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

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

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Rubi [A]  time = 0.11, antiderivative size = 132, normalized size of antiderivative = 1.00, 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} \begin {gather*} 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} \end {gather*}

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 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 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 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]

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]]

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*}

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Mathematica [A]  time = 0.12, size = 123, normalized size = 0.93 \begin {gather*} \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} \end {gather*}

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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IntegrateAlgebraic [A]  time = 0.12, size = 181, normalized size = 1.37 \begin {gather*} \frac {\sqrt {c+d x^2} \left (483 a^2 c^2 d^2+231 a^2 c d^3 x^2+63 a^2 d^4 x^4+90 a b c^3 d+270 a b c^2 d^2 x^2+270 a b c d^3 x^4+90 a b d^4 x^6-10 b^2 c^4+5 b^2 c^3 d x^2+75 b^2 c^2 d^2 x^4+95 b^2 c d^3 x^6+35 b^2 d^4 x^8\right )}{315 d^2}-a^2 c^{5/2} \tanh ^{-1}\left (\frac {\sqrt {c+d x^2}}{\sqrt {c}}\right ) \end {gather*}

Antiderivative was successfully verified.

[In]

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

[Out]

(Sqrt[c + d*x^2]*(-10*b^2*c^4 + 90*a*b*c^3*d + 483*a^2*c^2*d^2 + 5*b^2*c^3*d*x^2 + 270*a*b*c^2*d^2*x^2 + 231*a
^2*c*d^3*x^2 + 75*b^2*c^2*d^2*x^4 + 270*a*b*c*d^3*x^4 + 63*a^2*d^4*x^4 + 95*b^2*c*d^3*x^6 + 90*a*b*d^4*x^6 + 3
5*b^2*d^4*x^8))/(315*d^2) - a^2*c^(5/2)*ArcTanh[Sqrt[c + d*x^2]/Sqrt[c]]

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fricas [A]  time = 1.64, size = 360, normalized size = 2.73 \begin {gather*} \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 {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,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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giac [A]  time = 0.44, size = 141, normalized size = 1.07 \begin {gather*} \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 {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,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

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maple [A]  time = 0.01, size = 132, normalized size = 1.00 \begin {gather*} -a^{2} c^{\frac {5}{2}} \ln \left (\frac {2 c +2 \sqrt {d \,x^{2}+c}\, \sqrt {c}}{x}\right )+\sqrt {d \,x^{2}+c}\, a^{2} c^{2}+\frac {\left (d \,x^{2}+c \right )^{\frac {3}{2}} a^{2} c}{3}+\frac {\left (d \,x^{2}+c \right )^{\frac {7}{2}} b^{2} x^{2}}{9 d}+\frac {\left (d \,x^{2}+c \right )^{\frac {5}{2}} a^{2}}{5}+\frac {2 \left (d \,x^{2}+c \right )^{\frac {7}{2}} a b}{7 d}-\frac {2 \left (d \,x^{2}+c \right )^{\frac {7}{2}} b^{2} c}{63 d^{2}} \end {gather*}

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*(d*x^2+c)^(1/2)*c^(1/2))/x)+a^2*c^2*(d*x^2+c)^(1/2)

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maxima [A]  time = 0.89, size = 120, normalized size = 0.91 \begin {gather*} \frac {{\left (d x^{2} + c\right )}^{\frac {7}{2}} b^{2} x^{2}}{9 \, d} - a^{2} c^{\frac {5}{2}} \operatorname {arsinh}\left (\frac {c}{\sqrt {c d} {\left | x \right |}}\right ) + \frac {1}{5} \, {\left (d x^{2} + c\right )}^{\frac {5}{2}} a^{2} + \frac {1}{3} \, {\left (d x^{2} + c\right )}^{\frac {3}{2}} a^{2} c + \sqrt {d x^{2} + c} a^{2} c^{2} - \frac {2 \, {\left (d x^{2} + c\right )}^{\frac {7}{2}} b^{2} c}{63 \, d^{2}} + \frac {2 \, {\left (d x^{2} + c\right )}^{\frac {7}{2}} a b}{7 \, 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,x, algorithm="maxima")

[Out]

1/9*(d*x^2 + c)^(7/2)*b^2*x^2/d - a^2*c^(5/2)*arcsinh(c/(sqrt(c*d)*abs(x))) + 1/5*(d*x^2 + c)^(5/2)*a^2 + 1/3*
(d*x^2 + c)^(3/2)*a^2*c + sqrt(d*x^2 + c)*a^2*c^2 - 2/63*(d*x^2 + c)^(7/2)*b^2*c/d^2 + 2/7*(d*x^2 + c)^(7/2)*a
*b/d

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mupad [B]  time = 0.77, size = 249, normalized size = 1.89 \begin {gather*} {\left (d\,x^2+c\right )}^{5/2}\,\left (\frac {{\left (a\,d-b\,c\right )}^2}{5\,d^2}-\frac {c\,\left (\frac {2\,b^2\,c-2\,a\,b\,d}{d^2}-\frac {b^2\,c}{d^2}\right )}{5}\right )-\left (\frac {2\,b^2\,c-2\,a\,b\,d}{7\,d^2}-\frac {b^2\,c}{7\,d^2}\right )\,{\left (d\,x^2+c\right )}^{7/2}+c^2\,\sqrt {d\,x^2+c}\,\left (\frac {{\left (a\,d-b\,c\right )}^2}{d^2}-c\,\left (\frac {2\,b^2\,c-2\,a\,b\,d}{d^2}-\frac {b^2\,c}{d^2}\right )\right )+\frac {b^2\,{\left (d\,x^2+c\right )}^{9/2}}{9\,d^2}+\frac {c\,{\left (d\,x^2+c\right )}^{3/2}\,\left (\frac {{\left (a\,d-b\,c\right )}^2}{d^2}-c\,\left (\frac {2\,b^2\,c-2\,a\,b\,d}{d^2}-\frac {b^2\,c}{d^2}\right )\right )}{3}+a^2\,c^{5/2}\,\mathrm {atan}\left (\frac {\sqrt {d\,x^2+c}\,1{}\mathrm {i}}{\sqrt {c}}\right )\,1{}\mathrm {i} \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,x)

[Out]

(c + d*x^2)^(5/2)*((a*d - b*c)^2/(5*d^2) - (c*((2*b^2*c - 2*a*b*d)/d^2 - (b^2*c)/d^2))/5) - ((2*b^2*c - 2*a*b*
d)/(7*d^2) - (b^2*c)/(7*d^2))*(c + d*x^2)^(7/2) + a^2*c^(5/2)*atan(((c + d*x^2)^(1/2)*1i)/c^(1/2))*1i + c^2*(c
+ d*x^2)^(1/2)*((a*d - b*c)^2/d^2 - c*((2*b^2*c - 2*a*b*d)/d^2 - (b^2*c)/d^2)) + (b^2*(c + d*x^2)^(9/2))/(9*d
^2) + (c*(c + d*x^2)^(3/2)*((a*d - b*c)^2/d^2 - c*((2*b^2*c - 2*a*b*d)/d^2 - (b^2*c)/d^2)))/3

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sympy [A]  time = 128.71, size = 128, normalized size = 0.97 \begin {gather*} \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 {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,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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