\(\int \frac {(c+d x)^{5/4}}{(a+b x)^{9/4}} \, dx\) [1681]

Optimal result

Integrand size = 19, antiderivative size = 134 \[ \int \frac {(c+d x)^{5/4}}{(a+b x)^{9/4}} \, dx=-\frac {4 d \sqrt [4]{c+d x}}{b^2 \sqrt [4]{a+b x}}-\frac {4 (c+d x)^{5/4}}{5 b (a+b x)^{5/4}}-\frac {2 d^{5/4} \arctan \left (\frac {\sqrt [4]{d} \sqrt [4]{a+b x}}{\sqrt [4]{b} \sqrt [4]{c+d x}}\right )}{b^{9/4}}+\frac {2 d^{5/4} \text {arctanh}\left (\frac {\sqrt [4]{d} \sqrt [4]{a+b x}}{\sqrt [4]{b} \sqrt [4]{c+d x}}\right )}{b^{9/4}} \]

[Out]

-4*d*(d*x+c)^(1/4)/b^2/(b*x+a)^(1/4)-4/5*(d*x+c)^(5/4)/b/(b*x+a)^(5/4)-2*d^(5/4)*arctan(d^(1/4)*(b*x+a)^(1/4)/
b^(1/4)/(d*x+c)^(1/4))/b^(9/4)+2*d^(5/4)*arctanh(d^(1/4)*(b*x+a)^(1/4)/b^(1/4)/(d*x+c)^(1/4))/b^(9/4)

Rubi [A] (verified)

Time = 0.06 (sec) , antiderivative size = 134, normalized size of antiderivative = 1.00, number of steps used = 7, number of rules used = 6, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.316, Rules used = {49, 65, 338, 304, 211, 214} \[ \int \frac {(c+d x)^{5/4}}{(a+b x)^{9/4}} \, dx=-\frac {2 d^{5/4} \arctan \left (\frac {\sqrt [4]{d} \sqrt [4]{a+b x}}{\sqrt [4]{b} \sqrt [4]{c+d x}}\right )}{b^{9/4}}+\frac {2 d^{5/4} \text {arctanh}\left (\frac {\sqrt [4]{d} \sqrt [4]{a+b x}}{\sqrt [4]{b} \sqrt [4]{c+d x}}\right )}{b^{9/4}}-\frac {4 d \sqrt [4]{c+d x}}{b^2 \sqrt [4]{a+b x}}-\frac {4 (c+d x)^{5/4}}{5 b (a+b x)^{5/4}} \]

[In]

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

[Out]

(-4*d*(c + d*x)^(1/4))/(b^2*(a + b*x)^(1/4)) - (4*(c + d*x)^(5/4))/(5*b*(a + b*x)^(5/4)) - (2*d^(5/4)*ArcTan[(
d^(1/4)*(a + b*x)^(1/4))/(b^(1/4)*(c + d*x)^(1/4))])/b^(9/4) + (2*d^(5/4)*ArcTanh[(d^(1/4)*(a + b*x)^(1/4))/(b
^(1/4)*(c + d*x)^(1/4))])/b^(9/4)

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 304

Int[(x_)^2/((a_) + (b_.)*(x_)^4), x_Symbol] :> With[{r = Numerator[Rt[-a/b, 2]], s = Denominator[Rt[-a/b, 2]]}
, Dist[s/(2*b), Int[1/(r + s*x^2), x], x] - Dist[s/(2*b), Int[1/(r - s*x^2), x], x]] /; FreeQ[{a, b}, x] &&  !
GtQ[a/b, 0]

Rule 338

Int[(x_)^(m_.)*((a_) + (b_.)*(x_)^(n_))^(p_), x_Symbol] :> Dist[a^(p + (m + 1)/n), Subst[Int[x^m/(1 - b*x^n)^(
p + (m + 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)] && IntegersQ[m, p + (m + 1)/n]

Rubi steps \begin{align*} \text {integral}& = -\frac {4 (c+d x)^{5/4}}{5 b (a+b x)^{5/4}}+\frac {d \int \frac {\sqrt [4]{c+d x}}{(a+b x)^{5/4}} \, dx}{b} \\ & = -\frac {4 d \sqrt [4]{c+d x}}{b^2 \sqrt [4]{a+b x}}-\frac {4 (c+d x)^{5/4}}{5 b (a+b x)^{5/4}}+\frac {d^2 \int \frac {1}{\sqrt [4]{a+b x} (c+d x)^{3/4}} \, dx}{b^2} \\ & = -\frac {4 d \sqrt [4]{c+d x}}{b^2 \sqrt [4]{a+b x}}-\frac {4 (c+d x)^{5/4}}{5 b (a+b x)^{5/4}}+\frac {\left (4 d^2\right ) \text {Subst}\left (\int \frac {x^2}{\left (c-\frac {a d}{b}+\frac {d x^4}{b}\right )^{3/4}} \, dx,x,\sqrt [4]{a+b x}\right )}{b^3} \\ & = -\frac {4 d \sqrt [4]{c+d x}}{b^2 \sqrt [4]{a+b x}}-\frac {4 (c+d x)^{5/4}}{5 b (a+b x)^{5/4}}+\frac {\left (4 d^2\right ) \text {Subst}\left (\int \frac {x^2}{1-\frac {d x^4}{b}} \, dx,x,\frac {\sqrt [4]{a+b x}}{\sqrt [4]{c+d x}}\right )}{b^3} \\ & = -\frac {4 d \sqrt [4]{c+d x}}{b^2 \sqrt [4]{a+b x}}-\frac {4 (c+d x)^{5/4}}{5 b (a+b x)^{5/4}}+\frac {\left (2 d^{3/2}\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 )}{b^2}-\frac {\left (2 d^{3/2}\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 )}{b^2} \\ & = -\frac {4 d \sqrt [4]{c+d x}}{b^2 \sqrt [4]{a+b x}}-\frac {4 (c+d x)^{5/4}}{5 b (a+b x)^{5/4}}-\frac {2 d^{5/4} \tan ^{-1}\left (\frac {\sqrt [4]{d} \sqrt [4]{a+b x}}{\sqrt [4]{b} \sqrt [4]{c+d x}}\right )}{b^{9/4}}+\frac {2 d^{5/4} \tanh ^{-1}\left (\frac {\sqrt [4]{d} \sqrt [4]{a+b x}}{\sqrt [4]{b} \sqrt [4]{c+d x}}\right )}{b^{9/4}} \\ \end{align*}

Mathematica [A] (verified)

Time = 0.25 (sec) , antiderivative size = 123, normalized size of antiderivative = 0.92 \[ \int \frac {(c+d x)^{5/4}}{(a+b x)^{9/4}} \, dx=-\frac {4 \sqrt [4]{c+d x} (b c+5 a d+6 b d x)}{5 b^2 (a+b x)^{5/4}}+\frac {2 d^{5/4} \arctan \left (\frac {\sqrt [4]{b} \sqrt [4]{c+d x}}{\sqrt [4]{d} \sqrt [4]{a+b x}}\right )}{b^{9/4}}+\frac {2 d^{5/4} \text {arctanh}\left (\frac {\sqrt [4]{b} \sqrt [4]{c+d x}}{\sqrt [4]{d} \sqrt [4]{a+b x}}\right )}{b^{9/4}} \]

[In]

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

[Out]

(-4*(c + d*x)^(1/4)*(b*c + 5*a*d + 6*b*d*x))/(5*b^2*(a + b*x)^(5/4)) + (2*d^(5/4)*ArcTan[(b^(1/4)*(c + d*x)^(1
/4))/(d^(1/4)*(a + b*x)^(1/4))])/b^(9/4) + (2*d^(5/4)*ArcTanh[(b^(1/4)*(c + d*x)^(1/4))/(d^(1/4)*(a + b*x)^(1/
4))])/b^(9/4)

Maple [F]

\[\int \frac {\left (d x +c \right )^{\frac {5}{4}}}{\left (b x +a \right )^{\frac {9}{4}}}d x\]

[In]

int((d*x+c)^(5/4)/(b*x+a)^(9/4),x)

[Out]

int((d*x+c)^(5/4)/(b*x+a)^(9/4),x)

Fricas [C] (verification not implemented)

Result contains complex when optimal does not.

Time = 0.26 (sec) , antiderivative size = 385, normalized size of antiderivative = 2.87 \[ \int \frac {(c+d x)^{5/4}}{(a+b x)^{9/4}} \, dx=\frac {5 \, {\left (b^{4} x^{2} + 2 \, a b^{3} x + a^{2} b^{2}\right )} \left (\frac {d^{5}}{b^{9}}\right )^{\frac {1}{4}} \log \left (\frac {{\left (b x + a\right )}^{\frac {3}{4}} {\left (d x + c\right )}^{\frac {1}{4}} d + {\left (b^{3} x + a b^{2}\right )} \left (\frac {d^{5}}{b^{9}}\right )^{\frac {1}{4}}}{b x + a}\right ) - 5 \, {\left (b^{4} x^{2} + 2 \, a b^{3} x + a^{2} b^{2}\right )} \left (\frac {d^{5}}{b^{9}}\right )^{\frac {1}{4}} \log \left (\frac {{\left (b x + a\right )}^{\frac {3}{4}} {\left (d x + c\right )}^{\frac {1}{4}} d - {\left (b^{3} x + a b^{2}\right )} \left (\frac {d^{5}}{b^{9}}\right )^{\frac {1}{4}}}{b x + a}\right ) - 5 \, {\left (-i \, b^{4} x^{2} - 2 i \, a b^{3} x - i \, a^{2} b^{2}\right )} \left (\frac {d^{5}}{b^{9}}\right )^{\frac {1}{4}} \log \left (\frac {{\left (b x + a\right )}^{\frac {3}{4}} {\left (d x + c\right )}^{\frac {1}{4}} d + {\left (i \, b^{3} x + i \, a b^{2}\right )} \left (\frac {d^{5}}{b^{9}}\right )^{\frac {1}{4}}}{b x + a}\right ) - 5 \, {\left (i \, b^{4} x^{2} + 2 i \, a b^{3} x + i \, a^{2} b^{2}\right )} \left (\frac {d^{5}}{b^{9}}\right )^{\frac {1}{4}} \log \left (\frac {{\left (b x + a\right )}^{\frac {3}{4}} {\left (d x + c\right )}^{\frac {1}{4}} d + {\left (-i \, b^{3} x - i \, a b^{2}\right )} \left (\frac {d^{5}}{b^{9}}\right )^{\frac {1}{4}}}{b x + a}\right ) - 4 \, {\left (6 \, b d x + b c + 5 \, a d\right )} {\left (b x + a\right )}^{\frac {3}{4}} {\left (d x + c\right )}^{\frac {1}{4}}}{5 \, {\left (b^{4} x^{2} + 2 \, a b^{3} x + a^{2} b^{2}\right )}} \]

[In]

integrate((d*x+c)^(5/4)/(b*x+a)^(9/4),x, algorithm="fricas")

[Out]

1/5*(5*(b^4*x^2 + 2*a*b^3*x + a^2*b^2)*(d^5/b^9)^(1/4)*log(((b*x + a)^(3/4)*(d*x + c)^(1/4)*d + (b^3*x + a*b^2
)*(d^5/b^9)^(1/4))/(b*x + a)) - 5*(b^4*x^2 + 2*a*b^3*x + a^2*b^2)*(d^5/b^9)^(1/4)*log(((b*x + a)^(3/4)*(d*x +
c)^(1/4)*d - (b^3*x + a*b^2)*(d^5/b^9)^(1/4))/(b*x + a)) - 5*(-I*b^4*x^2 - 2*I*a*b^3*x - I*a^2*b^2)*(d^5/b^9)^
(1/4)*log(((b*x + a)^(3/4)*(d*x + c)^(1/4)*d + (I*b^3*x + I*a*b^2)*(d^5/b^9)^(1/4))/(b*x + a)) - 5*(I*b^4*x^2
+ 2*I*a*b^3*x + I*a^2*b^2)*(d^5/b^9)^(1/4)*log(((b*x + a)^(3/4)*(d*x + c)^(1/4)*d + (-I*b^3*x - I*a*b^2)*(d^5/
b^9)^(1/4))/(b*x + a)) - 4*(6*b*d*x + b*c + 5*a*d)*(b*x + a)^(3/4)*(d*x + c)^(1/4))/(b^4*x^2 + 2*a*b^3*x + a^2
*b^2)

Sympy [F]

\[ \int \frac {(c+d x)^{5/4}}{(a+b x)^{9/4}} \, dx=\int \frac {\left (c + d x\right )^{\frac {5}{4}}}{\left (a + b x\right )^{\frac {9}{4}}}\, dx \]

[In]

integrate((d*x+c)**(5/4)/(b*x+a)**(9/4),x)

[Out]

Integral((c + d*x)**(5/4)/(a + b*x)**(9/4), x)

Maxima [F]

\[ \int \frac {(c+d x)^{5/4}}{(a+b x)^{9/4}} \, dx=\int { \frac {{\left (d x + c\right )}^{\frac {5}{4}}}{{\left (b x + a\right )}^{\frac {9}{4}}} \,d x } \]

[In]

integrate((d*x+c)^(5/4)/(b*x+a)^(9/4),x, algorithm="maxima")

[Out]

integrate((d*x + c)^(5/4)/(b*x + a)^(9/4), x)

Giac [F]

\[ \int \frac {(c+d x)^{5/4}}{(a+b x)^{9/4}} \, dx=\int { \frac {{\left (d x + c\right )}^{\frac {5}{4}}}{{\left (b x + a\right )}^{\frac {9}{4}}} \,d x } \]

[In]

integrate((d*x+c)^(5/4)/(b*x+a)^(9/4),x, algorithm="giac")

[Out]

integrate((d*x + c)^(5/4)/(b*x + a)^(9/4), x)

Mupad [F(-1)]

Timed out. \[ \int \frac {(c+d x)^{5/4}}{(a+b x)^{9/4}} \, dx=\int \frac {{\left (c+d\,x\right )}^{5/4}}{{\left (a+b\,x\right )}^{9/4}} \,d x \]

[In]

int((c + d*x)^(5/4)/(a + b*x)^(9/4),x)

[Out]

int((c + d*x)^(5/4)/(a + b*x)^(9/4), x)