3.904 \(\int \frac{1}{x (a+b x)^{3/4} \sqrt [4]{c+d x}} \, dx\)

Optimal. Leaf size=85 \[ -\frac{2 \tan ^{-1}\left (\frac{\sqrt [4]{c} \sqrt [4]{a+b x}}{\sqrt [4]{a} \sqrt [4]{c+d x}}\right )}{a^{3/4} \sqrt [4]{c}}-\frac{2 \tanh ^{-1}\left (\frac{\sqrt [4]{c} \sqrt [4]{a+b x}}{\sqrt [4]{a} \sqrt [4]{c+d x}}\right )}{a^{3/4} \sqrt [4]{c}} \]

[Out]

(-2*ArcTan[(c^(1/4)*(a + b*x)^(1/4))/(a^(1/4)*(c + d*x)^(1/4))])/(a^(3/4)*c^(1/4)) - (2*ArcTanh[(c^(1/4)*(a +
b*x)^(1/4))/(a^(1/4)*(c + d*x)^(1/4))])/(a^(3/4)*c^(1/4))

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Rubi [A]  time = 0.0329855, antiderivative size = 85, normalized size of antiderivative = 1., number of steps used = 4, number of rules used = 4, integrand size = 22, \(\frac{\text{number of rules}}{\text{integrand size}}\) = 0.182, Rules used = {93, 212, 208, 205} \[ -\frac{2 \tan ^{-1}\left (\frac{\sqrt [4]{c} \sqrt [4]{a+b x}}{\sqrt [4]{a} \sqrt [4]{c+d x}}\right )}{a^{3/4} \sqrt [4]{c}}-\frac{2 \tanh ^{-1}\left (\frac{\sqrt [4]{c} \sqrt [4]{a+b x}}{\sqrt [4]{a} \sqrt [4]{c+d x}}\right )}{a^{3/4} \sqrt [4]{c}} \]

Antiderivative was successfully verified.

[In]

Int[1/(x*(a + b*x)^(3/4)*(c + d*x)^(1/4)),x]

[Out]

(-2*ArcTan[(c^(1/4)*(a + b*x)^(1/4))/(a^(1/4)*(c + d*x)^(1/4))])/(a^(3/4)*c^(1/4)) - (2*ArcTanh[(c^(1/4)*(a +
b*x)^(1/4))/(a^(1/4)*(c + d*x)^(1/4))])/(a^(3/4)*c^(1/4))

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]

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]

Rubi steps

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

Mathematica [C]  time = 0.0104104, size = 48, normalized size = 0.56 \[ -\frac{4 \sqrt [4]{a+b x} \, _2F_1\left (\frac{1}{4},1;\frac{5}{4};\frac{c (a+b x)}{a (c+d x)}\right )}{a \sqrt [4]{c+d x}} \]

Antiderivative was successfully verified.

[In]

Integrate[1/(x*(a + b*x)^(3/4)*(c + d*x)^(1/4)),x]

[Out]

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

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Maple [F]  time = 0.043, size = 0, normalized size = 0. \begin{align*} \int{\frac{1}{x} \left ( bx+a \right ) ^{-{\frac{3}{4}}}{\frac{1}{\sqrt [4]{dx+c}}}}\, dx \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

int(1/x/(b*x+a)^(3/4)/(d*x+c)^(1/4),x)

[Out]

int(1/x/(b*x+a)^(3/4)/(d*x+c)^(1/4),x)

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Maxima [F]  time = 0., size = 0, normalized size = 0. \begin{align*} \int \frac{1}{{\left (b x + a\right )}^{\frac{3}{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(1/x/(b*x+a)^(3/4)/(d*x+c)^(1/4),x, algorithm="maxima")

[Out]

integrate(1/((b*x + a)^(3/4)*(d*x + c)^(1/4)*x), x)

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Fricas [B]  time = 1.73982, size = 579, normalized size = 6.81 \begin{align*} 4 \, \left (\frac{1}{a^{3} c}\right )^{\frac{1}{4}} \arctan \left (-\frac{{\left (b x + a\right )}^{\frac{1}{4}}{\left (d x + c\right )}^{\frac{3}{4}} a^{2} c \left (\frac{1}{a^{3} c}\right )^{\frac{3}{4}} -{\left (a^{2} c d x + a^{2} c^{2}\right )} \sqrt{\frac{{\left (a^{2} d x + a^{2} c\right )} \sqrt{\frac{1}{a^{3} c}} + \sqrt{b x + a} \sqrt{d x + c}}{d x + c}} \left (\frac{1}{a^{3} c}\right )^{\frac{3}{4}}}{d x + c}\right ) - \left (\frac{1}{a^{3} c}\right )^{\frac{1}{4}} \log \left (\frac{{\left (a d x + a c\right )} \left (\frac{1}{a^{3} 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{1}{a^{3} c}\right )^{\frac{1}{4}} \log \left (-\frac{{\left (a d x + a c\right )} \left (\frac{1}{a^{3} 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 ) \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate(1/x/(b*x+a)^(3/4)/(d*x+c)^(1/4),x, algorithm="fricas")

[Out]

4*(1/(a^3*c))^(1/4)*arctan(-((b*x + a)^(1/4)*(d*x + c)^(3/4)*a^2*c*(1/(a^3*c))^(3/4) - (a^2*c*d*x + a^2*c^2)*s
qrt(((a^2*d*x + a^2*c)*sqrt(1/(a^3*c)) + sqrt(b*x + a)*sqrt(d*x + c))/(d*x + c))*(1/(a^3*c))^(3/4))/(d*x + c))
- (1/(a^3*c))^(1/4)*log(((a*d*x + a*c)*(1/(a^3*c))^(1/4) + (b*x + a)^(1/4)*(d*x + c)^(3/4))/(d*x + c)) + (1/(
a^3*c))^(1/4)*log(-((a*d*x + a*c)*(1/(a^3*c))^(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{1}{x \left (a + b x\right )^{\frac{3}{4}} \sqrt [4]{c + d x}}\, dx \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate(1/x/(b*x+a)**(3/4)/(d*x+c)**(1/4),x)

[Out]

Integral(1/(x*(a + b*x)**(3/4)*(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(1/x/(b*x+a)^(3/4)/(d*x+c)^(1/4),x, algorithm="giac")

[Out]

Timed out