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

Optimal. Leaf size=132 \[ a^2 c^2 \sqrt{c+d x^2}-a^2 c^{5/2} \tanh ^{-1}\left (\frac{\sqrt{c+d x^2}}{\sqrt{c}}\right )+\frac{1}{5} a^2 \left (c+d x^2\right )^{5/2}+\frac{1}{3} a^2 c \left (c+d x^2\right )^{3/2}-\frac{b \left (c+d x^2\right )^{7/2} (b c-2 a d)}{7 d^2}+\frac{b^2 \left (c+d x^2\right )^{9/2}}{9 d^2} \]

[Out]

a^2*c^2*Sqrt[c + d*x^2] + (a^2*c*(c + d*x^2)^(3/2))/3 + (a^2*(c + d*x^2)^(5/2))/5 - (b*(b*c - 2*a*d)*(c + d*x^
2)^(7/2))/(7*d^2) + (b^2*(c + d*x^2)^(9/2))/(9*d^2) - a^2*c^(5/2)*ArcTanh[Sqrt[c + d*x^2]/Sqrt[c]]

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Rubi [A]  time = 0.110641, antiderivative size = 132, normalized size of antiderivative = 1., number of steps used = 8, number of rules used = 5, integrand size = 24, \(\frac{\text{number of rules}}{\text{integrand size}}\) = 0.208, Rules used = {446, 88, 50, 63, 208} \[ a^2 c^2 \sqrt{c+d x^2}-a^2 c^{5/2} \tanh ^{-1}\left (\frac{\sqrt{c+d x^2}}{\sqrt{c}}\right )+\frac{1}{5} a^2 \left (c+d x^2\right )^{5/2}+\frac{1}{3} a^2 c \left (c+d x^2\right )^{3/2}-\frac{b \left (c+d x^2\right )^{7/2} (b c-2 a d)}{7 d^2}+\frac{b^2 \left (c+d x^2\right )^{9/2}}{9 d^2} \]

Antiderivative was successfully verified.

[In]

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

[Out]

a^2*c^2*Sqrt[c + d*x^2] + (a^2*c*(c + d*x^2)^(3/2))/3 + (a^2*(c + d*x^2)^(5/2))/5 - (b*(b*c - 2*a*d)*(c + d*x^
2)^(7/2))/(7*d^2) + (b^2*(c + d*x^2)^(9/2))/(9*d^2) - a^2*c^(5/2)*ArcTanh[Sqrt[c + d*x^2]/Sqrt[c]]

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 88

Int[((a_.) + (b_.)*(x_))^(m_.)*((c_.) + (d_.)*(x_))^(n_.)*((e_.) + (f_.)*(x_))^(p_.), x_Symbol] :> Int[ExpandI
ntegrand[(a + b*x)^m*(c + d*x)^n*(e + f*x)^p, x], x] /; FreeQ[{a, b, c, d, e, f, p}, x] && IntegersQ[m, n] &&
(IntegerQ[p] || (GtQ[m, 0] && GeQ[n, -1]))

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} \, dx &=\frac{1}{2} \operatorname{Subst}\left (\int \frac{(a+b x)^2 (c+d x)^{5/2}}{x} \, dx,x,x^2\right )\\ &=\frac{1}{2} \operatorname{Subst}\left (\int \left (-\frac{b (b c-2 a d) (c+d x)^{5/2}}{d}+\frac{a^2 (c+d x)^{5/2}}{x}+\frac{b^2 (c+d x)^{7/2}}{d}\right ) \, dx,x,x^2\right )\\ &=-\frac{b (b c-2 a d) \left (c+d x^2\right )^{7/2}}{7 d^2}+\frac{b^2 \left (c+d x^2\right )^{9/2}}{9 d^2}+\frac{1}{2} a^2 \operatorname{Subst}\left (\int \frac{(c+d x)^{5/2}}{x} \, dx,x,x^2\right )\\ &=\frac{1}{5} a^2 \left (c+d x^2\right )^{5/2}-\frac{b (b c-2 a d) \left (c+d x^2\right )^{7/2}}{7 d^2}+\frac{b^2 \left (c+d x^2\right )^{9/2}}{9 d^2}+\frac{1}{2} \left (a^2 c\right ) \operatorname{Subst}\left (\int \frac{(c+d x)^{3/2}}{x} \, dx,x,x^2\right )\\ &=\frac{1}{3} a^2 c \left (c+d x^2\right )^{3/2}+\frac{1}{5} a^2 \left (c+d x^2\right )^{5/2}-\frac{b (b c-2 a d) \left (c+d x^2\right )^{7/2}}{7 d^2}+\frac{b^2 \left (c+d x^2\right )^{9/2}}{9 d^2}+\frac{1}{2} \left (a^2 c^2\right ) \operatorname{Subst}\left (\int \frac{\sqrt{c+d x}}{x} \, dx,x,x^2\right )\\ &=a^2 c^2 \sqrt{c+d x^2}+\frac{1}{3} a^2 c \left (c+d x^2\right )^{3/2}+\frac{1}{5} a^2 \left (c+d x^2\right )^{5/2}-\frac{b (b c-2 a d) \left (c+d x^2\right )^{7/2}}{7 d^2}+\frac{b^2 \left (c+d x^2\right )^{9/2}}{9 d^2}+\frac{1}{2} \left (a^2 c^3\right ) \operatorname{Subst}\left (\int \frac{1}{x \sqrt{c+d x}} \, dx,x,x^2\right )\\ &=a^2 c^2 \sqrt{c+d x^2}+\frac{1}{3} a^2 c \left (c+d x^2\right )^{3/2}+\frac{1}{5} a^2 \left (c+d x^2\right )^{5/2}-\frac{b (b c-2 a d) \left (c+d x^2\right )^{7/2}}{7 d^2}+\frac{b^2 \left (c+d x^2\right )^{9/2}}{9 d^2}+\frac{\left (a^2 c^3\right ) \operatorname{Subst}\left (\int \frac{1}{-\frac{c}{d}+\frac{x^2}{d}} \, dx,x,\sqrt{c+d x^2}\right )}{d}\\ &=a^2 c^2 \sqrt{c+d x^2}+\frac{1}{3} a^2 c \left (c+d x^2\right )^{3/2}+\frac{1}{5} a^2 \left (c+d x^2\right )^{5/2}-\frac{b (b c-2 a d) \left (c+d x^2\right )^{7/2}}{7 d^2}+\frac{b^2 \left (c+d x^2\right )^{9/2}}{9 d^2}-a^2 c^{5/2} \tanh ^{-1}\left (\frac{\sqrt{c+d x^2}}{\sqrt{c}}\right )\\ \end{align*}

Mathematica [A]  time = 0.108364, size = 123, normalized size = 0.93 \[ \frac{1}{3} a^2 c \left (\sqrt{c+d x^2} \left (4 c+d x^2\right )-3 c^{3/2} \tanh ^{-1}\left (\frac{\sqrt{c+d x^2}}{\sqrt{c}}\right )\right )+\frac{1}{5} a^2 \left (c+d x^2\right )^{5/2}+\frac{b \left (c+d x^2\right )^{7/2} (2 a d-b c)}{7 d^2}+\frac{b^2 \left (c+d x^2\right )^{9/2}}{9 d^2} \]

Antiderivative was successfully verified.

[In]

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

[Out]

(a^2*(c + d*x^2)^(5/2))/5 + (b*(-(b*c) + 2*a*d)*(c + d*x^2)^(7/2))/(7*d^2) + (b^2*(c + d*x^2)^(9/2))/(9*d^2) +
(a^2*c*(Sqrt[c + d*x^2]*(4*c + d*x^2) - 3*c^(3/2)*ArcTanh[Sqrt[c + d*x^2]/Sqrt[c]]))/3

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Maple [A]  time = 0.009, size = 132, normalized size = 1. \begin{align*}{\frac{{b}^{2}{x}^{2}}{9\,d} \left ( d{x}^{2}+c \right ) ^{{\frac{7}{2}}}}-{\frac{2\,{b}^{2}c}{63\,{d}^{2}} \left ( d{x}^{2}+c \right ) ^{{\frac{7}{2}}}}+{\frac{2\,ab}{7\,d} \left ( d{x}^{2}+c \right ) ^{{\frac{7}{2}}}}+{\frac{{a}^{2}}{5} \left ( d{x}^{2}+c \right ) ^{{\frac{5}{2}}}}+{\frac{{a}^{2}c}{3} \left ( d{x}^{2}+c \right ) ^{{\frac{3}{2}}}}-{a}^{2}{c}^{{\frac{5}{2}}}\ln \left ({\frac{1}{x} \left ( 2\,c+2\,\sqrt{c}\sqrt{d{x}^{2}+c} \right ) } \right ) +{a}^{2}{c}^{2}\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,x)

[Out]

1/9*b^2*x^2*(d*x^2+c)^(7/2)/d-2/63*b^2*c/d^2*(d*x^2+c)^(7/2)+2/7*a*b*(d*x^2+c)^(7/2)/d+1/5*a^2*(d*x^2+c)^(5/2)
+1/3*a^2*c*(d*x^2+c)^(3/2)-a^2*c^(5/2)*ln((2*c+2*c^(1/2)*(d*x^2+c)^(1/2))/x)+a^2*c^2*(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,x, algorithm="maxima")

[Out]

Exception raised: ValueError

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Fricas [A]  time = 1.45754, size = 824, normalized size = 6.24 \begin{align*} \left [\frac{315 \, a^{2} c^{\frac{5}{2}} d^{2} \log \left (-\frac{d x^{2} - 2 \, \sqrt{d x^{2} + c} \sqrt{c} + 2 \, c}{x^{2}}\right ) + 2 \,{\left (35 \, b^{2} d^{4} x^{8} + 5 \,{\left (19 \, b^{2} c d^{3} + 18 \, a b d^{4}\right )} x^{6} - 10 \, b^{2} c^{4} + 90 \, a b c^{3} d + 483 \, a^{2} c^{2} d^{2} + 3 \,{\left (25 \, b^{2} c^{2} d^{2} + 90 \, a b c d^{3} + 21 \, a^{2} d^{4}\right )} x^{4} +{\left (5 \, b^{2} c^{3} d + 270 \, a b c^{2} d^{2} + 231 \, a^{2} c d^{3}\right )} x^{2}\right )} \sqrt{d x^{2} + c}}{630 \, d^{2}}, \frac{315 \, a^{2} \sqrt{-c} c^{2} d^{2} \arctan \left (\frac{\sqrt{-c}}{\sqrt{d x^{2} + c}}\right ) +{\left (35 \, b^{2} d^{4} x^{8} + 5 \,{\left (19 \, b^{2} c d^{3} + 18 \, a b d^{4}\right )} x^{6} - 10 \, b^{2} c^{4} + 90 \, a b c^{3} d + 483 \, a^{2} c^{2} d^{2} + 3 \,{\left (25 \, b^{2} c^{2} d^{2} + 90 \, a b c d^{3} + 21 \, a^{2} d^{4}\right )} x^{4} +{\left (5 \, b^{2} c^{3} d + 270 \, a b c^{2} d^{2} + 231 \, a^{2} c d^{3}\right )} x^{2}\right )} \sqrt{d x^{2} + c}}{315 \, d^{2}}\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,x, algorithm="fricas")

[Out]

[1/630*(315*a^2*c^(5/2)*d^2*log(-(d*x^2 - 2*sqrt(d*x^2 + c)*sqrt(c) + 2*c)/x^2) + 2*(35*b^2*d^4*x^8 + 5*(19*b^
2*c*d^3 + 18*a*b*d^4)*x^6 - 10*b^2*c^4 + 90*a*b*c^3*d + 483*a^2*c^2*d^2 + 3*(25*b^2*c^2*d^2 + 90*a*b*c*d^3 + 2
1*a^2*d^4)*x^4 + (5*b^2*c^3*d + 270*a*b*c^2*d^2 + 231*a^2*c*d^3)*x^2)*sqrt(d*x^2 + c))/d^2, 1/315*(315*a^2*sqr
t(-c)*c^2*d^2*arctan(sqrt(-c)/sqrt(d*x^2 + c)) + (35*b^2*d^4*x^8 + 5*(19*b^2*c*d^3 + 18*a*b*d^4)*x^6 - 10*b^2*
c^4 + 90*a*b*c^3*d + 483*a^2*c^2*d^2 + 3*(25*b^2*c^2*d^2 + 90*a*b*c*d^3 + 21*a^2*d^4)*x^4 + (5*b^2*c^3*d + 270
*a*b*c^2*d^2 + 231*a^2*c*d^3)*x^2)*sqrt(d*x^2 + c))/d^2]

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Sympy [A]  time = 85.5675, size = 128, normalized size = 0.97 \begin{align*} \frac{a^{2} c^{3} \operatorname{atan}{\left (\frac{\sqrt{c + d x^{2}}}{\sqrt{- c}} \right )}}{\sqrt{- c}} + a^{2} c^{2} \sqrt{c + d x^{2}} + \frac{a^{2} c \left (c + d x^{2}\right )^{\frac{3}{2}}}{3} + \frac{a^{2} \left (c + d x^{2}\right )^{\frac{5}{2}}}{5} + \frac{b^{2} \left (c + d x^{2}\right )^{\frac{9}{2}}}{9 d^{2}} + \frac{\left (c + d x^{2}\right )^{\frac{7}{2}} \left (4 a b d - 2 b^{2} c\right )}{14 d^{2}} \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,x)

[Out]

a**2*c**3*atan(sqrt(c + d*x**2)/sqrt(-c))/sqrt(-c) + a**2*c**2*sqrt(c + d*x**2) + a**2*c*(c + d*x**2)**(3/2)/3
+ a**2*(c + d*x**2)**(5/2)/5 + b**2*(c + d*x**2)**(9/2)/(9*d**2) + (c + d*x**2)**(7/2)*(4*a*b*d - 2*b**2*c)/(
14*d**2)

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Giac [A]  time = 1.14128, size = 190, normalized size = 1.44 \begin{align*} \frac{a^{2} c^{3} \arctan \left (\frac{\sqrt{d x^{2} + c}}{\sqrt{-c}}\right )}{\sqrt{-c}} + \frac{35 \,{\left (d x^{2} + c\right )}^{\frac{9}{2}} b^{2} d^{16} - 45 \,{\left (d x^{2} + c\right )}^{\frac{7}{2}} b^{2} c d^{16} + 90 \,{\left (d x^{2} + c\right )}^{\frac{7}{2}} a b d^{17} + 63 \,{\left (d x^{2} + c\right )}^{\frac{5}{2}} a^{2} d^{18} + 105 \,{\left (d x^{2} + c\right )}^{\frac{3}{2}} a^{2} c d^{18} + 315 \, \sqrt{d x^{2} + c} a^{2} c^{2} d^{18}}{315 \, d^{18}} \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,x, algorithm="giac")

[Out]

a^2*c^3*arctan(sqrt(d*x^2 + c)/sqrt(-c))/sqrt(-c) + 1/315*(35*(d*x^2 + c)^(9/2)*b^2*d^16 - 45*(d*x^2 + c)^(7/2
)*b^2*c*d^16 + 90*(d*x^2 + c)^(7/2)*a*b*d^17 + 63*(d*x^2 + c)^(5/2)*a^2*d^18 + 105*(d*x^2 + c)^(3/2)*a^2*c*d^1
8 + 315*sqrt(d*x^2 + c)*a^2*c^2*d^18)/d^18