3.1694 \(\int \frac {\sqrt [4]{a+b x}}{\sqrt [4]{c+d x}} \, dx\)

Optimal. Leaf size=127 \[ -\frac {(b c-a d) \tan ^{-1}\left (\frac {\sqrt [4]{d} \sqrt [4]{a+b x}}{\sqrt [4]{b} \sqrt [4]{c+d x}}\right )}{2 b^{3/4} d^{5/4}}-\frac {(b c-a d) \tanh ^{-1}\left (\frac {\sqrt [4]{d} \sqrt [4]{a+b x}}{\sqrt [4]{b} \sqrt [4]{c+d x}}\right )}{2 b^{3/4} d^{5/4}}+\frac {\sqrt [4]{a+b x} (c+d x)^{3/4}}{d} \]

[Out]

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

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Rubi [A]  time = 0.07, antiderivative size = 127, normalized size of antiderivative = 1.00, number of steps used = 6, number of rules used = 6, integrand size = 19, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.316, Rules used = {50, 63, 240, 212, 208, 205} \[ -\frac {(b c-a d) \tan ^{-1}\left (\frac {\sqrt [4]{d} \sqrt [4]{a+b x}}{\sqrt [4]{b} \sqrt [4]{c+d x}}\right )}{2 b^{3/4} d^{5/4}}-\frac {(b c-a d) \tanh ^{-1}\left (\frac {\sqrt [4]{d} \sqrt [4]{a+b x}}{\sqrt [4]{b} \sqrt [4]{c+d x}}\right )}{2 b^{3/4} d^{5/4}}+\frac {\sqrt [4]{a+b x} (c+d x)^{3/4}}{d} \]

Antiderivative was successfully verified.

[In]

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

[Out]

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

Rule 50

Int[((a_.) + (b_.)*(x_))^(m_)*((c_.) + (d_.)*(x_))^(n_), x_Symbol] :> Simp[((a + b*x)^(m + 1)*(c + d*x)^n)/(b*
(m + n + 1)), x] + Dist[(n*(b*c - a*d))/(b*(m + n + 1)), Int[(a + b*x)^m*(c + d*x)^(n - 1), x], x] /; FreeQ[{a
, b, c, d}, x] && NeQ[b*c - a*d, 0] && GtQ[n, 0] && NeQ[m + n + 1, 0] &&  !(IGtQ[m, 0] && ( !IntegerQ[n] || (G
tQ[m, 0] && LtQ[m - n, 0]))) &&  !ILtQ[m + n + 2, 0] && IntLinearQ[a, b, c, d, m, n, x]

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 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]

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]

Rubi steps

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

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Mathematica [C]  time = 0.03, size = 73, normalized size = 0.57 \[ \frac {4 (a+b x)^{5/4} \sqrt [4]{\frac {b (c+d x)}{b c-a d}} \, _2F_1\left (\frac {1}{4},\frac {5}{4};\frac {9}{4};\frac {d (a+b x)}{a d-b c}\right )}{5 b \sqrt [4]{c+d x}} \]

Antiderivative was successfully verified.

[In]

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

[Out]

(4*(a + b*x)^(5/4)*((b*(c + d*x))/(b*c - a*d))^(1/4)*Hypergeometric2F1[1/4, 5/4, 9/4, (d*(a + b*x))/(-(b*c) +
a*d)])/(5*b*(c + d*x)^(1/4))

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fricas [B]  time = 0.49, size = 814, normalized size = 6.41 \[ -\frac {4 \, d \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}}{b^{3} d^{5}}\right )^{\frac {1}{4}} \arctan \left (\frac {{\left (b^{3} c d^{4} - a b^{2} d^{5}\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}}{b^{3} d^{5}}\right )^{\frac {3}{4}} + {\left (b^{2} d^{5} x + b^{2} c d^{4}\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 (b^{2} d^{3} x + b^{2} c d^{2}\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}}{b^{3} d^{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}}{b^{3} d^{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 ) + d \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}}{b^{3} d^{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 (b d^{2} x + b c d\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}}{b^{3} d^{5}}\right )^{\frac {1}{4}}}{d x + c}\right ) - d \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}}{b^{3} d^{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 (b d^{2} x + b c d\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}}{b^{3} d^{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 \, d} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

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

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

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

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

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maple [F]  time = 0.04, size = 0, normalized size = 0.00 \[ \int \frac {\left (b x +a \right )^{\frac {1}{4}}}{\left (d x +c \right )^{\frac {1}{4}}}\, dx \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

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

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

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

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

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

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

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

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sympy [F]  time = 0.00, size = 0, normalized size = 0.00 \[ \int \frac {\sqrt [4]{a + b x}}{\sqrt [4]{c + d x}}\, dx \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

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

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