3.767 \(\int \frac {x^4}{(a+b x^2)^2 (c+d x^2)^{3/2}} \, dx\)

Optimal. Leaf size=130 \[ \frac {x (a d+2 b c)}{2 b \sqrt {c+d x^2} (b c-a d)^2}+\frac {a x}{2 b \left (a+b x^2\right ) \sqrt {c+d x^2} (b c-a d)}-\frac {3 \sqrt {a} c \tan ^{-1}\left (\frac {x \sqrt {b c-a d}}{\sqrt {a} \sqrt {c+d x^2}}\right )}{2 (b c-a d)^{5/2}} \]

[Out]

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

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Rubi [A]  time = 0.10, antiderivative size = 130, normalized size of antiderivative = 1.00, number of steps used = 5, number of rules used = 5, integrand size = 24, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.208, Rules used = {470, 527, 12, 377, 205} \[ \frac {x (a d+2 b c)}{2 b \sqrt {c+d x^2} (b c-a d)^2}+\frac {a x}{2 b \left (a+b x^2\right ) \sqrt {c+d x^2} (b c-a d)}-\frac {3 \sqrt {a} c \tan ^{-1}\left (\frac {x \sqrt {b c-a d}}{\sqrt {a} \sqrt {c+d x^2}}\right )}{2 (b c-a d)^{5/2}} \]

Antiderivative was successfully verified.

[In]

Int[x^4/((a + b*x^2)^2*(c + d*x^2)^(3/2)),x]

[Out]

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

Rule 12

Int[(a_)*(u_), x_Symbol] :> Dist[a, Int[u, x], x] /; FreeQ[a, x] &&  !MatchQ[u, (b_)*(v_) /; FreeQ[b, x]]

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 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 527

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

Rubi steps

\begin {align*} \int \frac {x^4}{\left (a+b x^2\right )^2 \left (c+d x^2\right )^{3/2}} \, dx &=\frac {a x}{2 b (b c-a d) \left (a+b x^2\right ) \sqrt {c+d x^2}}-\frac {\int \frac {a c-2 b c x^2}{\left (a+b x^2\right ) \left (c+d x^2\right )^{3/2}} \, dx}{2 b (b c-a d)}\\ &=\frac {(2 b c+a d) x}{2 b (b c-a d)^2 \sqrt {c+d x^2}}+\frac {a x}{2 b (b c-a d) \left (a+b x^2\right ) \sqrt {c+d x^2}}-\frac {\int \frac {3 a b c^2}{\left (a+b x^2\right ) \sqrt {c+d x^2}} \, dx}{2 b c (b c-a d)^2}\\ &=\frac {(2 b c+a d) x}{2 b (b c-a d)^2 \sqrt {c+d x^2}}+\frac {a x}{2 b (b c-a d) \left (a+b x^2\right ) \sqrt {c+d x^2}}-\frac {(3 a c) \int \frac {1}{\left (a+b x^2\right ) \sqrt {c+d x^2}} \, dx}{2 (b c-a d)^2}\\ &=\frac {(2 b c+a d) x}{2 b (b c-a d)^2 \sqrt {c+d x^2}}+\frac {a x}{2 b (b c-a d) \left (a+b x^2\right ) \sqrt {c+d x^2}}-\frac {(3 a c) \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 c-a d)^2}\\ &=\frac {(2 b c+a d) x}{2 b (b c-a d)^2 \sqrt {c+d x^2}}+\frac {a x}{2 b (b c-a d) \left (a+b x^2\right ) \sqrt {c+d x^2}}-\frac {3 \sqrt {a} c \tan ^{-1}\left (\frac {\sqrt {b c-a d} x}{\sqrt {a} \sqrt {c+d x^2}}\right )}{2 (b c-a d)^{5/2}}\\ \end {align*}

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Mathematica [C]  time = 0.02, size = 54, normalized size = 0.42 \[ \frac {c x^5 \, _2F_1\left (2,\frac {5}{2};\frac {7}{2};\frac {(a d-b c) x^2}{a \left (d x^2+c\right )}\right )}{5 a^2 \left (c+d x^2\right )^{5/2}} \]

Antiderivative was successfully verified.

[In]

Integrate[x^4/((a + b*x^2)^2*(c + d*x^2)^(3/2)),x]

[Out]

(c*x^5*Hypergeometric2F1[2, 5/2, 7/2, ((-(b*c) + a*d)*x^2)/(a*(c + d*x^2))])/(5*a^2*(c + d*x^2)^(5/2))

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fricas [B]  time = 1.31, size = 552, normalized size = 4.25 \[ \left [\frac {3 \, {\left (b c d x^{4} + a c^{2} + {\left (b c^{2} + a c d\right )} x^{2}\right )} \sqrt {-\frac {a}{b c - a d}} \log \left (\frac {{\left (b^{2} c^{2} - 8 \, a b c d + 8 \, a^{2} d^{2}\right )} x^{4} + a^{2} c^{2} - 2 \, {\left (3 \, a b c^{2} - 4 \, a^{2} c d\right )} x^{2} - 4 \, {\left ({\left (b^{2} c^{2} - 3 \, a b c d + 2 \, a^{2} d^{2}\right )} x^{3} - {\left (a b c^{2} - a^{2} c d\right )} x\right )} \sqrt {d x^{2} + c} \sqrt {-\frac {a}{b c - a d}}}{b^{2} x^{4} + 2 \, a b x^{2} + a^{2}}\right ) + 4 \, {\left ({\left (2 \, b c + a d\right )} x^{3} + 3 \, a c x\right )} \sqrt {d x^{2} + c}}{8 \, {\left (a b^{2} c^{3} - 2 \, a^{2} b c^{2} d + a^{3} c d^{2} + {\left (b^{3} c^{2} d - 2 \, a b^{2} c d^{2} + a^{2} b d^{3}\right )} x^{4} + {\left (b^{3} c^{3} - a b^{2} c^{2} d - a^{2} b c d^{2} + a^{3} d^{3}\right )} x^{2}\right )}}, \frac {3 \, {\left (b c d x^{4} + a c^{2} + {\left (b c^{2} + a c d\right )} x^{2}\right )} \sqrt {\frac {a}{b c - a d}} \arctan \left (-\frac {{\left ({\left (b c - 2 \, a d\right )} x^{2} - a c\right )} \sqrt {d x^{2} + c} \sqrt {\frac {a}{b c - a d}}}{2 \, {\left (a d x^{3} + a c x\right )}}\right ) + 2 \, {\left ({\left (2 \, b c + a d\right )} x^{3} + 3 \, a c x\right )} \sqrt {d x^{2} + c}}{4 \, {\left (a b^{2} c^{3} - 2 \, a^{2} b c^{2} d + a^{3} c d^{2} + {\left (b^{3} c^{2} d - 2 \, a b^{2} c d^{2} + a^{2} b d^{3}\right )} x^{4} + {\left (b^{3} c^{3} - a b^{2} c^{2} d - a^{2} b c d^{2} + a^{3} d^{3}\right )} x^{2}\right )}}\right ] \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate(x^4/(b*x^2+a)^2/(d*x^2+c)^(3/2),x, algorithm="fricas")

[Out]

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

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giac [B]  time = 4.33, size = 298, normalized size = 2.29 \[ \frac {3 \, a c \sqrt {d} \arctan \left (\frac {{\left (\sqrt {d} x - \sqrt {d x^{2} + c}\right )}^{2} b - b c + 2 \, a d}{2 \, \sqrt {a b c d - a^{2} d^{2}}}\right )}{2 \, {\left (b^{2} c^{2} - 2 \, a b c d + a^{2} d^{2}\right )} \sqrt {a b c d - a^{2} d^{2}}} + \frac {c x}{{\left (b^{2} c^{2} - 2 \, a b c d + a^{2} d^{2}\right )} \sqrt {d x^{2} + c}} - \frac {{\left (\sqrt {d} x - \sqrt {d x^{2} + c}\right )}^{2} a b c \sqrt {d} - 2 \, {\left (\sqrt {d} x - \sqrt {d x^{2} + c}\right )}^{2} a^{2} d^{\frac {3}{2}} - a b c^{2} \sqrt {d}}{{\left ({\left (\sqrt {d} x - \sqrt {d x^{2} + c}\right )}^{4} b - 2 \, {\left (\sqrt {d} x - \sqrt {d x^{2} + c}\right )}^{2} b c + 4 \, {\left (\sqrt {d} x - \sqrt {d x^{2} + c}\right )}^{2} a d + b c^{2}\right )} {\left (b^{3} c^{2} - 2 \, a b^{2} c d + a^{2} b d^{2}\right )}} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate(x^4/(b*x^2+a)^2/(d*x^2+c)^(3/2),x, algorithm="giac")

[Out]

3/2*a*c*sqrt(d)*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^2*c^
2 - 2*a*b*c*d + a^2*d^2)*sqrt(a*b*c*d - a^2*d^2)) + c*x/((b^2*c^2 - 2*a*b*c*d + a^2*d^2)*sqrt(d*x^2 + c)) - ((
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*(sqrt(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^2 - 2*a*b^2*c*d + a^2*b*d^2))

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maple [B]  time = 0.03, size = 1498, normalized size = 11.52 \[ \text {result too large to display} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

int(x^4/(b*x^2+a)^2/(d*x^2+c)^(3/2),x)

[Out]

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

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maxima [F]  time = 0.00, size = 0, normalized size = 0.00 \[ \int \frac {x^{4}}{{\left (b x^{2} + a\right )}^{2} {\left (d x^{2} + c\right )}^{\frac {3}{2}}}\,{d x} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate(x^4/(b*x^2+a)^2/(d*x^2+c)^(3/2),x, algorithm="maxima")

[Out]

integrate(x^4/((b*x^2 + a)^2*(d*x^2 + c)^(3/2)), x)

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mupad [F]  time = 0.00, size = -1, normalized size = -0.01 \[ \int \frac {x^4}{{\left (b\,x^2+a\right )}^2\,{\left (d\,x^2+c\right )}^{3/2}} \,d x \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

int(x^4/((a + b*x^2)^2*(c + d*x^2)^(3/2)),x)

[Out]

int(x^4/((a + b*x^2)^2*(c + d*x^2)^(3/2)), x)

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sympy [F]  time = 0.00, size = 0, normalized size = 0.00 \[ \int \frac {x^{4}}{\left (a + b x^{2}\right )^{2} \left (c + d x^{2}\right )^{\frac {3}{2}}}\, dx \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate(x**4/(b*x**2+a)**2/(d*x**2+c)**(3/2),x)

[Out]

Integral(x**4/((a + b*x**2)**2*(c + d*x**2)**(3/2)), x)

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