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

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

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Rubi [A]  time = 0.03, antiderivative size = 85, normalized size of antiderivative = 1.00, 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} \begin {gather*} -\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 {gather*}

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 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 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 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]

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*}

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Mathematica [C]  time = 0.01, size = 48, normalized size = 0.56 \begin {gather*} -\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}} \end {gather*}

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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IntegrateAlgebraic [A]  time = 0.13, size = 85, normalized size = 1.00 \begin {gather*} -\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 {gather*}

Antiderivative was successfully verified.

[In]

IntegrateAlgebraic[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))

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fricas [B]  time = 1.45, size = 234, normalized size = 2.75 \begin {gather*} 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 {gather*}

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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giac [F]  time = 0.00, size = 0, normalized size = 0.00 \begin {gather*} \int \frac {1}{{\left (b x + a\right )}^{\frac {3}{4}} {\left (d x + c\right )}^{\frac {1}{4}} x}\,{d x} \end {gather*}

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]

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

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maple [F]  time = 0.14, size = 0, normalized size = 0.00 \begin {gather*} \int \frac {1}{\left (b x +a \right )^{\frac {3}{4}} \left (d x +c \right )^{\frac {1}{4}} x}\, dx \end {gather*}

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 = 0.00, size = 0, normalized size = 0.00 \begin {gather*} \int \frac {1}{{\left (b x + a\right )}^{\frac {3}{4}} {\left (d x + c\right )}^{\frac {1}{4}} x}\,{d x} \end {gather*}

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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mupad [F]  time = 0.00, size = -1, normalized size = -0.01 \begin {gather*} \int \frac {1}{x\,{\left (a+b\,x\right )}^{3/4}\,{\left (c+d\,x\right )}^{1/4}} \,d x \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

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

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sympy [F]  time = 0.00, size = 0, normalized size = 0.00 \begin {gather*} \int \frac {1}{x \left (a + b x\right )^{\frac {3}{4}} \sqrt [4]{c + d x}}\, dx \end {gather*}

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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