**10)
- 12*A*a**2*c**(17/2)*x**8*sqrt(a/(c*x**2) + 1)/(105*a**5*c**4*x**6 + 210*a**4*c**5*x**8 + 105*a**3*c**6*x**10
) - 8*A*a*c**(19/2)*x**10*sqrt(a/(c*x**2) + 1)/(105*a**5*c**4*x**6 + 210*a**4*c**5*x**8 + 105*a**3*c**6*x**10)
- A*c**(3/2)*sqrt(a/(c*x**2) + 1)/(5*x**4) - A*c**(5/2)*sqrt(a/(c*x**2) + 1)/(15*a*x**2) + 2*A*c**(7/2)*sqrt(
a/(c*x**2) + 1)/(15*a**2) - B*a**2/(6*sqrt(c)*x**7*sqrt(a/(c*x**2) + 1)) - 11*B*a*sqrt(c)/(24*x**5*sqrt(a/(c*x
**2) + 1)) - 17*B*c**(3/2)/(48*x**3*sqrt(a/(c*x**2) + 1)) - B*c**(5/2)/(16*a*x*sqrt(a/(c*x**2) + 1)) + B*c**3*
asinh(sqrt(a)/(sqrt(c)*x))/(16*a**(3/2))

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