3.7.35 \(\int \frac {(a+b x^2)^2 (c+d x^2)^{5/2}}{x^7} \, dx\) [635]

Optimal. Leaf size=222 \[ \frac {5 d \left (8 b^2 c^2+a d (12 b c+a d)\right ) \sqrt {c+d x^2}}{16 c}+\frac {5 d \left (8 b^2 c^2+a d (12 b c+a d)\right ) \left (c+d x^2\right )^{3/2}}{48 c^2}-\frac {\left (8 b^2 c^2+a d (12 b c+a d)\right ) \left (c+d x^2\right )^{5/2}}{16 c^2 x^2}-\frac {a^2 \left (c+d x^2\right )^{7/2}}{6 c x^6}-\frac {a (12 b c+a d) \left (c+d x^2\right )^{7/2}}{24 c^2 x^4}-\frac {5 d \left (8 b^2 c^2+a d (12 b c+a d)\right ) \tanh ^{-1}\left (\frac {\sqrt {c+d x^2}}{\sqrt {c}}\right )}{16 \sqrt {c}} \]

[Out]

5/48*d*(8*b^2*c^2+a*d*(a*d+12*b*c))*(d*x^2+c)^(3/2)/c^2-1/16*(8*b^2*c^2+a*d*(a*d+12*b*c))*(d*x^2+c)^(5/2)/c^2/
x^2-1/6*a^2*(d*x^2+c)^(7/2)/c/x^6-1/24*a*(a*d+12*b*c)*(d*x^2+c)^(7/2)/c^2/x^4-5/16*d*(8*b^2*c^2+a*d*(a*d+12*b*
c))*arctanh((d*x^2+c)^(1/2)/c^(1/2))/c^(1/2)+5/16*d*(8*b^2*c^2+a*d*(a*d+12*b*c))*(d*x^2+c)^(1/2)/c

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Rubi [A]
time = 0.17, antiderivative size = 219, normalized size of antiderivative = 0.99, number of steps used = 8, number of rules used = 7, integrand size = 24, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.292, Rules used = {457, 91, 79, 43, 52, 65, 214} \begin {gather*} -\frac {a^2 \left (c+d x^2\right )^{7/2}}{6 c x^6}-\frac {\left (c+d x^2\right )^{5/2} \left (\frac {a d (a d+12 b c)}{c^2}+8 b^2\right )}{16 x^2}+\frac {5 d \left (c+d x^2\right )^{3/2} \left (a d (a d+12 b c)+8 b^2 c^2\right )}{48 c^2}+\frac {5 d \sqrt {c+d x^2} \left (a d (a d+12 b c)+8 b^2 c^2\right )}{16 c}-\frac {5 d \left (a d (a d+12 b c)+8 b^2 c^2\right ) \tanh ^{-1}\left (\frac {\sqrt {c+d x^2}}{\sqrt {c}}\right )}{16 \sqrt {c}}-\frac {a \left (c+d x^2\right )^{7/2} (a d+12 b c)}{24 c^2 x^4} \end {gather*}

Antiderivative was successfully verified.

[In]

Int[((a + b*x^2)^2*(c + d*x^2)^(5/2))/x^7,x]

[Out]

(5*d*(8*b^2*c^2 + a*d*(12*b*c + a*d))*Sqrt[c + d*x^2])/(16*c) + (5*d*(8*b^2*c^2 + a*d*(12*b*c + a*d))*(c + d*x
^2)^(3/2))/(48*c^2) - ((8*b^2 + (a*d*(12*b*c + a*d))/c^2)*(c + d*x^2)^(5/2))/(16*x^2) - (a^2*(c + d*x^2)^(7/2)
)/(6*c*x^6) - (a*(12*b*c + a*d)*(c + d*x^2)^(7/2))/(24*c^2*x^4) - (5*d*(8*b^2*c^2 + a*d*(12*b*c + a*d))*ArcTan
h[Sqrt[c + d*x^2]/Sqrt[c]])/(16*Sqrt[c])

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 52

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

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 \left (c+d x^2\right )^{5/2}}{x^7} \, dx &=\frac {1}{2} \text {Subst}\left (\int \frac {(a+b x)^2 (c+d x)^{5/2}}{x^4} \, dx,x,x^2\right )\\ &=-\frac {a^2 \left (c+d x^2\right )^{7/2}}{6 c x^6}+\frac {\text {Subst}\left (\int \frac {\left (\frac {1}{2} a (12 b c+a d)+3 b^2 c x\right ) (c+d x)^{5/2}}{x^3} \, dx,x,x^2\right )}{6 c}\\ &=-\frac {a^2 \left (c+d x^2\right )^{7/2}}{6 c x^6}-\frac {a (12 b c+a d) \left (c+d x^2\right )^{7/2}}{24 c^2 x^4}+\frac {1}{16} \left (8 b^2+\frac {a d (12 b c+a d)}{c^2}\right ) \text {Subst}\left (\int \frac {(c+d x)^{5/2}}{x^2} \, dx,x,x^2\right )\\ &=-\frac {\left (8 b^2+\frac {a d (12 b c+a d)}{c^2}\right ) \left (c+d x^2\right )^{5/2}}{16 x^2}-\frac {a^2 \left (c+d x^2\right )^{7/2}}{6 c x^6}-\frac {a (12 b c+a d) \left (c+d x^2\right )^{7/2}}{24 c^2 x^4}+\frac {1}{32} \left (5 d \left (8 b^2+\frac {a d (12 b c+a d)}{c^2}\right )\right ) \text {Subst}\left (\int \frac {(c+d x)^{3/2}}{x} \, dx,x,x^2\right )\\ &=\frac {5}{48} d \left (8 b^2+\frac {a d (12 b c+a d)}{c^2}\right ) \left (c+d x^2\right )^{3/2}-\frac {\left (8 b^2+\frac {a d (12 b c+a d)}{c^2}\right ) \left (c+d x^2\right )^{5/2}}{16 x^2}-\frac {a^2 \left (c+d x^2\right )^{7/2}}{6 c x^6}-\frac {a (12 b c+a d) \left (c+d x^2\right )^{7/2}}{24 c^2 x^4}+\frac {1}{32} \left (5 c d \left (8 b^2+\frac {a d (12 b c+a d)}{c^2}\right )\right ) \text {Subst}\left (\int \frac {\sqrt {c+d x}}{x} \, dx,x,x^2\right )\\ &=\frac {5}{16} c d \left (8 b^2+\frac {a d (12 b c+a d)}{c^2}\right ) \sqrt {c+d x^2}+\frac {5}{48} d \left (8 b^2+\frac {a d (12 b c+a d)}{c^2}\right ) \left (c+d x^2\right )^{3/2}-\frac {\left (8 b^2+\frac {a d (12 b c+a d)}{c^2}\right ) \left (c+d x^2\right )^{5/2}}{16 x^2}-\frac {a^2 \left (c+d x^2\right )^{7/2}}{6 c x^6}-\frac {a (12 b c+a d) \left (c+d x^2\right )^{7/2}}{24 c^2 x^4}+\frac {1}{32} \left (5 d \left (8 b^2 c^2+12 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 )\\ &=\frac {5}{16} c d \left (8 b^2+\frac {a d (12 b c+a d)}{c^2}\right ) \sqrt {c+d x^2}+\frac {5}{48} d \left (8 b^2+\frac {a d (12 b c+a d)}{c^2}\right ) \left (c+d x^2\right )^{3/2}-\frac {\left (8 b^2+\frac {a d (12 b c+a d)}{c^2}\right ) \left (c+d x^2\right )^{5/2}}{16 x^2}-\frac {a^2 \left (c+d x^2\right )^{7/2}}{6 c x^6}-\frac {a (12 b c+a d) \left (c+d x^2\right )^{7/2}}{24 c^2 x^4}+\frac {1}{16} \left (5 \left (8 b^2 c^2+12 a b c d+a^2 d^2\right )\right ) \text {Subst}\left (\int \frac {1}{-\frac {c}{d}+\frac {x^2}{d}} \, dx,x,\sqrt {c+d x^2}\right )\\ &=\frac {5}{16} c d \left (8 b^2+\frac {a d (12 b c+a d)}{c^2}\right ) \sqrt {c+d x^2}+\frac {5}{48} d \left (8 b^2+\frac {a d (12 b c+a d)}{c^2}\right ) \left (c+d x^2\right )^{3/2}-\frac {\left (8 b^2+\frac {a d (12 b c+a d)}{c^2}\right ) \left (c+d x^2\right )^{5/2}}{16 x^2}-\frac {a^2 \left (c+d x^2\right )^{7/2}}{6 c x^6}-\frac {a (12 b c+a d) \left (c+d x^2\right )^{7/2}}{24 c^2 x^4}-\frac {5 d \left (8 b^2 c^2+12 a b c d+a^2 d^2\right ) \tanh ^{-1}\left (\frac {\sqrt {c+d x^2}}{\sqrt {c}}\right )}{16 \sqrt {c}}\\ \end {align*}

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Mathematica [A]
time = 0.25, size = 152, normalized size = 0.68 \begin {gather*} -\frac {\sqrt {c+d x^2} \left (12 a b x^2 \left (2 c^2+9 c d x^2-8 d^2 x^4\right )-8 b^2 x^4 \left (-3 c^2+14 c d x^2+2 d^2 x^4\right )+a^2 \left (8 c^2+26 c d x^2+33 d^2 x^4\right )\right )}{48 x^6}-\frac {5 d \left (8 b^2 c^2+12 a b c d+a^2 d^2\right ) \tanh ^{-1}\left (\frac {\sqrt {c+d x^2}}{\sqrt {c}}\right )}{16 \sqrt {c}} \end {gather*}

Antiderivative was successfully verified.

[In]

Integrate[((a + b*x^2)^2*(c + d*x^2)^(5/2))/x^7,x]

[Out]

-1/48*(Sqrt[c + d*x^2]*(12*a*b*x^2*(2*c^2 + 9*c*d*x^2 - 8*d^2*x^4) - 8*b^2*x^4*(-3*c^2 + 14*c*d*x^2 + 2*d^2*x^
4) + a^2*(8*c^2 + 26*c*d*x^2 + 33*d^2*x^4)))/x^6 - (5*d*(8*b^2*c^2 + 12*a*b*c*d + a^2*d^2)*ArcTanh[Sqrt[c + d*
x^2]/Sqrt[c]])/(16*Sqrt[c])

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Maple [A]
time = 0.12, size = 356, normalized size = 1.60

method	result	size
risch	\(-\frac {\sqrt {d \,x^{2}+c}\, \left (33 a^{2} d^{2} x^{4}+108 a b c d \,x^{4}+24 b^{2} c^{2} x^{4}+26 a^{2} c d \,x^{2}+24 a b \,c^{2} x^{2}+8 a^{2} c^{2}\right )}{48 x^{6}}+\frac {b^{2} d^{2} x^{2} \sqrt {d \,x^{2}+c}}{3}+\frac {7 d \,b^{2} c \sqrt {d \,x^{2}+c}}{3}+2 a b \,d^{2} \sqrt {d \,x^{2}+c}-\frac {5 \ln \left (\frac {2 c +2 \sqrt {c}\, \sqrt {d \,x^{2}+c}}{x}\right ) a^{2} d^{3}}{16 \sqrt {c}}-\frac {15 \sqrt {c}\, \ln \left (\frac {2 c +2 \sqrt {c}\, \sqrt {d \,x^{2}+c}}{x}\right ) a b \,d^{2}}{4}-\frac {5 d \,c^{\frac {3}{2}} \ln \left (\frac {2 c +2 \sqrt {c}\, \sqrt {d \,x^{2}+c}}{x}\right ) b^{2}}{2}\)	\(227\)
default	\(2 a b \left (-\frac {\left (d \,x^{2}+c \right )^{\frac {7}{2}}}{4 c \,x^{4}}+\frac {3 d \left (-\frac {\left (d \,x^{2}+c \right )^{\frac {7}{2}}}{2 c \,x^{2}}+\frac {5 d \left (\frac {\left (d \,x^{2}+c \right )^{\frac {5}{2}}}{5}+c \left (\frac {\left (d \,x^{2}+c \right )^{\frac {3}{2}}}{3}+c \left (\sqrt {d \,x^{2}+c}-\sqrt {c}\, \ln \left (\frac {2 c +2 \sqrt {c}\, \sqrt {d \,x^{2}+c}}{x}\right )\right )\right )\right )}{2 c}\right )}{4 c}\right )+a^{2} \left (-\frac {\left (d \,x^{2}+c \right )^{\frac {7}{2}}}{6 c \,x^{6}}+\frac {d \left (-\frac {\left (d \,x^{2}+c \right )^{\frac {7}{2}}}{4 c \,x^{4}}+\frac {3 d \left (-\frac {\left (d \,x^{2}+c \right )^{\frac {7}{2}}}{2 c \,x^{2}}+\frac {5 d \left (\frac {\left (d \,x^{2}+c \right )^{\frac {5}{2}}}{5}+c \left (\frac {\left (d \,x^{2}+c \right )^{\frac {3}{2}}}{3}+c \left (\sqrt {d \,x^{2}+c}-\sqrt {c}\, \ln \left (\frac {2 c +2 \sqrt {c}\, \sqrt {d \,x^{2}+c}}{x}\right )\right )\right )\right )}{2 c}\right )}{4 c}\right )}{6 c}\right )+b^{2} \left (-\frac {\left (d \,x^{2}+c \right )^{\frac {7}{2}}}{2 c \,x^{2}}+\frac {5 d \left (\frac {\left (d \,x^{2}+c \right )^{\frac {5}{2}}}{5}+c \left (\frac {\left (d \,x^{2}+c \right )^{\frac {3}{2}}}{3}+c \left (\sqrt {d \,x^{2}+c}-\sqrt {c}\, \ln \left (\frac {2 c +2 \sqrt {c}\, \sqrt {d \,x^{2}+c}}{x}\right )\right )\right )\right )}{2 c}\right )\)	\(356\)

Verification of antiderivative is not currently implemented for this CAS.

[In]

int((b*x^2+a)^2*(d*x^2+c)^(5/2)/x^7,x,method=_RETURNVERBOSE)

[Out]

2*a*b*(-1/4/c/x^4*(d*x^2+c)^(7/2)+3/4*d/c*(-1/2/c/x^2*(d*x^2+c)^(7/2)+5/2*d/c*(1/5*(d*x^2+c)^(5/2)+c*(1/3*(d*x
^2+c)^(3/2)+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)^(7
/2)+1/6*d/c*(-1/4/c/x^4*(d*x^2+c)^(7/2)+3/4*d/c*(-1/2/c/x^2*(d*x^2+c)^(7/2)+5/2*d/c*(1/5*(d*x^2+c)^(5/2)+c*(1/
3*(d*x^2+c)^(3/2)+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)^(7/2)+5/2*d/c*(1/5*(d*x^2+c)^(5/2)+c*(1/3*(d*x^2+c)^(3/2)+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 = 0.31, size = 353, normalized size = 1.59 \begin {gather*} -\frac {5}{2} \, b^{2} c^{\frac {3}{2}} d \operatorname {arsinh}\left (\frac {c}{\sqrt {c d} {\left | x \right |}}\right ) - \frac {15}{4} \, a b \sqrt {c} d^{2} \operatorname {arsinh}\left (\frac {c}{\sqrt {c d} {\left | x \right |}}\right ) - \frac {5 \, a^{2} d^{3} \operatorname {arsinh}\left (\frac {c}{\sqrt {c d} {\left | x \right |}}\right )}{16 \, \sqrt {c}} + \frac {5}{6} \, {\left (d x^{2} + c\right )}^{\frac {3}{2}} b^{2} d + \frac {{\left (d x^{2} + c\right )}^{\frac {5}{2}} b^{2} d}{2 \, c} + \frac {5}{2} \, \sqrt {d x^{2} + c} b^{2} c d + \frac {15}{4} \, \sqrt {d x^{2} + c} a b d^{2} + \frac {3 \, {\left (d x^{2} + c\right )}^{\frac {5}{2}} a b d^{2}}{4 \, c^{2}} + \frac {5 \, {\left (d x^{2} + c\right )}^{\frac {3}{2}} a b d^{2}}{4 \, c} + \frac {{\left (d x^{2} + c\right )}^{\frac {5}{2}} a^{2} d^{3}}{16 \, c^{3}} + \frac {5 \, {\left (d x^{2} + c\right )}^{\frac {3}{2}} a^{2} d^{3}}{48 \, c^{2}} + \frac {5 \, \sqrt {d x^{2} + c} a^{2} d^{3}}{16 \, c} - \frac {{\left (d x^{2} + c\right )}^{\frac {7}{2}} b^{2}}{2 \, c x^{2}} - \frac {3 \, {\left (d x^{2} + c\right )}^{\frac {7}{2}} a b d}{4 \, c^{2} x^{2}} - \frac {{\left (d x^{2} + c\right )}^{\frac {7}{2}} a^{2} d^{2}}{16 \, c^{3} x^{2}} - \frac {{\left (d x^{2} + c\right )}^{\frac {7}{2}} a b}{2 \, c x^{4}} - \frac {{\left (d x^{2} + c\right )}^{\frac {7}{2}} a^{2} d}{24 \, c^{2} x^{4}} - \frac {{\left (d x^{2} + c\right )}^{\frac {7}{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)^(5/2)/x^7,x, algorithm="maxima")

[Out]

-5/2*b^2*c^(3/2)*d*arcsinh(c/(sqrt(c*d)*abs(x))) - 15/4*a*b*sqrt(c)*d^2*arcsinh(c/(sqrt(c*d)*abs(x))) - 5/16*a
^2*d^3*arcsinh(c/(sqrt(c*d)*abs(x)))/sqrt(c) + 5/6*(d*x^2 + c)^(3/2)*b^2*d + 1/2*(d*x^2 + c)^(5/2)*b^2*d/c + 5
/2*sqrt(d*x^2 + c)*b^2*c*d + 15/4*sqrt(d*x^2 + c)*a*b*d^2 + 3/4*(d*x^2 + c)^(5/2)*a*b*d^2/c^2 + 5/4*(d*x^2 + c
)^(3/2)*a*b*d^2/c + 1/16*(d*x^2 + c)^(5/2)*a^2*d^3/c^3 + 5/48*(d*x^2 + c)^(3/2)*a^2*d^3/c^2 + 5/16*sqrt(d*x^2
+ c)*a^2*d^3/c - 1/2*(d*x^2 + c)^(7/2)*b^2/(c*x^2) - 3/4*(d*x^2 + c)^(7/2)*a*b*d/(c^2*x^2) - 1/16*(d*x^2 + c)^
(7/2)*a^2*d^2/(c^3*x^2) - 1/2*(d*x^2 + c)^(7/2)*a*b/(c*x^4) - 1/24*(d*x^2 + c)^(7/2)*a^2*d/(c^2*x^4) - 1/6*(d*
x^2 + c)^(7/2)*a^2/(c*x^6)

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Fricas [A]
time = 1.84, size = 347, normalized size = 1.56 \begin {gather*} \left [\frac {15 \, {\left (8 \, b^{2} c^{2} d + 12 \, 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 (16 \, b^{2} c d^{2} x^{8} + 16 \, {\left (7 \, b^{2} c^{2} d + 6 \, a b c d^{2}\right )} x^{6} - 8 \, a^{2} c^{3} - 3 \, {\left (8 \, b^{2} c^{3} + 36 \, a b c^{2} d + 11 \, a^{2} c d^{2}\right )} x^{4} - 2 \, {\left (12 \, a b c^{3} + 13 \, a^{2} c^{2} d\right )} x^{2}\right )} \sqrt {d x^{2} + c}}{96 \, c x^{6}}, \frac {15 \, {\left (8 \, b^{2} c^{2} d + 12 \, 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 (16 \, b^{2} c d^{2} x^{8} + 16 \, {\left (7 \, b^{2} c^{2} d + 6 \, a b c d^{2}\right )} x^{6} - 8 \, a^{2} c^{3} - 3 \, {\left (8 \, b^{2} c^{3} + 36 \, a b c^{2} d + 11 \, a^{2} c d^{2}\right )} x^{4} - 2 \, {\left (12 \, a b c^{3} + 13 \, a^{2} c^{2} d\right )} x^{2}\right )} \sqrt {d x^{2} + c}}{48 \, c 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)^(5/2)/x^7,x, algorithm="fricas")

[Out]

[1/96*(15*(8*b^2*c^2*d + 12*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*(16*b^2*c*d^2*x^8 + 16*(7*b^2*c^2*d + 6*a*b*c*d^2)*x^6 - 8*a^2*c^3 - 3*(8*b^2*c^3 + 36*a*b*c^2*d + 11*a
^2*c*d^2)*x^4 - 2*(12*a*b*c^3 + 13*a^2*c^2*d)*x^2)*sqrt(d*x^2 + c))/(c*x^6), 1/48*(15*(8*b^2*c^2*d + 12*a*b*c*
d^2 + a^2*d^3)*sqrt(-c)*x^6*arctan(sqrt(-c)/sqrt(d*x^2 + c)) + (16*b^2*c*d^2*x^8 + 16*(7*b^2*c^2*d + 6*a*b*c*d
^2)*x^6 - 8*a^2*c^3 - 3*(8*b^2*c^3 + 36*a*b*c^2*d + 11*a^2*c*d^2)*x^4 - 2*(12*a*b*c^3 + 13*a^2*c^2*d)*x^2)*sqr
t(d*x^2 + c))/(c*x^6)]

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Sympy [A]
time = 117.70, size = 468, normalized size = 2.11 \begin {gather*} - \frac {a^{2} c^{3}}{6 \sqrt {d} x^{7} \sqrt {\frac {c}{d x^{2}} + 1}} - \frac {17 a^{2} c^{2} \sqrt {d}}{24 x^{5} \sqrt {\frac {c}{d x^{2}} + 1}} - \frac {35 a^{2} c d^{\frac {3}{2}}}{48 x^{3} \sqrt {\frac {c}{d x^{2}} + 1}} - \frac {a^{2} d^{\frac {5}{2}} \sqrt {\frac {c}{d x^{2}} + 1}}{2 x} - \frac {3 a^{2} d^{\frac {5}{2}}}{16 x \sqrt {\frac {c}{d x^{2}} + 1}} - \frac {5 a^{2} d^{3} \operatorname {asinh}{\left (\frac {\sqrt {c}}{\sqrt {d} x} \right )}}{16 \sqrt {c}} - \frac {15 a b \sqrt {c} d^{2} \operatorname {asinh}{\left (\frac {\sqrt {c}}{\sqrt {d} x} \right )}}{4} - \frac {a b c^{3}}{2 \sqrt {d} x^{5} \sqrt {\frac {c}{d x^{2}} + 1}} - \frac {3 a b c^{2} \sqrt {d}}{4 x^{3} \sqrt {\frac {c}{d x^{2}} + 1}} - \frac {2 a b c d^{\frac {3}{2}} \sqrt {\frac {c}{d x^{2}} + 1}}{x} + \frac {7 a b c d^{\frac {3}{2}}}{4 x \sqrt {\frac {c}{d x^{2}} + 1}} + \frac {2 a b d^{\frac {5}{2}} x}{\sqrt {\frac {c}{d x^{2}} + 1}} - \frac {5 b^{2} c^{\frac {3}{2}} d \operatorname {asinh}{\left (\frac {\sqrt {c}}{\sqrt {d} x} \right )}}{2} - \frac {b^{2} c^{2} \sqrt {d} \sqrt {\frac {c}{d x^{2}} + 1}}{2 x} + \frac {2 b^{2} c^{2} \sqrt {d}}{x \sqrt {\frac {c}{d x^{2}} + 1}} + \frac {2 b^{2} c d^{\frac {3}{2}} x}{\sqrt {\frac {c}{d x^{2}} + 1}} + b^{2} d^{2} \left (\begin {cases} \frac {\sqrt {c} x^{2}}{2} & \text {for}\: d = 0 \\\frac {\left (c + d x^{2}\right )^{\frac {3}{2}}}{3 d} & \text {otherwise} \end {cases}\right ) \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate((b*x**2+a)**2*(d*x**2+c)**(5/2)/x**7,x)

[Out]

-a**2*c**3/(6*sqrt(d)*x**7*sqrt(c/(d*x**2) + 1)) - 17*a**2*c**2*sqrt(d)/(24*x**5*sqrt(c/(d*x**2) + 1)) - 35*a*
*2*c*d**(3/2)/(48*x**3*sqrt(c/(d*x**2) + 1)) - a**2*d**(5/2)*sqrt(c/(d*x**2) + 1)/(2*x) - 3*a**2*d**(5/2)/(16*
x*sqrt(c/(d*x**2) + 1)) - 5*a**2*d**3*asinh(sqrt(c)/(sqrt(d)*x))/(16*sqrt(c)) - 15*a*b*sqrt(c)*d**2*asinh(sqrt
(c)/(sqrt(d)*x))/4 - a*b*c**3/(2*sqrt(d)*x**5*sqrt(c/(d*x**2) + 1)) - 3*a*b*c**2*sqrt(d)/(4*x**3*sqrt(c/(d*x**
2) + 1)) - 2*a*b*c*d**(3/2)*sqrt(c/(d*x**2) + 1)/x + 7*a*b*c*d**(3/2)/(4*x*sqrt(c/(d*x**2) + 1)) + 2*a*b*d**(5
/2)*x/sqrt(c/(d*x**2) + 1) - 5*b**2*c**(3/2)*d*asinh(sqrt(c)/(sqrt(d)*x))/2 - b**2*c**2*sqrt(d)*sqrt(c/(d*x**2
) + 1)/(2*x) + 2*b**2*c**2*sqrt(d)/(x*sqrt(c/(d*x**2) + 1)) + 2*b**2*c*d**(3/2)*x/sqrt(c/(d*x**2) + 1) + b**2*
d**2*Piecewise((sqrt(c)*x**2/2, Eq(d, 0)), ((c + d*x**2)**(3/2)/(3*d), True))

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Giac [A]
time = 0.57, size = 286, normalized size = 1.29 \begin {gather*} \frac {16 \, {\left (d x^{2} + c\right )}^{\frac {3}{2}} b^{2} d^{2} + 96 \, \sqrt {d x^{2} + c} b^{2} c d^{2} + 96 \, \sqrt {d x^{2} + c} a b d^{3} + \frac {15 \, {\left (8 \, b^{2} c^{2} d^{2} + 12 \, a b c d^{3} + a^{2} d^{4}\right )} \arctan \left (\frac {\sqrt {d x^{2} + c}}{\sqrt {-c}}\right )}{\sqrt {-c}} - \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} + 108 \, {\left (d x^{2} + c\right )}^{\frac {5}{2}} a b c d^{3} - 192 \, {\left (d x^{2} + c\right )}^{\frac {3}{2}} a b c^{2} d^{3} + 84 \, \sqrt {d x^{2} + c} a b c^{3} d^{3} + 33 \, {\left (d x^{2} + c\right )}^{\frac {5}{2}} a^{2} d^{4} - 40 \, {\left (d x^{2} + c\right )}^{\frac {3}{2}} a^{2} c d^{4} + 15 \, \sqrt {d x^{2} + c} a^{2} c^{2} d^{4}}{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)^(5/2)/x^7,x, algorithm="giac")

[Out]

1/48*(16*(d*x^2 + c)^(3/2)*b^2*d^2 + 96*sqrt(d*x^2 + c)*b^2*c*d^2 + 96*sqrt(d*x^2 + c)*a*b*d^3 + 15*(8*b^2*c^2
*d^2 + 12*a*b*c*d^3 + a^2*d^4)*arctan(sqrt(d*x^2 + c)/sqrt(-c))/sqrt(-c) - (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 + 108*(d*x^2 + c)^(5/2)*a*b*c*d^3 - 192*(d*
x^2 + c)^(3/2)*a*b*c^2*d^3 + 84*sqrt(d*x^2 + c)*a*b*c^3*d^3 + 33*(d*x^2 + c)^(5/2)*a^2*d^4 - 40*(d*x^2 + c)^(3
/2)*a^2*c*d^4 + 15*sqrt(d*x^2 + c)*a^2*c^2*d^4)/(d^3*x^6))/d

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Mupad [B]
time = 1.31, size = 301, normalized size = 1.36 \begin {gather*} \frac {\sqrt {d\,x^2+c}\,\left (\frac {5\,a^2\,c^2\,d^3}{16}+\frac {7\,a\,b\,c^3\,d^2}{4}+\frac {b^2\,c^4\,d}{2}\right )-{\left (d\,x^2+c\right )}^{3/2}\,\left (\frac {5\,a^2\,c\,d^3}{6}+4\,a\,b\,c^2\,d^2+b^2\,c^3\,d\right )+{\left (d\,x^2+c\right )}^{5/2}\,\left (\frac {11\,a^2\,d^3}{16}+\frac {9\,a\,b\,c\,d^2}{4}+\frac {b^2\,c^2\,d}{2}\right )}{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}+\left (2\,b\,d\,\left (a\,d-b\,c\right )+4\,b^2\,c\,d\right )\,\sqrt {d\,x^2+c}+\frac {b^2\,d\,{\left (d\,x^2+c\right )}^{3/2}}{3}+\frac {d\,\mathrm {atan}\left (\frac {d\,\sqrt {d\,x^2+c}\,\left (a^2\,d^2+12\,a\,b\,c\,d+8\,b^2\,c^2\right )\,5{}\mathrm {i}}{8\,\sqrt {c}\,\left (\frac {5\,a^2\,d^3}{8}+\frac {15\,a\,b\,c\,d^2}{2}+5\,b^2\,c^2\,d\right )}\right )\,\left (a^2\,d^2+12\,a\,b\,c\,d+8\,b^2\,c^2\right )\,5{}\mathrm {i}}{16\,\sqrt {c}} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

int(((a + b*x^2)^2*(c + d*x^2)^(5/2))/x^7,x)

[Out]

((c + d*x^2)^(1/2)*((b^2*c^4*d)/2 + (5*a^2*c^2*d^3)/16 + (7*a*b*c^3*d^2)/4) - (c + d*x^2)^(3/2)*((5*a^2*c*d^3)
/6 + b^2*c^3*d + 4*a*b*c^2*d^2) + (c + d*x^2)^(5/2)*((11*a^2*d^3)/16 + (b^2*c^2*d)/2 + (9*a*b*c*d^2)/4))/(3*c*
(c + d*x^2)^2 - 3*c^2*(c + d*x^2) - (c + d*x^2)^3 + c^3) + (2*b*d*(a*d - b*c) + 4*b^2*c*d)*(c + d*x^2)^(1/2) +
(b^2*d*(c + d*x^2)^(3/2))/3 + (d*atan((d*(c + d*x^2)^(1/2)*(a^2*d^2 + 8*b^2*c^2 + 12*a*b*c*d)*5i)/(8*c^(1/2)*
((5*a^2*d^3)/8 + 5*b^2*c^2*d + (15*a*b*c*d^2)/2)))*(a^2*d^2 + 8*b^2*c^2 + 12*a*b*c*d)*5i)/(16*c^(1/2))

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