3.6.33 \(\int \frac {x^7}{(a+b x^4)^2 \sqrt {c+d x^4}} \, dx\)

Optimal. Leaf size=99 \[ \frac {a \sqrt {c+d x^4}}{4 b \left (a+b x^4\right ) (b c-a d)}-\frac {(2 b c-a d) \tanh ^{-1}\left (\frac {\sqrt {b} \sqrt {c+d x^4}}{\sqrt {b c-a d}}\right )}{4 b^{3/2} (b c-a d)^{3/2}} \]

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Rubi [A]  time = 0.09, antiderivative size = 99, normalized size of antiderivative = 1.00, number of steps used = 4, number of rules used = 4, integrand size = 24, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.167, Rules used = {446, 78, 63, 208} \begin {gather*} \frac {a \sqrt {c+d x^4}}{4 b \left (a+b x^4\right ) (b c-a d)}-\frac {(2 b c-a d) \tanh ^{-1}\left (\frac {\sqrt {b} \sqrt {c+d x^4}}{\sqrt {b c-a d}}\right )}{4 b^{3/2} (b c-a d)^{3/2}} \end {gather*}

Antiderivative was successfully verified.

[In]

Int[x^7/((a + b*x^4)^2*Sqrt[c + d*x^4]),x]

[Out]

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

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 78

Int[((a_.) + (b_.)*(x_))*((c_.) + (d_.)*(x_))^(n_.)*((e_.) + (f_.)*(x_))^(p_.), x_Symbol] :> -Simp[((b*e - a*f
)*(c + d*x)^(n + 1)*(e + f*x)^(p + 1))/(f*(p + 1)*(c*f - d*e)), x] - Dist[(a*d*f*(n + p + 2) - b*(d*e*(n + 1)
+ c*f*(p + 1)))/(f*(p + 1)*(c*f - d*e)), Int[(c + d*x)^n*(e + f*x)^(p + 1), x], x] /; FreeQ[{a, b, c, d, e, f,
 n}, x] && LtQ[p, -1] && ( !LtQ[n, -1] || IntegerQ[p] ||  !(IntegerQ[n] ||  !(EqQ[e, 0] ||  !(EqQ[c, 0] || LtQ
[p, n]))))

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

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Mathematica [A]  time = 0.11, size = 98, normalized size = 0.99 \begin {gather*} \frac {\frac {a \sqrt {b} \sqrt {c+d x^4}}{\left (a+b x^4\right ) (b c-a d)}+\frac {(a d-2 b c) \tanh ^{-1}\left (\frac {\sqrt {b} \sqrt {c+d x^4}}{\sqrt {b c-a d}}\right )}{(b c-a d)^{3/2}}}{4 b^{3/2}} \end {gather*}

Antiderivative was successfully verified.

[In]

Integrate[x^7/((a + b*x^4)^2*Sqrt[c + d*x^4]),x]

[Out]

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

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IntegrateAlgebraic [A]  time = 0.17, size = 109, normalized size = 1.10 \begin {gather*} \frac {(2 b c-a d) \tan ^{-1}\left (\frac {\sqrt {b} \sqrt {c+d x^4} \sqrt {a d-b c}}{b c-a d}\right )}{4 b^{3/2} (a d-b c)^{3/2}}+\frac {a \sqrt {c+d x^4}}{4 b \left (a+b x^4\right ) (b c-a d)} \end {gather*}

Antiderivative was successfully verified.

[In]

IntegrateAlgebraic[x^7/((a + b*x^4)^2*Sqrt[c + d*x^4]),x]

[Out]

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

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fricas [A]  time = 0.71, size = 348, normalized size = 3.52 \begin {gather*} \left [\frac {{\left ({\left (2 \, b^{2} c - a b d\right )} x^{4} + 2 \, a b c - a^{2} d\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 \, \sqrt {d x^{4} + c} {\left (a b^{2} c - a^{2} b d\right )}}{8 \, {\left (a b^{4} c^{2} - 2 \, a^{2} b^{3} c d + a^{3} b^{2} d^{2} + {\left (b^{5} c^{2} - 2 \, a b^{4} c d + a^{2} b^{3} d^{2}\right )} x^{4}\right )}}, \frac {{\left ({\left (2 \, b^{2} c - a b d\right )} x^{4} + 2 \, a b c - a^{2} d\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 ) + \sqrt {d x^{4} + c} {\left (a b^{2} c - a^{2} b d\right )}}{4 \, {\left (a b^{4} c^{2} - 2 \, a^{2} b^{3} c d + a^{3} b^{2} d^{2} + {\left (b^{5} c^{2} - 2 \, a b^{4} c d + a^{2} b^{3} d^{2}\right )} x^{4}\right )}}\right ] \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate(x^7/(b*x^4+a)^2/(d*x^4+c)^(1/2),x, algorithm="fricas")

[Out]

[1/8*(((2*b^2*c - a*b*d)*x^4 + 2*a*b*c - a^2*d)*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*sqrt(d*x^4 + c)*(a*b^2*c - a^2*b*d))/(a*b^4*c^2 - 2*a^2*b^3*c*d + a
^3*b^2*d^2 + (b^5*c^2 - 2*a*b^4*c*d + a^2*b^3*d^2)*x^4), 1/4*(((2*b^2*c - a*b*d)*x^4 + 2*a*b*c - a^2*d)*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)) + sqrt(d*x^4 + c)*(a*b^2*c - a^2*b
*d))/(a*b^4*c^2 - 2*a^2*b^3*c*d + a^3*b^2*d^2 + (b^5*c^2 - 2*a*b^4*c*d + a^2*b^3*d^2)*x^4)]

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giac [A]  time = 0.17, size = 116, normalized size = 1.17 \begin {gather*} \frac {\frac {\sqrt {d x^{4} + c} a d^{2}}{{\left (b^{2} c - a b d\right )} {\left ({\left (d x^{4} + c\right )} b - b c + a d\right )}} + \frac {{\left (2 \, b c d - a d^{2}\right )} \arctan \left (\frac {\sqrt {d x^{4} + c} b}{\sqrt {-b^{2} c + a b d}}\right )}{{\left (b^{2} c - a b d\right )} \sqrt {-b^{2} c + a b d}}}{4 \, d} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate(x^7/(b*x^4+a)^2/(d*x^4+c)^(1/2),x, algorithm="giac")

[Out]

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

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maple [B]  time = 0.20, size = 851, normalized size = 8.60 \begin {gather*} \frac {a d \ln \left (\frac {\frac {2 \sqrt {-a b}\, \left (x^{2}-\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^{2}-\frac {\sqrt {-a b}}{b}\right )^{2} d +\frac {2 \sqrt {-a b}\, \left (x^{2}-\frac {\sqrt {-a b}}{b}\right ) d}{b}-\frac {a d -b c}{b}}}{x^{2}-\frac {\sqrt {-a b}}{b}}\right )}{8 \left (a d -b c \right ) \sqrt {-\frac {a d -b c}{b}}\, b^{2}}+\frac {a d \ln \left (\frac {-\frac {2 \sqrt {-a b}\, \left (x^{2}+\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^{2}+\frac {\sqrt {-a b}}{b}\right )^{2} d -\frac {2 \sqrt {-a b}\, \left (x^{2}+\frac {\sqrt {-a b}}{b}\right ) d}{b}-\frac {a d -b c}{b}}}{x^{2}+\frac {\sqrt {-a b}}{b}}\right )}{8 \left (a d -b c \right ) \sqrt {-\frac {a d -b c}{b}}\, b^{2}}-\frac {\ln \left (\frac {\frac {2 \sqrt {-a b}\, \left (x^{2}-\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^{2}-\frac {\sqrt {-a b}}{b}\right )^{2} d +\frac {2 \sqrt {-a b}\, \left (x^{2}-\frac {\sqrt {-a b}}{b}\right ) d}{b}-\frac {a d -b c}{b}}}{x^{2}-\frac {\sqrt {-a b}}{b}}\right )}{4 \sqrt {-\frac {a d -b c}{b}}\, b^{2}}-\frac {\ln \left (\frac {-\frac {2 \sqrt {-a b}\, \left (x^{2}+\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^{2}+\frac {\sqrt {-a b}}{b}\right )^{2} d -\frac {2 \sqrt {-a b}\, \left (x^{2}+\frac {\sqrt {-a b}}{b}\right ) d}{b}-\frac {a d -b c}{b}}}{x^{2}+\frac {\sqrt {-a b}}{b}}\right )}{4 \sqrt {-\frac {a d -b c}{b}}\, b^{2}}-\frac {\sqrt {-a b}\, \sqrt {\left (x^{2}+\frac {\sqrt {-a b}}{b}\right )^{2} d -\frac {2 \sqrt {-a b}\, \left (x^{2}+\frac {\sqrt {-a b}}{b}\right ) d}{b}-\frac {a d -b c}{b}}}{8 \left (a d -b c \right ) \left (x^{2}+\frac {\sqrt {-a b}}{b}\right ) b^{2}}+\frac {\sqrt {-a b}\, \sqrt {\left (x^{2}-\frac {\sqrt {-a b}}{b}\right )^{2} d +\frac {2 \sqrt {-a b}\, \left (x^{2}-\frac {\sqrt {-a b}}{b}\right ) d}{b}-\frac {a d -b c}{b}}}{8 \left (a d -b c \right ) \left (x^{2}-\frac {\sqrt {-a b}}{b}\right ) b^{2}} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

int(x^7/(b*x^4+a)^2/(d*x^4+c)^(1/2),x)

[Out]

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

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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^7/(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 details)Is a*d-b*c positive or negative?

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mupad [B]  time = 4.93, size = 95, normalized size = 0.96 \begin {gather*} \frac {\mathrm {atan}\left (\frac {\sqrt {b}\,\sqrt {d\,x^4+c}}{\sqrt {a\,d-b\,c}}\right )\,\left (a\,d-2\,b\,c\right )}{4\,b^{3/2}\,{\left (a\,d-b\,c\right )}^{3/2}}-\frac {a\,d\,\sqrt {d\,x^4+c}}{2\,b\,\left (a\,d-b\,c\right )\,\left (2\,b\,\left (d\,x^4+c\right )+2\,a\,d-2\,b\,c\right )} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

int(x^7/((a + b*x^4)^2*(c + d*x^4)^(1/2)),x)

[Out]

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

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sympy [F(-1)]  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**7/(b*x**4+a)**2/(d*x**4+c)**(1/2),x)

[Out]

Timed out

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