3.758 \(\int \frac {x^4}{(a+b x^2)^2 \sqrt {c+d x^2}} \, dx\)

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

[Out]

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

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Rubi [A]  time = 0.11, antiderivative size = 132, normalized size of antiderivative = 1.00, number of steps used = 6, number of rules used = 6, integrand size = 24, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.250, Rules used = {470, 523, 217, 206, 377, 205} \[ -\frac {\sqrt {a} (3 b c-2 a d) \tan ^{-1}\left (\frac {x \sqrt {b c-a d}}{\sqrt {a} \sqrt {c+d x^2}}\right )}{2 b^2 (b c-a d)^{3/2}}+\frac {a x \sqrt {c+d x^2}}{2 b \left (a+b x^2\right ) (b c-a d)}+\frac {\tanh ^{-1}\left (\frac {\sqrt {d} x}{\sqrt {c+d x^2}}\right )}{b^2 \sqrt {d}} \]

Antiderivative was successfully verified.

[In]

Int[x^4/((a + b*x^2)^2*Sqrt[c + d*x^2]),x]

[Out]

(a*x*Sqrt[c + d*x^2])/(2*b*(b*c - a*d)*(a + b*x^2)) - (Sqrt[a]*(3*b*c - 2*a*d)*ArcTan[(Sqrt[b*c - a*d]*x)/(Sqr
t[a]*Sqrt[c + d*x^2])])/(2*b^2*(b*c - a*d)^(3/2)) + ArcTanh[(Sqrt[d]*x)/Sqrt[c + d*x^2]]/(b^2*Sqrt[d])

Rule 205

Int[((a_) + (b_.)*(x_)^2)^(-1), x_Symbol] :> Simp[(Rt[a/b, 2]*ArcTan[x/Rt[a/b, 2]])/a, x] /; FreeQ[{a, b}, x]
&& PosQ[a/b]

Rule 206

Int[((a_) + (b_.)*(x_)^2)^(-1), x_Symbol] :> Simp[(1*ArcTanh[(Rt[-b, 2]*x)/Rt[a, 2]])/(Rt[a, 2]*Rt[-b, 2]), x]
/; FreeQ[{a, b}, x] && NegQ[a/b] && (GtQ[a, 0] || LtQ[b, 0])

Rule 217

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 377

Int[((a_) + (b_.)*(x_)^(n_))^(p_)/((c_) + (d_.)*(x_)^(n_)), x_Symbol] :> Subst[Int[1/(c - (b*c - a*d)*x^n), x]
, x, x/(a + b*x^n)^(1/n)] /; FreeQ[{a, b, c, d}, x] && NeQ[b*c - a*d, 0] && EqQ[n*p + 1, 0] && IntegerQ[n]

Rule 470

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

Rule 523

Int[((e_) + (f_.)*(x_)^(n_))/(((a_) + (b_.)*(x_)^(n_))*Sqrt[(c_) + (d_.)*(x_)^(n_)]), x_Symbol] :> Dist[f/b, I
nt[1/Sqrt[c + d*x^n], x], x] + Dist[(b*e - a*f)/b, Int[1/((a + b*x^n)*Sqrt[c + d*x^n]), x], x] /; FreeQ[{a, b,
c, d, e, f, n}, x]

Rubi steps

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

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Mathematica [A]  time = 0.19, size = 129, normalized size = 0.98 \[ \frac {\frac {a b x \sqrt {c+d x^2}}{\left (a+b x^2\right ) (b c-a d)}+\frac {\sqrt {a} (2 a d-3 b c) \tan ^{-1}\left (\frac {x \sqrt {b c-a d}}{\sqrt {a} \sqrt {c+d x^2}}\right )}{(b c-a d)^{3/2}}+\frac {2 \log \left (\sqrt {d} \sqrt {c+d x^2}+d x\right )}{\sqrt {d}}}{2 b^2} \]

Antiderivative was successfully verified.

[In]

Integrate[x^4/((a + b*x^2)^2*Sqrt[c + d*x^2]),x]

[Out]

((a*b*x*Sqrt[c + d*x^2])/((b*c - a*d)*(a + b*x^2)) + (Sqrt[a]*(-3*b*c + 2*a*d)*ArcTan[(Sqrt[b*c - a*d]*x)/(Sqr
t[a]*Sqrt[c + d*x^2])])/(b*c - a*d)^(3/2) + (2*Log[d*x + Sqrt[d]*Sqrt[c + d*x^2]])/Sqrt[d])/(2*b^2)

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fricas [A]  time = 1.28, size = 1053, normalized size = 7.98 \[ \left [\frac {4 \, \sqrt {d x^{2} + c} a b d x + 4 \, {\left (a b c - a^{2} d + {\left (b^{2} c - a b d\right )} x^{2}\right )} \sqrt {d} \log \left (-2 \, d x^{2} - 2 \, \sqrt {d x^{2} + c} \sqrt {d} x - c\right ) + {\left (3 \, a b c d - 2 \, a^{2} d^{2} + {\left (3 \, b^{2} c d - 2 \, a b d^{2}\right )} x^{2}\right )} \sqrt {-\frac {a}{b c - a d}} \log \left (\frac {{\left (b^{2} c^{2} - 8 \, a b c d + 8 \, a^{2} d^{2}\right )} x^{4} + a^{2} c^{2} - 2 \, {\left (3 \, a b c^{2} - 4 \, a^{2} c d\right )} x^{2} - 4 \, {\left ({\left (b^{2} c^{2} - 3 \, a b c d + 2 \, a^{2} d^{2}\right )} x^{3} - {\left (a b c^{2} - a^{2} c d\right )} x\right )} \sqrt {d x^{2} + c} \sqrt {-\frac {a}{b c - a d}}}{b^{2} x^{4} + 2 \, a b x^{2} + a^{2}}\right )}{8 \, {\left (a b^{3} c d - a^{2} b^{2} d^{2} + {\left (b^{4} c d - a b^{3} d^{2}\right )} x^{2}\right )}}, \frac {4 \, \sqrt {d x^{2} + c} a b d x - 8 \, {\left (a b c - a^{2} d + {\left (b^{2} c - a b d\right )} x^{2}\right )} \sqrt {-d} \arctan \left (\frac {\sqrt {-d} x}{\sqrt {d x^{2} + c}}\right ) + {\left (3 \, a b c d - 2 \, a^{2} d^{2} + {\left (3 \, b^{2} c d - 2 \, a b d^{2}\right )} x^{2}\right )} \sqrt {-\frac {a}{b c - a d}} \log \left (\frac {{\left (b^{2} c^{2} - 8 \, a b c d + 8 \, a^{2} d^{2}\right )} x^{4} + a^{2} c^{2} - 2 \, {\left (3 \, a b c^{2} - 4 \, a^{2} c d\right )} x^{2} - 4 \, {\left ({\left (b^{2} c^{2} - 3 \, a b c d + 2 \, a^{2} d^{2}\right )} x^{3} - {\left (a b c^{2} - a^{2} c d\right )} x\right )} \sqrt {d x^{2} + c} \sqrt {-\frac {a}{b c - a d}}}{b^{2} x^{4} + 2 \, a b x^{2} + a^{2}}\right )}{8 \, {\left (a b^{3} c d - a^{2} b^{2} d^{2} + {\left (b^{4} c d - a b^{3} d^{2}\right )} x^{2}\right )}}, \frac {2 \, \sqrt {d x^{2} + c} a b d x + {\left (3 \, a b c d - 2 \, a^{2} d^{2} + {\left (3 \, b^{2} c d - 2 \, a b d^{2}\right )} x^{2}\right )} \sqrt {\frac {a}{b c - a d}} \arctan \left (-\frac {{\left ({\left (b c - 2 \, a d\right )} x^{2} - a c\right )} \sqrt {d x^{2} + c} \sqrt {\frac {a}{b c - a d}}}{2 \, {\left (a d x^{3} + a c x\right )}}\right ) + 2 \, {\left (a b c - a^{2} d + {\left (b^{2} c - a b d\right )} x^{2}\right )} \sqrt {d} \log \left (-2 \, d x^{2} - 2 \, \sqrt {d x^{2} + c} \sqrt {d} x - c\right )}{4 \, {\left (a b^{3} c d - a^{2} b^{2} d^{2} + {\left (b^{4} c d - a b^{3} d^{2}\right )} x^{2}\right )}}, \frac {2 \, \sqrt {d x^{2} + c} a b d x - 4 \, {\left (a b c - a^{2} d + {\left (b^{2} c - a b d\right )} x^{2}\right )} \sqrt {-d} \arctan \left (\frac {\sqrt {-d} x}{\sqrt {d x^{2} + c}}\right ) + {\left (3 \, a b c d - 2 \, a^{2} d^{2} + {\left (3 \, b^{2} c d - 2 \, a b d^{2}\right )} x^{2}\right )} \sqrt {\frac {a}{b c - a d}} \arctan \left (-\frac {{\left ({\left (b c - 2 \, a d\right )} x^{2} - a c\right )} \sqrt {d x^{2} + c} \sqrt {\frac {a}{b c - a d}}}{2 \, {\left (a d x^{3} + a c x\right )}}\right )}{4 \, {\left (a b^{3} c d - a^{2} b^{2} d^{2} + {\left (b^{4} c d - a b^{3} d^{2}\right )} x^{2}\right )}}\right ] \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate(x^4/(b*x^2+a)^2/(d*x^2+c)^(1/2),x, algorithm="fricas")

[Out]

[1/8*(4*sqrt(d*x^2 + c)*a*b*d*x + 4*(a*b*c - a^2*d + (b^2*c - a*b*d)*x^2)*sqrt(d)*log(-2*d*x^2 - 2*sqrt(d*x^2
+ c)*sqrt(d)*x - c) + (3*a*b*c*d - 2*a^2*d^2 + (3*b^2*c*d - 2*a*b*d^2)*x^2)*sqrt(-a/(b*c - a*d))*log(((b^2*c^2
- 8*a*b*c*d + 8*a^2*d^2)*x^4 + a^2*c^2 - 2*(3*a*b*c^2 - 4*a^2*c*d)*x^2 - 4*((b^2*c^2 - 3*a*b*c*d + 2*a^2*d^2)
*x^3 - (a*b*c^2 - a^2*c*d)*x)*sqrt(d*x^2 + c)*sqrt(-a/(b*c - a*d)))/(b^2*x^4 + 2*a*b*x^2 + a^2)))/(a*b^3*c*d -
a^2*b^2*d^2 + (b^4*c*d - a*b^3*d^2)*x^2), 1/8*(4*sqrt(d*x^2 + c)*a*b*d*x - 8*(a*b*c - a^2*d + (b^2*c - a*b*d)
*x^2)*sqrt(-d)*arctan(sqrt(-d)*x/sqrt(d*x^2 + c)) + (3*a*b*c*d - 2*a^2*d^2 + (3*b^2*c*d - 2*a*b*d^2)*x^2)*sqrt
(-a/(b*c - a*d))*log(((b^2*c^2 - 8*a*b*c*d + 8*a^2*d^2)*x^4 + a^2*c^2 - 2*(3*a*b*c^2 - 4*a^2*c*d)*x^2 - 4*((b^
2*c^2 - 3*a*b*c*d + 2*a^2*d^2)*x^3 - (a*b*c^2 - a^2*c*d)*x)*sqrt(d*x^2 + c)*sqrt(-a/(b*c - a*d)))/(b^2*x^4 + 2
*a*b*x^2 + a^2)))/(a*b^3*c*d - a^2*b^2*d^2 + (b^4*c*d - a*b^3*d^2)*x^2), 1/4*(2*sqrt(d*x^2 + c)*a*b*d*x + (3*a
*b*c*d - 2*a^2*d^2 + (3*b^2*c*d - 2*a*b*d^2)*x^2)*sqrt(a/(b*c - a*d))*arctan(-1/2*((b*c - 2*a*d)*x^2 - a*c)*sq
rt(d*x^2 + c)*sqrt(a/(b*c - a*d))/(a*d*x^3 + a*c*x)) + 2*(a*b*c - a^2*d + (b^2*c - a*b*d)*x^2)*sqrt(d)*log(-2*
d*x^2 - 2*sqrt(d*x^2 + c)*sqrt(d)*x - c))/(a*b^3*c*d - a^2*b^2*d^2 + (b^4*c*d - a*b^3*d^2)*x^2), 1/4*(2*sqrt(d
*x^2 + c)*a*b*d*x - 4*(a*b*c - a^2*d + (b^2*c - a*b*d)*x^2)*sqrt(-d)*arctan(sqrt(-d)*x/sqrt(d*x^2 + c)) + (3*a
*b*c*d - 2*a^2*d^2 + (3*b^2*c*d - 2*a*b*d^2)*x^2)*sqrt(a/(b*c - a*d))*arctan(-1/2*((b*c - 2*a*d)*x^2 - a*c)*sq
rt(d*x^2 + c)*sqrt(a/(b*c - a*d))/(a*d*x^3 + a*c*x)))/(a*b^3*c*d - a^2*b^2*d^2 + (b^4*c*d - a*b^3*d^2)*x^2)]

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giac [B]  time = 0.57, size = 284, normalized size = 2.15 \[ -\frac {{\left (3 \, a b c \sqrt {d} - 2 \, a^{2} d^{\frac {3}{2}}\right )} \arctan \left (-\frac {{\left (\sqrt {d} x - \sqrt {d x^{2} + c}\right )}^{2} b - b c + 2 \, a d}{2 \, \sqrt {a b c d - a^{2} d^{2}}}\right )}{2 \, {\left (b^{3} c - a b^{2} d\right )} \sqrt {a b c d - a^{2} d^{2}}} - \frac {{\left (\sqrt {d} x - \sqrt {d x^{2} + c}\right )}^{2} a b c \sqrt {d} - 2 \, {\left (\sqrt {d} x - \sqrt {d x^{2} + c}\right )}^{2} a^{2} d^{\frac {3}{2}} - a b c^{2} \sqrt {d}}{{\left ({\left (\sqrt {d} x - \sqrt {d x^{2} + c}\right )}^{4} b - 2 \, {\left (\sqrt {d} x - \sqrt {d x^{2} + c}\right )}^{2} b c + 4 \, {\left (\sqrt {d} x - \sqrt {d x^{2} + c}\right )}^{2} a d + b c^{2}\right )} {\left (b^{3} c - a b^{2} d\right )}} - \frac {\log \left ({\left (\sqrt {d} x - \sqrt {d x^{2} + c}\right )}^{2}\right )}{2 \, b^{2} \sqrt {d}} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate(x^4/(b*x^2+a)^2/(d*x^2+c)^(1/2),x, algorithm="giac")

[Out]

-1/2*(3*a*b*c*sqrt(d) - 2*a^2*d^(3/2))*arctan(-1/2*((sqrt(d)*x - sqrt(d*x^2 + c))^2*b - b*c + 2*a*d)/sqrt(a*b*
c*d - a^2*d^2))/((b^3*c - a*b^2*d)*sqrt(a*b*c*d - a^2*d^2)) - ((sqrt(d)*x - sqrt(d*x^2 + c))^2*a*b*c*sqrt(d) -
2*(sqrt(d)*x - sqrt(d*x^2 + c))^2*a^2*d^(3/2) - a*b*c^2*sqrt(d))/(((sqrt(d)*x - sqrt(d*x^2 + c))^4*b - 2*(sqr
t(d)*x - sqrt(d*x^2 + c))^2*b*c + 4*(sqrt(d)*x - sqrt(d*x^2 + c))^2*a*d + b*c^2)*(b^3*c - a*b^2*d)) - 1/2*log(
(sqrt(d)*x - sqrt(d*x^2 + c))^2)/(b^2*sqrt(d))

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maple [B]  time = 0.02, size = 846, normalized size = 6.41 \[ \frac {3 a \ln \left (\frac {\frac {2 \sqrt {-a b}\, \left (x -\frac {\sqrt {-a b}}{b}\right ) d}{b}-\frac {2 \left (a d -b c \right )}{b}+2 \sqrt {-\frac {a d -b c}{b}}\, \sqrt {\left (x -\frac {\sqrt {-a b}}{b}\right )^{2} d +\frac {2 \sqrt {-a b}\, \left (x -\frac {\sqrt {-a b}}{b}\right ) d}{b}-\frac {a d -b c}{b}}}{x -\frac {\sqrt {-a b}}{b}}\right )}{4 \sqrt {-a b}\, \sqrt {-\frac {a d -b c}{b}}\, b^{2}}-\frac {3 a \ln \left (\frac {-\frac {2 \sqrt {-a b}\, \left (x +\frac {\sqrt {-a b}}{b}\right ) d}{b}-\frac {2 \left (a d -b c \right )}{b}+2 \sqrt {-\frac {a d -b c}{b}}\, \sqrt {\left (x +\frac {\sqrt {-a b}}{b}\right )^{2} d -\frac {2 \sqrt {-a b}\, \left (x +\frac {\sqrt {-a b}}{b}\right ) d}{b}-\frac {a d -b c}{b}}}{x +\frac {\sqrt {-a b}}{b}}\right )}{4 \sqrt {-a b}\, \sqrt {-\frac {a d -b c}{b}}\, b^{2}}+\frac {\sqrt {-a b}\, a d \ln \left (\frac {\frac {2 \sqrt {-a b}\, \left (x -\frac {\sqrt {-a b}}{b}\right ) d}{b}-\frac {2 \left (a d -b c \right )}{b}+2 \sqrt {-\frac {a d -b c}{b}}\, \sqrt {\left (x -\frac {\sqrt {-a b}}{b}\right )^{2} d +\frac {2 \sqrt {-a b}\, \left (x -\frac {\sqrt {-a b}}{b}\right ) d}{b}-\frac {a d -b c}{b}}}{x -\frac {\sqrt {-a b}}{b}}\right )}{4 \left (a d -b c \right ) \sqrt {-\frac {a d -b c}{b}}\, b^{3}}-\frac {\sqrt {-a b}\, a d \ln \left (\frac {-\frac {2 \sqrt {-a b}\, \left (x +\frac {\sqrt {-a b}}{b}\right ) d}{b}-\frac {2 \left (a d -b c \right )}{b}+2 \sqrt {-\frac {a d -b c}{b}}\, \sqrt {\left (x +\frac {\sqrt {-a b}}{b}\right )^{2} d -\frac {2 \sqrt {-a b}\, \left (x +\frac {\sqrt {-a b}}{b}\right ) d}{b}-\frac {a d -b c}{b}}}{x +\frac {\sqrt {-a b}}{b}}\right )}{4 \left (a d -b c \right ) \sqrt {-\frac {a d -b c}{b}}\, b^{3}}-\frac {\sqrt {\left (x -\frac {\sqrt {-a b}}{b}\right )^{2} d +\frac {2 \sqrt {-a b}\, \left (x -\frac {\sqrt {-a b}}{b}\right ) d}{b}-\frac {a d -b c}{b}}\, a}{4 \left (a d -b c \right ) \left (x -\frac {\sqrt {-a b}}{b}\right ) b^{2}}-\frac {\sqrt {\left (x +\frac {\sqrt {-a b}}{b}\right )^{2} d -\frac {2 \sqrt {-a b}\, \left (x +\frac {\sqrt {-a b}}{b}\right ) d}{b}-\frac {a d -b c}{b}}\, a}{4 \left (a d -b c \right ) \left (x +\frac {\sqrt {-a b}}{b}\right ) b^{2}}+\frac {\ln \left (\sqrt {d}\, x +\sqrt {d \,x^{2}+c}\right )}{b^{2} \sqrt {d}} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

int(x^4/(b*x^2+a)^2/(d*x^2+c)^(1/2),x)

[Out]

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

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maxima [F]  time = 0.00, size = 0, normalized size = 0.00 \[ \int \frac {x^{4}}{{\left (b x^{2} + a\right )}^{2} \sqrt {d x^{2} + c}}\,{d x} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate(x^4/(b*x^2+a)^2/(d*x^2+c)^(1/2),x, algorithm="maxima")

[Out]

integrate(x^4/((b*x^2 + a)^2*sqrt(d*x^2 + c)), x)

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mupad [F]  time = 0.00, size = -1, normalized size = -0.01 \[ \int \frac {x^4}{{\left (b\,x^2+a\right )}^2\,\sqrt {d\,x^2+c}} \,d x \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

int(x^4/((a + b*x^2)^2*(c + d*x^2)^(1/2)),x)

[Out]

int(x^4/((a + b*x^2)^2*(c + d*x^2)^(1/2)), x)

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sympy [F]  time = 0.00, size = 0, normalized size = 0.00 \[ \int \frac {x^{4}}{\left (a + b x^{2}\right )^{2} \sqrt {c + d x^{2}}}\, dx \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate(x**4/(b*x**2+a)**2/(d*x**2+c)**(1/2),x)

[Out]

Integral(x**4/((a + b*x**2)**2*sqrt(c + d*x**2)), x)

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