3.4.49 \(\int \frac {(A+B x) (a+c x^2)^{5/2}}{x^6} \, dx\) [349]

Optimal. Leaf size=140 \[ -\frac {c^2 (8 A-15 B x) \sqrt {a+c x^2}}{8 x}-\frac {c (8 A+15 B x) \left (a+c x^2\right )^{3/2}}{24 x^3}-\frac {(4 A+5 B x) \left (a+c x^2\right )^{5/2}}{20 x^5}+A c^{5/2} \tanh ^{-1}\left (\frac {\sqrt {c} x}{\sqrt {a+c x^2}}\right )-\frac {15}{8} \sqrt {a} B c^2 \tanh ^{-1}\left (\frac {\sqrt {a+c x^2}}{\sqrt {a}}\right ) \]

[Out]

-1/24*c*(15*B*x+8*A)*(c*x^2+a)^(3/2)/x^3-1/20*(5*B*x+4*A)*(c*x^2+a)^(5/2)/x^5+A*c^(5/2)*arctanh(x*c^(1/2)/(c*x
^2+a)^(1/2))-15/8*B*c^2*arctanh((c*x^2+a)^(1/2)/a^(1/2))*a^(1/2)-1/8*c^2*(-15*B*x+8*A)*(c*x^2+a)^(1/2)/x

________________________________________________________________________________________

Rubi [A]
time = 0.08, antiderivative size = 140, normalized size of antiderivative = 1.00, number of steps used = 9, number of rules used = 8, integrand size = 20, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.400, Rules used = {825, 827, 858, 223, 212, 272, 65, 214} \begin {gather*} -\frac {c^2 \sqrt {a+c x^2} (8 A-15 B x)}{8 x}-\frac {\left (a+c x^2\right )^{5/2} (4 A+5 B x)}{20 x^5}-\frac {c \left (a+c x^2\right )^{3/2} (8 A+15 B x)}{24 x^3}+A c^{5/2} \tanh ^{-1}\left (\frac {\sqrt {c} x}{\sqrt {a+c x^2}}\right )-\frac {15}{8} \sqrt {a} B c^2 \tanh ^{-1}\left (\frac {\sqrt {a+c x^2}}{\sqrt {a}}\right ) \end {gather*}

Antiderivative was successfully verified.

[In]

Int[((A + B*x)*(a + c*x^2)^(5/2))/x^6,x]

[Out]

-1/8*(c^2*(8*A - 15*B*x)*Sqrt[a + c*x^2])/x - (c*(8*A + 15*B*x)*(a + c*x^2)^(3/2))/(24*x^3) - ((4*A + 5*B*x)*(
a + c*x^2)^(5/2))/(20*x^5) + A*c^(5/2)*ArcTanh[(Sqrt[c]*x)/Sqrt[a + c*x^2]] - (15*Sqrt[a]*B*c^2*ArcTanh[Sqrt[a
 + c*x^2]/Sqrt[a]])/8

Rule 65

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 212

Int[((a_) + (b_.)*(x_)^2)^(-1), x_Symbol] :> Simp[(1/(Rt[a, 2]*Rt[-b, 2]))*ArcTanh[Rt[-b, 2]*(x/Rt[a, 2])], x]
 /; FreeQ[{a, b}, x] && NegQ[a/b] && (GtQ[a, 0] || LtQ[b, 0])

Rule 214

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

Rule 223

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

Rule 272

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 825

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

Rule 827

Int[((d_.) + (e_.)*(x_))^(m_)*((f_.) + (g_.)*(x_))*((a_) + (c_.)*(x_)^2)^(p_.), x_Symbol] :> Simp[(d + e*x)^(m
 + 1)*(e*f*(m + 2*p + 2) - d*g*(2*p + 1) + e*g*(m + 1)*x)*((a + c*x^2)^p/(e^2*(m + 1)*(m + 2*p + 2))), x] + Di
st[p/(e^2*(m + 1)*(m + 2*p + 2)), Int[(d + e*x)^(m + 1)*(a + c*x^2)^(p - 1)*Simp[g*(2*a*e + 2*a*e*m) + (g*(2*c
*d + 4*c*d*p) - 2*c*e*f*(m + 2*p + 2))*x, x], x], x] /; FreeQ[{a, c, d, e, f, g, m}, x] && NeQ[c*d^2 + a*e^2,
0] && RationalQ[p] && p > 0 && (LtQ[m, -1] || EqQ[p, 1] || (IntegerQ[p] &&  !RationalQ[m])) && NeQ[m, -1] &&
!ILtQ[m + 2*p + 1, 0] && (IntegerQ[m] || IntegerQ[p] || IntegersQ[2*m, 2*p])

Rule 858

Int[((d_.) + (e_.)*(x_))^(m_)*((f_.) + (g_.)*(x_))*((a_) + (c_.)*(x_)^2)^(p_.), x_Symbol] :> Dist[g/e, Int[(d
+ e*x)^(m + 1)*(a + c*x^2)^p, x], x] + Dist[(e*f - d*g)/e, Int[(d + e*x)^m*(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] &&  !IGtQ[m, 0]

Rubi steps

\begin {align*} \int \frac {(A+B x) \left (a+c x^2\right )^{5/2}}{x^6} \, dx &=-\frac {(4 A+5 B x) \left (a+c x^2\right )^{5/2}}{20 x^5}-\frac {\int \frac {(-8 a A c-10 a B c x) \left (a+c x^2\right )^{3/2}}{x^4} \, dx}{8 a}\\ &=-\frac {c (8 A+15 B x) \left (a+c x^2\right )^{3/2}}{24 x^3}-\frac {(4 A+5 B x) \left (a+c x^2\right )^{5/2}}{20 x^5}+\frac {\int \frac {\left (32 a^2 A c^2+60 a^2 B c^2 x\right ) \sqrt {a+c x^2}}{x^2} \, dx}{32 a^2}\\ &=-\frac {c^2 (8 A-15 B x) \sqrt {a+c x^2}}{8 x}-\frac {c (8 A+15 B x) \left (a+c x^2\right )^{3/2}}{24 x^3}-\frac {(4 A+5 B x) \left (a+c x^2\right )^{5/2}}{20 x^5}-\frac {\int \frac {-120 a^3 B c^2-64 a^2 A c^3 x}{x \sqrt {a+c x^2}} \, dx}{64 a^2}\\ &=-\frac {c^2 (8 A-15 B x) \sqrt {a+c x^2}}{8 x}-\frac {c (8 A+15 B x) \left (a+c x^2\right )^{3/2}}{24 x^3}-\frac {(4 A+5 B x) \left (a+c x^2\right )^{5/2}}{20 x^5}+\frac {1}{8} \left (15 a B c^2\right ) \int \frac {1}{x \sqrt {a+c x^2}} \, dx+\left (A c^3\right ) \int \frac {1}{\sqrt {a+c x^2}} \, dx\\ &=-\frac {c^2 (8 A-15 B x) \sqrt {a+c x^2}}{8 x}-\frac {c (8 A+15 B x) \left (a+c x^2\right )^{3/2}}{24 x^3}-\frac {(4 A+5 B x) \left (a+c x^2\right )^{5/2}}{20 x^5}+\frac {1}{16} \left (15 a B c^2\right ) \text {Subst}\left (\int \frac {1}{x \sqrt {a+c x}} \, dx,x,x^2\right )+\left (A c^3\right ) \text {Subst}\left (\int \frac {1}{1-c x^2} \, dx,x,\frac {x}{\sqrt {a+c x^2}}\right )\\ &=-\frac {c^2 (8 A-15 B x) \sqrt {a+c x^2}}{8 x}-\frac {c (8 A+15 B x) \left (a+c x^2\right )^{3/2}}{24 x^3}-\frac {(4 A+5 B x) \left (a+c x^2\right )^{5/2}}{20 x^5}+A c^{5/2} \tanh ^{-1}\left (\frac {\sqrt {c} x}{\sqrt {a+c x^2}}\right )+\frac {1}{8} (15 a B c) \text {Subst}\left (\int \frac {1}{-\frac {a}{c}+\frac {x^2}{c}} \, dx,x,\sqrt {a+c x^2}\right )\\ &=-\frac {c^2 (8 A-15 B x) \sqrt {a+c x^2}}{8 x}-\frac {c (8 A+15 B x) \left (a+c x^2\right )^{3/2}}{24 x^3}-\frac {(4 A+5 B x) \left (a+c x^2\right )^{5/2}}{20 x^5}+A c^{5/2} \tanh ^{-1}\left (\frac {\sqrt {c} x}{\sqrt {a+c x^2}}\right )-\frac {15}{8} \sqrt {a} B c^2 \tanh ^{-1}\left (\frac {\sqrt {a+c x^2}}{\sqrt {a}}\right )\\ \end {align*}

________________________________________________________________________________________

Mathematica [A]
time = 0.62, size = 133, normalized size = 0.95 \begin {gather*} -\frac {\sqrt {a+c x^2} \left (8 c^2 x^4 (23 A-15 B x)+6 a^2 (4 A+5 B x)+a c x^2 (88 A+135 B x)\right )}{120 x^5}+\frac {15}{4} \sqrt {a} B c^2 \tanh ^{-1}\left (\frac {\sqrt {c} x-\sqrt {a+c x^2}}{\sqrt {a}}\right )-A c^{5/2} \log \left (-\sqrt {c} x+\sqrt {a+c x^2}\right ) \end {gather*}

Antiderivative was successfully verified.

[In]

Integrate[((A + B*x)*(a + c*x^2)^(5/2))/x^6,x]

[Out]

-1/120*(Sqrt[a + c*x^2]*(8*c^2*x^4*(23*A - 15*B*x) + 6*a^2*(4*A + 5*B*x) + a*c*x^2*(88*A + 135*B*x)))/x^5 + (1
5*Sqrt[a]*B*c^2*ArcTanh[(Sqrt[c]*x - Sqrt[a + c*x^2])/Sqrt[a]])/4 - A*c^(5/2)*Log[-(Sqrt[c]*x) + Sqrt[a + c*x^
2]]

________________________________________________________________________________________

Maple [B] Leaf count of result is larger than twice the leaf count of optimal. \(258\) vs. \(2(114)=228\).
time = 0.60, size = 259, normalized size = 1.85

method result size
risch \(-\frac {\sqrt {c \,x^{2}+a}\, \left (184 A \,c^{2} x^{4}+135 a B c \,x^{3}+88 a A c \,x^{2}+30 a^{2} B x +24 a^{2} A \right )}{120 x^{5}}+A \,c^{\frac {5}{2}} \ln \left (\sqrt {c}\, x +\sqrt {c \,x^{2}+a}\right )-\frac {15 B \sqrt {a}\, \ln \left (\frac {2 a +2 \sqrt {a}\, \sqrt {c \,x^{2}+a}}{x}\right ) c^{2}}{8}+B \sqrt {c \,x^{2}+a}\, c^{2}\) \(122\)
default \(A \left (-\frac {\left (c \,x^{2}+a \right )^{\frac {7}{2}}}{5 a \,x^{5}}+\frac {2 c \left (-\frac {\left (c \,x^{2}+a \right )^{\frac {7}{2}}}{3 a \,x^{3}}+\frac {4 c \left (-\frac {\left (c \,x^{2}+a \right )^{\frac {7}{2}}}{a x}+\frac {6 c \left (\frac {x \left (c \,x^{2}+a \right )^{\frac {5}{2}}}{6}+\frac {5 a \left (\frac {x \left (c \,x^{2}+a \right )^{\frac {3}{2}}}{4}+\frac {3 a \left (\frac {x \sqrt {c \,x^{2}+a}}{2}+\frac {a \ln \left (\sqrt {c}\, x +\sqrt {c \,x^{2}+a}\right )}{2 \sqrt {c}}\right )}{4}\right )}{6}\right )}{a}\right )}{3 a}\right )}{5 a}\right )+B \left (-\frac {\left (c \,x^{2}+a \right )^{\frac {7}{2}}}{4 a \,x^{4}}+\frac {3 c \left (-\frac {\left (c \,x^{2}+a \right )^{\frac {7}{2}}}{2 a \,x^{2}}+\frac {5 c \left (\frac {\left (c \,x^{2}+a \right )^{\frac {5}{2}}}{5}+a \left (\frac {\left (c \,x^{2}+a \right )^{\frac {3}{2}}}{3}+a \left (\sqrt {c \,x^{2}+a}-\sqrt {a}\, \ln \left (\frac {2 a +2 \sqrt {a}\, \sqrt {c \,x^{2}+a}}{x}\right )\right )\right )\right )}{2 a}\right )}{4 a}\right )\) \(259\)

Verification of antiderivative is not currently implemented for this CAS.

[In]

int((B*x+A)*(c*x^2+a)^(5/2)/x^6,x,method=_RETURNVERBOSE)

[Out]

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

________________________________________________________________________________________

Maxima [A]
time = 0.27, size = 219, normalized size = 1.56 \begin {gather*} \frac {2 \, {\left (c x^{2} + a\right )}^{\frac {3}{2}} A c^{3} x}{3 \, a^{2}} + \frac {\sqrt {c x^{2} + a} A c^{3} x}{a} + A c^{\frac {5}{2}} \operatorname {arsinh}\left (\frac {c x}{\sqrt {a c}}\right ) - \frac {15}{8} \, B \sqrt {a} c^{2} \operatorname {arsinh}\left (\frac {a}{\sqrt {a c} {\left | x \right |}}\right ) + \frac {15}{8} \, \sqrt {c x^{2} + a} B c^{2} + \frac {3 \, {\left (c x^{2} + a\right )}^{\frac {5}{2}} B c^{2}}{8 \, a^{2}} + \frac {5 \, {\left (c x^{2} + a\right )}^{\frac {3}{2}} B c^{2}}{8 \, a} - \frac {8 \, {\left (c x^{2} + a\right )}^{\frac {5}{2}} A c^{2}}{15 \, a^{2} x} - \frac {3 \, {\left (c x^{2} + a\right )}^{\frac {7}{2}} B c}{8 \, a^{2} x^{2}} - \frac {2 \, {\left (c x^{2} + a\right )}^{\frac {7}{2}} A c}{15 \, a^{2} x^{3}} - \frac {{\left (c x^{2} + a\right )}^{\frac {7}{2}} B}{4 \, a x^{4}} - \frac {{\left (c x^{2} + a\right )}^{\frac {7}{2}} A}{5 \, a x^{5}} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate((B*x+A)*(c*x^2+a)^(5/2)/x^6,x, algorithm="maxima")

[Out]

2/3*(c*x^2 + a)^(3/2)*A*c^3*x/a^2 + sqrt(c*x^2 + a)*A*c^3*x/a + A*c^(5/2)*arcsinh(c*x/sqrt(a*c)) - 15/8*B*sqrt
(a)*c^2*arcsinh(a/(sqrt(a*c)*abs(x))) + 15/8*sqrt(c*x^2 + a)*B*c^2 + 3/8*(c*x^2 + a)^(5/2)*B*c^2/a^2 + 5/8*(c*
x^2 + a)^(3/2)*B*c^2/a - 8/15*(c*x^2 + a)^(5/2)*A*c^2/(a^2*x) - 3/8*(c*x^2 + a)^(7/2)*B*c/(a^2*x^2) - 2/15*(c*
x^2 + a)^(7/2)*A*c/(a^2*x^3) - 1/4*(c*x^2 + a)^(7/2)*B/(a*x^4) - 1/5*(c*x^2 + a)^(7/2)*A/(a*x^5)

________________________________________________________________________________________

Fricas [A]
time = 4.02, size = 534, normalized size = 3.81 \begin {gather*} \left [\frac {120 \, A c^{\frac {5}{2}} x^{5} \log \left (-2 \, c x^{2} - 2 \, \sqrt {c x^{2} + a} \sqrt {c} x - a\right ) + 225 \, B \sqrt {a} c^{2} x^{5} \log \left (-\frac {c x^{2} - 2 \, \sqrt {c x^{2} + a} \sqrt {a} + 2 \, a}{x^{2}}\right ) + 2 \, {\left (120 \, B c^{2} x^{5} - 184 \, A c^{2} x^{4} - 135 \, B a c x^{3} - 88 \, A a c x^{2} - 30 \, B a^{2} x - 24 \, A a^{2}\right )} \sqrt {c x^{2} + a}}{240 \, x^{5}}, -\frac {240 \, A \sqrt {-c} c^{2} x^{5} \arctan \left (\frac {\sqrt {-c} x}{\sqrt {c x^{2} + a}}\right ) - 225 \, B \sqrt {a} c^{2} x^{5} \log \left (-\frac {c x^{2} - 2 \, \sqrt {c x^{2} + a} \sqrt {a} + 2 \, a}{x^{2}}\right ) - 2 \, {\left (120 \, B c^{2} x^{5} - 184 \, A c^{2} x^{4} - 135 \, B a c x^{3} - 88 \, A a c x^{2} - 30 \, B a^{2} x - 24 \, A a^{2}\right )} \sqrt {c x^{2} + a}}{240 \, x^{5}}, \frac {225 \, B \sqrt {-a} c^{2} x^{5} \arctan \left (\frac {\sqrt {-a}}{\sqrt {c x^{2} + a}}\right ) + 60 \, A c^{\frac {5}{2}} x^{5} \log \left (-2 \, c x^{2} - 2 \, \sqrt {c x^{2} + a} \sqrt {c} x - a\right ) + {\left (120 \, B c^{2} x^{5} - 184 \, A c^{2} x^{4} - 135 \, B a c x^{3} - 88 \, A a c x^{2} - 30 \, B a^{2} x - 24 \, A a^{2}\right )} \sqrt {c x^{2} + a}}{120 \, x^{5}}, -\frac {120 \, A \sqrt {-c} c^{2} x^{5} \arctan \left (\frac {\sqrt {-c} x}{\sqrt {c x^{2} + a}}\right ) - 225 \, B \sqrt {-a} c^{2} x^{5} \arctan \left (\frac {\sqrt {-a}}{\sqrt {c x^{2} + a}}\right ) - {\left (120 \, B c^{2} x^{5} - 184 \, A c^{2} x^{4} - 135 \, B a c x^{3} - 88 \, A a c x^{2} - 30 \, B a^{2} x - 24 \, A a^{2}\right )} \sqrt {c x^{2} + a}}{120 \, x^{5}}\right ] \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate((B*x+A)*(c*x^2+a)^(5/2)/x^6,x, algorithm="fricas")

[Out]

[1/240*(120*A*c^(5/2)*x^5*log(-2*c*x^2 - 2*sqrt(c*x^2 + a)*sqrt(c)*x - a) + 225*B*sqrt(a)*c^2*x^5*log(-(c*x^2
- 2*sqrt(c*x^2 + a)*sqrt(a) + 2*a)/x^2) + 2*(120*B*c^2*x^5 - 184*A*c^2*x^4 - 135*B*a*c*x^3 - 88*A*a*c*x^2 - 30
*B*a^2*x - 24*A*a^2)*sqrt(c*x^2 + a))/x^5, -1/240*(240*A*sqrt(-c)*c^2*x^5*arctan(sqrt(-c)*x/sqrt(c*x^2 + a)) -
 225*B*sqrt(a)*c^2*x^5*log(-(c*x^2 - 2*sqrt(c*x^2 + a)*sqrt(a) + 2*a)/x^2) - 2*(120*B*c^2*x^5 - 184*A*c^2*x^4
- 135*B*a*c*x^3 - 88*A*a*c*x^2 - 30*B*a^2*x - 24*A*a^2)*sqrt(c*x^2 + a))/x^5, 1/120*(225*B*sqrt(-a)*c^2*x^5*ar
ctan(sqrt(-a)/sqrt(c*x^2 + a)) + 60*A*c^(5/2)*x^5*log(-2*c*x^2 - 2*sqrt(c*x^2 + a)*sqrt(c)*x - a) + (120*B*c^2
*x^5 - 184*A*c^2*x^4 - 135*B*a*c*x^3 - 88*A*a*c*x^2 - 30*B*a^2*x - 24*A*a^2)*sqrt(c*x^2 + a))/x^5, -1/120*(120
*A*sqrt(-c)*c^2*x^5*arctan(sqrt(-c)*x/sqrt(c*x^2 + a)) - 225*B*sqrt(-a)*c^2*x^5*arctan(sqrt(-a)/sqrt(c*x^2 + a
)) - (120*B*c^2*x^5 - 184*A*c^2*x^4 - 135*B*a*c*x^3 - 88*A*a*c*x^2 - 30*B*a^2*x - 24*A*a^2)*sqrt(c*x^2 + a))/x
^5]

________________________________________________________________________________________

Sympy [B] Leaf count of result is larger than twice the leaf count of optimal. 294 vs. \(2 (128) = 256\).
time = 6.27, size = 294, normalized size = 2.10 \begin {gather*} - \frac {A \sqrt {a} c^{2}}{x \sqrt {1 + \frac {c x^{2}}{a}}} - \frac {A a^{2} \sqrt {c} \sqrt {\frac {a}{c x^{2}} + 1}}{5 x^{4}} - \frac {11 A a c^{\frac {3}{2}} \sqrt {\frac {a}{c x^{2}} + 1}}{15 x^{2}} - \frac {8 A c^{\frac {5}{2}} \sqrt {\frac {a}{c x^{2}} + 1}}{15} + A c^{\frac {5}{2}} \operatorname {asinh}{\left (\frac {\sqrt {c} x}{\sqrt {a}} \right )} - \frac {A c^{3} x}{\sqrt {a} \sqrt {1 + \frac {c x^{2}}{a}}} - \frac {15 B \sqrt {a} c^{2} \operatorname {asinh}{\left (\frac {\sqrt {a}}{\sqrt {c} x} \right )}}{8} - \frac {B a^{3}}{4 \sqrt {c} x^{5} \sqrt {\frac {a}{c x^{2}} + 1}} - \frac {3 B a^{2} \sqrt {c}}{8 x^{3} \sqrt {\frac {a}{c x^{2}} + 1}} - \frac {B a c^{\frac {3}{2}} \sqrt {\frac {a}{c x^{2}} + 1}}{x} + \frac {7 B a c^{\frac {3}{2}}}{8 x \sqrt {\frac {a}{c x^{2}} + 1}} + \frac {B c^{\frac {5}{2}} x}{\sqrt {\frac {a}{c x^{2}} + 1}} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate((B*x+A)*(c*x**2+a)**(5/2)/x**6,x)

[Out]

-A*sqrt(a)*c**2/(x*sqrt(1 + c*x**2/a)) - A*a**2*sqrt(c)*sqrt(a/(c*x**2) + 1)/(5*x**4) - 11*A*a*c**(3/2)*sqrt(a
/(c*x**2) + 1)/(15*x**2) - 8*A*c**(5/2)*sqrt(a/(c*x**2) + 1)/15 + A*c**(5/2)*asinh(sqrt(c)*x/sqrt(a)) - A*c**3
*x/(sqrt(a)*sqrt(1 + c*x**2/a)) - 15*B*sqrt(a)*c**2*asinh(sqrt(a)/(sqrt(c)*x))/8 - B*a**3/(4*sqrt(c)*x**5*sqrt
(a/(c*x**2) + 1)) - 3*B*a**2*sqrt(c)/(8*x**3*sqrt(a/(c*x**2) + 1)) - B*a*c**(3/2)*sqrt(a/(c*x**2) + 1)/x + 7*B
*a*c**(3/2)/(8*x*sqrt(a/(c*x**2) + 1)) + B*c**(5/2)*x/sqrt(a/(c*x**2) + 1)

________________________________________________________________________________________

Giac [B] Leaf count of result is larger than twice the leaf count of optimal. 331 vs. \(2 (114) = 228\).
time = 1.42, size = 331, normalized size = 2.36 \begin {gather*} \frac {15 \, B a c^{2} \arctan \left (-\frac {\sqrt {c} x - \sqrt {c x^{2} + a}}{\sqrt {-a}}\right )}{4 \, \sqrt {-a}} - A c^{\frac {5}{2}} \log \left ({\left | -\sqrt {c} x + \sqrt {c x^{2} + a} \right |}\right ) + \sqrt {c x^{2} + a} B c^{2} + \frac {135 \, {\left (\sqrt {c} x - \sqrt {c x^{2} + a}\right )}^{9} B a c^{2} + 360 \, {\left (\sqrt {c} x - \sqrt {c x^{2} + a}\right )}^{8} A a c^{\frac {5}{2}} - 150 \, {\left (\sqrt {c} x - \sqrt {c x^{2} + a}\right )}^{7} B a^{2} c^{2} - 720 \, {\left (\sqrt {c} x - \sqrt {c x^{2} + a}\right )}^{6} A a^{2} c^{\frac {5}{2}} + 1120 \, {\left (\sqrt {c} x - \sqrt {c x^{2} + a}\right )}^{4} A a^{3} c^{\frac {5}{2}} + 150 \, {\left (\sqrt {c} x - \sqrt {c x^{2} + a}\right )}^{3} B a^{4} c^{2} - 560 \, {\left (\sqrt {c} x - \sqrt {c x^{2} + a}\right )}^{2} A a^{4} c^{\frac {5}{2}} - 135 \, {\left (\sqrt {c} x - \sqrt {c x^{2} + a}\right )} B a^{5} c^{2} + 184 \, A a^{5} c^{\frac {5}{2}}}{60 \, {\left ({\left (\sqrt {c} x - \sqrt {c x^{2} + a}\right )}^{2} - a\right )}^{5}} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate((B*x+A)*(c*x^2+a)^(5/2)/x^6,x, algorithm="giac")

[Out]

15/4*B*a*c^2*arctan(-(sqrt(c)*x - sqrt(c*x^2 + a))/sqrt(-a))/sqrt(-a) - A*c^(5/2)*log(abs(-sqrt(c)*x + sqrt(c*
x^2 + a))) + sqrt(c*x^2 + a)*B*c^2 + 1/60*(135*(sqrt(c)*x - sqrt(c*x^2 + a))^9*B*a*c^2 + 360*(sqrt(c)*x - sqrt
(c*x^2 + a))^8*A*a*c^(5/2) - 150*(sqrt(c)*x - sqrt(c*x^2 + a))^7*B*a^2*c^2 - 720*(sqrt(c)*x - sqrt(c*x^2 + a))
^6*A*a^2*c^(5/2) + 1120*(sqrt(c)*x - sqrt(c*x^2 + a))^4*A*a^3*c^(5/2) + 150*(sqrt(c)*x - sqrt(c*x^2 + a))^3*B*
a^4*c^2 - 560*(sqrt(c)*x - sqrt(c*x^2 + a))^2*A*a^4*c^(5/2) - 135*(sqrt(c)*x - sqrt(c*x^2 + a))*B*a^5*c^2 + 18
4*A*a^5*c^(5/2))/((sqrt(c)*x - sqrt(c*x^2 + a))^2 - a)^5

________________________________________________________________________________________

Mupad [F]
time = 0.00, size = -1, normalized size = -0.01 \begin {gather*} \int \frac {{\left (c\,x^2+a\right )}^{5/2}\,\left (A+B\,x\right )}{x^6} \,d x \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

int(((a + c*x^2)^(5/2)*(A + B*x))/x^6,x)

[Out]

int(((a + c*x^2)^(5/2)*(A + B*x))/x^6, x)

________________________________________________________________________________________