3.18.69 \(\int \sqrt [4]{b x^7+a x^8} \, dx\)

Optimal. Leaf size=119 \[ \frac {7 b^3 \tan ^{-1}\left (\frac {\sqrt [4]{a x^8+b x^7}}{\sqrt [4]{a} x^2}\right )}{64 a^{11/4}}+\frac {7 b^3 \tanh ^{-1}\left (\frac {\sqrt [4]{a x^8+b x^7}}{\sqrt [4]{a} x^2}\right )}{64 a^{11/4}}+\frac {\left (32 a^2 x^2+4 a b x-7 b^2\right ) \sqrt [4]{a x^8+b x^7}}{96 a^2 x} \]

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Rubi [A]  time = 0.25, antiderivative size = 196, normalized size of antiderivative = 1.65, number of steps used = 9, number of rules used = 8, integrand size = 15, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.533, Rules used = {2004, 2024, 2032, 63, 331, 298, 203, 206} \begin {gather*} -\frac {7 b^3 x^{21/4} (a x+b)^{3/4} \tan ^{-1}\left (\frac {\sqrt [4]{a} \sqrt [4]{x}}{\sqrt [4]{a x+b}}\right )}{64 a^{11/4} \left (a x^8+b x^7\right )^{3/4}}+\frac {7 b^3 x^{21/4} (a x+b)^{3/4} \tanh ^{-1}\left (\frac {\sqrt [4]{a} \sqrt [4]{x}}{\sqrt [4]{a x+b}}\right )}{64 a^{11/4} \left (a x^8+b x^7\right )^{3/4}}-\frac {7 b^2 \sqrt [4]{a x^8+b x^7}}{96 a^2 x}+\frac {1}{3} x \sqrt [4]{a x^8+b x^7}+\frac {b \sqrt [4]{a x^8+b x^7}}{24 a} \end {gather*}

Antiderivative was successfully verified.

[In]

Int[(b*x^7 + a*x^8)^(1/4),x]

[Out]

(b*(b*x^7 + a*x^8)^(1/4))/(24*a) - (7*b^2*(b*x^7 + a*x^8)^(1/4))/(96*a^2*x) + (x*(b*x^7 + a*x^8)^(1/4))/3 - (7
*b^3*x^(21/4)*(b + a*x)^(3/4)*ArcTan[(a^(1/4)*x^(1/4))/(b + a*x)^(1/4)])/(64*a^(11/4)*(b*x^7 + a*x^8)^(3/4)) +
(7*b^3*x^(21/4)*(b + a*x)^(3/4)*ArcTanh[(a^(1/4)*x^(1/4))/(b + a*x)^(1/4)])/(64*a^(11/4)*(b*x^7 + a*x^8)^(3/4
))

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 203

Int[((a_) + (b_.)*(x_)^2)^(-1), x_Symbol] :> Simp[(1*ArcTan[(Rt[b, 2]*x)/Rt[a, 2]])/(Rt[a, 2]*Rt[b, 2]), x] /;
FreeQ[{a, b}, x] && PosQ[a/b] && (GtQ[a, 0] || GtQ[b, 0])

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 298

Int[(x_)^2/((a_) + (b_.)*(x_)^4), x_Symbol] :> With[{r = Numerator[Rt[-(a/b), 2]], s = Denominator[Rt[-(a/b),
2]]}, Dist[s/(2*b), Int[1/(r + s*x^2), x], x] - Dist[s/(2*b), Int[1/(r - s*x^2), x], x]] /; FreeQ[{a, b}, x] &
&  !GtQ[a/b, 0]

Rule 331

Int[(x_)^(m_.)*((a_) + (b_.)*(x_)^(n_))^(p_), x_Symbol] :> Dist[a^(p + (m + 1)/n), Subst[Int[x^m/(1 - b*x^n)^(
p + (m + 1)/n + 1), x], x, x/(a + b*x^n)^(1/n)], x] /; FreeQ[{a, b}, x] && IGtQ[n, 0] && LtQ[-1, p, 0] && NeQ[
p, -2^(-1)] && IntegersQ[m, p + (m + 1)/n]

Rule 2004

Int[((a_.)*(x_)^(j_.) + (b_.)*(x_)^(n_.))^(p_), x_Symbol] :> Simp[(x*(a*x^j + b*x^n)^p)/(n*p + 1), x] + Dist[(
a*(n - j)*p)/(n*p + 1), Int[x^j*(a*x^j + b*x^n)^(p - 1), x], x] /; FreeQ[{a, b}, x] &&  !IntegerQ[p] && LtQ[0,
j, n] && GtQ[p, 0] && NeQ[n*p + 1, 0]

Rule 2024

Int[((c_.)*(x_))^(m_.)*((a_.)*(x_)^(j_.) + (b_.)*(x_)^(n_.))^(p_), x_Symbol] :> Simp[(c^(n - 1)*(c*x)^(m - n +
1)*(a*x^j + b*x^n)^(p + 1))/(b*(m + n*p + 1)), x] - Dist[(a*c^(n - j)*(m + j*p - n + j + 1))/(b*(m + n*p + 1)
), Int[(c*x)^(m - (n - j))*(a*x^j + b*x^n)^p, x], x] /; FreeQ[{a, b, c, m, p}, x] &&  !IntegerQ[p] && LtQ[0, j
, n] && (IntegersQ[j, n] || GtQ[c, 0]) && GtQ[m + j*p + 1 - n + j, 0] && NeQ[m + n*p + 1, 0]

Rule 2032

Int[((c_.)*(x_))^(m_.)*((a_.)*(x_)^(j_.) + (b_.)*(x_)^(n_.))^(p_), x_Symbol] :> Dist[(c^IntPart[m]*(c*x)^FracP
art[m]*(a*x^j + b*x^n)^FracPart[p])/(x^(FracPart[m] + j*FracPart[p])*(a + b*x^(n - j))^FracPart[p]), Int[x^(m
+ j*p)*(a + b*x^(n - j))^p, x], x] /; FreeQ[{a, b, c, j, m, n, p}, x] &&  !IntegerQ[p] && NeQ[n, j] && PosQ[n
- j]

Rubi steps

\begin {align*} \int \sqrt [4]{b x^7+a x^8} \, dx &=\frac {1}{3} x \sqrt [4]{b x^7+a x^8}+\frac {1}{12} b \int \frac {x^7}{\left (b x^7+a x^8\right )^{3/4}} \, dx\\ &=\frac {b \sqrt [4]{b x^7+a x^8}}{24 a}+\frac {1}{3} x \sqrt [4]{b x^7+a x^8}-\frac {\left (7 b^2\right ) \int \frac {x^6}{\left (b x^7+a x^8\right )^{3/4}} \, dx}{96 a}\\ &=\frac {b \sqrt [4]{b x^7+a x^8}}{24 a}-\frac {7 b^2 \sqrt [4]{b x^7+a x^8}}{96 a^2 x}+\frac {1}{3} x \sqrt [4]{b x^7+a x^8}+\frac {\left (7 b^3\right ) \int \frac {x^5}{\left (b x^7+a x^8\right )^{3/4}} \, dx}{128 a^2}\\ &=\frac {b \sqrt [4]{b x^7+a x^8}}{24 a}-\frac {7 b^2 \sqrt [4]{b x^7+a x^8}}{96 a^2 x}+\frac {1}{3} x \sqrt [4]{b x^7+a x^8}+\frac {\left (7 b^3 x^{21/4} (b+a x)^{3/4}\right ) \int \frac {1}{\sqrt [4]{x} (b+a x)^{3/4}} \, dx}{128 a^2 \left (b x^7+a x^8\right )^{3/4}}\\ &=\frac {b \sqrt [4]{b x^7+a x^8}}{24 a}-\frac {7 b^2 \sqrt [4]{b x^7+a x^8}}{96 a^2 x}+\frac {1}{3} x \sqrt [4]{b x^7+a x^8}+\frac {\left (7 b^3 x^{21/4} (b+a x)^{3/4}\right ) \operatorname {Subst}\left (\int \frac {x^2}{\left (b+a x^4\right )^{3/4}} \, dx,x,\sqrt [4]{x}\right )}{32 a^2 \left (b x^7+a x^8\right )^{3/4}}\\ &=\frac {b \sqrt [4]{b x^7+a x^8}}{24 a}-\frac {7 b^2 \sqrt [4]{b x^7+a x^8}}{96 a^2 x}+\frac {1}{3} x \sqrt [4]{b x^7+a x^8}+\frac {\left (7 b^3 x^{21/4} (b+a x)^{3/4}\right ) \operatorname {Subst}\left (\int \frac {x^2}{1-a x^4} \, dx,x,\frac {\sqrt [4]{x}}{\sqrt [4]{b+a x}}\right )}{32 a^2 \left (b x^7+a x^8\right )^{3/4}}\\ &=\frac {b \sqrt [4]{b x^7+a x^8}}{24 a}-\frac {7 b^2 \sqrt [4]{b x^7+a x^8}}{96 a^2 x}+\frac {1}{3} x \sqrt [4]{b x^7+a x^8}+\frac {\left (7 b^3 x^{21/4} (b+a x)^{3/4}\right ) \operatorname {Subst}\left (\int \frac {1}{1-\sqrt {a} x^2} \, dx,x,\frac {\sqrt [4]{x}}{\sqrt [4]{b+a x}}\right )}{64 a^{5/2} \left (b x^7+a x^8\right )^{3/4}}-\frac {\left (7 b^3 x^{21/4} (b+a x)^{3/4}\right ) \operatorname {Subst}\left (\int \frac {1}{1+\sqrt {a} x^2} \, dx,x,\frac {\sqrt [4]{x}}{\sqrt [4]{b+a x}}\right )}{64 a^{5/2} \left (b x^7+a x^8\right )^{3/4}}\\ &=\frac {b \sqrt [4]{b x^7+a x^8}}{24 a}-\frac {7 b^2 \sqrt [4]{b x^7+a x^8}}{96 a^2 x}+\frac {1}{3} x \sqrt [4]{b x^7+a x^8}-\frac {7 b^3 x^{21/4} (b+a x)^{3/4} \tan ^{-1}\left (\frac {\sqrt [4]{a} \sqrt [4]{x}}{\sqrt [4]{b+a x}}\right )}{64 a^{11/4} \left (b x^7+a x^8\right )^{3/4}}+\frac {7 b^3 x^{21/4} (b+a x)^{3/4} \tanh ^{-1}\left (\frac {\sqrt [4]{a} \sqrt [4]{x}}{\sqrt [4]{b+a x}}\right )}{64 a^{11/4} \left (b x^7+a x^8\right )^{3/4}}\\ \end {align*}

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Mathematica [C]  time = 0.02, size = 47, normalized size = 0.39 \begin {gather*} \frac {4 x \sqrt [4]{x^7 (a x+b)} \, _2F_1\left (-\frac {1}{4},\frac {11}{4};\frac {15}{4};-\frac {a x}{b}\right )}{11 \sqrt [4]{\frac {a x}{b}+1}} \end {gather*}

Antiderivative was successfully verified.

[In]

Integrate[(b*x^7 + a*x^8)^(1/4),x]

[Out]

(4*x*(x^7*(b + a*x))^(1/4)*Hypergeometric2F1[-1/4, 11/4, 15/4, -((a*x)/b)])/(11*(1 + (a*x)/b)^(1/4))

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IntegrateAlgebraic [A]  time = 0.58, size = 119, normalized size = 1.00 \begin {gather*} \frac {\left (-7 b^2+4 a b x+32 a^2 x^2\right ) \sqrt [4]{b x^7+a x^8}}{96 a^2 x}+\frac {7 b^3 \tan ^{-1}\left (\frac {\sqrt [4]{b x^7+a x^8}}{\sqrt [4]{a} x^2}\right )}{64 a^{11/4}}+\frac {7 b^3 \tanh ^{-1}\left (\frac {\sqrt [4]{b x^7+a x^8}}{\sqrt [4]{a} x^2}\right )}{64 a^{11/4}} \end {gather*}

Antiderivative was successfully verified.

[In]

IntegrateAlgebraic[(b*x^7 + a*x^8)^(1/4),x]

[Out]

((-7*b^2 + 4*a*b*x + 32*a^2*x^2)*(b*x^7 + a*x^8)^(1/4))/(96*a^2*x) + (7*b^3*ArcTan[(b*x^7 + a*x^8)^(1/4)/(a^(1
/4)*x^2)])/(64*a^(11/4)) + (7*b^3*ArcTanh[(b*x^7 + a*x^8)^(1/4)/(a^(1/4)*x^2)])/(64*a^(11/4))

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fricas [B]  time = 0.47, size = 265, normalized size = 2.23 \begin {gather*} -\frac {84 \, a^{2} \left (\frac {b^{12}}{a^{11}}\right )^{\frac {1}{4}} x \arctan \left (-\frac {{\left (a x^{8} + b x^{7}\right )}^{\frac {1}{4}} a^{8} b^{3} \left (\frac {b^{12}}{a^{11}}\right )^{\frac {3}{4}} - a^{8} \left (\frac {b^{12}}{a^{11}}\right )^{\frac {3}{4}} x^{2} \sqrt {\frac {a^{6} \sqrt {\frac {b^{12}}{a^{11}}} x^{4} + \sqrt {a x^{8} + b x^{7}} b^{6}}{x^{4}}}}{b^{12} x^{2}}\right ) - 21 \, a^{2} \left (\frac {b^{12}}{a^{11}}\right )^{\frac {1}{4}} x \log \left (\frac {7 \, {\left (a^{3} \left (\frac {b^{12}}{a^{11}}\right )^{\frac {1}{4}} x^{2} + {\left (a x^{8} + b x^{7}\right )}^{\frac {1}{4}} b^{3}\right )}}{x^{2}}\right ) + 21 \, a^{2} \left (\frac {b^{12}}{a^{11}}\right )^{\frac {1}{4}} x \log \left (-\frac {7 \, {\left (a^{3} \left (\frac {b^{12}}{a^{11}}\right )^{\frac {1}{4}} x^{2} - {\left (a x^{8} + b x^{7}\right )}^{\frac {1}{4}} b^{3}\right )}}{x^{2}}\right ) - 4 \, {\left (a x^{8} + b x^{7}\right )}^{\frac {1}{4}} {\left (32 \, a^{2} x^{2} + 4 \, a b x - 7 \, b^{2}\right )}}{384 \, a^{2} x} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate((a*x^8+b*x^7)^(1/4),x, algorithm="fricas")

[Out]

-1/384*(84*a^2*(b^12/a^11)^(1/4)*x*arctan(-((a*x^8 + b*x^7)^(1/4)*a^8*b^3*(b^12/a^11)^(3/4) - a^8*(b^12/a^11)^
(3/4)*x^2*sqrt((a^6*sqrt(b^12/a^11)*x^4 + sqrt(a*x^8 + b*x^7)*b^6)/x^4))/(b^12*x^2)) - 21*a^2*(b^12/a^11)^(1/4
)*x*log(7*(a^3*(b^12/a^11)^(1/4)*x^2 + (a*x^8 + b*x^7)^(1/4)*b^3)/x^2) + 21*a^2*(b^12/a^11)^(1/4)*x*log(-7*(a^
3*(b^12/a^11)^(1/4)*x^2 - (a*x^8 + b*x^7)^(1/4)*b^3)/x^2) - 4*(a*x^8 + b*x^7)^(1/4)*(32*a^2*x^2 + 4*a*b*x - 7*
b^2))/(a^2*x)

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giac [B]  time = 0.34, size = 261, normalized size = 2.19 \begin {gather*} \frac {\frac {42 \, \sqrt {2} \left (-a\right )^{\frac {1}{4}} b^{4} \arctan \left (\frac {\sqrt {2} {\left (\sqrt {2} \left (-a\right )^{\frac {1}{4}} + 2 \, {\left (a + \frac {b}{x}\right )}^{\frac {1}{4}}\right )}}{2 \, \left (-a\right )^{\frac {1}{4}}}\right )}{a^{3}} + \frac {42 \, \sqrt {2} \left (-a\right )^{\frac {1}{4}} b^{4} \arctan \left (-\frac {\sqrt {2} {\left (\sqrt {2} \left (-a\right )^{\frac {1}{4}} - 2 \, {\left (a + \frac {b}{x}\right )}^{\frac {1}{4}}\right )}}{2 \, \left (-a\right )^{\frac {1}{4}}}\right )}{a^{3}} + \frac {21 \, \sqrt {2} \left (-a\right )^{\frac {1}{4}} b^{4} \log \left (\sqrt {2} \left (-a\right )^{\frac {1}{4}} {\left (a + \frac {b}{x}\right )}^{\frac {1}{4}} + \sqrt {-a} + \sqrt {a + \frac {b}{x}}\right )}{a^{3}} + \frac {21 \, \sqrt {2} b^{4} \log \left (-\sqrt {2} \left (-a\right )^{\frac {1}{4}} {\left (a + \frac {b}{x}\right )}^{\frac {1}{4}} + \sqrt {-a} + \sqrt {a + \frac {b}{x}}\right )}{\left (-a\right )^{\frac {3}{4}} a^{2}} - \frac {8 \, {\left (7 \, {\left (a + \frac {b}{x}\right )}^{\frac {9}{4}} b^{4} - 18 \, {\left (a + \frac {b}{x}\right )}^{\frac {5}{4}} a b^{4} - 21 \, {\left (a + \frac {b}{x}\right )}^{\frac {1}{4}} a^{2} b^{4}\right )} x^{3}}{a^{2} b^{3}}}{768 \, b} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate((a*x^8+b*x^7)^(1/4),x, algorithm="giac")

[Out]

1/768*(42*sqrt(2)*(-a)^(1/4)*b^4*arctan(1/2*sqrt(2)*(sqrt(2)*(-a)^(1/4) + 2*(a + b/x)^(1/4))/(-a)^(1/4))/a^3 +
42*sqrt(2)*(-a)^(1/4)*b^4*arctan(-1/2*sqrt(2)*(sqrt(2)*(-a)^(1/4) - 2*(a + b/x)^(1/4))/(-a)^(1/4))/a^3 + 21*s
qrt(2)*(-a)^(1/4)*b^4*log(sqrt(2)*(-a)^(1/4)*(a + b/x)^(1/4) + sqrt(-a) + sqrt(a + b/x))/a^3 + 21*sqrt(2)*b^4*
log(-sqrt(2)*(-a)^(1/4)*(a + b/x)^(1/4) + sqrt(-a) + sqrt(a + b/x))/((-a)^(3/4)*a^2) - 8*(7*(a + b/x)^(9/4)*b^
4 - 18*(a + b/x)^(5/4)*a*b^4 - 21*(a + b/x)^(1/4)*a^2*b^4)*x^3/(a^2*b^3))/b

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maple [F]  time = 0.02, size = 0, normalized size = 0.00 \[\int \left (a \,x^{8}+b \,x^{7}\right )^{\frac {1}{4}}\, dx\]

Verification of antiderivative is not currently implemented for this CAS.

[In]

int((a*x^8+b*x^7)^(1/4),x)

[Out]

int((a*x^8+b*x^7)^(1/4),x)

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maxima [F]  time = 0.00, size = 0, normalized size = 0.00 \begin {gather*} \int {\left (a x^{8} + b x^{7}\right )}^{\frac {1}{4}}\,{d x} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate((a*x^8+b*x^7)^(1/4),x, algorithm="maxima")

[Out]

integrate((a*x^8 + b*x^7)^(1/4), x)

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mupad [B]  time = 0.93, size = 38, normalized size = 0.32 \begin {gather*} \frac {4\,x\,{\left (a\,x^8+b\,x^7\right )}^{1/4}\,{{}}_2{\mathrm {F}}_1\left (-\frac {1}{4},\frac {11}{4};\ \frac {15}{4};\ -\frac {a\,x}{b}\right )}{11\,{\left (\frac {a\,x}{b}+1\right )}^{1/4}} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

int((a*x^8 + b*x^7)^(1/4),x)

[Out]

(4*x*(a*x^8 + b*x^7)^(1/4)*hypergeom([-1/4, 11/4], 15/4, -(a*x)/b))/(11*((a*x)/b + 1)^(1/4))

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sympy [F]  time = 0.00, size = 0, normalized size = 0.00 \begin {gather*} \int \sqrt [4]{a x^{8} + b x^{7}}\, dx \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate((a*x**8+b*x**7)**(1/4),x)

[Out]

Integral((a*x**8 + b*x**7)**(1/4), x)

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