3.887 \(\int \frac{\sqrt [4]{a+b x}}{x \sqrt [4]{c+d x}} \, dx\)

Optimal. Leaf size=169 \[ -\frac{2 \sqrt [4]{a} \tan ^{-1}\left (\frac{\sqrt [4]{c} \sqrt [4]{a+b x}}{\sqrt [4]{a} \sqrt [4]{c+d x}}\right )}{\sqrt [4]{c}}+\frac{2 \sqrt [4]{b} \tan ^{-1}\left (\frac{\sqrt [4]{d} \sqrt [4]{a+b x}}{\sqrt [4]{b} \sqrt [4]{c+d x}}\right )}{\sqrt [4]{d}}-\frac{2 \sqrt [4]{a} \tanh ^{-1}\left (\frac{\sqrt [4]{c} \sqrt [4]{a+b x}}{\sqrt [4]{a} \sqrt [4]{c+d x}}\right )}{\sqrt [4]{c}}+\frac{2 \sqrt [4]{b} \tanh ^{-1}\left (\frac{\sqrt [4]{d} \sqrt [4]{a+b x}}{\sqrt [4]{b} \sqrt [4]{c+d x}}\right )}{\sqrt [4]{d}} \]

[Out]

(-2*a^(1/4)*ArcTan[(c^(1/4)*(a + b*x)^(1/4))/(a^(1/4)*(c + d*x)^(1/4))])/c^(1/4) + (2*b^(1/4)*ArcTan[(d^(1/4)*
(a + b*x)^(1/4))/(b^(1/4)*(c + d*x)^(1/4))])/d^(1/4) - (2*a^(1/4)*ArcTanh[(c^(1/4)*(a + b*x)^(1/4))/(a^(1/4)*(
c + d*x)^(1/4))])/c^(1/4) + (2*b^(1/4)*ArcTanh[(d^(1/4)*(a + b*x)^(1/4))/(b^(1/4)*(c + d*x)^(1/4))])/d^(1/4)

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Rubi [A]  time = 0.11753, antiderivative size = 169, normalized size of antiderivative = 1., number of steps used = 10, number of rules used = 7, integrand size = 22, \(\frac{\text{number of rules}}{\text{integrand size}}\) = 0.318, Rules used = {105, 63, 240, 212, 208, 205, 93} \[ -\frac{2 \sqrt [4]{a} \tan ^{-1}\left (\frac{\sqrt [4]{c} \sqrt [4]{a+b x}}{\sqrt [4]{a} \sqrt [4]{c+d x}}\right )}{\sqrt [4]{c}}+\frac{2 \sqrt [4]{b} \tan ^{-1}\left (\frac{\sqrt [4]{d} \sqrt [4]{a+b x}}{\sqrt [4]{b} \sqrt [4]{c+d x}}\right )}{\sqrt [4]{d}}-\frac{2 \sqrt [4]{a} \tanh ^{-1}\left (\frac{\sqrt [4]{c} \sqrt [4]{a+b x}}{\sqrt [4]{a} \sqrt [4]{c+d x}}\right )}{\sqrt [4]{c}}+\frac{2 \sqrt [4]{b} \tanh ^{-1}\left (\frac{\sqrt [4]{d} \sqrt [4]{a+b x}}{\sqrt [4]{b} \sqrt [4]{c+d x}}\right )}{\sqrt [4]{d}} \]

Antiderivative was successfully verified.

[In]

Int[(a + b*x)^(1/4)/(x*(c + d*x)^(1/4)),x]

[Out]

(-2*a^(1/4)*ArcTan[(c^(1/4)*(a + b*x)^(1/4))/(a^(1/4)*(c + d*x)^(1/4))])/c^(1/4) + (2*b^(1/4)*ArcTan[(d^(1/4)*
(a + b*x)^(1/4))/(b^(1/4)*(c + d*x)^(1/4))])/d^(1/4) - (2*a^(1/4)*ArcTanh[(c^(1/4)*(a + b*x)^(1/4))/(a^(1/4)*(
c + d*x)^(1/4))])/c^(1/4) + (2*b^(1/4)*ArcTanh[(d^(1/4)*(a + b*x)^(1/4))/(b^(1/4)*(c + d*x)^(1/4))])/d^(1/4)

Rule 105

Int[(((a_.) + (b_.)*(x_))^(m_)*((c_.) + (d_.)*(x_))^(n_))/((e_.) + (f_.)*(x_)), x_Symbol] :> Dist[b/f, Int[(a
+ b*x)^(m - 1)*(c + d*x)^n, x], x] - Dist[(b*e - a*f)/f, Int[((a + b*x)^(m - 1)*(c + d*x)^n)/(e + f*x), x], x]
/; FreeQ[{a, b, c, d, e, f, m, n}, x] && IGtQ[Simplify[m + n + 1], 0] && (GtQ[m, 0] || ( !RationalQ[m] && (Su
mSimplerQ[m, -1] ||  !SumSimplerQ[n, -1])))

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 240

Int[((a_) + (b_.)*(x_)^(n_))^(p_), x_Symbol] :> Dist[a^(p + 1/n), Subst[Int[1/(1 - b*x^n)^(p + 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)] && IntegerQ[p
+ 1/n]

Rule 212

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

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 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 93

Int[(((a_.) + (b_.)*(x_))^(m_)*((c_.) + (d_.)*(x_))^(n_))/((e_.) + (f_.)*(x_)), x_Symbol] :> With[{q = Denomin
ator[m]}, Dist[q, Subst[Int[x^(q*(m + 1) - 1)/(b*e - a*f - (d*e - c*f)*x^q), x], x, (a + b*x)^(1/q)/(c + d*x)^
(1/q)], x]] /; FreeQ[{a, b, c, d, e, f}, x] && EqQ[m + n + 1, 0] && RationalQ[n] && LtQ[-1, m, 0] && SimplerQ[
a + b*x, c + d*x]

Rubi steps

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

Mathematica [C]  time = 0.0514882, size = 97, normalized size = 0.57 \[ \frac{4 \sqrt [4]{a+b x} \left (\sqrt [4]{\frac{b (c+d x)}{b c-a d}} \, _2F_1\left (\frac{1}{4},\frac{1}{4};\frac{5}{4};\frac{d (a+b x)}{a d-b c}\right )-\, _2F_1\left (\frac{1}{4},1;\frac{5}{4};\frac{c (a+b x)}{a (c+d x)}\right )\right )}{\sqrt [4]{c+d x}} \]

Antiderivative was successfully verified.

[In]

Integrate[(a + b*x)^(1/4)/(x*(c + d*x)^(1/4)),x]

[Out]

(4*(a + b*x)^(1/4)*(((b*(c + d*x))/(b*c - a*d))^(1/4)*Hypergeometric2F1[1/4, 1/4, 5/4, (d*(a + b*x))/(-(b*c) +
a*d)] - Hypergeometric2F1[1/4, 1, 5/4, (c*(a + b*x))/(a*(c + d*x))]))/(c + d*x)^(1/4)

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Maple [F]  time = 0.039, size = 0, normalized size = 0. \begin{align*} \int{\frac{1}{x}\sqrt [4]{bx+a}{\frac{1}{\sqrt [4]{dx+c}}}}\, dx \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

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

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Maxima [F]  time = 0., size = 0, normalized size = 0. \begin{align*} \int \frac{{\left (b x + a\right )}^{\frac{1}{4}}}{{\left (d x + c\right )}^{\frac{1}{4}} x}\,{d x} \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

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

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Fricas [B]  time = 2.2348, size = 965, normalized size = 5.71 \begin{align*} 4 \, \left (\frac{a}{c}\right )^{\frac{1}{4}} \arctan \left (-\frac{{\left (b x + a\right )}^{\frac{1}{4}}{\left (d x + c\right )}^{\frac{3}{4}} c \left (\frac{a}{c}\right )^{\frac{3}{4}} -{\left (c d x + c^{2}\right )} \sqrt{\frac{{\left (d x + c\right )} \sqrt{\frac{a}{c}} + \sqrt{b x + a} \sqrt{d x + c}}{d x + c}} \left (\frac{a}{c}\right )^{\frac{3}{4}}}{a d x + a c}\right ) - 4 \, \left (\frac{b}{d}\right )^{\frac{1}{4}} \arctan \left (-\frac{{\left (b x + a\right )}^{\frac{1}{4}}{\left (d x + c\right )}^{\frac{3}{4}} d \left (\frac{b}{d}\right )^{\frac{3}{4}} -{\left (d^{2} x + c d\right )} \sqrt{\frac{{\left (d x + c\right )} \sqrt{\frac{b}{d}} + \sqrt{b x + a} \sqrt{d x + c}}{d x + c}} \left (\frac{b}{d}\right )^{\frac{3}{4}}}{b d x + b c}\right ) - \left (\frac{a}{c}\right )^{\frac{1}{4}} \log \left (\frac{{\left (d x + c\right )} \left (\frac{a}{c}\right )^{\frac{1}{4}} +{\left (b x + a\right )}^{\frac{1}{4}}{\left (d x + c\right )}^{\frac{3}{4}}}{d x + c}\right ) + \left (\frac{a}{c}\right )^{\frac{1}{4}} \log \left (-\frac{{\left (d x + c\right )} \left (\frac{a}{c}\right )^{\frac{1}{4}} -{\left (b x + a\right )}^{\frac{1}{4}}{\left (d x + c\right )}^{\frac{3}{4}}}{d x + c}\right ) + \left (\frac{b}{d}\right )^{\frac{1}{4}} \log \left (\frac{{\left (d x + c\right )} \left (\frac{b}{d}\right )^{\frac{1}{4}} +{\left (b x + a\right )}^{\frac{1}{4}}{\left (d x + c\right )}^{\frac{3}{4}}}{d x + c}\right ) - \left (\frac{b}{d}\right )^{\frac{1}{4}} \log \left (-\frac{{\left (d x + c\right )} \left (\frac{b}{d}\right )^{\frac{1}{4}} -{\left (b x + a\right )}^{\frac{1}{4}}{\left (d x + c\right )}^{\frac{3}{4}}}{d x + c}\right ) \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

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

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Sympy [F]  time = 0., size = 0, normalized size = 0. \begin{align*} \int \frac{\sqrt [4]{a + b x}}{x \sqrt [4]{c + d x}}\, dx \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

Integral((a + b*x)**(1/4)/(x*(c + d*x)**(1/4)), x)

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Giac [F(-1)]  time = 0., size = 0, normalized size = 0. \begin{align*} \text{Timed out} \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

Timed out