\(\int \frac {\sqrt [4]{a+b x}}{(c+d x)^{5/4}} \, dx\) [1718]

Optimal result

Integrand size = 19, antiderivative size = 108 \[ \int \frac {\sqrt [4]{a+b x}}{(c+d x)^{5/4}} \, dx=-\frac {4 \sqrt [4]{a+b x}}{d \sqrt [4]{c+d x}}+\frac {2 \sqrt [4]{b} \arctan \left (\frac {\sqrt [4]{d} \sqrt [4]{a+b x}}{\sqrt [4]{b} \sqrt [4]{c+d x}}\right )}{d^{5/4}}+\frac {2 \sqrt [4]{b} \text {arctanh}\left (\frac {\sqrt [4]{d} \sqrt [4]{a+b x}}{\sqrt [4]{b} \sqrt [4]{c+d x}}\right )}{d^{5/4}} \]

[Out]

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

Rubi [A] (verified)

Time = 0.05 (sec) , antiderivative size = 108, normalized size of antiderivative = 1.00, number of steps used = 6, number of rules used = 6, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.316, Rules used = {49, 65, 246, 218, 214, 211} \[ \int \frac {\sqrt [4]{a+b x}}{(c+d x)^{5/4}} \, dx=\frac {2 \sqrt [4]{b} \arctan \left (\frac {\sqrt [4]{d} \sqrt [4]{a+b x}}{\sqrt [4]{b} \sqrt [4]{c+d x}}\right )}{d^{5/4}}+\frac {2 \sqrt [4]{b} \text {arctanh}\left (\frac {\sqrt [4]{d} \sqrt [4]{a+b x}}{\sqrt [4]{b} \sqrt [4]{c+d x}}\right )}{d^{5/4}}-\frac {4 \sqrt [4]{a+b x}}{d \sqrt [4]{c+d x}} \]

[In]

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

[Out]

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

Rule 49

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

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

Mathematica [A] (verified)

Time = 0.15 (sec) , antiderivative size = 108, normalized size of antiderivative = 1.00 \[ \int \frac {\sqrt [4]{a+b x}}{(c+d x)^{5/4}} \, dx=-\frac {4 \sqrt [4]{a+b x}}{d \sqrt [4]{c+d x}}-\frac {2 \sqrt [4]{b} \arctan \left (\frac {\sqrt [4]{b} \sqrt [4]{c+d x}}{\sqrt [4]{d} \sqrt [4]{a+b x}}\right )}{d^{5/4}}+\frac {2 \sqrt [4]{b} \text {arctanh}\left (\frac {\sqrt [4]{b} \sqrt [4]{c+d x}}{\sqrt [4]{d} \sqrt [4]{a+b x}}\right )}{d^{5/4}} \]

[In]

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

[Out]

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

Maple [F]

\[\int \frac {\left (b x +a \right )^{\frac {1}{4}}}{\left (d x +c \right )^{\frac {5}{4}}}d x\]

[In]

int((b*x+a)^(1/4)/(d*x+c)^(5/4),x)

[Out]

int((b*x+a)^(1/4)/(d*x+c)^(5/4),x)

Fricas [C] (verification not implemented)

Result contains complex when optimal does not.

Time = 0.24 (sec) , antiderivative size = 276, normalized size of antiderivative = 2.56 \[ \int \frac {\sqrt [4]{a+b x}}{(c+d x)^{5/4}} \, dx=\frac {{\left (d^{2} x + c d\right )} \left (\frac {b}{d^{5}}\right )^{\frac {1}{4}} \log \left (\frac {{\left (d^{2} x + c d\right )} \left (\frac {b}{d^{5}}\right )^{\frac {1}{4}} + {\left (b x + a\right )}^{\frac {1}{4}} {\left (d x + c\right )}^{\frac {3}{4}}}{d x + c}\right ) - {\left (d^{2} x + c d\right )} \left (\frac {b}{d^{5}}\right )^{\frac {1}{4}} \log \left (-\frac {{\left (d^{2} x + c d\right )} \left (\frac {b}{d^{5}}\right )^{\frac {1}{4}} - {\left (b x + a\right )}^{\frac {1}{4}} {\left (d x + c\right )}^{\frac {3}{4}}}{d x + c}\right ) + {\left (i \, d^{2} x + i \, c d\right )} \left (\frac {b}{d^{5}}\right )^{\frac {1}{4}} \log \left (\frac {{\left (i \, d^{2} x + i \, c d\right )} \left (\frac {b}{d^{5}}\right )^{\frac {1}{4}} + {\left (b x + a\right )}^{\frac {1}{4}} {\left (d x + c\right )}^{\frac {3}{4}}}{d x + c}\right ) + {\left (-i \, d^{2} x - i \, c d\right )} \left (\frac {b}{d^{5}}\right )^{\frac {1}{4}} \log \left (\frac {{\left (-i \, d^{2} x - i \, c d\right )} \left (\frac {b}{d^{5}}\right )^{\frac {1}{4}} + {\left (b x + a\right )}^{\frac {1}{4}} {\left (d x + c\right )}^{\frac {3}{4}}}{d x + c}\right ) - 4 \, {\left (b x + a\right )}^{\frac {1}{4}} {\left (d x + c\right )}^{\frac {3}{4}}}{d^{2} x + c d} \]

[In]

integrate((b*x+a)^(1/4)/(d*x+c)^(5/4),x, algorithm="fricas")

[Out]

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

Sympy [F]

\[ \int \frac {\sqrt [4]{a+b x}}{(c+d x)^{5/4}} \, dx=\int \frac {\sqrt [4]{a + b x}}{\left (c + d x\right )^{\frac {5}{4}}}\, dx \]

[In]

integrate((b*x+a)**(1/4)/(d*x+c)**(5/4),x)

[Out]

Integral((a + b*x)**(1/4)/(c + d*x)**(5/4), x)

Maxima [F]

\[ \int \frac {\sqrt [4]{a+b x}}{(c+d x)^{5/4}} \, dx=\int { \frac {{\left (b x + a\right )}^{\frac {1}{4}}}{{\left (d x + c\right )}^{\frac {5}{4}}} \,d x } \]

[In]

integrate((b*x+a)^(1/4)/(d*x+c)^(5/4),x, algorithm="maxima")

[Out]

integrate((b*x + a)^(1/4)/(d*x + c)^(5/4), x)

Giac [F]

\[ \int \frac {\sqrt [4]{a+b x}}{(c+d x)^{5/4}} \, dx=\int { \frac {{\left (b x + a\right )}^{\frac {1}{4}}}{{\left (d x + c\right )}^{\frac {5}{4}}} \,d x } \]

[In]

integrate((b*x+a)^(1/4)/(d*x+c)^(5/4),x, algorithm="giac")

[Out]

integrate((b*x + a)^(1/4)/(d*x + c)^(5/4), x)

Mupad [F(-1)]

Timed out. \[ \int \frac {\sqrt [4]{a+b x}}{(c+d x)^{5/4}} \, dx=\int \frac {{\left (a+b\,x\right )}^{1/4}}{{\left (c+d\,x\right )}^{5/4}} \,d x \]

[In]

int((a + b*x)^(1/4)/(c + d*x)^(5/4),x)

[Out]

int((a + b*x)^(1/4)/(c + d*x)^(5/4), x)