3.601 \(\int \frac {(a+b x^2)^2 \sqrt {c+d x^2}}{x} \, dx\)

Optimal. Leaf size=92 \[ a^2 \sqrt {c+d x^2}-a^2 \sqrt {c} \tanh ^{-1}\left (\frac {\sqrt {c+d x^2}}{\sqrt {c}}\right )-\frac {b \left (c+d x^2\right )^{3/2} (b c-2 a d)}{3 d^2}+\frac {b^2 \left (c+d x^2\right )^{5/2}}{5 d^2} \]

[Out]

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

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Rubi [A]  time = 0.08, antiderivative size = 92, normalized size of antiderivative = 1.00, number of steps used = 6, 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} \[ a^2 \sqrt {c+d x^2}-a^2 \sqrt {c} \tanh ^{-1}\left (\frac {\sqrt {c+d x^2}}{\sqrt {c}}\right )-\frac {b \left (c+d x^2\right )^{3/2} (b c-2 a d)}{3 d^2}+\frac {b^2 \left (c+d x^2\right )^{5/2}}{5 d^2} \]

Antiderivative was successfully verified.

[In]

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

[Out]

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

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Mathematica [A]  time = 0.14, size = 93, normalized size = 1.01 \[ a^2 \sqrt {c+d x^2}-a^2 \sqrt {c} \tanh ^{-1}\left (\frac {\sqrt {c+d x^2}}{\sqrt {c}}\right )+\frac {b \left (c+d x^2\right )^{3/2} (2 a d-b c)}{3 d^2}+\frac {b^2 \left (c+d x^2\right )^{5/2}}{5 d^2} \]

Antiderivative was successfully verified.

[In]

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

[Out]

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

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fricas [A]  time = 0.63, size = 207, normalized size = 2.25 \[ \left [\frac {15 \, a^{2} \sqrt {c} d^{2} \log \left (-\frac {d x^{2} - 2 \, \sqrt {d x^{2} + c} \sqrt {c} + 2 \, c}{x^{2}}\right ) + 2 \, {\left (3 \, b^{2} d^{2} x^{4} - 2 \, b^{2} c^{2} + 10 \, a b c d + 15 \, a^{2} d^{2} + {\left (b^{2} c d + 10 \, a b d^{2}\right )} x^{2}\right )} \sqrt {d x^{2} + c}}{30 \, d^{2}}, \frac {15 \, a^{2} \sqrt {-c} d^{2} \arctan \left (\frac {\sqrt {-c}}{\sqrt {d x^{2} + c}}\right ) + {\left (3 \, b^{2} d^{2} x^{4} - 2 \, b^{2} c^{2} + 10 \, a b c d + 15 \, a^{2} d^{2} + {\left (b^{2} c d + 10 \, a b d^{2}\right )} x^{2}\right )} \sqrt {d x^{2} + c}}{15 \, d^{2}}\right ] \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate((b*x^2+a)^2*(d*x^2+c)^(1/2)/x,x, algorithm="fricas")

[Out]

[1/30*(15*a^2*sqrt(c)*d^2*log(-(d*x^2 - 2*sqrt(d*x^2 + c)*sqrt(c) + 2*c)/x^2) + 2*(3*b^2*d^2*x^4 - 2*b^2*c^2 +
10*a*b*c*d + 15*a^2*d^2 + (b^2*c*d + 10*a*b*d^2)*x^2)*sqrt(d*x^2 + c))/d^2, 1/15*(15*a^2*sqrt(-c)*d^2*arctan(
sqrt(-c)/sqrt(d*x^2 + c)) + (3*b^2*d^2*x^4 - 2*b^2*c^2 + 10*a*b*c*d + 15*a^2*d^2 + (b^2*c*d + 10*a*b*d^2)*x^2)
*sqrt(d*x^2 + c))/d^2]

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giac [A]  time = 0.32, size = 101, normalized size = 1.10 \[ \frac {a^{2} c \arctan \left (\frac {\sqrt {d x^{2} + c}}{\sqrt {-c}}\right )}{\sqrt {-c}} + \frac {3 \, {\left (d x^{2} + c\right )}^{\frac {5}{2}} b^{2} d^{8} - 5 \, {\left (d x^{2} + c\right )}^{\frac {3}{2}} b^{2} c d^{8} + 10 \, {\left (d x^{2} + c\right )}^{\frac {3}{2}} a b d^{9} + 15 \, \sqrt {d x^{2} + c} a^{2} d^{10}}{15 \, d^{10}} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate((b*x^2+a)^2*(d*x^2+c)^(1/2)/x,x, algorithm="giac")

[Out]

a^2*c*arctan(sqrt(d*x^2 + c)/sqrt(-c))/sqrt(-c) + 1/15*(3*(d*x^2 + c)^(5/2)*b^2*d^8 - 5*(d*x^2 + c)^(3/2)*b^2*
c*d^8 + 10*(d*x^2 + c)^(3/2)*a*b*d^9 + 15*sqrt(d*x^2 + c)*a^2*d^10)/d^10

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maple [A]  time = 0.01, size = 100, normalized size = 1.09 \[ -a^{2} \sqrt {c}\, \ln \left (\frac {2 c +2 \sqrt {d \,x^{2}+c}\, \sqrt {c}}{x}\right )+\frac {\left (d \,x^{2}+c \right )^{\frac {3}{2}} b^{2} x^{2}}{5 d}+\sqrt {d \,x^{2}+c}\, a^{2}+\frac {2 \left (d \,x^{2}+c \right )^{\frac {3}{2}} a b}{3 d}-\frac {2 \left (d \,x^{2}+c \right )^{\frac {3}{2}} b^{2} c}{15 d^{2}} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

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

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maxima [A]  time = 1.03, size = 88, normalized size = 0.96 \[ \frac {{\left (d x^{2} + c\right )}^{\frac {3}{2}} b^{2} x^{2}}{5 \, d} - a^{2} \sqrt {c} \operatorname {arsinh}\left (\frac {c}{\sqrt {c d} {\left | x \right |}}\right ) + \sqrt {d x^{2} + c} a^{2} - \frac {2 \, {\left (d x^{2} + c\right )}^{\frac {3}{2}} b^{2} c}{15 \, d^{2}} + \frac {2 \, {\left (d x^{2} + c\right )}^{\frac {3}{2}} a b}{3 \, d} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate((b*x^2+a)^2*(d*x^2+c)^(1/2)/x,x, algorithm="maxima")

[Out]

1/5*(d*x^2 + c)^(3/2)*b^2*x^2/d - a^2*sqrt(c)*arcsinh(c/(sqrt(c*d)*abs(x))) + sqrt(d*x^2 + c)*a^2 - 2/15*(d*x^
2 + c)^(3/2)*b^2*c/d^2 + 2/3*(d*x^2 + c)^(3/2)*a*b/d

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mupad [B]  time = 0.69, size = 135, normalized size = 1.47 \[ \sqrt {d\,x^2+c}\,\left (\frac {{\left (a\,d-b\,c\right )}^2}{d^2}-c\,\left (\frac {2\,b^2\,c-2\,a\,b\,d}{d^2}-\frac {b^2\,c}{d^2}\right )\right )-\left (\frac {2\,b^2\,c-2\,a\,b\,d}{3\,d^2}-\frac {b^2\,c}{3\,d^2}\right )\,{\left (d\,x^2+c\right )}^{3/2}+\frac {b^2\,{\left (d\,x^2+c\right )}^{5/2}}{5\,d^2}+a^2\,\sqrt {c}\,\mathrm {atan}\left (\frac {\sqrt {d\,x^2+c}\,1{}\mathrm {i}}{\sqrt {c}}\right )\,1{}\mathrm {i} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

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

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sympy [A]  time = 72.46, size = 90, normalized size = 0.98 \[ \frac {a^{2} c \operatorname {atan}{\left (\frac {\sqrt {c + d x^{2}}}{\sqrt {- c}} \right )}}{\sqrt {- c}} + a^{2} \sqrt {c + d x^{2}} + \frac {b^{2} \left (c + d x^{2}\right )^{\frac {5}{2}}}{5 d^{2}} + \frac {\left (c + d x^{2}\right )^{\frac {3}{2}} \left (4 a b d - 2 b^{2} c\right )}{6 d^{2}} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate((b*x**2+a)**2*(d*x**2+c)**(1/2)/x,x)

[Out]

a**2*c*atan(sqrt(c + d*x**2)/sqrt(-c))/sqrt(-c) + a**2*sqrt(c + d*x**2) + b**2*(c + d*x**2)**(5/2)/(5*d**2) +
(c + d*x**2)**(3/2)*(4*a*b*d - 2*b**2*c)/(6*d**2)

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