3.8.4 \(\int \frac {x^4}{(a+b x^2) (c+d x^2)^{5/2}} \, dx\)

Optimal. Leaf size=117 \[ \frac {a^{3/2} \tan ^{-1}\left (\frac {x \sqrt {b c-a d}}{\sqrt {a} \sqrt {c+d x^2}}\right )}{(b c-a d)^{5/2}}+\frac {x (b c-4 a d)}{3 d \sqrt {c+d x^2} (b c-a d)^2}-\frac {c x}{3 d \left (c+d x^2\right )^{3/2} (b c-a d)} \]

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Rubi [A]  time = 0.11, antiderivative size = 117, 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} \begin {gather*} \frac {a^{3/2} \tan ^{-1}\left (\frac {x \sqrt {b c-a d}}{\sqrt {a} \sqrt {c+d x^2}}\right )}{(b c-a d)^{5/2}}+\frac {x (b c-4 a d)}{3 d \sqrt {c+d x^2} (b c-a d)^2}-\frac {c x}{3 d \left (c+d x^2\right )^{3/2} (b c-a d)} \end {gather*}

Antiderivative was successfully verified.

[In]

Int[x^4/((a + b*x^2)*(c + d*x^2)^(5/2)),x]

[Out]

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

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Mathematica [A]  time = 0.29, size = 160, normalized size = 1.37 \begin {gather*} \frac {x^2 \left (a^2 d \left (3 c+4 d x^2\right )-a b c \left (3 c+5 d x^2\right )+b^2 c^2 x^2\right )-\frac {3 a^2 \left (c+d x^2\right )^2 \sqrt {\frac {x^2 (a d-b c)}{a c}} \tanh ^{-1}\left (\frac {\sqrt {x^2 \left (\frac {d}{c}-\frac {b}{a}\right )}}{\sqrt {\frac {d x^2}{c}+1}}\right )}{\sqrt {\frac {d x^2}{c}+1}}}{3 x \left (c+d x^2\right )^{3/2} (b c-a d)^3} \end {gather*}

Antiderivative was successfully verified.

[In]

Integrate[x^4/((a + b*x^2)*(c + d*x^2)^(5/2)),x]

[Out]

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

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IntegrateAlgebraic [A]  time = 0.44, size = 148, normalized size = 1.26 \begin {gather*} \frac {-3 a c x-4 a d x^3+b c x^3}{3 \left (c+d x^2\right )^{3/2} (b c-a d)^2}-\frac {a^{3/2} \tan ^{-1}\left (\frac {b \sqrt {d} x^2}{\sqrt {a} \sqrt {b c-a d}}-\frac {b x \sqrt {c+d x^2}}{\sqrt {a} \sqrt {b c-a d}}+\frac {\sqrt {a} \sqrt {d}}{\sqrt {b c-a d}}\right )}{(b c-a d)^{5/2}} \end {gather*}

Antiderivative was successfully verified.

[In]

IntegrateAlgebraic[x^4/((a + b*x^2)*(c + d*x^2)^(5/2)),x]

[Out]

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

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fricas [B]  time = 1.94, size = 524, normalized size = 4.48 \begin {gather*} \left [\frac {3 \, {\left (a d^{2} x^{4} + 2 \, a c d x^{2} + a c^{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 (b c - 4 \, a d\right )} x^{3} - 3 \, a c x\right )} \sqrt {d x^{2} + c}}{12 \, {\left (b^{2} c^{4} - 2 \, a b c^{3} d + a^{2} c^{2} d^{2} + {\left (b^{2} c^{2} d^{2} - 2 \, a b c d^{3} + a^{2} d^{4}\right )} x^{4} + 2 \, {\left (b^{2} c^{3} d - 2 \, a b c^{2} d^{2} + a^{2} c d^{3}\right )} x^{2}\right )}}, -\frac {3 \, {\left (a d^{2} x^{4} + 2 \, a c d x^{2} + a c^{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 (b c - 4 \, a d\right )} x^{3} - 3 \, a c x\right )} \sqrt {d x^{2} + c}}{6 \, {\left (b^{2} c^{4} - 2 \, a b c^{3} d + a^{2} c^{2} d^{2} + {\left (b^{2} c^{2} d^{2} - 2 \, a b c d^{3} + a^{2} d^{4}\right )} x^{4} + 2 \, {\left (b^{2} c^{3} d - 2 \, a b c^{2} d^{2} + a^{2} c d^{3}\right )} x^{2}\right )}}\right ] \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate(x^4/(b*x^2+a)/(d*x^2+c)^(5/2),x, algorithm="fricas")

[Out]

[1/12*(3*(a*d^2*x^4 + 2*a*c*d*x^2 + a*c^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)*sqr
t(d*x^2 + c)*sqrt(-a/(b*c - a*d)))/(b^2*x^4 + 2*a*b*x^2 + a^2)) + 4*((b*c - 4*a*d)*x^3 - 3*a*c*x)*sqrt(d*x^2 +
c))/(b^2*c^4 - 2*a*b*c^3*d + a^2*c^2*d^2 + (b^2*c^2*d^2 - 2*a*b*c*d^3 + a^2*d^4)*x^4 + 2*(b^2*c^3*d - 2*a*b*c
^2*d^2 + a^2*c*d^3)*x^2), -1/6*(3*(a*d^2*x^4 + 2*a*c*d*x^2 + a*c^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*((b*c - 4*a*d)*x^3 - 3*a*c*x)*sqrt(
d*x^2 + c))/(b^2*c^4 - 2*a*b*c^3*d + a^2*c^2*d^2 + (b^2*c^2*d^2 - 2*a*b*c*d^3 + a^2*d^4)*x^4 + 2*(b^2*c^3*d -
2*a*b*c^2*d^2 + a^2*c*d^3)*x^2)]

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giac [B]  time = 0.45, size = 304, normalized size = 2.60 \begin {gather*} -\frac {a^{2} \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 )}{{\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 {{\left (\frac {{\left (b^{3} c^{4} d - 6 \, a b^{2} c^{3} d^{2} + 9 \, a^{2} b c^{2} d^{3} - 4 \, a^{3} c d^{4}\right )} x^{2}}{b^{4} c^{5} d - 4 \, a b^{3} c^{4} d^{2} + 6 \, a^{2} b^{2} c^{3} d^{3} - 4 \, a^{3} b c^{2} d^{4} + a^{4} c d^{5}} - \frac {3 \, {\left (a b^{2} c^{4} d - 2 \, a^{2} b c^{3} d^{2} + a^{3} c^{2} d^{3}\right )}}{b^{4} c^{5} d - 4 \, a b^{3} c^{4} d^{2} + 6 \, a^{2} b^{2} c^{3} d^{3} - 4 \, a^{3} b c^{2} d^{4} + a^{4} c d^{5}}\right )} x}{3 \, {\left (d x^{2} + c\right )}^{\frac {3}{2}}} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate(x^4/(b*x^2+a)/(d*x^2+c)^(5/2),x, algorithm="giac")

[Out]

-a^2*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)) + 1/3*((b^3*c^4*d - 6*a*b^2*c^3*d^2 + 9*a^2*b*c^2*d^3 - 4*a^3*c
*d^4)*x^2/(b^4*c^5*d - 4*a*b^3*c^4*d^2 + 6*a^2*b^2*c^3*d^3 - 4*a^3*b*c^2*d^4 + a^4*c*d^5) - 3*(a*b^2*c^4*d - 2
*a^2*b*c^3*d^2 + a^3*c^2*d^3)/(b^4*c^5*d - 4*a*b^3*c^4*d^2 + 6*a^2*b^2*c^3*d^3 - 4*a^3*b*c^2*d^4 + a^4*c*d^5))
*x/(d*x^2 + c)^(3/2)

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maple [B]  time = 0.03, size = 1207, normalized size = 10.32

result too large to display

Verification of antiderivative is not currently implemented for this CAS.

[In]

int(x^4/(b*x^2+a)/(d*x^2+c)^(5/2),x)

[Out]

-1/3/b/d*x/(d*x^2+c)^(3/2)+1/3/b/c/d*x/(d*x^2+c)^(1/2)-1/3/b^2*a*x/c/(d*x^2+c)^(3/2)-2/3/b^2*a/c^2*x/(d*x^2+c)
^(1/2)+1/6/b*a^2/(-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)^(3/2)+1/6/b^2*a^2*d/(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
)^(3/2)*x+1/3/b^2*a^2*d/(a*d-b*c)/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)*x-1/2*a^2/(-a*b)^(1/2)/(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)-1/2/b*a^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)*d*x+1/2*a^2/(-a*b)^(1/2)/(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))-1/6/b*a^2/(-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)^(3/2)+1/6/b^2*a^2*d/(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)^(3/2)*x+1/3/b^2*a^2*d/(a*d-b*c)/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)*x+1/2*a^2/(-a*b)^(1/2)/(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)-1/2/b*a^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)*d*x-1/2*a^2/(-a*b)^(1/2)/(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))

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maxima [F]  time = 0.00, size = 0, normalized size = 0.00 \begin {gather*} \int \frac {x^{4}}{{\left (b x^{2} + a\right )} {\left (d x^{2} + c\right )}^{\frac {5}{2}}}\,{d x} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate(x^4/(b*x^2+a)/(d*x^2+c)^(5/2),x, algorithm="maxima")

[Out]

integrate(x^4/((b*x^2 + a)*(d*x^2 + c)^(5/2)), x)

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mupad [F]  time = 0.00, size = -1, normalized size = -0.01 \begin {gather*} \int \frac {x^4}{\left (b\,x^2+a\right )\,{\left (d\,x^2+c\right )}^{5/2}} \,d x \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

int(x^4/((a + b*x^2)*(c + d*x^2)^(5/2)),x)

[Out]

int(x^4/((a + b*x^2)*(c + d*x^2)^(5/2)), x)

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sympy [F]  time = 0.00, size = 0, normalized size = 0.00 \begin {gather*} \int \frac {x^{4}}{\left (a + b x^{2}\right ) \left (c + d x^{2}\right )^{\frac {5}{2}}}\, dx \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate(x**4/(b*x**2+a)/(d*x**2+c)**(5/2),x)

[Out]

Integral(x**4/((a + b*x**2)*(c + d*x**2)**(5/2)), x)

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