\(\int \frac {(a+b x^2)^2 (c+d x^2)^{5/2}}{x^6} \, dx\) [634]

Optimal result

Integrand size = 24, antiderivative size = 228 \[ \int \frac {\left (a+b x^2\right )^2 \left (c+d x^2\right )^{5/2}}{x^6} \, dx=\frac {d \left (15 b^2 c^2+8 a d (5 b c+a d)\right ) x \sqrt {c+d x^2}}{8 c}+\frac {d \left (15 b^2 c^2+8 a d (5 b c+a d)\right ) x \left (c+d x^2\right )^{3/2}}{12 c^2}-\frac {\left (15 b^2 c^2+8 a d (5 b c+a d)\right ) \left (c+d x^2\right )^{5/2}}{15 c^2 x}-\frac {a^2 \left (c+d x^2\right )^{7/2}}{5 c x^5}-\frac {2 a (5 b c+a d) \left (c+d x^2\right )^{7/2}}{15 c^2 x^3}+\frac {1}{8} \sqrt {d} \left (15 b^2 c^2+8 a d (5 b c+a d)\right ) \text {arctanh}\left (\frac {\sqrt {d} x}{\sqrt {c+d x^2}}\right ) \]

[Out]

1/12*d*(15*b^2*c^2+8*a*d*(a*d+5*b*c))*x*(d*x^2+c)^(3/2)/c^2-1/15*(15*b^2*c^2+8*a*d*(a*d+5*b*c))*(d*x^2+c)^(5/2
)/c^2/x-1/5*a^2*(d*x^2+c)^(7/2)/c/x^5-2/15*a*(a*d+5*b*c)*(d*x^2+c)^(7/2)/c^2/x^3+1/8*(15*b^2*c^2+8*a*d*(a*d+5*
b*c))*arctanh(x*d^(1/2)/(d*x^2+c)^(1/2))*d^(1/2)+1/8*d*(15*b^2*c^2+8*a*d*(a*d+5*b*c))*x*(d*x^2+c)^(1/2)/c

Rubi [A] (verified)

Time = 0.13 (sec) , antiderivative size = 225, normalized size of antiderivative = 0.99, number of steps used = 7, number of rules used = 6, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.250, Rules used = {473, 464, 283, 201, 223, 212} \[ \int \frac {\left (a+b x^2\right )^2 \left (c+d x^2\right )^{5/2}}{x^6} \, dx=-\frac {a^2 \left (c+d x^2\right )^{7/2}}{5 c x^5}+\frac {1}{8} \sqrt {d} \left (8 a d (a d+5 b c)+15 b^2 c^2\right ) \text {arctanh}\left (\frac {\sqrt {d} x}{\sqrt {c+d x^2}}\right )-\frac {\left (c+d x^2\right )^{5/2} \left (\frac {8 a d (a d+5 b c)}{c^2}+15 b^2\right )}{15 x}+\frac {d x \left (c+d x^2\right )^{3/2} \left (8 a d (a d+5 b c)+15 b^2 c^2\right )}{12 c^2}+\frac {d x \sqrt {c+d x^2} \left (8 a d (a d+5 b c)+15 b^2 c^2\right )}{8 c}-\frac {2 a \left (c+d x^2\right )^{7/2} (a d+5 b c)}{15 c^2 x^3} \]

[In]

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

[Out]

(d*(15*b^2*c^2 + 8*a*d*(5*b*c + a*d))*x*Sqrt[c + d*x^2])/(8*c) + (d*(15*b^2*c^2 + 8*a*d*(5*b*c + a*d))*x*(c +
d*x^2)^(3/2))/(12*c^2) - ((15*b^2 + (8*a*d*(5*b*c + a*d))/c^2)*(c + d*x^2)^(5/2))/(15*x) - (a^2*(c + d*x^2)^(7
/2))/(5*c*x^5) - (2*a*(5*b*c + a*d)*(c + d*x^2)^(7/2))/(15*c^2*x^3) + (Sqrt[d]*(15*b^2*c^2 + 8*a*d*(5*b*c + a*
d))*ArcTanh[(Sqrt[d]*x)/Sqrt[c + d*x^2]])/8

Rule 201

Int[((a_) + (b_.)*(x_)^(n_))^(p_), x_Symbol] :> Simp[x*((a + b*x^n)^p/(n*p + 1)), x] + Dist[a*n*(p/(n*p + 1)),
Int[(a + b*x^n)^(p - 1), x], x] /; FreeQ[{a, b}, x] && IGtQ[n, 0] && GtQ[p, 0] && (IntegerQ[2*p] || (EqQ[n, 2
] && IntegerQ[4*p]) || (EqQ[n, 2] && IntegerQ[3*p]) || LtQ[Denominator[p + 1/n], Denominator[p]])

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 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 283

Int[((c_.)*(x_))^(m_.)*((a_) + (b_.)*(x_)^(n_))^(p_), x_Symbol] :> Simp[(c*x)^(m + 1)*((a + b*x^n)^p/(c*(m + 1
))), x] - Dist[b*n*(p/(c^n*(m + 1))), Int[(c*x)^(m + n)*(a + b*x^n)^(p - 1), x], x] /; FreeQ[{a, b, c}, x] &&
IGtQ[n, 0] && GtQ[p, 0] && LtQ[m, -1] &&  !ILtQ[(m + n*p + n + 1)/n, 0] && IntBinomialQ[a, b, c, n, m, p, x]

Rule 464

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

Rule 473

Int[((e_.)*(x_))^(m_)*((a_) + (b_.)*(x_)^(n_))^(p_)*((c_) + (d_.)*(x_)^(n_))^2, x_Symbol] :> Simp[c^2*(e*x)^(m
+ 1)*((a + b*x^n)^(p + 1)/(a*e*(m + 1))), x] - Dist[1/(a*e^n*(m + 1)), Int[(e*x)^(m + n)*(a + b*x^n)^p*Simp[b
*c^2*n*(p + 1) + c*(b*c - 2*a*d)*(m + 1) - a*(m + 1)*d^2*x^n, x], x], x] /; FreeQ[{a, b, c, d, e, p}, x] && Ne
Q[b*c - a*d, 0] && IGtQ[n, 0] && LtQ[m, -1] && GtQ[n, 0]

Rubi steps \begin{align*} \text {integral}& = -\frac {a^2 \left (c+d x^2\right )^{7/2}}{5 c x^5}+\frac {\int \frac {\left (2 a (5 b c+a d)+5 b^2 c x^2\right ) \left (c+d x^2\right )^{5/2}}{x^4} \, dx}{5 c} \\ & = -\frac {a^2 \left (c+d x^2\right )^{7/2}}{5 c x^5}-\frac {2 a (5 b c+a d) \left (c+d x^2\right )^{7/2}}{15 c^2 x^3}-\frac {1}{15} \left (-15 b^2-\frac {8 a d (5 b c+a d)}{c^2}\right ) \int \frac {\left (c+d x^2\right )^{5/2}}{x^2} \, dx \\ & = -\frac {\left (15 b^2+\frac {8 a d (5 b c+a d)}{c^2}\right ) \left (c+d x^2\right )^{5/2}}{15 x}-\frac {a^2 \left (c+d x^2\right )^{7/2}}{5 c x^5}-\frac {2 a (5 b c+a d) \left (c+d x^2\right )^{7/2}}{15 c^2 x^3}+\frac {1}{3} \left (d \left (15 b^2+\frac {8 a d (5 b c+a d)}{c^2}\right )\right ) \int \left (c+d x^2\right )^{3/2} \, dx \\ & = \frac {1}{12} d \left (15 b^2+\frac {8 a d (5 b c+a d)}{c^2}\right ) x \left (c+d x^2\right )^{3/2}-\frac {\left (15 b^2+\frac {8 a d (5 b c+a d)}{c^2}\right ) \left (c+d x^2\right )^{5/2}}{15 x}-\frac {a^2 \left (c+d x^2\right )^{7/2}}{5 c x^5}-\frac {2 a (5 b c+a d) \left (c+d x^2\right )^{7/2}}{15 c^2 x^3}+\frac {1}{4} \left (c d \left (15 b^2+\frac {8 a d (5 b c+a d)}{c^2}\right )\right ) \int \sqrt {c+d x^2} \, dx \\ & = \frac {1}{8} c d \left (15 b^2+\frac {8 a d (5 b c+a d)}{c^2}\right ) x \sqrt {c+d x^2}+\frac {1}{12} d \left (15 b^2+\frac {8 a d (5 b c+a d)}{c^2}\right ) x \left (c+d x^2\right )^{3/2}-\frac {\left (15 b^2+\frac {8 a d (5 b c+a d)}{c^2}\right ) \left (c+d x^2\right )^{5/2}}{15 x}-\frac {a^2 \left (c+d x^2\right )^{7/2}}{5 c x^5}-\frac {2 a (5 b c+a d) \left (c+d x^2\right )^{7/2}}{15 c^2 x^3}+\frac {1}{8} \left (d \left (15 b^2 c^2+40 a b c d+8 a^2 d^2\right )\right ) \int \frac {1}{\sqrt {c+d x^2}} \, dx \\ & = \frac {1}{8} c d \left (15 b^2+\frac {8 a d (5 b c+a d)}{c^2}\right ) x \sqrt {c+d x^2}+\frac {1}{12} d \left (15 b^2+\frac {8 a d (5 b c+a d)}{c^2}\right ) x \left (c+d x^2\right )^{3/2}-\frac {\left (15 b^2+\frac {8 a d (5 b c+a d)}{c^2}\right ) \left (c+d x^2\right )^{5/2}}{15 x}-\frac {a^2 \left (c+d x^2\right )^{7/2}}{5 c x^5}-\frac {2 a (5 b c+a d) \left (c+d x^2\right )^{7/2}}{15 c^2 x^3}+\frac {1}{8} \left (d \left (15 b^2 c^2+40 a b c d+8 a^2 d^2\right )\right ) \text {Subst}\left (\int \frac {1}{1-d x^2} \, dx,x,\frac {x}{\sqrt {c+d x^2}}\right ) \\ & = \frac {1}{8} c d \left (15 b^2+\frac {8 a d (5 b c+a d)}{c^2}\right ) x \sqrt {c+d x^2}+\frac {1}{12} d \left (15 b^2+\frac {8 a d (5 b c+a d)}{c^2}\right ) x \left (c+d x^2\right )^{3/2}-\frac {\left (15 b^2+\frac {8 a d (5 b c+a d)}{c^2}\right ) \left (c+d x^2\right )^{5/2}}{15 x}-\frac {a^2 \left (c+d x^2\right )^{7/2}}{5 c x^5}-\frac {2 a (5 b c+a d) \left (c+d x^2\right )^{7/2}}{15 c^2 x^3}+\frac {1}{8} \sqrt {d} \left (15 b^2 c^2+40 a b c d+8 a^2 d^2\right ) \tanh ^{-1}\left (\frac {\sqrt {d} x}{\sqrt {c+d x^2}}\right ) \\ \end{align*}

Mathematica [A] (verified)

Time = 0.40 (sec) , antiderivative size = 164, normalized size of antiderivative = 0.72 \[ \int \frac {\left (a+b x^2\right )^2 \left (c+d x^2\right )^{5/2}}{x^6} \, dx=\frac {\sqrt {c+d x^2} \left (15 b^2 x^4 \left (-8 c^2+9 c d x^2+2 d^2 x^4\right )+40 a b x^2 \left (-2 c^2-14 c d x^2+3 d^2 x^4\right )-8 a^2 \left (3 c^2+11 c d x^2+23 d^2 x^4\right )\right )}{120 x^5}+\frac {1}{4} \sqrt {d} \left (15 b^2 c^2+40 a b c d+8 a^2 d^2\right ) \text {arctanh}\left (\frac {\sqrt {d} x}{-\sqrt {c}+\sqrt {c+d x^2}}\right ) \]

[In]

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

[Out]

(Sqrt[c + d*x^2]*(15*b^2*x^4*(-8*c^2 + 9*c*d*x^2 + 2*d^2*x^4) + 40*a*b*x^2*(-2*c^2 - 14*c*d*x^2 + 3*d^2*x^4) -
8*a^2*(3*c^2 + 11*c*d*x^2 + 23*d^2*x^4)))/(120*x^5) + (Sqrt[d]*(15*b^2*c^2 + 40*a*b*c*d + 8*a^2*d^2)*ArcTanh[
(Sqrt[d]*x)/(-Sqrt[c] + Sqrt[c + d*x^2])])/4

Maple [A] (verified)

Time = 2.98 (sec) , antiderivative size = 148, normalized size of antiderivative = 0.65

method	result	size
pseudoelliptic	\(-\frac {-5 x^{5} \left (a^{2} d^{2}+5 a b c d +\frac {15}{8} b^{2} c^{2}\right ) d \,\operatorname {arctanh}\left (\frac {\sqrt {d \,x^{2}+c}}{x \sqrt {d}}\right )+\sqrt {d \,x^{2}+c}\, \left (\frac {11 x^{2} \left (-\frac {135}{88} b^{2} x^{4}+\frac {70}{11} a b \,x^{2}+a^{2}\right ) c \,d^{\frac {3}{2}}}{3}+\left (-\frac {5}{4} b^{2} x^{8}-5 a b \,x^{6}+\frac {23}{3} a^{2} x^{4}\right ) d^{\frac {5}{2}}+c^{2} \sqrt {d}\, \left (5 b^{2} x^{4}+\frac {10}{3} a b \,x^{2}+a^{2}\right )\right )}{5 \sqrt {d}\, x^{5}}\)	\(148\)
risch	\(-\frac {\sqrt {d \,x^{2}+c}\, \left (-30 b^{2} d^{2} x^{8}-120 a b \,d^{2} x^{6}-135 b^{2} c d \,x^{6}+184 a^{2} d^{2} x^{4}+560 x^{4} a b c d +120 b^{2} c^{2} x^{4}+88 a^{2} c d \,x^{2}+80 a b \,c^{2} x^{2}+24 a^{2} c^{2}\right )}{120 x^{5}}+\frac {\sqrt {d}\, \left (8 a^{2} d^{2}+40 a b c d +15 b^{2} c^{2}\right ) \ln \left (x \sqrt {d}+\sqrt {d \,x^{2}+c}\right )}{8}\)	\(151\)
default	\(b^{2} \left (-\frac {\left (d \,x^{2}+c \right )^{\frac {7}{2}}}{c x}+\frac {6 d \left (\frac {x \left (d \,x^{2}+c \right )^{\frac {5}{2}}}{6}+\frac {5 c \left (\frac {x \left (d \,x^{2}+c \right )^{\frac {3}{2}}}{4}+\frac {3 c \left (\frac {x \sqrt {d \,x^{2}+c}}{2}+\frac {c \ln \left (x \sqrt {d}+\sqrt {d \,x^{2}+c}\right )}{2 \sqrt {d}}\right )}{4}\right )}{6}\right )}{c}\right )+a^{2} \left (-\frac {\left (d \,x^{2}+c \right )^{\frac {7}{2}}}{5 c \,x^{5}}+\frac {2 d \left (-\frac {\left (d \,x^{2}+c \right )^{\frac {7}{2}}}{3 c \,x^{3}}+\frac {4 d \left (-\frac {\left (d \,x^{2}+c \right )^{\frac {7}{2}}}{c x}+\frac {6 d \left (\frac {x \left (d \,x^{2}+c \right )^{\frac {5}{2}}}{6}+\frac {5 c \left (\frac {x \left (d \,x^{2}+c \right )^{\frac {3}{2}}}{4}+\frac {3 c \left (\frac {x \sqrt {d \,x^{2}+c}}{2}+\frac {c \ln \left (x \sqrt {d}+\sqrt {d \,x^{2}+c}\right )}{2 \sqrt {d}}\right )}{4}\right )}{6}\right )}{c}\right )}{3 c}\right )}{5 c}\right )+2 a b \left (-\frac {\left (d \,x^{2}+c \right )^{\frac {7}{2}}}{3 c \,x^{3}}+\frac {4 d \left (-\frac {\left (d \,x^{2}+c \right )^{\frac {7}{2}}}{c x}+\frac {6 d \left (\frac {x \left (d \,x^{2}+c \right )^{\frac {5}{2}}}{6}+\frac {5 c \left (\frac {x \left (d \,x^{2}+c \right )^{\frac {3}{2}}}{4}+\frac {3 c \left (\frac {x \sqrt {d \,x^{2}+c}}{2}+\frac {c \ln \left (x \sqrt {d}+\sqrt {d \,x^{2}+c}\right )}{2 \sqrt {d}}\right )}{4}\right )}{6}\right )}{c}\right )}{3 c}\right )\)	\(359\)

[In]

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

[Out]

-1/5/d^(1/2)*(-5*x^5*(a^2*d^2+5*a*b*c*d+15/8*b^2*c^2)*d*arctanh((d*x^2+c)^(1/2)/x/d^(1/2))+(d*x^2+c)^(1/2)*(11
/3*x^2*(-135/88*b^2*x^4+70/11*a*b*x^2+a^2)*c*d^(3/2)+(-5/4*b^2*x^8-5*a*b*x^6+23/3*a^2*x^4)*d^(5/2)+c^2*d^(1/2)
*(5*b^2*x^4+10/3*a*b*x^2+a^2)))/x^5

Fricas [A] (verification not implemented)

none

Time = 0.30 (sec) , antiderivative size = 318, normalized size of antiderivative = 1.39 \[ \int \frac {\left (a+b x^2\right )^2 \left (c+d x^2\right )^{5/2}}{x^6} \, dx=\left [\frac {15 \, {\left (15 \, b^{2} c^{2} + 40 \, a b c d + 8 \, a^{2} d^{2}\right )} \sqrt {d} x^{5} \log \left (-2 \, d x^{2} - 2 \, \sqrt {d x^{2} + c} \sqrt {d} x - c\right ) + 2 \, {\left (30 \, b^{2} d^{2} x^{8} + 15 \, {\left (9 \, b^{2} c d + 8 \, a b d^{2}\right )} x^{6} - 8 \, {\left (15 \, b^{2} c^{2} + 70 \, a b c d + 23 \, a^{2} d^{2}\right )} x^{4} - 24 \, a^{2} c^{2} - 8 \, {\left (10 \, a b c^{2} + 11 \, a^{2} c d\right )} x^{2}\right )} \sqrt {d x^{2} + c}}{240 \, x^{5}}, -\frac {15 \, {\left (15 \, b^{2} c^{2} + 40 \, a b c d + 8 \, a^{2} d^{2}\right )} \sqrt {-d} x^{5} \arctan \left (\frac {\sqrt {-d} x}{\sqrt {d x^{2} + c}}\right ) - {\left (30 \, b^{2} d^{2} x^{8} + 15 \, {\left (9 \, b^{2} c d + 8 \, a b d^{2}\right )} x^{6} - 8 \, {\left (15 \, b^{2} c^{2} + 70 \, a b c d + 23 \, a^{2} d^{2}\right )} x^{4} - 24 \, a^{2} c^{2} - 8 \, {\left (10 \, a b c^{2} + 11 \, a^{2} c d\right )} x^{2}\right )} \sqrt {d x^{2} + c}}{120 \, x^{5}}\right ] \]

[In]

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

[Out]

[1/240*(15*(15*b^2*c^2 + 40*a*b*c*d + 8*a^2*d^2)*sqrt(d)*x^5*log(-2*d*x^2 - 2*sqrt(d*x^2 + c)*sqrt(d)*x - c) +
2*(30*b^2*d^2*x^8 + 15*(9*b^2*c*d + 8*a*b*d^2)*x^6 - 8*(15*b^2*c^2 + 70*a*b*c*d + 23*a^2*d^2)*x^4 - 24*a^2*c^
2 - 8*(10*a*b*c^2 + 11*a^2*c*d)*x^2)*sqrt(d*x^2 + c))/x^5, -1/120*(15*(15*b^2*c^2 + 40*a*b*c*d + 8*a^2*d^2)*sq
rt(-d)*x^5*arctan(sqrt(-d)*x/sqrt(d*x^2 + c)) - (30*b^2*d^2*x^8 + 15*(9*b^2*c*d + 8*a*b*d^2)*x^6 - 8*(15*b^2*c
^2 + 70*a*b*c*d + 23*a^2*d^2)*x^4 - 24*a^2*c^2 - 8*(10*a*b*c^2 + 11*a^2*c*d)*x^2)*sqrt(d*x^2 + c))/x^5]

Sympy [A] (verification not implemented)

Time = 4.01 (sec) , antiderivative size = 607, normalized size of antiderivative = 2.66 \[ \int \frac {\left (a+b x^2\right )^2 \left (c+d x^2\right )^{5/2}}{x^6} \, dx=- \frac {a^{2} \sqrt {c} d^{2}}{x \sqrt {1 + \frac {d x^{2}}{c}}} - \frac {a^{2} c^{2} \sqrt {d} \sqrt {\frac {c}{d x^{2}} + 1}}{5 x^{4}} - \frac {11 a^{2} c d^{\frac {3}{2}} \sqrt {\frac {c}{d x^{2}} + 1}}{15 x^{2}} - \frac {8 a^{2} d^{\frac {5}{2}} \sqrt {\frac {c}{d x^{2}} + 1}}{15} + a^{2} d^{\frac {5}{2}} \operatorname {asinh}{\left (\frac {\sqrt {d} x}{\sqrt {c}} \right )} - \frac {a^{2} d^{3} x}{\sqrt {c} \sqrt {1 + \frac {d x^{2}}{c}}} - \frac {4 a b c^{\frac {3}{2}} d}{x \sqrt {1 + \frac {d x^{2}}{c}}} - \frac {4 a b \sqrt {c} d^{2} x}{\sqrt {1 + \frac {d x^{2}}{c}}} - \frac {2 a b c^{2} \sqrt {d} \sqrt {\frac {c}{d x^{2}} + 1}}{3 x^{2}} - \frac {2 a b c d^{\frac {3}{2}} \sqrt {\frac {c}{d x^{2}} + 1}}{3} + 4 a b c d^{\frac {3}{2}} \operatorname {asinh}{\left (\frac {\sqrt {d} x}{\sqrt {c}} \right )} + 2 a b d^{2} \left (\begin {cases} \frac {c \left (\begin {cases} \frac {\log {\left (2 \sqrt {d} \sqrt {c + d x^{2}} + 2 d x \right )}}{\sqrt {d}} & \text {for}\: c \neq 0 \\\frac {x \log {\left (x \right )}}{\sqrt {d x^{2}}} & \text {otherwise} \end {cases}\right )}{2} + \frac {x \sqrt {c + d x^{2}}}{2} & \text {for}\: d \neq 0 \\\sqrt {c} x & \text {otherwise} \end {cases}\right ) - \frac {b^{2} c^{\frac {5}{2}}}{x \sqrt {1 + \frac {d x^{2}}{c}}} - \frac {b^{2} c^{\frac {3}{2}} d x}{\sqrt {1 + \frac {d x^{2}}{c}}} + b^{2} c^{2} \sqrt {d} \operatorname {asinh}{\left (\frac {\sqrt {d} x}{\sqrt {c}} \right )} + 2 b^{2} c d \left (\begin {cases} \frac {c \left (\begin {cases} \frac {\log {\left (2 \sqrt {d} \sqrt {c + d x^{2}} + 2 d x \right )}}{\sqrt {d}} & \text {for}\: c \neq 0 \\\frac {x \log {\left (x \right )}}{\sqrt {d x^{2}}} & \text {otherwise} \end {cases}\right )}{2} + \frac {x \sqrt {c + d x^{2}}}{2} & \text {for}\: d \neq 0 \\\sqrt {c} x & \text {otherwise} \end {cases}\right ) + b^{2} d^{2} \left (\begin {cases} - \frac {c^{2} \left (\begin {cases} \frac {\log {\left (2 \sqrt {d} \sqrt {c + d x^{2}} + 2 d x \right )}}{\sqrt {d}} & \text {for}\: c \neq 0 \\\frac {x \log {\left (x \right )}}{\sqrt {d x^{2}}} & \text {otherwise} \end {cases}\right )}{8 d} + \frac {c x \sqrt {c + d x^{2}}}{8 d} + \frac {x^{3} \sqrt {c + d x^{2}}}{4} & \text {for}\: d \neq 0 \\\frac {\sqrt {c} x^{3}}{3} & \text {otherwise} \end {cases}\right ) \]

[In]

integrate((b*x**2+a)**2*(d*x**2+c)**(5/2)/x**6,x)

[Out]

-a**2*sqrt(c)*d**2/(x*sqrt(1 + d*x**2/c)) - a**2*c**2*sqrt(d)*sqrt(c/(d*x**2) + 1)/(5*x**4) - 11*a**2*c*d**(3/
2)*sqrt(c/(d*x**2) + 1)/(15*x**2) - 8*a**2*d**(5/2)*sqrt(c/(d*x**2) + 1)/15 + a**2*d**(5/2)*asinh(sqrt(d)*x/sq
rt(c)) - a**2*d**3*x/(sqrt(c)*sqrt(1 + d*x**2/c)) - 4*a*b*c**(3/2)*d/(x*sqrt(1 + d*x**2/c)) - 4*a*b*sqrt(c)*d*
*2*x/sqrt(1 + d*x**2/c) - 2*a*b*c**2*sqrt(d)*sqrt(c/(d*x**2) + 1)/(3*x**2) - 2*a*b*c*d**(3/2)*sqrt(c/(d*x**2)
+ 1)/3 + 4*a*b*c*d**(3/2)*asinh(sqrt(d)*x/sqrt(c)) + 2*a*b*d**2*Piecewise((c*Piecewise((log(2*sqrt(d)*sqrt(c +
d*x**2) + 2*d*x)/sqrt(d), Ne(c, 0)), (x*log(x)/sqrt(d*x**2), True))/2 + x*sqrt(c + d*x**2)/2, Ne(d, 0)), (sqr
t(c)*x, True)) - b**2*c**(5/2)/(x*sqrt(1 + d*x**2/c)) - b**2*c**(3/2)*d*x/sqrt(1 + d*x**2/c) + b**2*c**2*sqrt(
d)*asinh(sqrt(d)*x/sqrt(c)) + 2*b**2*c*d*Piecewise((c*Piecewise((log(2*sqrt(d)*sqrt(c + d*x**2) + 2*d*x)/sqrt(
d), Ne(c, 0)), (x*log(x)/sqrt(d*x**2), True))/2 + x*sqrt(c + d*x**2)/2, Ne(d, 0)), (sqrt(c)*x, True)) + b**2*d
**2*Piecewise((-c**2*Piecewise((log(2*sqrt(d)*sqrt(c + d*x**2) + 2*d*x)/sqrt(d), Ne(c, 0)), (x*log(x)/sqrt(d*x
**2), True))/(8*d) + c*x*sqrt(c + d*x**2)/(8*d) + x**3*sqrt(c + d*x**2)/4, Ne(d, 0)), (sqrt(c)*x**3/3, True))

Maxima [A] (verification not implemented)

none

Time = 0.21 (sec) , antiderivative size = 285, normalized size of antiderivative = 1.25 \[ \int \frac {\left (a+b x^2\right )^2 \left (c+d x^2\right )^{5/2}}{x^6} \, dx=\frac {5}{4} \, {\left (d x^{2} + c\right )}^{\frac {3}{2}} b^{2} d x + \frac {15}{8} \, \sqrt {d x^{2} + c} b^{2} c d x + 5 \, \sqrt {d x^{2} + c} a b d^{2} x + \frac {10 \, {\left (d x^{2} + c\right )}^{\frac {3}{2}} a b d^{2} x}{3 \, c} + \frac {2 \, {\left (d x^{2} + c\right )}^{\frac {3}{2}} a^{2} d^{3} x}{3 \, c^{2}} + \frac {\sqrt {d x^{2} + c} a^{2} d^{3} x}{c} + \frac {15}{8} \, b^{2} c^{2} \sqrt {d} \operatorname {arsinh}\left (\frac {d x}{\sqrt {c d}}\right ) + 5 \, a b c d^{\frac {3}{2}} \operatorname {arsinh}\left (\frac {d x}{\sqrt {c d}}\right ) + a^{2} d^{\frac {5}{2}} \operatorname {arsinh}\left (\frac {d x}{\sqrt {c d}}\right ) - \frac {{\left (d x^{2} + c\right )}^{\frac {5}{2}} b^{2}}{x} - \frac {8 \, {\left (d x^{2} + c\right )}^{\frac {5}{2}} a b d}{3 \, c x} - \frac {8 \, {\left (d x^{2} + c\right )}^{\frac {5}{2}} a^{2} d^{2}}{15 \, c^{2} x} - \frac {2 \, {\left (d x^{2} + c\right )}^{\frac {7}{2}} a b}{3 \, c x^{3}} - \frac {2 \, {\left (d x^{2} + c\right )}^{\frac {7}{2}} a^{2} d}{15 \, c^{2} x^{3}} - \frac {{\left (d x^{2} + c\right )}^{\frac {7}{2}} a^{2}}{5 \, c x^{5}} \]

[In]

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

[Out]

5/4*(d*x^2 + c)^(3/2)*b^2*d*x + 15/8*sqrt(d*x^2 + c)*b^2*c*d*x + 5*sqrt(d*x^2 + c)*a*b*d^2*x + 10/3*(d*x^2 + c
)^(3/2)*a*b*d^2*x/c + 2/3*(d*x^2 + c)^(3/2)*a^2*d^3*x/c^2 + sqrt(d*x^2 + c)*a^2*d^3*x/c + 15/8*b^2*c^2*sqrt(d)
*arcsinh(d*x/sqrt(c*d)) + 5*a*b*c*d^(3/2)*arcsinh(d*x/sqrt(c*d)) + a^2*d^(5/2)*arcsinh(d*x/sqrt(c*d)) - (d*x^2
+ c)^(5/2)*b^2/x - 8/3*(d*x^2 + c)^(5/2)*a*b*d/(c*x) - 8/15*(d*x^2 + c)^(5/2)*a^2*d^2/(c^2*x) - 2/3*(d*x^2 +
c)^(7/2)*a*b/(c*x^3) - 2/15*(d*x^2 + c)^(7/2)*a^2*d/(c^2*x^3) - 1/5*(d*x^2 + c)^(7/2)*a^2/(c*x^5)

Giac [B] (verification not implemented)

Leaf count of result is larger than twice the leaf count of optimal. 510 vs. \(2 (200) = 400\).

Time = 0.36 (sec) , antiderivative size = 510, normalized size of antiderivative = 2.24 \[ \int \frac {\left (a+b x^2\right )^2 \left (c+d x^2\right )^{5/2}}{x^6} \, dx=\frac {1}{8} \, {\left (2 \, b^{2} d^{2} x^{2} + \frac {9 \, b^{2} c d^{3} + 8 \, a b d^{4}}{d^{2}}\right )} \sqrt {d x^{2} + c} x - \frac {1}{16} \, {\left (15 \, b^{2} c^{2} \sqrt {d} + 40 \, a b c d^{\frac {3}{2}} + 8 \, a^{2} d^{\frac {5}{2}}\right )} \log \left ({\left (\sqrt {d} x - \sqrt {d x^{2} + c}\right )}^{2}\right ) + \frac {2 \, {\left (15 \, {\left (\sqrt {d} x - \sqrt {d x^{2} + c}\right )}^{8} b^{2} c^{3} \sqrt {d} + 90 \, {\left (\sqrt {d} x - \sqrt {d x^{2} + c}\right )}^{8} a b c^{2} d^{\frac {3}{2}} + 45 \, {\left (\sqrt {d} x - \sqrt {d x^{2} + c}\right )}^{8} a^{2} c d^{\frac {5}{2}} - 60 \, {\left (\sqrt {d} x - \sqrt {d x^{2} + c}\right )}^{6} b^{2} c^{4} \sqrt {d} - 300 \, {\left (\sqrt {d} x - \sqrt {d x^{2} + c}\right )}^{6} a b c^{3} d^{\frac {3}{2}} - 90 \, {\left (\sqrt {d} x - \sqrt {d x^{2} + c}\right )}^{6} a^{2} c^{2} d^{\frac {5}{2}} + 90 \, {\left (\sqrt {d} x - \sqrt {d x^{2} + c}\right )}^{4} b^{2} c^{5} \sqrt {d} + 400 \, {\left (\sqrt {d} x - \sqrt {d x^{2} + c}\right )}^{4} a b c^{4} d^{\frac {3}{2}} + 140 \, {\left (\sqrt {d} x - \sqrt {d x^{2} + c}\right )}^{4} a^{2} c^{3} d^{\frac {5}{2}} - 60 \, {\left (\sqrt {d} x - \sqrt {d x^{2} + c}\right )}^{2} b^{2} c^{6} \sqrt {d} - 260 \, {\left (\sqrt {d} x - \sqrt {d x^{2} + c}\right )}^{2} a b c^{5} d^{\frac {3}{2}} - 70 \, {\left (\sqrt {d} x - \sqrt {d x^{2} + c}\right )}^{2} a^{2} c^{4} d^{\frac {5}{2}} + 15 \, b^{2} c^{7} \sqrt {d} + 70 \, a b c^{6} d^{\frac {3}{2}} + 23 \, a^{2} c^{5} d^{\frac {5}{2}}\right )}}{15 \, {\left ({\left (\sqrt {d} x - \sqrt {d x^{2} + c}\right )}^{2} - c\right )}^{5}} \]

[In]

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

[Out]

1/8*(2*b^2*d^2*x^2 + (9*b^2*c*d^3 + 8*a*b*d^4)/d^2)*sqrt(d*x^2 + c)*x - 1/16*(15*b^2*c^2*sqrt(d) + 40*a*b*c*d^
(3/2) + 8*a^2*d^(5/2))*log((sqrt(d)*x - sqrt(d*x^2 + c))^2) + 2/15*(15*(sqrt(d)*x - sqrt(d*x^2 + c))^8*b^2*c^3
*sqrt(d) + 90*(sqrt(d)*x - sqrt(d*x^2 + c))^8*a*b*c^2*d^(3/2) + 45*(sqrt(d)*x - sqrt(d*x^2 + c))^8*a^2*c*d^(5/
2) - 60*(sqrt(d)*x - sqrt(d*x^2 + c))^6*b^2*c^4*sqrt(d) - 300*(sqrt(d)*x - sqrt(d*x^2 + c))^6*a*b*c^3*d^(3/2)
- 90*(sqrt(d)*x - sqrt(d*x^2 + c))^6*a^2*c^2*d^(5/2) + 90*(sqrt(d)*x - sqrt(d*x^2 + c))^4*b^2*c^5*sqrt(d) + 40
0*(sqrt(d)*x - sqrt(d*x^2 + c))^4*a*b*c^4*d^(3/2) + 140*(sqrt(d)*x - sqrt(d*x^2 + c))^4*a^2*c^3*d^(5/2) - 60*(
sqrt(d)*x - sqrt(d*x^2 + c))^2*b^2*c^6*sqrt(d) - 260*(sqrt(d)*x - sqrt(d*x^2 + c))^2*a*b*c^5*d^(3/2) - 70*(sqr
t(d)*x - sqrt(d*x^2 + c))^2*a^2*c^4*d^(5/2) + 15*b^2*c^7*sqrt(d) + 70*a*b*c^6*d^(3/2) + 23*a^2*c^5*d^(5/2))/((
sqrt(d)*x - sqrt(d*x^2 + c))^2 - c)^5

Mupad [F(-1)]

Timed out. \[ \int \frac {\left (a+b x^2\right )^2 \left (c+d x^2\right )^{5/2}}{x^6} \, dx=\int \frac {{\left (b\,x^2+a\right )}^2\,{\left (d\,x^2+c\right )}^{5/2}}{x^6} \,d x \]

[In]

int(((a + b*x^2)^2*(c + d*x^2)^(5/2))/x^6,x)

[Out]

int(((a + b*x^2)^2*(c + d*x^2)^(5/2))/x^6, x)