3.9.22 \(\int \frac {x^{15}}{(a+b x^4)^2 \sqrt {c+d x^4}} \, dx\) [822]

Optimal. Leaf size=175 \[ \frac {a x^8 \sqrt {c+d x^4}}{4 b (b c-a d) \left (a+b x^4\right )}-\frac {\sqrt {c+d x^4} \left (4 b^2 c^2+8 a b c d-15 a^2 d^2-b d (2 b c-5 a d) x^4\right )}{12 b^3 d^2 (b c-a d)}-\frac {a^2 (6 b c-5 a d) \tanh ^{-1}\left (\frac {\sqrt {b} \sqrt {c+d x^4}}{\sqrt {b c-a d}}\right )}{4 b^{7/2} (b c-a d)^{3/2}} \]

[Out]

-1/4*a^2*(-5*a*d+6*b*c)*arctanh(b^(1/2)*(d*x^4+c)^(1/2)/(-a*d+b*c)^(1/2))/b^(7/2)/(-a*d+b*c)^(3/2)+1/4*a*x^8*(
d*x^4+c)^(1/2)/b/(-a*d+b*c)/(b*x^4+a)-1/12*(4*b^2*c^2+8*a*b*c*d-15*a^2*d^2-b*d*(-5*a*d+2*b*c)*x^4)*(d*x^4+c)^(
1/2)/b^3/d^2/(-a*d+b*c)

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Rubi [A]
time = 0.12, antiderivative size = 175, 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 = {457, 100, 152, 65, 214} \begin {gather*} -\frac {a^2 (6 b c-5 a d) \tanh ^{-1}\left (\frac {\sqrt {b} \sqrt {c+d x^4}}{\sqrt {b c-a d}}\right )}{4 b^{7/2} (b c-a d)^{3/2}}-\frac {\sqrt {c+d x^4} \left (-15 a^2 d^2-b d x^4 (2 b c-5 a d)+8 a b c d+4 b^2 c^2\right )}{12 b^3 d^2 (b c-a d)}+\frac {a x^8 \sqrt {c+d x^4}}{4 b \left (a+b x^4\right ) (b c-a d)} \end {gather*}

Antiderivative was successfully verified.

[In]

Int[x^15/((a + b*x^4)^2*Sqrt[c + d*x^4]),x]

[Out]

(a*x^8*Sqrt[c + d*x^4])/(4*b*(b*c - a*d)*(a + b*x^4)) - (Sqrt[c + d*x^4]*(4*b^2*c^2 + 8*a*b*c*d - 15*a^2*d^2 -
b*d*(2*b*c - 5*a*d)*x^4))/(12*b^3*d^2*(b*c - a*d)) - (a^2*(6*b*c - 5*a*d)*ArcTanh[(Sqrt[b]*Sqrt[c + d*x^4])/S
qrt[b*c - a*d]])/(4*b^(7/2)*(b*c - a*d)^(3/2))

Rule 65

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 100

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

Rule 152

Int[((a_.) + (b_.)*(x_))^(m_.)*((c_.) + (d_.)*(x_))^(n_.)*((e_) + (f_.)*(x_))*((g_.) + (h_.)*(x_)), x_Symbol]
:> Simp[(-(a*d*f*h*(n + 2) + b*c*f*h*(m + 2) - b*d*(f*g + e*h)*(m + n + 3) - b*d*f*h*(m + n + 2)*x))*(a + b*x)
^(m + 1)*((c + d*x)^(n + 1)/(b^2*d^2*(m + n + 2)*(m + n + 3))), x] + Dist[(a^2*d^2*f*h*(n + 1)*(n + 2) + a*b*d
*(n + 1)*(2*c*f*h*(m + 1) - d*(f*g + e*h)*(m + n + 3)) + b^2*(c^2*f*h*(m + 1)*(m + 2) - c*d*(f*g + e*h)*(m + 1
)*(m + n + 3) + d^2*e*g*(m + n + 2)*(m + n + 3)))/(b^2*d^2*(m + n + 2)*(m + n + 3)), Int[(a + b*x)^m*(c + d*x)
^n, x], x] /; FreeQ[{a, b, c, d, e, f, g, h, m, n}, x] && NeQ[m + n + 2, 0] && NeQ[m + n + 3, 0]

Rule 214

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

Rule 457

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 {x^{15}}{\left (a+b x^4\right )^2 \sqrt {c+d x^4}} \, dx &=\frac {1}{4} \text {Subst}\left (\int \frac {x^3}{(a+b x)^2 \sqrt {c+d x}} \, dx,x,x^4\right )\\ &=\frac {a x^8 \sqrt {c+d x^4}}{4 b (b c-a d) \left (a+b x^4\right )}-\frac {\text {Subst}\left (\int \frac {x \left (2 a c+\frac {1}{2} (-2 b c+5 a d) x\right )}{(a+b x) \sqrt {c+d x}} \, dx,x,x^4\right )}{4 b (b c-a d)}\\ &=\frac {a x^8 \sqrt {c+d x^4}}{4 b (b c-a d) \left (a+b x^4\right )}-\frac {\sqrt {c+d x^4} \left (4 b^2 c^2+8 a b c d-15 a^2 d^2-b d (2 b c-5 a d) x^4\right )}{12 b^3 d^2 (b c-a d)}+\frac {\left (a^2 (6 b c-5 a d)\right ) \text {Subst}\left (\int \frac {1}{(a+b x) \sqrt {c+d x}} \, dx,x,x^4\right )}{8 b^3 (b c-a d)}\\ &=\frac {a x^8 \sqrt {c+d x^4}}{4 b (b c-a d) \left (a+b x^4\right )}-\frac {\sqrt {c+d x^4} \left (4 b^2 c^2+8 a b c d-15 a^2 d^2-b d (2 b c-5 a d) x^4\right )}{12 b^3 d^2 (b c-a d)}+\frac {\left (a^2 (6 b c-5 a d)\right ) \text {Subst}\left (\int \frac {1}{a-\frac {b c}{d}+\frac {b x^2}{d}} \, dx,x,\sqrt {c+d x^4}\right )}{4 b^3 d (b c-a d)}\\ &=\frac {a x^8 \sqrt {c+d x^4}}{4 b (b c-a d) \left (a+b x^4\right )}-\frac {\sqrt {c+d x^4} \left (4 b^2 c^2+8 a b c d-15 a^2 d^2-b d (2 b c-5 a d) x^4\right )}{12 b^3 d^2 (b c-a d)}-\frac {a^2 (6 b c-5 a d) \tanh ^{-1}\left (\frac {\sqrt {b} \sqrt {c+d x^4}}{\sqrt {b c-a d}}\right )}{4 b^{7/2} (b c-a d)^{3/2}}\\ \end {align*}

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Mathematica [A]
time = 0.44, size = 175, normalized size = 1.00 \begin {gather*} -\frac {\sqrt {c+d x^4} \left (-15 a^3 d^2+2 a^2 b d \left (4 c-5 d x^4\right )+2 b^3 c x^4 \left (2 c-d x^4\right )+2 a b^2 \left (2 c^2+3 c d x^4+d^2 x^8\right )\right )}{12 b^3 d^2 (b c-a d) \left (a+b x^4\right )}+\frac {a^2 (-6 b c+5 a d) \tan ^{-1}\left (\frac {\sqrt {b} \sqrt {c+d x^4}}{\sqrt {-b c+a d}}\right )}{4 b^{7/2} (-b c+a d)^{3/2}} \end {gather*}

Antiderivative was successfully verified.

[In]

Integrate[x^15/((a + b*x^4)^2*Sqrt[c + d*x^4]),x]

[Out]

-1/12*(Sqrt[c + d*x^4]*(-15*a^3*d^2 + 2*a^2*b*d*(4*c - 5*d*x^4) + 2*b^3*c*x^4*(2*c - d*x^4) + 2*a*b^2*(2*c^2 +
3*c*d*x^4 + d^2*x^8)))/(b^3*d^2*(b*c - a*d)*(a + b*x^4)) + (a^2*(-6*b*c + 5*a*d)*ArcTan[(Sqrt[b]*Sqrt[c + d*x
^4])/Sqrt[-(b*c) + a*d]])/(4*b^(7/2)*(-(b*c) + a*d)^(3/2))

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Maple [B] Leaf count of result is larger than twice the leaf count of optimal. \(917\) vs. \(2(155)=310\).
time = 0.39, size = 918, normalized size = 5.25 Too large to display

Verification of antiderivative is not currently implemented for this CAS.

[In]

int(x^15/(b*x^4+a)^2/(d*x^4+c)^(1/2),x,method=_RETURNVERBOSE)

[Out]

-1/6/b^2*(d*x^4+c)^(1/2)*(-d*x^4+2*c)/d^2-a/b^3/d*(d*x^4+c)^(1/2)+3*a^2/b^3*(-1/4/b/(-(a*d-b*c)/b)^(1/2)*ln((-
2*(a*d-b*c)/b+2*d*(-a*b)^(1/2)/b*(x^2-1/b*(-a*b)^(1/2))+2*(-(a*d-b*c)/b)^(1/2)*((x^2-1/b*(-a*b)^(1/2))^2*d+2*d
*(-a*b)^(1/2)/b*(x^2-1/b*(-a*b)^(1/2))-(a*d-b*c)/b)^(1/2))/(x^2-1/b*(-a*b)^(1/2)))-1/4/b/(-(a*d-b*c)/b)^(1/2)*
ln((-2*(a*d-b*c)/b-2*d*(-a*b)^(1/2)/b*(x^2+1/b*(-a*b)^(1/2))+2*(-(a*d-b*c)/b)^(1/2)*((x^2+1/b*(-a*b)^(1/2))^2*
d-2*d*(-a*b)^(1/2)/b*(x^2+1/b*(-a*b)^(1/2))-(a*d-b*c)/b)^(1/2))/(x^2+1/b*(-a*b)^(1/2))))-a^3/b^3*(-1/8/a*(-a*b
)^(1/2)/b/(a*d-b*c)/(x^2-1/b*(-a*b)^(1/2))*((x^2-1/b*(-a*b)^(1/2))^2*d+2*d*(-a*b)^(1/2)/b*(x^2-1/b*(-a*b)^(1/2
))-(a*d-b*c)/b)^(1/2)-1/8/b*d/(a*d-b*c)/(-(a*d-b*c)/b)^(1/2)*ln((-2*(a*d-b*c)/b+2*d*(-a*b)^(1/2)/b*(x^2-1/b*(-
a*b)^(1/2))+2*(-(a*d-b*c)/b)^(1/2)*((x^2-1/b*(-a*b)^(1/2))^2*d+2*d*(-a*b)^(1/2)/b*(x^2-1/b*(-a*b)^(1/2))-(a*d-
b*c)/b)^(1/2))/(x^2-1/b*(-a*b)^(1/2)))+1/8/a*(-a*b)^(1/2)/b/(a*d-b*c)/(x^2+1/b*(-a*b)^(1/2))*((x^2+1/b*(-a*b)^
(1/2))^2*d-2*d*(-a*b)^(1/2)/b*(x^2+1/b*(-a*b)^(1/2))-(a*d-b*c)/b)^(1/2)-1/8/b*d/(a*d-b*c)/(-(a*d-b*c)/b)^(1/2)
*ln((-2*(a*d-b*c)/b-2*d*(-a*b)^(1/2)/b*(x^2+1/b*(-a*b)^(1/2))+2*(-(a*d-b*c)/b)^(1/2)*((x^2+1/b*(-a*b)^(1/2))^2
*d-2*d*(-a*b)^(1/2)/b*(x^2+1/b*(-a*b)^(1/2))-(a*d-b*c)/b)^(1/2))/(x^2+1/b*(-a*b)^(1/2))))

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Maxima [F(-2)]
time = 0.00, size = 0, normalized size = 0.00 \begin {gather*} \text {Exception raised: ValueError} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate(x^15/(b*x^4+a)^2/(d*x^4+c)^(1/2),x, algorithm="maxima")

[Out]

Exception raised: ValueError >> Computation failed since Maxima requested additional constraints; using the 'a
ssume' command before evaluation *may* help (example of legal syntax is 'assume(a*d-b*c>0)', see `assume?` for
more detail

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Fricas [A]
time = 3.17, size = 622, normalized size = 3.55 \begin {gather*} \left [\frac {3 \, {\left (6 \, a^{3} b c d^{2} - 5 \, a^{4} d^{3} + {\left (6 \, a^{2} b^{2} c d^{2} - 5 \, a^{3} b d^{3}\right )} x^{4}\right )} \sqrt {b^{2} c - a b d} \log \left (\frac {b d x^{4} + 2 \, b c - a d - 2 \, \sqrt {d x^{4} + c} \sqrt {b^{2} c - a b d}}{b x^{4} + a}\right ) + 2 \, {\left (2 \, {\left (b^{5} c^{2} d - 2 \, a b^{4} c d^{2} + a^{2} b^{3} d^{3}\right )} x^{8} - 4 \, a b^{4} c^{3} - 4 \, a^{2} b^{3} c^{2} d + 23 \, a^{3} b^{2} c d^{2} - 15 \, a^{4} b d^{3} - 2 \, {\left (2 \, b^{5} c^{3} + a b^{4} c^{2} d - 8 \, a^{2} b^{3} c d^{2} + 5 \, a^{3} b^{2} d^{3}\right )} x^{4}\right )} \sqrt {d x^{4} + c}}{24 \, {\left (a b^{6} c^{2} d^{2} - 2 \, a^{2} b^{5} c d^{3} + a^{3} b^{4} d^{4} + {\left (b^{7} c^{2} d^{2} - 2 \, a b^{6} c d^{3} + a^{2} b^{5} d^{4}\right )} x^{4}\right )}}, \frac {3 \, {\left (6 \, a^{3} b c d^{2} - 5 \, a^{4} d^{3} + {\left (6 \, a^{2} b^{2} c d^{2} - 5 \, a^{3} b d^{3}\right )} x^{4}\right )} \sqrt {-b^{2} c + a b d} \arctan \left (\frac {\sqrt {d x^{4} + c} \sqrt {-b^{2} c + a b d}}{b d x^{4} + b c}\right ) + {\left (2 \, {\left (b^{5} c^{2} d - 2 \, a b^{4} c d^{2} + a^{2} b^{3} d^{3}\right )} x^{8} - 4 \, a b^{4} c^{3} - 4 \, a^{2} b^{3} c^{2} d + 23 \, a^{3} b^{2} c d^{2} - 15 \, a^{4} b d^{3} - 2 \, {\left (2 \, b^{5} c^{3} + a b^{4} c^{2} d - 8 \, a^{2} b^{3} c d^{2} + 5 \, a^{3} b^{2} d^{3}\right )} x^{4}\right )} \sqrt {d x^{4} + c}}{12 \, {\left (a b^{6} c^{2} d^{2} - 2 \, a^{2} b^{5} c d^{3} + a^{3} b^{4} d^{4} + {\left (b^{7} c^{2} d^{2} - 2 \, a b^{6} c d^{3} + a^{2} b^{5} d^{4}\right )} x^{4}\right )}}\right ] \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate(x^15/(b*x^4+a)^2/(d*x^4+c)^(1/2),x, algorithm="fricas")

[Out]

[1/24*(3*(6*a^3*b*c*d^2 - 5*a^4*d^3 + (6*a^2*b^2*c*d^2 - 5*a^3*b*d^3)*x^4)*sqrt(b^2*c - a*b*d)*log((b*d*x^4 +
2*b*c - a*d - 2*sqrt(d*x^4 + c)*sqrt(b^2*c - a*b*d))/(b*x^4 + a)) + 2*(2*(b^5*c^2*d - 2*a*b^4*c*d^2 + a^2*b^3*
d^3)*x^8 - 4*a*b^4*c^3 - 4*a^2*b^3*c^2*d + 23*a^3*b^2*c*d^2 - 15*a^4*b*d^3 - 2*(2*b^5*c^3 + a*b^4*c^2*d - 8*a^
2*b^3*c*d^2 + 5*a^3*b^2*d^3)*x^4)*sqrt(d*x^4 + c))/(a*b^6*c^2*d^2 - 2*a^2*b^5*c*d^3 + a^3*b^4*d^4 + (b^7*c^2*d
^2 - 2*a*b^6*c*d^3 + a^2*b^5*d^4)*x^4), 1/12*(3*(6*a^3*b*c*d^2 - 5*a^4*d^3 + (6*a^2*b^2*c*d^2 - 5*a^3*b*d^3)*x
^4)*sqrt(-b^2*c + a*b*d)*arctan(sqrt(d*x^4 + c)*sqrt(-b^2*c + a*b*d)/(b*d*x^4 + b*c)) + (2*(b^5*c^2*d - 2*a*b^
4*c*d^2 + a^2*b^3*d^3)*x^8 - 4*a*b^4*c^3 - 4*a^2*b^3*c^2*d + 23*a^3*b^2*c*d^2 - 15*a^4*b*d^3 - 2*(2*b^5*c^3 +
a*b^4*c^2*d - 8*a^2*b^3*c*d^2 + 5*a^3*b^2*d^3)*x^4)*sqrt(d*x^4 + c))/(a*b^6*c^2*d^2 - 2*a^2*b^5*c*d^3 + a^3*b^
4*d^4 + (b^7*c^2*d^2 - 2*a*b^6*c*d^3 + a^2*b^5*d^4)*x^4)]

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Sympy [F(-1)] Timed out
time = 0.00, size = 0, normalized size = 0.00 \begin {gather*} \text {Timed out} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate(x**15/(b*x**4+a)**2/(d*x**4+c)**(1/2),x)

[Out]

Timed out

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Giac [A]
time = 2.22, size = 180, normalized size = 1.03 \begin {gather*} \frac {\sqrt {d x^{4} + c} a^{3} d}{4 \, {\left (b^{4} c - a b^{3} d\right )} {\left ({\left (d x^{4} + c\right )} b - b c + a d\right )}} + \frac {{\left (6 \, a^{2} b c - 5 \, a^{3} d\right )} \arctan \left (\frac {\sqrt {d x^{4} + c} b}{\sqrt {-b^{2} c + a b d}}\right )}{4 \, {\left (b^{4} c - a b^{3} d\right )} \sqrt {-b^{2} c + a b d}} + \frac {{\left (d x^{4} + c\right )}^{\frac {3}{2}} b^{4} d^{4} - 3 \, \sqrt {d x^{4} + c} b^{4} c d^{4} - 6 \, \sqrt {d x^{4} + c} a b^{3} d^{5}}{6 \, b^{6} d^{6}} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate(x^15/(b*x^4+a)^2/(d*x^4+c)^(1/2),x, algorithm="giac")

[Out]

1/4*sqrt(d*x^4 + c)*a^3*d/((b^4*c - a*b^3*d)*((d*x^4 + c)*b - b*c + a*d)) + 1/4*(6*a^2*b*c - 5*a^3*d)*arctan(s
qrt(d*x^4 + c)*b/sqrt(-b^2*c + a*b*d))/((b^4*c - a*b^3*d)*sqrt(-b^2*c + a*b*d)) + 1/6*((d*x^4 + c)^(3/2)*b^4*d
^4 - 3*sqrt(d*x^4 + c)*b^4*c*d^4 - 6*sqrt(d*x^4 + c)*a*b^3*d^5)/(b^6*d^6)

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Mupad [B]
time = 5.19, size = 186, normalized size = 1.06 \begin {gather*} \frac {{\left (d\,x^4+c\right )}^{3/2}}{6\,b^2\,d^2}-\left (\frac {3\,c}{2\,b^2\,d^2}+\frac {a\,d-b\,c}{b^3\,d^2}\right )\,\sqrt {d\,x^4+c}+\frac {a^2\,\mathrm {atan}\left (\frac {a^2\,\sqrt {b}\,\sqrt {d\,x^4+c}\,\left (5\,a\,d-6\,b\,c\right )}{\sqrt {a\,d-b\,c}\,\left (5\,a^3\,d-6\,a^2\,b\,c\right )}\right )\,\left (5\,a\,d-6\,b\,c\right )}{4\,b^{7/2}\,{\left (a\,d-b\,c\right )}^{3/2}}-\frac {a^3\,d\,\sqrt {d\,x^4+c}}{2\,\left (a\,d-b\,c\right )\,\left (2\,b^4\,\left (d\,x^4+c\right )-2\,b^4\,c+2\,a\,b^3\,d\right )} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

int(x^15/((a + b*x^4)^2*(c + d*x^4)^(1/2)),x)

[Out]

(c + d*x^4)^(3/2)/(6*b^2*d^2) - ((3*c)/(2*b^2*d^2) + (a*d - b*c)/(b^3*d^2))*(c + d*x^4)^(1/2) + (a^2*atan((a^2
*b^(1/2)*(c + d*x^4)^(1/2)*(5*a*d - 6*b*c))/((a*d - b*c)^(1/2)*(5*a^3*d - 6*a^2*b*c)))*(5*a*d - 6*b*c))/(4*b^(
7/2)*(a*d - b*c)^(3/2)) - (a^3*d*(c + d*x^4)^(1/2))/(2*(a*d - b*c)*(2*b^4*(c + d*x^4) - 2*b^4*c + 2*a*b^3*d))

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