3.892 \(\int \frac {x^{19}}{(a+b x^8) \sqrt {c+d x^8}} \, dx\)

Optimal. Leaf size=123 \[ \frac {a^{3/2} \tan ^{-1}\left (\frac {x^4 \sqrt {b c-a d}}{\sqrt {a} \sqrt {c+d x^8}}\right )}{4 b^2 \sqrt {b c-a d}}-\frac {(2 a d+b c) \tanh ^{-1}\left (\frac {\sqrt {d} x^4}{\sqrt {c+d x^8}}\right )}{8 b^2 d^{3/2}}+\frac {x^4 \sqrt {c+d x^8}}{8 b d} \]

[Out]

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

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Rubi [A]  time = 0.13, antiderivative size = 123, normalized size of antiderivative = 1.00, number of steps used = 7, number of rules used = 7, integrand size = 24, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.292, Rules used = {465, 479, 523, 217, 206, 377, 205} \[ \frac {a^{3/2} \tan ^{-1}\left (\frac {x^4 \sqrt {b c-a d}}{\sqrt {a} \sqrt {c+d x^8}}\right )}{4 b^2 \sqrt {b c-a d}}-\frac {(2 a d+b c) \tanh ^{-1}\left (\frac {\sqrt {d} x^4}{\sqrt {c+d x^8}}\right )}{8 b^2 d^{3/2}}+\frac {x^4 \sqrt {c+d x^8}}{8 b d} \]

Antiderivative was successfully verified.

[In]

Int[x^19/((a + b*x^8)*Sqrt[c + d*x^8]),x]

[Out]

(x^4*Sqrt[c + d*x^8])/(8*b*d) + (a^(3/2)*ArcTan[(Sqrt[b*c - a*d]*x^4)/(Sqrt[a]*Sqrt[c + d*x^8])])/(4*b^2*Sqrt[
b*c - a*d]) - ((b*c + 2*a*d)*ArcTanh[(Sqrt[d]*x^4)/Sqrt[c + d*x^8]])/(8*b^2*d^(3/2))

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 479

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

Rule 523

Int[((e_) + (f_.)*(x_)^(n_))/(((a_) + (b_.)*(x_)^(n_))*Sqrt[(c_) + (d_.)*(x_)^(n_)]), x_Symbol] :> Dist[f/b, I
nt[1/Sqrt[c + d*x^n], x], x] + Dist[(b*e - a*f)/b, Int[1/((a + b*x^n)*Sqrt[c + d*x^n]), x], x] /; FreeQ[{a, b,
c, d, e, f, n}, x]

Rubi steps

\begin {align*} \int \frac {x^{19}}{\left (a+b x^8\right ) \sqrt {c+d x^8}} \, dx &=\frac {1}{4} \operatorname {Subst}\left (\int \frac {x^4}{\left (a+b x^2\right ) \sqrt {c+d x^2}} \, dx,x,x^4\right )\\ &=\frac {x^4 \sqrt {c+d x^8}}{8 b d}-\frac {\operatorname {Subst}\left (\int \frac {a c+(b c+2 a d) x^2}{\left (a+b x^2\right ) \sqrt {c+d x^2}} \, dx,x,x^4\right )}{8 b d}\\ &=\frac {x^4 \sqrt {c+d x^8}}{8 b d}+\frac {a^2 \operatorname {Subst}\left (\int \frac {1}{\left (a+b x^2\right ) \sqrt {c+d x^2}} \, dx,x,x^4\right )}{4 b^2}-\frac {(b c+2 a d) \operatorname {Subst}\left (\int \frac {1}{\sqrt {c+d x^2}} \, dx,x,x^4\right )}{8 b^2 d}\\ &=\frac {x^4 \sqrt {c+d x^8}}{8 b d}+\frac {a^2 \operatorname {Subst}\left (\int \frac {1}{a-(-b c+a d) x^2} \, dx,x,\frac {x^4}{\sqrt {c+d x^8}}\right )}{4 b^2}-\frac {(b c+2 a d) \operatorname {Subst}\left (\int \frac {1}{1-d x^2} \, dx,x,\frac {x^4}{\sqrt {c+d x^8}}\right )}{8 b^2 d}\\ &=\frac {x^4 \sqrt {c+d x^8}}{8 b d}+\frac {a^{3/2} \tan ^{-1}\left (\frac {\sqrt {b c-a d} x^4}{\sqrt {a} \sqrt {c+d x^8}}\right )}{4 b^2 \sqrt {b c-a d}}-\frac {(b c+2 a d) \tanh ^{-1}\left (\frac {\sqrt {d} x^4}{\sqrt {c+d x^8}}\right )}{8 b^2 d^{3/2}}\\ \end {align*}

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Mathematica [A]  time = 0.24, size = 118, normalized size = 0.96 \[ \frac {\frac {2 a^{3/2} \tan ^{-1}\left (\frac {x^4 \sqrt {b c-a d}}{\sqrt {a} \sqrt {c+d x^8}}\right )}{\sqrt {b c-a d}}-\frac {(2 a d+b c) \log \left (\sqrt {d} \sqrt {c+d x^8}+d x^4\right )}{d^{3/2}}+\frac {b x^4 \sqrt {c+d x^8}}{d}}{8 b^2} \]

Antiderivative was successfully verified.

[In]

Integrate[x^19/((a + b*x^8)*Sqrt[c + d*x^8]),x]

[Out]

((b*x^4*Sqrt[c + d*x^8])/d + (2*a^(3/2)*ArcTan[(Sqrt[b*c - a*d]*x^4)/(Sqrt[a]*Sqrt[c + d*x^8])])/Sqrt[b*c - a*
d] - ((b*c + 2*a*d)*Log[d*x^4 + Sqrt[d]*Sqrt[c + d*x^8]])/d^(3/2))/(8*b^2)

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fricas [A]  time = 0.92, size = 739, normalized size = 6.01 \[ \left [\frac {2 \, \sqrt {d x^{8} + c} b d x^{4} + a d^{2} \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^{16} - 2 \, {\left (3 \, a b c^{2} - 4 \, a^{2} c d\right )} x^{8} + a^{2} c^{2} + 4 \, {\left ({\left (b^{2} c^{2} - 3 \, a b c d + 2 \, a^{2} d^{2}\right )} x^{12} - {\left (a b c^{2} - a^{2} c d\right )} x^{4}\right )} \sqrt {d x^{8} + c} \sqrt {-\frac {a}{b c - a d}}}{b^{2} x^{16} + 2 \, a b x^{8} + a^{2}}\right ) + {\left (b c + 2 \, a d\right )} \sqrt {d} \log \left (-2 \, d x^{8} + 2 \, \sqrt {d x^{8} + c} \sqrt {d} x^{4} - c\right )}{16 \, b^{2} d^{2}}, \frac {2 \, \sqrt {d x^{8} + c} b d x^{4} + a d^{2} \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^{16} - 2 \, {\left (3 \, a b c^{2} - 4 \, a^{2} c d\right )} x^{8} + a^{2} c^{2} + 4 \, {\left ({\left (b^{2} c^{2} - 3 \, a b c d + 2 \, a^{2} d^{2}\right )} x^{12} - {\left (a b c^{2} - a^{2} c d\right )} x^{4}\right )} \sqrt {d x^{8} + c} \sqrt {-\frac {a}{b c - a d}}}{b^{2} x^{16} + 2 \, a b x^{8} + a^{2}}\right ) + 2 \, {\left (b c + 2 \, a d\right )} \sqrt {-d} \arctan \left (\frac {\sqrt {-d} x^{4}}{\sqrt {d x^{8} + c}}\right )}{16 \, b^{2} d^{2}}, \frac {2 \, \sqrt {d x^{8} + c} b d x^{4} - 2 \, a d^{2} \sqrt {\frac {a}{b c - a d}} \arctan \left (-\frac {{\left ({\left (b c - 2 \, a d\right )} x^{8} - a c\right )} \sqrt {d x^{8} + c} \sqrt {\frac {a}{b c - a d}}}{2 \, {\left (a d x^{12} + a c x^{4}\right )}}\right ) + {\left (b c + 2 \, a d\right )} \sqrt {d} \log \left (-2 \, d x^{8} + 2 \, \sqrt {d x^{8} + c} \sqrt {d} x^{4} - c\right )}{16 \, b^{2} d^{2}}, \frac {\sqrt {d x^{8} + c} b d x^{4} - a d^{2} \sqrt {\frac {a}{b c - a d}} \arctan \left (-\frac {{\left ({\left (b c - 2 \, a d\right )} x^{8} - a c\right )} \sqrt {d x^{8} + c} \sqrt {\frac {a}{b c - a d}}}{2 \, {\left (a d x^{12} + a c x^{4}\right )}}\right ) + {\left (b c + 2 \, a d\right )} \sqrt {-d} \arctan \left (\frac {\sqrt {-d} x^{4}}{\sqrt {d x^{8} + c}}\right )}{8 \, b^{2} d^{2}}\right ] \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate(x^19/(b*x^8+a)/(d*x^8+c)^(1/2),x, algorithm="fricas")

[Out]

[1/16*(2*sqrt(d*x^8 + c)*b*d*x^4 + a*d^2*sqrt(-a/(b*c - a*d))*log(((b^2*c^2 - 8*a*b*c*d + 8*a^2*d^2)*x^16 - 2*
(3*a*b*c^2 - 4*a^2*c*d)*x^8 + a^2*c^2 + 4*((b^2*c^2 - 3*a*b*c*d + 2*a^2*d^2)*x^12 - (a*b*c^2 - a^2*c*d)*x^4)*s
qrt(d*x^8 + c)*sqrt(-a/(b*c - a*d)))/(b^2*x^16 + 2*a*b*x^8 + a^2)) + (b*c + 2*a*d)*sqrt(d)*log(-2*d*x^8 + 2*sq
rt(d*x^8 + c)*sqrt(d)*x^4 - c))/(b^2*d^2), 1/16*(2*sqrt(d*x^8 + c)*b*d*x^4 + a*d^2*sqrt(-a/(b*c - a*d))*log(((
b^2*c^2 - 8*a*b*c*d + 8*a^2*d^2)*x^16 - 2*(3*a*b*c^2 - 4*a^2*c*d)*x^8 + a^2*c^2 + 4*((b^2*c^2 - 3*a*b*c*d + 2*
a^2*d^2)*x^12 - (a*b*c^2 - a^2*c*d)*x^4)*sqrt(d*x^8 + c)*sqrt(-a/(b*c - a*d)))/(b^2*x^16 + 2*a*b*x^8 + a^2)) +
2*(b*c + 2*a*d)*sqrt(-d)*arctan(sqrt(-d)*x^4/sqrt(d*x^8 + c)))/(b^2*d^2), 1/16*(2*sqrt(d*x^8 + c)*b*d*x^4 - 2
*a*d^2*sqrt(a/(b*c - a*d))*arctan(-1/2*((b*c - 2*a*d)*x^8 - a*c)*sqrt(d*x^8 + c)*sqrt(a/(b*c - a*d))/(a*d*x^12
+ a*c*x^4)) + (b*c + 2*a*d)*sqrt(d)*log(-2*d*x^8 + 2*sqrt(d*x^8 + c)*sqrt(d)*x^4 - c))/(b^2*d^2), 1/8*(sqrt(d
*x^8 + c)*b*d*x^4 - a*d^2*sqrt(a/(b*c - a*d))*arctan(-1/2*((b*c - 2*a*d)*x^8 - a*c)*sqrt(d*x^8 + c)*sqrt(a/(b*
c - a*d))/(a*d*x^12 + a*c*x^4)) + (b*c + 2*a*d)*sqrt(-d)*arctan(sqrt(-d)*x^4/sqrt(d*x^8 + c)))/(b^2*d^2)]

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

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate(x^19/(b*x^8+a)/(d*x^8+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 [F]  time = 0.58, size = 0, normalized size = 0.00 \[ \int \frac {x^{19}}{\left (b \,x^{8}+a \right ) \sqrt {d \,x^{8}+c}}\, dx \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

int(x^19/(b*x^8+a)/(d*x^8+c)^(1/2),x)

[Out]

int(x^19/(b*x^8+a)/(d*x^8+c)^(1/2),x)

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maxima [F]  time = 0.00, size = 0, normalized size = 0.00 \[ \int \frac {x^{19}}{{\left (b x^{8} + a\right )} \sqrt {d x^{8} + c}}\,{d x} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate(x^19/(b*x^8+a)/(d*x^8+c)^(1/2),x, algorithm="maxima")

[Out]

integrate(x^19/((b*x^8 + a)*sqrt(d*x^8 + c)), x)

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mupad [F]  time = 0.00, size = -1, normalized size = -0.01 \[ \int \frac {x^{19}}{\left (b\,x^8+a\right )\,\sqrt {d\,x^8+c}} \,d x \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

int(x^19/((a + b*x^8)*(c + d*x^8)^(1/2)),x)

[Out]

int(x^19/((a + b*x^8)*(c + d*x^8)^(1/2)), x)

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sympy [F]  time = 0.00, size = 0, normalized size = 0.00 \[ \int \frac {x^{19}}{\left (a + b x^{8}\right ) \sqrt {c + d x^{8}}}\, dx \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate(x**19/(b*x**8+a)/(d*x**8+c)**(1/2),x)

[Out]

Integral(x**19/((a + b*x**8)*sqrt(c + d*x**8)), x)

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