\(\int \frac {1}{(a+b x)^{3/4} \sqrt [4]{c+d x}} \, dx\) [1695]

Optimal result

Integrand size = 19, antiderivative size = 85 \[ \int \frac {1}{(a+b x)^{3/4} \sqrt [4]{c+d x}} \, dx=\frac {2 \arctan \left (\frac {\sqrt [4]{d} \sqrt [4]{a+b x}}{\sqrt [4]{b} \sqrt [4]{c+d x}}\right )}{b^{3/4} \sqrt [4]{d}}+\frac {2 \text {arctanh}\left (\frac {\sqrt [4]{d} \sqrt [4]{a+b x}}{\sqrt [4]{b} \sqrt [4]{c+d x}}\right )}{b^{3/4} \sqrt [4]{d}} \]

[Out]

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

Rubi [A] (verified)

Time = 0.04 (sec) , antiderivative size = 85, normalized size of antiderivative = 1.00, number of steps used = 5, number of rules used = 5, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.263, Rules used = {65, 246, 218, 214, 211} \[ \int \frac {1}{(a+b x)^{3/4} \sqrt [4]{c+d x}} \, dx=\frac {2 \arctan \left (\frac {\sqrt [4]{d} \sqrt [4]{a+b x}}{\sqrt [4]{b} \sqrt [4]{c+d x}}\right )}{b^{3/4} \sqrt [4]{d}}+\frac {2 \text {arctanh}\left (\frac {\sqrt [4]{d} \sqrt [4]{a+b x}}{\sqrt [4]{b} \sqrt [4]{c+d x}}\right )}{b^{3/4} \sqrt [4]{d}} \]

[In]

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

[Out]

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

Rule 65

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 211

Int[((a_) + (b_.)*(x_)^2)^(-1), x_Symbol] :> Simp[(Rt[a/b, 2]/a)*ArcTan[x/Rt[a/b, 2]], x] /; FreeQ[{a, b}, x]
&& PosQ[a/b]

Rule 214

Int[((a_) + (b_.)*(x_)^2)^(-1), x_Symbol] :> Simp[(Rt[-a/b, 2]/a)*ArcTanh[x/Rt[-a/b, 2]], x] /; FreeQ[{a, b},
x] && NegQ[a/b]

Rule 218

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] &&  !Gt
Q[a/b, 0]

Rule 246

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

Mathematica [A] (verified)

Time = 0.09 (sec) , antiderivative size = 75, normalized size of antiderivative = 0.88 \[ \int \frac {1}{(a+b x)^{3/4} \sqrt [4]{c+d x}} \, dx=-\frac {2 \left (\arctan \left (\frac {\sqrt [4]{b} \sqrt [4]{c+d x}}{\sqrt [4]{d} \sqrt [4]{a+b x}}\right )-\text {arctanh}\left (\frac {\sqrt [4]{b} \sqrt [4]{c+d x}}{\sqrt [4]{d} \sqrt [4]{a+b x}}\right )\right )}{b^{3/4} \sqrt [4]{d}} \]

[In]

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

[Out]

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

Maple [F]

\[\int \frac {1}{\left (b x +a \right )^{\frac {3}{4}} \left (d x +c \right )^{\frac {1}{4}}}d x\]

[In]

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

[Out]

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

Fricas [C] (verification not implemented)

Result contains complex when optimal does not.

Time = 0.25 (sec) , antiderivative size = 222, normalized size of antiderivative = 2.61 \[ \int \frac {1}{(a+b x)^{3/4} \sqrt [4]{c+d x}} \, dx=\left (\frac {1}{b^{3} d}\right )^{\frac {1}{4}} \log \left (\frac {{\left (b d x + b c\right )} \left (\frac {1}{b^{3} d}\right )^{\frac {1}{4}} + {\left (b x + a\right )}^{\frac {1}{4}} {\left (d x + c\right )}^{\frac {3}{4}}}{d x + c}\right ) - \left (\frac {1}{b^{3} d}\right )^{\frac {1}{4}} \log \left (-\frac {{\left (b d x + b c\right )} \left (\frac {1}{b^{3} d}\right )^{\frac {1}{4}} - {\left (b x + a\right )}^{\frac {1}{4}} {\left (d x + c\right )}^{\frac {3}{4}}}{d x + c}\right ) + i \, \left (\frac {1}{b^{3} d}\right )^{\frac {1}{4}} \log \left (\frac {{\left (i \, b d x + i \, b c\right )} \left (\frac {1}{b^{3} d}\right )^{\frac {1}{4}} + {\left (b x + a\right )}^{\frac {1}{4}} {\left (d x + c\right )}^{\frac {3}{4}}}{d x + c}\right ) - i \, \left (\frac {1}{b^{3} d}\right )^{\frac {1}{4}} \log \left (\frac {{\left (-i \, b d x - i \, b c\right )} \left (\frac {1}{b^{3} d}\right )^{\frac {1}{4}} + {\left (b x + a\right )}^{\frac {1}{4}} {\left (d x + c\right )}^{\frac {3}{4}}}{d x + c}\right ) \]

[In]

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

[Out]

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

Sympy [F]

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

[In]

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

[Out]

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

Maxima [F]

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

[In]

integrate(1/(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)

Giac [F]

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

[In]

integrate(1/(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)

Mupad [F(-1)]

Timed out. \[ \int \frac {1}{(a+b x)^{3/4} \sqrt [4]{c+d x}} \, dx=\int \frac {1}{{\left (a+b\,x\right )}^{3/4}\,{\left (c+d\,x\right )}^{1/4}} \,d x \]

[In]

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

[Out]

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