3.9.33 \(\int \frac {\sqrt [4]{a+b x}}{x^2 \sqrt [4]{c+d x}} \, dx\)

Optimal. Leaf size=131 \[ -\frac {(b c-a d) \tan ^{-1}\left (\frac {\sqrt [4]{c} \sqrt [4]{a+b x}}{\sqrt [4]{a} \sqrt [4]{c+d x}}\right )}{2 a^{3/4} c^{5/4}}-\frac {(b c-a d) \tanh ^{-1}\left (\frac {\sqrt [4]{c} \sqrt [4]{a+b x}}{\sqrt [4]{a} \sqrt [4]{c+d x}}\right )}{2 a^{3/4} c^{5/4}}-\frac {\sqrt [4]{a+b x} (c+d x)^{3/4}}{c x} \]

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Rubi [A]  time = 0.06, antiderivative size = 131, normalized size of antiderivative = 1.00, number of steps used = 5, number of rules used = 5, integrand size = 22, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.227, Rules used = {94, 93, 212, 208, 205} \begin {gather*} -\frac {(b c-a d) \tan ^{-1}\left (\frac {\sqrt [4]{c} \sqrt [4]{a+b x}}{\sqrt [4]{a} \sqrt [4]{c+d x}}\right )}{2 a^{3/4} c^{5/4}}-\frac {(b c-a d) \tanh ^{-1}\left (\frac {\sqrt [4]{c} \sqrt [4]{a+b x}}{\sqrt [4]{a} \sqrt [4]{c+d x}}\right )}{2 a^{3/4} c^{5/4}}-\frac {\sqrt [4]{a+b x} (c+d x)^{3/4}}{c x} \end {gather*}

Antiderivative was successfully verified.

[In]

Int[(a + b*x)^(1/4)/(x^2*(c + d*x)^(1/4)),x]

[Out]

-(((a + b*x)^(1/4)*(c + d*x)^(3/4))/(c*x)) - ((b*c - a*d)*ArcTan[(c^(1/4)*(a + b*x)^(1/4))/(a^(1/4)*(c + d*x)^
(1/4))])/(2*a^(3/4)*c^(5/4)) - ((b*c - a*d)*ArcTanh[(c^(1/4)*(a + b*x)^(1/4))/(a^(1/4)*(c + d*x)^(1/4))])/(2*a
^(3/4)*c^(5/4))

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 94

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

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 {\sqrt [4]{a+b x}}{x^2 \sqrt [4]{c+d x}} \, dx &=-\frac {\sqrt [4]{a+b x} (c+d x)^{3/4}}{c x}+\frac {(b c-a d) \int \frac {1}{x (a+b x)^{3/4} \sqrt [4]{c+d x}} \, dx}{4 c}\\ &=-\frac {\sqrt [4]{a+b x} (c+d x)^{3/4}}{c x}+\frac {(b c-a d) \operatorname {Subst}\left (\int \frac {1}{-a+c x^4} \, dx,x,\frac {\sqrt [4]{a+b x}}{\sqrt [4]{c+d x}}\right )}{c}\\ &=-\frac {\sqrt [4]{a+b x} (c+d x)^{3/4}}{c x}-\frac {(b c-a d) \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 )}{2 \sqrt {a} c}-\frac {(b c-a d) \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 )}{2 \sqrt {a} c}\\ &=-\frac {\sqrt [4]{a+b x} (c+d x)^{3/4}}{c x}-\frac {(b c-a d) \tan ^{-1}\left (\frac {\sqrt [4]{c} \sqrt [4]{a+b x}}{\sqrt [4]{a} \sqrt [4]{c+d x}}\right )}{2 a^{3/4} c^{5/4}}-\frac {(b c-a d) \tanh ^{-1}\left (\frac {\sqrt [4]{c} \sqrt [4]{a+b x}}{\sqrt [4]{a} \sqrt [4]{c+d x}}\right )}{2 a^{3/4} c^{5/4}}\\ \end {align*}

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Mathematica [C]  time = 0.02, size = 73, normalized size = 0.56 \begin {gather*} \frac {\sqrt [4]{a+b x} \left ((a d x-b c x) \, _2F_1\left (\frac {1}{4},1;\frac {5}{4};\frac {c (a+b x)}{a (c+d x)}\right )-a (c+d x)\right )}{a c x \sqrt [4]{c+d x}} \end {gather*}

Antiderivative was successfully verified.

[In]

Integrate[(a + b*x)^(1/4)/(x^2*(c + d*x)^(1/4)),x]

[Out]

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

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IntegrateAlgebraic [A]  time = 5.31, size = 173, normalized size = 1.32 \begin {gather*} \frac {(b c-a d) \tan ^{-1}\left (\frac {\sqrt [4]{a} \sqrt [4]{c+d x}}{\sqrt [4]{c} \sqrt [4]{a+\frac {b (c+d x)}{d}-\frac {b c}{d}}}\right )}{2 a^{3/4} c^{5/4}}+\frac {(a d-b c) \tanh ^{-1}\left (\frac {\sqrt [4]{c} \sqrt [4]{a+\frac {b (c+d x)}{d}-\frac {b c}{d}}}{\sqrt [4]{a} \sqrt [4]{c+d x}}\right )}{2 a^{3/4} c^{5/4}}-\frac {(c+d x)^{3/4} \sqrt [4]{a+\frac {b (c+d x)}{d}-\frac {b c}{d}}}{c x} \end {gather*}

Antiderivative was successfully verified.

[In]

IntegrateAlgebraic[(a + b*x)^(1/4)/(x^2*(c + d*x)^(1/4)),x]

[Out]

-(((c + d*x)^(3/4)*(a - (b*c)/d + (b*(c + d*x))/d)^(1/4))/(c*x)) + ((b*c - a*d)*ArcTan[(a^(1/4)*(c + d*x)^(1/4
))/(c^(1/4)*(a - (b*c)/d + (b*(c + d*x))/d)^(1/4))])/(2*a^(3/4)*c^(5/4)) + ((-(b*c) + a*d)*ArcTanh[(c^(1/4)*(a
- (b*c)/d + (b*(c + d*x))/d)^(1/4))/(a^(1/4)*(c + d*x)^(1/4))])/(2*a^(3/4)*c^(5/4))

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fricas [B]  time = 1.57, size = 820, normalized size = 6.26 \begin {gather*} -\frac {4 \, c x \left (\frac {b^{4} c^{4} - 4 \, a b^{3} c^{3} d + 6 \, a^{2} b^{2} c^{2} d^{2} - 4 \, a^{3} b c d^{3} + a^{4} d^{4}}{a^{3} c^{5}}\right )^{\frac {1}{4}} \arctan \left (\frac {{\left (a^{2} b c^{5} - a^{3} c^{4} d\right )} {\left (b x + a\right )}^{\frac {1}{4}} {\left (d x + c\right )}^{\frac {3}{4}} \left (\frac {b^{4} c^{4} - 4 \, a b^{3} c^{3} d + 6 \, a^{2} b^{2} c^{2} d^{2} - 4 \, a^{3} b c d^{3} + a^{4} d^{4}}{a^{3} c^{5}}\right )^{\frac {3}{4}} + {\left (a^{2} c^{4} d x + a^{2} c^{5}\right )} \sqrt {\frac {{\left (b^{2} c^{2} - 2 \, a b c d + a^{2} d^{2}\right )} \sqrt {b x + a} \sqrt {d x + c} + {\left (a^{2} c^{2} d x + a^{2} c^{3}\right )} \sqrt {\frac {b^{4} c^{4} - 4 \, a b^{3} c^{3} d + 6 \, a^{2} b^{2} c^{2} d^{2} - 4 \, a^{3} b c d^{3} + a^{4} d^{4}}{a^{3} c^{5}}}}{d x + c}} \left (\frac {b^{4} c^{4} - 4 \, a b^{3} c^{3} d + 6 \, a^{2} b^{2} c^{2} d^{2} - 4 \, a^{3} b c d^{3} + a^{4} d^{4}}{a^{3} c^{5}}\right )^{\frac {3}{4}}}{b^{4} c^{5} - 4 \, a b^{3} c^{4} d + 6 \, a^{2} b^{2} c^{3} d^{2} - 4 \, a^{3} b c^{2} d^{3} + a^{4} c d^{4} + {\left (b^{4} c^{4} d - 4 \, a b^{3} c^{3} d^{2} + 6 \, a^{2} b^{2} c^{2} d^{3} - 4 \, a^{3} b c d^{4} + a^{4} d^{5}\right )} x}\right ) + c x \left (\frac {b^{4} c^{4} - 4 \, a b^{3} c^{3} d + 6 \, a^{2} b^{2} c^{2} d^{2} - 4 \, a^{3} b c d^{3} + a^{4} d^{4}}{a^{3} c^{5}}\right )^{\frac {1}{4}} \log \left (-\frac {{\left (b c - a d\right )} {\left (b x + a\right )}^{\frac {1}{4}} {\left (d x + c\right )}^{\frac {3}{4}} + {\left (a c d x + a c^{2}\right )} \left (\frac {b^{4} c^{4} - 4 \, a b^{3} c^{3} d + 6 \, a^{2} b^{2} c^{2} d^{2} - 4 \, a^{3} b c d^{3} + a^{4} d^{4}}{a^{3} c^{5}}\right )^{\frac {1}{4}}}{d x + c}\right ) - c x \left (\frac {b^{4} c^{4} - 4 \, a b^{3} c^{3} d + 6 \, a^{2} b^{2} c^{2} d^{2} - 4 \, a^{3} b c d^{3} + a^{4} d^{4}}{a^{3} c^{5}}\right )^{\frac {1}{4}} \log \left (-\frac {{\left (b c - a d\right )} {\left (b x + a\right )}^{\frac {1}{4}} {\left (d x + c\right )}^{\frac {3}{4}} - {\left (a c d x + a c^{2}\right )} \left (\frac {b^{4} c^{4} - 4 \, a b^{3} c^{3} d + 6 \, a^{2} b^{2} c^{2} d^{2} - 4 \, a^{3} b c d^{3} + a^{4} d^{4}}{a^{3} c^{5}}\right )^{\frac {1}{4}}}{d x + c}\right ) + 4 \, {\left (b x + a\right )}^{\frac {1}{4}} {\left (d x + c\right )}^{\frac {3}{4}}}{4 \, c x} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate((b*x+a)^(1/4)/x^2/(d*x+c)^(1/4),x, algorithm="fricas")

[Out]

-1/4*(4*c*x*((b^4*c^4 - 4*a*b^3*c^3*d + 6*a^2*b^2*c^2*d^2 - 4*a^3*b*c*d^3 + a^4*d^4)/(a^3*c^5))^(1/4)*arctan((
(a^2*b*c^5 - a^3*c^4*d)*(b*x + a)^(1/4)*(d*x + c)^(3/4)*((b^4*c^4 - 4*a*b^3*c^3*d + 6*a^2*b^2*c^2*d^2 - 4*a^3*
b*c*d^3 + a^4*d^4)/(a^3*c^5))^(3/4) + (a^2*c^4*d*x + a^2*c^5)*sqrt(((b^2*c^2 - 2*a*b*c*d + a^2*d^2)*sqrt(b*x +
a)*sqrt(d*x + c) + (a^2*c^2*d*x + a^2*c^3)*sqrt((b^4*c^4 - 4*a*b^3*c^3*d + 6*a^2*b^2*c^2*d^2 - 4*a^3*b*c*d^3
+ a^4*d^4)/(a^3*c^5)))/(d*x + c))*((b^4*c^4 - 4*a*b^3*c^3*d + 6*a^2*b^2*c^2*d^2 - 4*a^3*b*c*d^3 + a^4*d^4)/(a^
3*c^5))^(3/4))/(b^4*c^5 - 4*a*b^3*c^4*d + 6*a^2*b^2*c^3*d^2 - 4*a^3*b*c^2*d^3 + a^4*c*d^4 + (b^4*c^4*d - 4*a*b
^3*c^3*d^2 + 6*a^2*b^2*c^2*d^3 - 4*a^3*b*c*d^4 + a^4*d^5)*x)) + c*x*((b^4*c^4 - 4*a*b^3*c^3*d + 6*a^2*b^2*c^2*
d^2 - 4*a^3*b*c*d^3 + a^4*d^4)/(a^3*c^5))^(1/4)*log(-((b*c - a*d)*(b*x + a)^(1/4)*(d*x + c)^(3/4) + (a*c*d*x +
a*c^2)*((b^4*c^4 - 4*a*b^3*c^3*d + 6*a^2*b^2*c^2*d^2 - 4*a^3*b*c*d^3 + a^4*d^4)/(a^3*c^5))^(1/4))/(d*x + c))
- c*x*((b^4*c^4 - 4*a*b^3*c^3*d + 6*a^2*b^2*c^2*d^2 - 4*a^3*b*c*d^3 + a^4*d^4)/(a^3*c^5))^(1/4)*log(-((b*c - a
*d)*(b*x + a)^(1/4)*(d*x + c)^(3/4) - (a*c*d*x + a*c^2)*((b^4*c^4 - 4*a*b^3*c^3*d + 6*a^2*b^2*c^2*d^2 - 4*a^3*
b*c*d^3 + a^4*d^4)/(a^3*c^5))^(1/4))/(d*x + c)) + 4*(b*x + a)^(1/4)*(d*x + c)^(3/4))/(c*x)

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giac [F]  time = 0.00, size = 0, normalized size = 0.00 \begin {gather*} \int \frac {{\left (b x + a\right )}^{\frac {1}{4}}}{{\left (d x + c\right )}^{\frac {1}{4}} x^{2}}\,{d x} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate((b*x+a)^(1/4)/x^2/(d*x+c)^(1/4),x, algorithm="giac")

[Out]

integrate((b*x + a)^(1/4)/((d*x + c)^(1/4)*x^2), x)

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maple [F]  time = 0.10, size = 0, normalized size = 0.00 \begin {gather*} \int \frac {\left (b x +a \right )^{\frac {1}{4}}}{\left (d x +c \right )^{\frac {1}{4}} x^{2}}\, dx \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

int((b*x+a)^(1/4)/x^2/(d*x+c)^(1/4),x)

[Out]

int((b*x+a)^(1/4)/x^2/(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 {{\left (b x + a\right )}^{\frac {1}{4}}}{{\left (d x + c\right )}^{\frac {1}{4}} x^{2}}\,{d x} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate((b*x+a)^(1/4)/x^2/(d*x+c)^(1/4),x, algorithm="maxima")

[Out]

integrate((b*x + a)^(1/4)/((d*x + c)^(1/4)*x^2), x)

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mupad [F]  time = 0.00, size = -1, normalized size = -0.01 \begin {gather*} \int \frac {{\left (a+b\,x\right )}^{1/4}}{x^2\,{\left (c+d\,x\right )}^{1/4}} \,d x \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

int((a + b*x)^(1/4)/(x^2*(c + d*x)^(1/4)),x)

[Out]

int((a + b*x)^(1/4)/(x^2*(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 {\sqrt [4]{a + b x}}{x^{2} \sqrt [4]{c + d x}}\, dx \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate((b*x+a)**(1/4)/x**2/(d*x+c)**(1/4),x)

[Out]

Integral((a + b*x)**(1/4)/(x**2*(c + d*x)**(1/4)), x)

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