3.6.27 \(\int \frac {x^5}{(a+b x^4) \sqrt {c+d x^4}} \, dx\)

Optimal. Leaf size=91 \[ \frac {\tanh ^{-1}\left (\frac {\sqrt {d} x^2}{\sqrt {c+d x^4}}\right )}{2 b \sqrt {d}}-\frac {\sqrt {a} \tan ^{-1}\left (\frac {x^2 \sqrt {b c-a d}}{\sqrt {a} \sqrt {c+d x^4}}\right )}{2 b \sqrt {b c-a d}} \]

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Rubi [A]  time = 0.09, antiderivative size = 91, 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 = {465, 483, 217, 206, 377, 205} \begin {gather*} \frac {\tanh ^{-1}\left (\frac {\sqrt {d} x^2}{\sqrt {c+d x^4}}\right )}{2 b \sqrt {d}}-\frac {\sqrt {a} \tan ^{-1}\left (\frac {x^2 \sqrt {b c-a d}}{\sqrt {a} \sqrt {c+d x^4}}\right )}{2 b \sqrt {b c-a d}} \end {gather*}

Antiderivative was successfully verified.

[In]

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

[Out]

-(Sqrt[a]*ArcTan[(Sqrt[b*c - a*d]*x^2)/(Sqrt[a]*Sqrt[c + d*x^4])])/(2*b*Sqrt[b*c - a*d]) + ArcTanh[(Sqrt[d]*x^
2)/Sqrt[c + d*x^4]]/(2*b*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 465

Int[(x_)^(m_.)*((a_) + (b_.)*(x_)^(n_))^(p_)*((c_) + (d_.)*(x_)^(n_))^(q_), x_Symbol] :> With[{k = GCD[m + 1,
n]}, Dist[1/k, Subst[Int[x^((m + 1)/k - 1)*(a + b*x^(n/k))^p*(c + d*x^(n/k))^q, x], x, x^k], x] /; k != 1] /;
FreeQ[{a, b, c, d, p, q}, x] && NeQ[b*c - a*d, 0] && IGtQ[n, 0] && IntegerQ[m]

Rule 483

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

Rubi steps

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

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Mathematica [A]  time = 0.07, size = 90, normalized size = 0.99 \begin {gather*} \frac {\frac {\log \left (\sqrt {d} \sqrt {c+d x^4}+d x^2\right )}{\sqrt {d}}-\frac {\sqrt {a} \tan ^{-1}\left (\frac {x^2 \sqrt {b c-a d}}{\sqrt {a} \sqrt {c+d x^4}}\right )}{\sqrt {b c-a d}}}{2 b} \end {gather*}

Antiderivative was successfully verified.

[In]

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

[Out]

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

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IntegrateAlgebraic [A]  time = 0.46, size = 144, normalized size = 1.58 \begin {gather*} \frac {\log \left (\sqrt {c+d x^4}+\sqrt {d} x^2\right )}{2 b \sqrt {d}}-\frac {\sqrt {a} \tan ^{-1}\left (\frac {b \sqrt {d} x^4}{\sqrt {a} \sqrt {b c-a d}}+\frac {b x^2 \sqrt {c+d x^4}}{\sqrt {a} \sqrt {b c-a d}}+\frac {\sqrt {a} \sqrt {d}}{\sqrt {b c-a d}}\right )}{2 b \sqrt {b c-a d}} \end {gather*}

Antiderivative was successfully verified.

[In]

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

[Out]

-1/2*(Sqrt[a]*ArcTan[(Sqrt[a]*Sqrt[d])/Sqrt[b*c - a*d] + (b*Sqrt[d]*x^4)/(Sqrt[a]*Sqrt[b*c - a*d]) + (b*x^2*Sq
rt[c + d*x^4])/(Sqrt[a]*Sqrt[b*c - a*d])])/(b*Sqrt[b*c - a*d]) + Log[Sqrt[d]*x^2 + Sqrt[c + d*x^4]]/(2*b*Sqrt[
d])

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fricas [A]  time = 0.67, size = 632, normalized size = 6.95 \begin {gather*} \left [\frac {d \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^{8} - 2 \, {\left (3 \, a b c^{2} - 4 \, a^{2} c d\right )} x^{4} + a^{2} c^{2} - 4 \, {\left ({\left (b^{2} c^{2} - 3 \, a b c d + 2 \, a^{2} d^{2}\right )} x^{6} - {\left (a b c^{2} - a^{2} c d\right )} x^{2}\right )} \sqrt {d x^{4} + c} \sqrt {-\frac {a}{b c - a d}}}{b^{2} x^{8} + 2 \, a b x^{4} + a^{2}}\right ) + 2 \, \sqrt {d} \log \left (-2 \, d x^{4} - 2 \, \sqrt {d x^{4} + c} \sqrt {d} x^{2} - c\right )}{8 \, b d}, \frac {d \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^{8} - 2 \, {\left (3 \, a b c^{2} - 4 \, a^{2} c d\right )} x^{4} + a^{2} c^{2} - 4 \, {\left ({\left (b^{2} c^{2} - 3 \, a b c d + 2 \, a^{2} d^{2}\right )} x^{6} - {\left (a b c^{2} - a^{2} c d\right )} x^{2}\right )} \sqrt {d x^{4} + c} \sqrt {-\frac {a}{b c - a d}}}{b^{2} x^{8} + 2 \, a b x^{4} + a^{2}}\right ) - 4 \, \sqrt {-d} \arctan \left (\frac {\sqrt {-d} x^{2}}{\sqrt {d x^{4} + c}}\right )}{8 \, b d}, \frac {d \sqrt {\frac {a}{b c - a d}} \arctan \left (-\frac {{\left ({\left (b c - 2 \, a d\right )} x^{4} - a c\right )} \sqrt {d x^{4} + c} \sqrt {\frac {a}{b c - a d}}}{2 \, {\left (a d x^{6} + a c x^{2}\right )}}\right ) + \sqrt {d} \log \left (-2 \, d x^{4} - 2 \, \sqrt {d x^{4} + c} \sqrt {d} x^{2} - c\right )}{4 \, b d}, \frac {d \sqrt {\frac {a}{b c - a d}} \arctan \left (-\frac {{\left ({\left (b c - 2 \, a d\right )} x^{4} - a c\right )} \sqrt {d x^{4} + c} \sqrt {\frac {a}{b c - a d}}}{2 \, {\left (a d x^{6} + a c x^{2}\right )}}\right ) - 2 \, \sqrt {-d} \arctan \left (\frac {\sqrt {-d} x^{2}}{\sqrt {d x^{4} + c}}\right )}{4 \, b d}\right ] \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate(x^5/(b*x^4+a)/(d*x^4+c)^(1/2),x, algorithm="fricas")

[Out]

[1/8*(d*sqrt(-a/(b*c - a*d))*log(((b^2*c^2 - 8*a*b*c*d + 8*a^2*d^2)*x^8 - 2*(3*a*b*c^2 - 4*a^2*c*d)*x^4 + a^2*
c^2 - 4*((b^2*c^2 - 3*a*b*c*d + 2*a^2*d^2)*x^6 - (a*b*c^2 - a^2*c*d)*x^2)*sqrt(d*x^4 + c)*sqrt(-a/(b*c - a*d))
)/(b^2*x^8 + 2*a*b*x^4 + a^2)) + 2*sqrt(d)*log(-2*d*x^4 - 2*sqrt(d*x^4 + c)*sqrt(d)*x^2 - c))/(b*d), 1/8*(d*sq
rt(-a/(b*c - a*d))*log(((b^2*c^2 - 8*a*b*c*d + 8*a^2*d^2)*x^8 - 2*(3*a*b*c^2 - 4*a^2*c*d)*x^4 + a^2*c^2 - 4*((
b^2*c^2 - 3*a*b*c*d + 2*a^2*d^2)*x^6 - (a*b*c^2 - a^2*c*d)*x^2)*sqrt(d*x^4 + c)*sqrt(-a/(b*c - a*d)))/(b^2*x^8
+ 2*a*b*x^4 + a^2)) - 4*sqrt(-d)*arctan(sqrt(-d)*x^2/sqrt(d*x^4 + c)))/(b*d), 1/4*(d*sqrt(a/(b*c - a*d))*arct
an(-1/2*((b*c - 2*a*d)*x^4 - a*c)*sqrt(d*x^4 + c)*sqrt(a/(b*c - a*d))/(a*d*x^6 + a*c*x^2)) + sqrt(d)*log(-2*d*
x^4 - 2*sqrt(d*x^4 + c)*sqrt(d)*x^2 - c))/(b*d), 1/4*(d*sqrt(a/(b*c - a*d))*arctan(-1/2*((b*c - 2*a*d)*x^4 - a
*c)*sqrt(d*x^4 + c)*sqrt(a/(b*c - a*d))/(a*d*x^6 + a*c*x^2)) - 2*sqrt(-d)*arctan(sqrt(-d)*x^2/sqrt(d*x^4 + c))
)/(b*d)]

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

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate(x^5/(b*x^4+a)/(d*x^4+c)^(1/2),x, algorithm="giac")

[Out]

Exception raised: TypeError >> An error occurred running a Giac command:INPUT:sage2:=int(sage0,x):;OUTPUT:inde
x.cc index_m i_lex_is_greater Error: Bad Argument Value

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maple [B]  time = 0.33, size = 356, normalized size = 3.91 \begin {gather*} \frac {a \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 {-a b}\, \sqrt {-\frac {a d -b c}{b}}\, b}-\frac {a \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 {-a b}\, \sqrt {-\frac {a d -b c}{b}}\, b}+\frac {\ln \left (\sqrt {d}\, x^{2}+\sqrt {d \,x^{4}+c}\right )}{2 b \sqrt {d}} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

int(x^5/(b*x^4+a)/(d*x^4+c)^(1/2),x)

[Out]

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

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

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate(x^5/(b*x^4+a)/(d*x^4+c)^(1/2),x, algorithm="maxima")

[Out]

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

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

Verification of antiderivative is not currently implemented for this CAS.

[In]

int(x^5/((a + b*x^4)*(c + d*x^4)^(1/2)),x)

[Out]

int(x^5/((a + b*x^4)*(c + d*x^4)^(1/2)), x)

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

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate(x**5/(b*x**4+a)/(d*x**4+c)**(1/2),x)

[Out]

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

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