3.633 \(\int \frac{(a+b x^2)^2 (c+d x^2)^{5/2}}{x^5} \, dx\)

Optimal. Leaf size=222 \[ -\frac{a^2 \left (c+d x^2\right )^{7/2}}{4 c x^4}+\frac{\left (c+d x^2\right )^{5/2} \left (5 a d (3 a d+8 b c)+8 b^2 c^2\right )}{40 c^2}+\frac{\left (c+d x^2\right )^{3/2} \left (5 a d (3 a d+8 b c)+8 b^2 c^2\right )}{24 c}+\frac{1}{8} \sqrt{c+d x^2} \left (5 a d (3 a d+8 b c)+8 b^2 c^2\right )-\frac{1}{8} \sqrt{c} \left (5 a d (3 a d+8 b c)+8 b^2 c^2\right ) \tanh ^{-1}\left (\frac{\sqrt{c+d x^2}}{\sqrt{c}}\right )-\frac{a \left (c+d x^2\right )^{7/2} (3 a d+8 b c)}{8 c^2 x^2} \]

[Out]

((8*b^2*c^2 + 5*a*d*(8*b*c + 3*a*d))*Sqrt[c + d*x^2])/8 + ((8*b^2*c^2 + 5*a*d*(8*b*c + 3*a*d))*(c + d*x^2)^(3/
2))/(24*c) + ((8*b^2*c^2 + 5*a*d*(8*b*c + 3*a*d))*(c + d*x^2)^(5/2))/(40*c^2) - (a^2*(c + d*x^2)^(7/2))/(4*c*x
^4) - (a*(8*b*c + 3*a*d)*(c + d*x^2)^(7/2))/(8*c^2*x^2) - (Sqrt[c]*(8*b^2*c^2 + 5*a*d*(8*b*c + 3*a*d))*ArcTanh
[Sqrt[c + d*x^2]/Sqrt[c]])/8

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Rubi [A]  time = 0.253353, antiderivative size = 219, normalized size of antiderivative = 0.99, number of steps used = 8, number of rules used = 6, integrand size = 24, \(\frac{\text{number of rules}}{\text{integrand size}}\) = 0.25, Rules used = {446, 89, 78, 50, 63, 208} \[ -\frac{a^2 \left (c+d x^2\right )^{7/2}}{4 c x^4}+\frac{1}{40} \left (c+d x^2\right )^{5/2} \left (\frac{5 a d (3 a d+8 b c)}{c^2}+8 b^2\right )+\frac{\left (c+d x^2\right )^{3/2} \left (5 a d (3 a d+8 b c)+8 b^2 c^2\right )}{24 c}+\frac{1}{8} \sqrt{c+d x^2} \left (5 a d (3 a d+8 b c)+8 b^2 c^2\right )-\frac{1}{8} \sqrt{c} \left (5 a d (3 a d+8 b c)+8 b^2 c^2\right ) \tanh ^{-1}\left (\frac{\sqrt{c+d x^2}}{\sqrt{c}}\right )-\frac{a \left (c+d x^2\right )^{7/2} (3 a d+8 b c)}{8 c^2 x^2} \]

Antiderivative was successfully verified.

[In]

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

[Out]

((8*b^2*c^2 + 5*a*d*(8*b*c + 3*a*d))*Sqrt[c + d*x^2])/8 + ((8*b^2*c^2 + 5*a*d*(8*b*c + 3*a*d))*(c + d*x^2)^(3/
2))/(24*c) + ((8*b^2 + (5*a*d*(8*b*c + 3*a*d))/c^2)*(c + d*x^2)^(5/2))/40 - (a^2*(c + d*x^2)^(7/2))/(4*c*x^4)
- (a*(8*b*c + 3*a*d)*(c + d*x^2)^(7/2))/(8*c^2*x^2) - (Sqrt[c]*(8*b^2*c^2 + 5*a*d*(8*b*c + 3*a*d))*ArcTanh[Sqr
t[c + d*x^2]/Sqrt[c]])/8

Rule 446

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]]

Rule 89

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 78

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] || LtQ
[p, n]))))

Rule 50

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 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]

Rubi steps

\begin{align*} \int \frac{\left (a+b x^2\right )^2 \left (c+d x^2\right )^{5/2}}{x^5} \, dx &=\frac{1}{2} \operatorname{Subst}\left (\int \frac{(a+b x)^2 (c+d x)^{5/2}}{x^3} \, dx,x,x^2\right )\\ &=-\frac{a^2 \left (c+d x^2\right )^{7/2}}{4 c x^4}+\frac{\operatorname{Subst}\left (\int \frac{\left (\frac{1}{2} a (8 b c+3 a d)+2 b^2 c x\right ) (c+d x)^{5/2}}{x^2} \, dx,x,x^2\right )}{4 c}\\ &=-\frac{a^2 \left (c+d x^2\right )^{7/2}}{4 c x^4}-\frac{a (8 b c+3 a d) \left (c+d x^2\right )^{7/2}}{8 c^2 x^2}+\frac{1}{16} \left (8 b^2+\frac{5 a d (8 b c+3 a d)}{c^2}\right ) \operatorname{Subst}\left (\int \frac{(c+d x)^{5/2}}{x} \, dx,x,x^2\right )\\ &=\frac{1}{40} \left (8 b^2+\frac{5 a d (8 b c+3 a d)}{c^2}\right ) \left (c+d x^2\right )^{5/2}-\frac{a^2 \left (c+d x^2\right )^{7/2}}{4 c x^4}-\frac{a (8 b c+3 a d) \left (c+d x^2\right )^{7/2}}{8 c^2 x^2}+\frac{1}{16} \left (c \left (8 b^2+\frac{5 a d (8 b c+3 a d)}{c^2}\right )\right ) \operatorname{Subst}\left (\int \frac{(c+d x)^{3/2}}{x} \, dx,x,x^2\right )\\ &=\frac{1}{24} c \left (8 b^2+\frac{5 a d (8 b c+3 a d)}{c^2}\right ) \left (c+d x^2\right )^{3/2}+\frac{1}{40} \left (8 b^2+\frac{5 a d (8 b c+3 a d)}{c^2}\right ) \left (c+d x^2\right )^{5/2}-\frac{a^2 \left (c+d x^2\right )^{7/2}}{4 c x^4}-\frac{a (8 b c+3 a d) \left (c+d x^2\right )^{7/2}}{8 c^2 x^2}+\frac{1}{16} \left (8 b^2 c^2+40 a b c d+15 a^2 d^2\right ) \operatorname{Subst}\left (\int \frac{\sqrt{c+d x}}{x} \, dx,x,x^2\right )\\ &=\frac{1}{8} \left (8 b^2 c^2+40 a b c d+15 a^2 d^2\right ) \sqrt{c+d x^2}+\frac{1}{24} c \left (8 b^2+\frac{5 a d (8 b c+3 a d)}{c^2}\right ) \left (c+d x^2\right )^{3/2}+\frac{1}{40} \left (8 b^2+\frac{5 a d (8 b c+3 a d)}{c^2}\right ) \left (c+d x^2\right )^{5/2}-\frac{a^2 \left (c+d x^2\right )^{7/2}}{4 c x^4}-\frac{a (8 b c+3 a d) \left (c+d x^2\right )^{7/2}}{8 c^2 x^2}+\frac{1}{16} \left (c \left (8 b^2 c^2+40 a b c d+15 a^2 d^2\right )\right ) \operatorname{Subst}\left (\int \frac{1}{x \sqrt{c+d x}} \, dx,x,x^2\right )\\ &=\frac{1}{8} \left (8 b^2 c^2+40 a b c d+15 a^2 d^2\right ) \sqrt{c+d x^2}+\frac{1}{24} c \left (8 b^2+\frac{5 a d (8 b c+3 a d)}{c^2}\right ) \left (c+d x^2\right )^{3/2}+\frac{1}{40} \left (8 b^2+\frac{5 a d (8 b c+3 a d)}{c^2}\right ) \left (c+d x^2\right )^{5/2}-\frac{a^2 \left (c+d x^2\right )^{7/2}}{4 c x^4}-\frac{a (8 b c+3 a d) \left (c+d x^2\right )^{7/2}}{8 c^2 x^2}+\frac{\left (c \left (8 b^2 c^2+40 a b c d+15 a^2 d^2\right )\right ) \operatorname{Subst}\left (\int \frac{1}{-\frac{c}{d}+\frac{x^2}{d}} \, dx,x,\sqrt{c+d x^2}\right )}{8 d}\\ &=\frac{1}{8} \left (8 b^2 c^2+40 a b c d+15 a^2 d^2\right ) \sqrt{c+d x^2}+\frac{1}{24} c \left (8 b^2+\frac{5 a d (8 b c+3 a d)}{c^2}\right ) \left (c+d x^2\right )^{3/2}+\frac{1}{40} \left (8 b^2+\frac{5 a d (8 b c+3 a d)}{c^2}\right ) \left (c+d x^2\right )^{5/2}-\frac{a^2 \left (c+d x^2\right )^{7/2}}{4 c x^4}-\frac{a (8 b c+3 a d) \left (c+d x^2\right )^{7/2}}{8 c^2 x^2}-\frac{1}{8} \sqrt{c} \left (8 b^2 c^2+40 a b c d+15 a^2 d^2\right ) \tanh ^{-1}\left (\frac{\sqrt{c+d x^2}}{\sqrt{c}}\right )\\ \end{align*}

Mathematica [A]  time = 0.103731, size = 153, normalized size = 0.69 \[ \frac{\sqrt{c+d x^2} \left (-15 a^2 \left (2 c^2+9 c d x^2-8 d^2 x^4\right )+40 a b x^2 \left (-3 c^2+14 c d x^2+2 d^2 x^4\right )+8 b^2 x^4 \left (23 c^2+11 c d x^2+3 d^2 x^4\right )\right )}{120 x^4}-\frac{1}{8} \sqrt{c} \left (15 a^2 d^2+40 a b c d+8 b^2 c^2\right ) \tanh ^{-1}\left (\frac{\sqrt{c+d x^2}}{\sqrt{c}}\right ) \]

Antiderivative was successfully verified.

[In]

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

[Out]

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

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Maple [A]  time = 0.01, size = 305, normalized size = 1.4 \begin{align*}{\frac{{b}^{2}}{5} \left ( d{x}^{2}+c \right ) ^{{\frac{5}{2}}}}+{\frac{{b}^{2}c}{3} \left ( d{x}^{2}+c \right ) ^{{\frac{3}{2}}}}-{b}^{2}{c}^{{\frac{5}{2}}}\ln \left ({\frac{1}{x} \left ( 2\,c+2\,\sqrt{c}\sqrt{d{x}^{2}+c} \right ) } \right ) +{b}^{2}\sqrt{d{x}^{2}+c}{c}^{2}-{\frac{{a}^{2}}{4\,c{x}^{4}} \left ( d{x}^{2}+c \right ) ^{{\frac{7}{2}}}}-{\frac{3\,{a}^{2}d}{8\,{c}^{2}{x}^{2}} \left ( d{x}^{2}+c \right ) ^{{\frac{7}{2}}}}+{\frac{3\,{a}^{2}{d}^{2}}{8\,{c}^{2}} \left ( d{x}^{2}+c \right ) ^{{\frac{5}{2}}}}+{\frac{5\,{a}^{2}{d}^{2}}{8\,c} \left ( d{x}^{2}+c \right ) ^{{\frac{3}{2}}}}-{\frac{15\,{a}^{2}{d}^{2}}{8}\sqrt{c}\ln \left ({\frac{1}{x} \left ( 2\,c+2\,\sqrt{c}\sqrt{d{x}^{2}+c} \right ) } \right ) }+{\frac{15\,{a}^{2}{d}^{2}}{8}\sqrt{d{x}^{2}+c}}-{\frac{ab}{c{x}^{2}} \left ( d{x}^{2}+c \right ) ^{{\frac{7}{2}}}}+{\frac{abd}{c} \left ( d{x}^{2}+c \right ) ^{{\frac{5}{2}}}}+{\frac{5\,abd}{3} \left ( d{x}^{2}+c \right ) ^{{\frac{3}{2}}}}-5\,abd{c}^{3/2}\ln \left ({\frac{2\,c+2\,\sqrt{c}\sqrt{d{x}^{2}+c}}{x}} \right ) +5\,abdc\sqrt{d{x}^{2}+c} \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

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

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Maxima [F(-2)]  time = 0., size = 0, normalized size = 0. \begin{align*} \text{Exception raised: ValueError} \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

Exception raised: ValueError

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Fricas [A]  time = 1.44407, size = 740, normalized size = 3.33 \begin{align*} \left [\frac{15 \,{\left (8 \, b^{2} c^{2} + 40 \, a b c d + 15 \, a^{2} d^{2}\right )} \sqrt{c} x^{4} \log \left (-\frac{d x^{2} - 2 \, \sqrt{d x^{2} + c} \sqrt{c} + 2 \, c}{x^{2}}\right ) + 2 \,{\left (24 \, b^{2} d^{2} x^{8} + 8 \,{\left (11 \, b^{2} c d + 10 \, a b d^{2}\right )} x^{6} + 8 \,{\left (23 \, b^{2} c^{2} + 70 \, a b c d + 15 \, a^{2} d^{2}\right )} x^{4} - 30 \, a^{2} c^{2} - 15 \,{\left (8 \, a b c^{2} + 9 \, a^{2} c d\right )} x^{2}\right )} \sqrt{d x^{2} + c}}{240 \, x^{4}}, \frac{15 \,{\left (8 \, b^{2} c^{2} + 40 \, a b c d + 15 \, a^{2} d^{2}\right )} \sqrt{-c} x^{4} \arctan \left (\frac{\sqrt{-c}}{\sqrt{d x^{2} + c}}\right ) +{\left (24 \, b^{2} d^{2} x^{8} + 8 \,{\left (11 \, b^{2} c d + 10 \, a b d^{2}\right )} x^{6} + 8 \,{\left (23 \, b^{2} c^{2} + 70 \, a b c d + 15 \, a^{2} d^{2}\right )} x^{4} - 30 \, a^{2} c^{2} - 15 \,{\left (8 \, a b c^{2} + 9 \, a^{2} c d\right )} x^{2}\right )} \sqrt{d x^{2} + c}}{120 \, x^{4}}\right ] \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

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

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Sympy [A]  time = 98.836, size = 473, normalized size = 2.13 \begin{align*} - \frac{15 a^{2} \sqrt{c} d^{2} \operatorname{asinh}{\left (\frac{\sqrt{c}}{\sqrt{d} x} \right )}}{8} - \frac{a^{2} c^{3}}{4 \sqrt{d} x^{5} \sqrt{\frac{c}{d x^{2}} + 1}} - \frac{3 a^{2} c^{2} \sqrt{d}}{8 x^{3} \sqrt{\frac{c}{d x^{2}} + 1}} - \frac{a^{2} c d^{\frac{3}{2}} \sqrt{\frac{c}{d x^{2}} + 1}}{x} + \frac{7 a^{2} c d^{\frac{3}{2}}}{8 x \sqrt{\frac{c}{d x^{2}} + 1}} + \frac{a^{2} d^{\frac{5}{2}} x}{\sqrt{\frac{c}{d x^{2}} + 1}} - 5 a b c^{\frac{3}{2}} d \operatorname{asinh}{\left (\frac{\sqrt{c}}{\sqrt{d} x} \right )} - \frac{a b c^{2} \sqrt{d} \sqrt{\frac{c}{d x^{2}} + 1}}{x} + \frac{4 a b c^{2} \sqrt{d}}{x \sqrt{\frac{c}{d x^{2}} + 1}} + \frac{4 a b c d^{\frac{3}{2}} x}{\sqrt{\frac{c}{d x^{2}} + 1}} + 2 a b 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 ) - b^{2} c^{\frac{5}{2}} \operatorname{asinh}{\left (\frac{\sqrt{c}}{\sqrt{d} x} \right )} + \frac{b^{2} c^{3}}{\sqrt{d} x \sqrt{\frac{c}{d x^{2}} + 1}} + \frac{b^{2} c^{2} \sqrt{d} x}{\sqrt{\frac{c}{d x^{2}} + 1}} + 2 b^{2} c d \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 ) + b^{2} d^{2} \left (\begin{cases} - \frac{2 c^{2} \sqrt{c + d x^{2}}}{15 d^{2}} + \frac{c x^{2} \sqrt{c + d x^{2}}}{15 d} + \frac{x^{4} \sqrt{c + d x^{2}}}{5} & \text{for}\: d \neq 0 \\\frac{\sqrt{c} x^{4}}{4} & \text{otherwise} \end{cases}\right ) \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

-15*a**2*sqrt(c)*d**2*asinh(sqrt(c)/(sqrt(d)*x))/8 - a**2*c**3/(4*sqrt(d)*x**5*sqrt(c/(d*x**2) + 1)) - 3*a**2*
c**2*sqrt(d)/(8*x**3*sqrt(c/(d*x**2) + 1)) - a**2*c*d**(3/2)*sqrt(c/(d*x**2) + 1)/x + 7*a**2*c*d**(3/2)/(8*x*s
qrt(c/(d*x**2) + 1)) + a**2*d**(5/2)*x/sqrt(c/(d*x**2) + 1) - 5*a*b*c**(3/2)*d*asinh(sqrt(c)/(sqrt(d)*x)) - a*
b*c**2*sqrt(d)*sqrt(c/(d*x**2) + 1)/x + 4*a*b*c**2*sqrt(d)/(x*sqrt(c/(d*x**2) + 1)) + 4*a*b*c*d**(3/2)*x/sqrt(
c/(d*x**2) + 1) + 2*a*b*d**2*Piecewise((sqrt(c)*x**2/2, Eq(d, 0)), ((c + d*x**2)**(3/2)/(3*d), True)) - b**2*c
**(5/2)*asinh(sqrt(c)/(sqrt(d)*x)) + b**2*c**3/(sqrt(d)*x*sqrt(c/(d*x**2) + 1)) + b**2*c**2*sqrt(d)*x/sqrt(c/(
d*x**2) + 1) + 2*b**2*c*d*Piecewise((sqrt(c)*x**2/2, Eq(d, 0)), ((c + d*x**2)**(3/2)/(3*d), True)) + b**2*d**2
*Piecewise((-2*c**2*sqrt(c + d*x**2)/(15*d**2) + c*x**2*sqrt(c + d*x**2)/(15*d) + x**4*sqrt(c + d*x**2)/5, Ne(
d, 0)), (sqrt(c)*x**4/4, True))

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Giac [A]  time = 1.14426, size = 327, normalized size = 1.47 \begin{align*} \frac{24 \,{\left (d x^{2} + c\right )}^{\frac{5}{2}} b^{2} d + 40 \,{\left (d x^{2} + c\right )}^{\frac{3}{2}} b^{2} c d + 120 \, \sqrt{d x^{2} + c} b^{2} c^{2} d + 80 \,{\left (d x^{2} + c\right )}^{\frac{3}{2}} a b d^{2} + 480 \, \sqrt{d x^{2} + c} a b c d^{2} + 120 \, \sqrt{d x^{2} + c} a^{2} d^{3} + \frac{15 \,{\left (8 \, b^{2} c^{3} d + 40 \, a b c^{2} d^{2} + 15 \, a^{2} c d^{3}\right )} \arctan \left (\frac{\sqrt{d x^{2} + c}}{\sqrt{-c}}\right )}{\sqrt{-c}} - \frac{15 \,{\left (8 \,{\left (d x^{2} + c\right )}^{\frac{3}{2}} a b c^{2} d^{2} - 8 \, \sqrt{d x^{2} + c} a b c^{3} d^{2} + 9 \,{\left (d x^{2} + c\right )}^{\frac{3}{2}} a^{2} c d^{3} - 7 \, \sqrt{d x^{2} + c} a^{2} c^{2} d^{3}\right )}}{d^{2} x^{4}}}{120 \, d} \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

1/120*(24*(d*x^2 + c)^(5/2)*b^2*d + 40*(d*x^2 + c)^(3/2)*b^2*c*d + 120*sqrt(d*x^2 + c)*b^2*c^2*d + 80*(d*x^2 +
c)^(3/2)*a*b*d^2 + 480*sqrt(d*x^2 + c)*a*b*c*d^2 + 120*sqrt(d*x^2 + c)*a^2*d^3 + 15*(8*b^2*c^3*d + 40*a*b*c^2
*d^2 + 15*a^2*c*d^3)*arctan(sqrt(d*x^2 + c)/sqrt(-c))/sqrt(-c) - 15*(8*(d*x^2 + c)^(3/2)*a*b*c^2*d^2 - 8*sqrt(
d*x^2 + c)*a*b*c^3*d^2 + 9*(d*x^2 + c)^(3/2)*a^2*c*d^3 - 7*sqrt(d*x^2 + c)*a^2*c^2*d^3)/(d^2*x^4))/d