3.7.4 \(\int \frac {(a+b x^2)^2 \sqrt {c+d x^2}}{x^7} \, dx\) [604]

Optimal. Leaf size=149 \[ -\frac {\left (8 b^2 c^2-4 a b c d+a^2 d^2\right ) \sqrt {c+d x^2}}{16 c^2 x^2}-\frac {a^2 \left (c+d x^2\right )^{3/2}}{6 c x^6}-\frac {a (4 b c-a d) \left (c+d x^2\right )^{3/2}}{8 c^2 x^4}-\frac {d \left (8 b^2 c^2-4 a b c d+a^2 d^2\right ) \tanh ^{-1}\left (\frac {\sqrt {c+d x^2}}{\sqrt {c}}\right )}{16 c^{5/2}} \]

[Out]

-1/6*a^2*(d*x^2+c)^(3/2)/c/x^6-1/8*a*(-a*d+4*b*c)*(d*x^2+c)^(3/2)/c^2/x^4-1/16*d*(a^2*d^2-4*a*b*c*d+8*b^2*c^2)
*arctanh((d*x^2+c)^(1/2)/c^(1/2))/c^(5/2)-1/16*(a^2*d^2-4*a*b*c*d+8*b^2*c^2)*(d*x^2+c)^(1/2)/c^2/x^2

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Rubi [A]
time = 0.10, antiderivative size = 149, normalized size of antiderivative = 1.00, number of steps used = 6, number of rules used = 6, integrand size = 24, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.250, Rules used = {457, 91, 79, 43, 65, 214} \begin {gather*} -\frac {\sqrt {c+d x^2} \left (a^2 d^2-4 a b c d+8 b^2 c^2\right )}{16 c^2 x^2}-\frac {d \left (a^2 d^2-4 a b c d+8 b^2 c^2\right ) \tanh ^{-1}\left (\frac {\sqrt {c+d x^2}}{\sqrt {c}}\right )}{16 c^{5/2}}-\frac {a^2 \left (c+d x^2\right )^{3/2}}{6 c x^6}-\frac {a \left (c+d x^2\right )^{3/2} (4 b c-a d)}{8 c^2 x^4} \end {gather*}

Antiderivative was successfully verified.

[In]

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

[Out]

-1/16*((8*b^2*c^2 - 4*a*b*c*d + a^2*d^2)*Sqrt[c + d*x^2])/(c^2*x^2) - (a^2*(c + d*x^2)^(3/2))/(6*c*x^6) - (a*(
4*b*c - a*d)*(c + d*x^2)^(3/2))/(8*c^2*x^4) - (d*(8*b^2*c^2 - 4*a*b*c*d + a^2*d^2)*ArcTanh[Sqrt[c + d*x^2]/Sqr
t[c]])/(16*c^(5/2))

Rule 43

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, n
}, x] && NeQ[b*c - a*d, 0] && ILtQ[m, -1] &&  !IntegerQ[n] && GtQ[n, 0]

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 79

Int[((a_.) + (b_.)*(x_))*((c_.) + (d_.)*(x_))^(n_.)*((e_.) + (f_.)*(x_))^(p_.), x_Symbol] :> Simp[(-(b*e - a*f
))*(c + d*x)^(n + 1)*((e + f*x)^(p + 1)/(f*(p + 1)*(c*f - d*e))), x] - Dist[(a*d*f*(n + p + 2) - b*(d*e*(n + 1
) + c*f*(p + 1)))/(f*(p + 1)*(c*f - d*e)), Int[(c + d*x)^n*(e + f*x)^(p + 1), x], x] /; FreeQ[{a, b, c, d, e,
f, n}, x] && LtQ[p, -1] && ( !LtQ[n, -1] || IntegerQ[p] ||  !(IntegerQ[n] ||  !(EqQ[e, 0] ||  !(EqQ[c, 0] || L
tQ[p, n]))))

Rule 91

Int[((a_.) + (b_.)*(x_))^2*((c_.) + (d_.)*(x_))^(n_.)*((e_.) + (f_.)*(x_))^(p_.), x_Symbol] :> Simp[(b*c - a*d
)^2*(c + d*x)^(n + 1)*((e + f*x)^(p + 1)/(d^2*(d*e - c*f)*(n + 1))), x] - Dist[1/(d^2*(d*e - c*f)*(n + 1)), In
t[(c + d*x)^(n + 1)*(e + f*x)^p*Simp[a^2*d^2*f*(n + p + 2) + b^2*c*(d*e*(n + 1) + c*f*(p + 1)) - 2*a*b*d*(d*e*
(n + 1) + c*f*(p + 1)) - b^2*d*(d*e - c*f)*(n + 1)*x, x], x], x] /; FreeQ[{a, b, c, d, e, f, n, p}, x] && (LtQ
[n, -1] || (EqQ[n + p + 3, 0] && NeQ[n, -1] && (SumSimplerQ[n, 1] ||  !SumSimplerQ[p, 1])))

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 457

Int[(x_)^(m_.)*((a_) + (b_.)*(x_)^(n_))^(p_.)*((c_) + (d_.)*(x_)^(n_))^(q_.), x_Symbol] :> Dist[1/n, Subst[Int
[x^(Simplify[(m + 1)/n] - 1)*(a + b*x)^p*(c + d*x)^q, x], x, x^n], x] /; FreeQ[{a, b, c, d, m, n, p, q}, x] &&
NeQ[b*c - a*d, 0] && IntegerQ[Simplify[(m + 1)/n]]

Rubi steps

\begin {align*} \int \frac {\left (a+b x^2\right )^2 \sqrt {c+d x^2}}{x^7} \, dx &=\frac {1}{2} \text {Subst}\left (\int \frac {(a+b x)^2 \sqrt {c+d x}}{x^4} \, dx,x,x^2\right )\\ &=-\frac {a^2 \left (c+d x^2\right )^{3/2}}{6 c x^6}+\frac {\text {Subst}\left (\int \frac {\left (\frac {3}{2} a (4 b c-a d)+3 b^2 c x\right ) \sqrt {c+d x}}{x^3} \, dx,x,x^2\right )}{6 c}\\ &=-\frac {a^2 \left (c+d x^2\right )^{3/2}}{6 c x^6}-\frac {a (4 b c-a d) \left (c+d x^2\right )^{3/2}}{8 c^2 x^4}+\frac {\left (8 b^2 c^2-4 a b c d+a^2 d^2\right ) \text {Subst}\left (\int \frac {\sqrt {c+d x}}{x^2} \, dx,x,x^2\right )}{16 c^2}\\ &=-\frac {\left (8 b^2 c^2-4 a b c d+a^2 d^2\right ) \sqrt {c+d x^2}}{16 c^2 x^2}-\frac {a^2 \left (c+d x^2\right )^{3/2}}{6 c x^6}-\frac {a (4 b c-a d) \left (c+d x^2\right )^{3/2}}{8 c^2 x^4}+\frac {\left (d \left (8 b^2 c^2-4 a b c d+a^2 d^2\right )\right ) \text {Subst}\left (\int \frac {1}{x \sqrt {c+d x}} \, dx,x,x^2\right )}{32 c^2}\\ &=-\frac {\left (8 b^2 c^2-4 a b c d+a^2 d^2\right ) \sqrt {c+d x^2}}{16 c^2 x^2}-\frac {a^2 \left (c+d x^2\right )^{3/2}}{6 c x^6}-\frac {a (4 b c-a d) \left (c+d x^2\right )^{3/2}}{8 c^2 x^4}+\frac {\left (8 b^2 c^2-4 a b c d+a^2 d^2\right ) \text {Subst}\left (\int \frac {1}{-\frac {c}{d}+\frac {x^2}{d}} \, dx,x,\sqrt {c+d x^2}\right )}{16 c^2}\\ &=-\frac {\left (8 b^2 c^2-4 a b c d+a^2 d^2\right ) \sqrt {c+d x^2}}{16 c^2 x^2}-\frac {a^2 \left (c+d x^2\right )^{3/2}}{6 c x^6}-\frac {a (4 b c-a d) \left (c+d x^2\right )^{3/2}}{8 c^2 x^4}-\frac {d \left (8 b^2 c^2-4 a b c d+a^2 d^2\right ) \tanh ^{-1}\left (\frac {\sqrt {c+d x^2}}{\sqrt {c}}\right )}{16 c^{5/2}}\\ \end {align*}

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Mathematica [A]
time = 0.22, size = 126, normalized size = 0.85 \begin {gather*} -\frac {\sqrt {c+d x^2} \left (24 b^2 c^2 x^4+12 a b c x^2 \left (2 c+d x^2\right )+a^2 \left (8 c^2+2 c d x^2-3 d^2 x^4\right )\right )}{48 c^2 x^6}-\frac {d \left (8 b^2 c^2-4 a b c d+a^2 d^2\right ) \tanh ^{-1}\left (\frac {\sqrt {c+d x^2}}{\sqrt {c}}\right )}{16 c^{5/2}} \end {gather*}

Antiderivative was successfully verified.

[In]

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

[Out]

-1/48*(Sqrt[c + d*x^2]*(24*b^2*c^2*x^4 + 12*a*b*c*x^2*(2*c + d*x^2) + a^2*(8*c^2 + 2*c*d*x^2 - 3*d^2*x^4)))/(c
^2*x^6) - (d*(8*b^2*c^2 - 4*a*b*c*d + a^2*d^2)*ArcTanh[Sqrt[c + d*x^2]/Sqrt[c]])/(16*c^(5/2))

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Maple [B] Leaf count of result is larger than twice the leaf count of optimal. \(271\) vs. \(2(129)=258\).
time = 0.10, size = 272, normalized size = 1.83

method	result	size
risch	\(-\frac {\sqrt {d \,x^{2}+c}\, \left (-3 a^{2} d^{2} x^{4}+12 a b c d \,x^{4}+24 b^{2} c^{2} x^{4}+2 a^{2} c d \,x^{2}+24 a b \,c^{2} x^{2}+8 a^{2} c^{2}\right )}{48 x^{6} c^{2}}-\frac {d^{3} \ln \left (\frac {2 c +2 \sqrt {c}\, \sqrt {d \,x^{2}+c}}{x}\right ) a^{2}}{16 c^{\frac {5}{2}}}+\frac {d^{2} \ln \left (\frac {2 c +2 \sqrt {c}\, \sqrt {d \,x^{2}+c}}{x}\right ) a b}{4 c^{\frac {3}{2}}}-\frac {d \ln \left (\frac {2 c +2 \sqrt {c}\, \sqrt {d \,x^{2}+c}}{x}\right ) b^{2}}{2 \sqrt {c}}\)	\(178\)
default	\(2 a b \left (-\frac {\left (d \,x^{2}+c \right )^{\frac {3}{2}}}{4 c \,x^{4}}-\frac {d \left (-\frac {\left (d \,x^{2}+c \right )^{\frac {3}{2}}}{2 c \,x^{2}}+\frac {d \left (\sqrt {d \,x^{2}+c}-\sqrt {c}\, \ln \left (\frac {2 c +2 \sqrt {c}\, \sqrt {d \,x^{2}+c}}{x}\right )\right )}{2 c}\right )}{4 c}\right )+a^{2} \left (-\frac {\left (d \,x^{2}+c \right )^{\frac {3}{2}}}{6 c \,x^{6}}-\frac {d \left (-\frac {\left (d \,x^{2}+c \right )^{\frac {3}{2}}}{4 c \,x^{4}}-\frac {d \left (-\frac {\left (d \,x^{2}+c \right )^{\frac {3}{2}}}{2 c \,x^{2}}+\frac {d \left (\sqrt {d \,x^{2}+c}-\sqrt {c}\, \ln \left (\frac {2 c +2 \sqrt {c}\, \sqrt {d \,x^{2}+c}}{x}\right )\right )}{2 c}\right )}{4 c}\right )}{2 c}\right )+b^{2} \left (-\frac {\left (d \,x^{2}+c \right )^{\frac {3}{2}}}{2 c \,x^{2}}+\frac {d \left (\sqrt {d \,x^{2}+c}-\sqrt {c}\, \ln \left (\frac {2 c +2 \sqrt {c}\, \sqrt {d \,x^{2}+c}}{x}\right )\right )}{2 c}\right )\)	\(272\)

Verification of antiderivative is not currently implemented for this CAS.

[In]

int((b*x^2+a)^2*(d*x^2+c)^(1/2)/x^7,x,method=_RETURNVERBOSE)

[Out]

2*a*b*(-1/4/c/x^4*(d*x^2+c)^(3/2)-1/4*d/c*(-1/2/c/x^2*(d*x^2+c)^(3/2)+1/2*d/c*((d*x^2+c)^(1/2)-c^(1/2)*ln((2*c
+2*c^(1/2)*(d*x^2+c)^(1/2))/x))))+a^2*(-1/6/c/x^6*(d*x^2+c)^(3/2)-1/2*d/c*(-1/4/c/x^4*(d*x^2+c)^(3/2)-1/4*d/c*
(-1/2/c/x^2*(d*x^2+c)^(3/2)+1/2*d/c*((d*x^2+c)^(1/2)-c^(1/2)*ln((2*c+2*c^(1/2)*(d*x^2+c)^(1/2))/x)))))+b^2*(-1
/2/c/x^2*(d*x^2+c)^(3/2)+1/2*d/c*((d*x^2+c)^(1/2)-c^(1/2)*ln((2*c+2*c^(1/2)*(d*x^2+c)^(1/2))/x)))

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Maxima [A]
time = 0.28, size = 247, normalized size = 1.66 \begin {gather*} -\frac {b^{2} d \operatorname {arsinh}\left (\frac {c}{\sqrt {c d} {\left | x \right |}}\right )}{2 \, \sqrt {c}} + \frac {a b d^{2} \operatorname {arsinh}\left (\frac {c}{\sqrt {c d} {\left | x \right |}}\right )}{4 \, c^{\frac {3}{2}}} - \frac {a^{2} d^{3} \operatorname {arsinh}\left (\frac {c}{\sqrt {c d} {\left | x \right |}}\right )}{16 \, c^{\frac {5}{2}}} + \frac {\sqrt {d x^{2} + c} b^{2} d}{2 \, c} - \frac {\sqrt {d x^{2} + c} a b d^{2}}{4 \, c^{2}} + \frac {\sqrt {d x^{2} + c} a^{2} d^{3}}{16 \, c^{3}} - \frac {{\left (d x^{2} + c\right )}^{\frac {3}{2}} b^{2}}{2 \, c x^{2}} + \frac {{\left (d x^{2} + c\right )}^{\frac {3}{2}} a b d}{4 \, c^{2} x^{2}} - \frac {{\left (d x^{2} + c\right )}^{\frac {3}{2}} a^{2} d^{2}}{16 \, c^{3} x^{2}} - \frac {{\left (d x^{2} + c\right )}^{\frac {3}{2}} a b}{2 \, c x^{4}} + \frac {{\left (d x^{2} + c\right )}^{\frac {3}{2}} a^{2} d}{8 \, c^{2} x^{4}} - \frac {{\left (d x^{2} + c\right )}^{\frac {3}{2}} a^{2}}{6 \, c x^{6}} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate((b*x^2+a)^2*(d*x^2+c)^(1/2)/x^7,x, algorithm="maxima")

[Out]

-1/2*b^2*d*arcsinh(c/(sqrt(c*d)*abs(x)))/sqrt(c) + 1/4*a*b*d^2*arcsinh(c/(sqrt(c*d)*abs(x)))/c^(3/2) - 1/16*a^
2*d^3*arcsinh(c/(sqrt(c*d)*abs(x)))/c^(5/2) + 1/2*sqrt(d*x^2 + c)*b^2*d/c - 1/4*sqrt(d*x^2 + c)*a*b*d^2/c^2 +
1/16*sqrt(d*x^2 + c)*a^2*d^3/c^3 - 1/2*(d*x^2 + c)^(3/2)*b^2/(c*x^2) + 1/4*(d*x^2 + c)^(3/2)*a*b*d/(c^2*x^2) -
1/16*(d*x^2 + c)^(3/2)*a^2*d^2/(c^3*x^2) - 1/2*(d*x^2 + c)^(3/2)*a*b/(c*x^4) + 1/8*(d*x^2 + c)^(3/2)*a^2*d/(c
^2*x^4) - 1/6*(d*x^2 + c)^(3/2)*a^2/(c*x^6)

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Fricas [A]
time = 1.50, size = 276, normalized size = 1.85 \begin {gather*} \left [\frac {3 \, {\left (8 \, b^{2} c^{2} d - 4 \, a b c d^{2} + a^{2} d^{3}\right )} \sqrt {c} x^{6} \log \left (-\frac {d x^{2} - 2 \, \sqrt {d x^{2} + c} \sqrt {c} + 2 \, c}{x^{2}}\right ) - 2 \, {\left (8 \, a^{2} c^{3} + 3 \, {\left (8 \, b^{2} c^{3} + 4 \, a b c^{2} d - a^{2} c d^{2}\right )} x^{4} + 2 \, {\left (12 \, a b c^{3} + a^{2} c^{2} d\right )} x^{2}\right )} \sqrt {d x^{2} + c}}{96 \, c^{3} x^{6}}, \frac {3 \, {\left (8 \, b^{2} c^{2} d - 4 \, a b c d^{2} + a^{2} d^{3}\right )} \sqrt {-c} x^{6} \arctan \left (\frac {\sqrt {-c}}{\sqrt {d x^{2} + c}}\right ) - {\left (8 \, a^{2} c^{3} + 3 \, {\left (8 \, b^{2} c^{3} + 4 \, a b c^{2} d - a^{2} c d^{2}\right )} x^{4} + 2 \, {\left (12 \, a b c^{3} + a^{2} c^{2} d\right )} x^{2}\right )} \sqrt {d x^{2} + c}}{48 \, c^{3} x^{6}}\right ] \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate((b*x^2+a)^2*(d*x^2+c)^(1/2)/x^7,x, algorithm="fricas")

[Out]

[1/96*(3*(8*b^2*c^2*d - 4*a*b*c*d^2 + a^2*d^3)*sqrt(c)*x^6*log(-(d*x^2 - 2*sqrt(d*x^2 + c)*sqrt(c) + 2*c)/x^2)
- 2*(8*a^2*c^3 + 3*(8*b^2*c^3 + 4*a*b*c^2*d - a^2*c*d^2)*x^4 + 2*(12*a*b*c^3 + a^2*c^2*d)*x^2)*sqrt(d*x^2 + c
))/(c^3*x^6), 1/48*(3*(8*b^2*c^2*d - 4*a*b*c*d^2 + a^2*d^3)*sqrt(-c)*x^6*arctan(sqrt(-c)/sqrt(d*x^2 + c)) - (8
*a^2*c^3 + 3*(8*b^2*c^3 + 4*a*b*c^2*d - a^2*c*d^2)*x^4 + 2*(12*a*b*c^3 + a^2*c^2*d)*x^2)*sqrt(d*x^2 + c))/(c^3
*x^6)]

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Sympy [B] Leaf count of result is larger than twice the leaf count of optimal. 291 vs. \(2 (141) = 282\).
time = 80.62, size = 291, normalized size = 1.95 \begin {gather*} - \frac {a^{2} c}{6 \sqrt {d} x^{7} \sqrt {\frac {c}{d x^{2}} + 1}} - \frac {5 a^{2} \sqrt {d}}{24 x^{5} \sqrt {\frac {c}{d x^{2}} + 1}} + \frac {a^{2} d^{\frac {3}{2}}}{48 c x^{3} \sqrt {\frac {c}{d x^{2}} + 1}} + \frac {a^{2} d^{\frac {5}{2}}}{16 c^{2} x \sqrt {\frac {c}{d x^{2}} + 1}} - \frac {a^{2} d^{3} \operatorname {asinh}{\left (\frac {\sqrt {c}}{\sqrt {d} x} \right )}}{16 c^{\frac {5}{2}}} - \frac {a b c}{2 \sqrt {d} x^{5} \sqrt {\frac {c}{d x^{2}} + 1}} - \frac {3 a b \sqrt {d}}{4 x^{3} \sqrt {\frac {c}{d x^{2}} + 1}} - \frac {a b d^{\frac {3}{2}}}{4 c x \sqrt {\frac {c}{d x^{2}} + 1}} + \frac {a b d^{2} \operatorname {asinh}{\left (\frac {\sqrt {c}}{\sqrt {d} x} \right )}}{4 c^{\frac {3}{2}}} - \frac {b^{2} \sqrt {d} \sqrt {\frac {c}{d x^{2}} + 1}}{2 x} - \frac {b^{2} d \operatorname {asinh}{\left (\frac {\sqrt {c}}{\sqrt {d} x} \right )}}{2 \sqrt {c}} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate((b*x**2+a)**2*(d*x**2+c)**(1/2)/x**7,x)

[Out]

-a**2*c/(6*sqrt(d)*x**7*sqrt(c/(d*x**2) + 1)) - 5*a**2*sqrt(d)/(24*x**5*sqrt(c/(d*x**2) + 1)) + a**2*d**(3/2)/
(48*c*x**3*sqrt(c/(d*x**2) + 1)) + a**2*d**(5/2)/(16*c**2*x*sqrt(c/(d*x**2) + 1)) - a**2*d**3*asinh(sqrt(c)/(s
qrt(d)*x))/(16*c**(5/2)) - a*b*c/(2*sqrt(d)*x**5*sqrt(c/(d*x**2) + 1)) - 3*a*b*sqrt(d)/(4*x**3*sqrt(c/(d*x**2)
+ 1)) - a*b*d**(3/2)/(4*c*x*sqrt(c/(d*x**2) + 1)) + a*b*d**2*asinh(sqrt(c)/(sqrt(d)*x))/(4*c**(3/2)) - b**2*s
qrt(d)*sqrt(c/(d*x**2) + 1)/(2*x) - b**2*d*asinh(sqrt(c)/(sqrt(d)*x))/(2*sqrt(c))

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Giac [A]
time = 0.94, size = 222, normalized size = 1.49 \begin {gather*} \frac {\frac {3 \, {\left (8 \, b^{2} c^{2} d^{2} - 4 \, a b c d^{3} + a^{2} d^{4}\right )} \arctan \left (\frac {\sqrt {d x^{2} + c}}{\sqrt {-c}}\right )}{\sqrt {-c} c^{2}} - \frac {24 \, {\left (d x^{2} + c\right )}^{\frac {5}{2}} b^{2} c^{2} d^{2} - 48 \, {\left (d x^{2} + c\right )}^{\frac {3}{2}} b^{2} c^{3} d^{2} + 24 \, \sqrt {d x^{2} + c} b^{2} c^{4} d^{2} + 12 \, {\left (d x^{2} + c\right )}^{\frac {5}{2}} a b c d^{3} - 12 \, \sqrt {d x^{2} + c} a b c^{3} d^{3} - 3 \, {\left (d x^{2} + c\right )}^{\frac {5}{2}} a^{2} d^{4} + 8 \, {\left (d x^{2} + c\right )}^{\frac {3}{2}} a^{2} c d^{4} + 3 \, \sqrt {d x^{2} + c} a^{2} c^{2} d^{4}}{c^{2} d^{3} x^{6}}}{48 \, d} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate((b*x^2+a)^2*(d*x^2+c)^(1/2)/x^7,x, algorithm="giac")

[Out]

1/48*(3*(8*b^2*c^2*d^2 - 4*a*b*c*d^3 + a^2*d^4)*arctan(sqrt(d*x^2 + c)/sqrt(-c))/(sqrt(-c)*c^2) - (24*(d*x^2 +
c)^(5/2)*b^2*c^2*d^2 - 48*(d*x^2 + c)^(3/2)*b^2*c^3*d^2 + 24*sqrt(d*x^2 + c)*b^2*c^4*d^2 + 12*(d*x^2 + c)^(5/
2)*a*b*c*d^3 - 12*sqrt(d*x^2 + c)*a*b*c^3*d^3 - 3*(d*x^2 + c)^(5/2)*a^2*d^4 + 8*(d*x^2 + c)^(3/2)*a^2*c*d^4 +
3*sqrt(d*x^2 + c)*a^2*c^2*d^4)/(c^2*d^3*x^6))/d

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Mupad [B]
time = 0.86, size = 193, normalized size = 1.30 \begin {gather*} \frac {\sqrt {d\,x^2+c}\,\left (\frac {a^2\,d^3}{16}-\frac {a\,b\,c\,d^2}{4}+\frac {b^2\,c^2\,d}{2}\right )+\frac {{\left (d\,x^2+c\right )}^{3/2}\,\left (a^2\,d^3-6\,b^2\,c^2\,d\right )}{6\,c}+\frac {{\left (d\,x^2+c\right )}^{5/2}\,\left (-a^2\,d^3+4\,a\,b\,c\,d^2+8\,b^2\,c^2\,d\right )}{16\,c^2}}{3\,c\,{\left (d\,x^2+c\right )}^2-3\,c^2\,\left (d\,x^2+c\right )-{\left (d\,x^2+c\right )}^3+c^3}-\frac {d\,\mathrm {atanh}\left (\frac {\sqrt {d\,x^2+c}}{\sqrt {c}}\right )\,\left (a^2\,d^2-4\,a\,b\,c\,d+8\,b^2\,c^2\right )}{16\,c^{5/2}} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

int(((a + b*x^2)^2*(c + d*x^2)^(1/2))/x^7,x)

[Out]

((c + d*x^2)^(1/2)*((a^2*d^3)/16 + (b^2*c^2*d)/2 - (a*b*c*d^2)/4) + ((c + d*x^2)^(3/2)*(a^2*d^3 - 6*b^2*c^2*d)
)/(6*c) + ((c + d*x^2)^(5/2)*(8*b^2*c^2*d - a^2*d^3 + 4*a*b*c*d^2))/(16*c^2))/(3*c*(c + d*x^2)^2 - 3*c^2*(c +
d*x^2) - (c + d*x^2)^3 + c^3) - (d*atanh((c + d*x^2)^(1/2)/c^(1/2))*(a^2*d^2 + 8*b^2*c^2 - 4*a*b*c*d))/(16*c^(
5/2))

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