\(\int \frac {(c+d x)^{5/4}}{(a+b x)^{5/4}} \, dx\) [575]

Optimal result

Integrand size = 19, antiderivative size = 152 \[ \int \frac {(c+d x)^{5/4}}{(a+b x)^{5/4}} \, dx=\frac {5 d (a+b x)^{3/4} \sqrt [4]{c+d x}}{b^2}-\frac {4 (c+d x)^{5/4}}{b \sqrt [4]{a+b x}}-\frac {5 \sqrt [4]{d} (b c-a d) \arctan \left (\frac {\sqrt [4]{d} \sqrt [4]{a+b x}}{\sqrt [4]{b} \sqrt [4]{c+d x}}\right )}{2 b^{9/4}}+\frac {5 \sqrt [4]{d} (b c-a d) \text {arctanh}\left (\frac {\sqrt [4]{d} \sqrt [4]{a+b x}}{\sqrt [4]{b} \sqrt [4]{c+d x}}\right )}{2 b^{9/4}} \] Output:

5*d*(b*x+a)^(3/4)*(d*x+c)^(1/4)/b^2-4*(d*x+c)^(5/4)/b/(b*x+a)^(1/4)-5/2*d^ 
(1/4)*(-a*d+b*c)*arctan(d^(1/4)*(b*x+a)^(1/4)/b^(1/4)/(d*x+c)^(1/4))/b^(9/ 
4)+5/2*d^(1/4)*(-a*d+b*c)*arctanh(d^(1/4)*(b*x+a)^(1/4)/b^(1/4)/(d*x+c)^(1 
/4))/b^(9/4)
 

Mathematica [A] (verified)

Time = 0.30 (sec) , antiderivative size = 140, normalized size of antiderivative = 0.92 \[ \int \frac {(c+d x)^{5/4}}{(a+b x)^{5/4}} \, dx=\frac {\sqrt [4]{c+d x} (-4 b c+5 a d+b d x)}{b^2 \sqrt [4]{a+b x}}+\frac {5 \sqrt [4]{d} (b c-a d) \arctan \left (\frac {\sqrt [4]{b} \sqrt [4]{c+d x}}{\sqrt [4]{d} \sqrt [4]{a+b x}}\right )}{2 b^{9/4}}+\frac {5 \sqrt [4]{d} (b c-a d) \text {arctanh}\left (\frac {\sqrt [4]{b} \sqrt [4]{c+d x}}{\sqrt [4]{d} \sqrt [4]{a+b x}}\right )}{2 b^{9/4}} \] Input:

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

Output:

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

Rubi [A] (verified)

Time = 0.25 (sec) , antiderivative size = 182, normalized size of antiderivative = 1.20, number of steps used = 9, number of rules used = 8, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.421, Rules used = {57, 60, 73, 854, 27, 827, 218, 221}

Below are the steps used by Rubi to obtain the solution. The rule number used for the transformation is given above next to the arrow. The rules definitions used are listed below.

\(\displaystyle \int \frac {(c+d x)^{5/4}}{(a+b x)^{5/4}} \, dx\)

\(\Big \downarrow \) 57

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

\(\Big \downarrow \) 60

\(\displaystyle \frac {5 d \left (\frac {(b c-a d) \int \frac {1}{\sqrt [4]{a+b x} (c+d x)^{3/4}}dx}{4 b}+\frac {(a+b x)^{3/4} \sqrt [4]{c+d x}}{b}\right )}{b}-\frac {4 (c+d x)^{5/4}}{b \sqrt [4]{a+b x}}\)

\(\Big \downarrow \) 73

\(\displaystyle \frac {5 d \left (\frac {(b c-a d) \int \frac {\sqrt {a+b x}}{\left (c-\frac {a d}{b}+\frac {d (a+b x)}{b}\right )^{3/4}}d\sqrt [4]{a+b x}}{b^2}+\frac {(a+b x)^{3/4} \sqrt [4]{c+d x}}{b}\right )}{b}-\frac {4 (c+d x)^{5/4}}{b \sqrt [4]{a+b x}}\)

\(\Big \downarrow \) 854

\(\displaystyle \frac {5 d \left (\frac {(b c-a d) \int \frac {b \sqrt {a+b x}}{b-d (a+b x)}d\frac {\sqrt [4]{a+b x}}{\sqrt [4]{c-\frac {a d}{b}+\frac {d (a+b x)}{b}}}}{b^2}+\frac {(a+b x)^{3/4} \sqrt [4]{c+d x}}{b}\right )}{b}-\frac {4 (c+d x)^{5/4}}{b \sqrt [4]{a+b x}}\)

\(\Big \downarrow \) 27

\(\displaystyle \frac {5 d \left (\frac {(b c-a d) \int \frac {\sqrt {a+b x}}{b-d (a+b x)}d\frac {\sqrt [4]{a+b x}}{\sqrt [4]{c-\frac {a d}{b}+\frac {d (a+b x)}{b}}}}{b}+\frac {(a+b x)^{3/4} \sqrt [4]{c+d x}}{b}\right )}{b}-\frac {4 (c+d x)^{5/4}}{b \sqrt [4]{a+b x}}\)

\(\Big \downarrow \) 827

\(\displaystyle \frac {5 d \left (\frac {(b c-a d) \left (\frac {\int \frac {1}{\sqrt {b}-\sqrt {d} \sqrt {a+b x}}d\frac {\sqrt [4]{a+b x}}{\sqrt [4]{c-\frac {a d}{b}+\frac {d (a+b x)}{b}}}}{2 \sqrt {d}}-\frac {\int \frac {1}{\sqrt {b}+\sqrt {d} \sqrt {a+b x}}d\frac {\sqrt [4]{a+b x}}{\sqrt [4]{c-\frac {a d}{b}+\frac {d (a+b x)}{b}}}}{2 \sqrt {d}}\right )}{b}+\frac {(a+b x)^{3/4} \sqrt [4]{c+d x}}{b}\right )}{b}-\frac {4 (c+d x)^{5/4}}{b \sqrt [4]{a+b x}}\)

\(\Big \downarrow \) 218

\(\displaystyle \frac {5 d \left (\frac {(b c-a d) \left (\frac {\int \frac {1}{\sqrt {b}-\sqrt {d} \sqrt {a+b x}}d\frac {\sqrt [4]{a+b x}}{\sqrt [4]{c-\frac {a d}{b}+\frac {d (a+b x)}{b}}}}{2 \sqrt {d}}-\frac {\arctan \left (\frac {\sqrt [4]{d} \sqrt [4]{a+b x}}{\sqrt [4]{b} \sqrt [4]{\frac {d (a+b x)}{b}-\frac {a d}{b}+c}}\right )}{2 \sqrt [4]{b} d^{3/4}}\right )}{b}+\frac {(a+b x)^{3/4} \sqrt [4]{c+d x}}{b}\right )}{b}-\frac {4 (c+d x)^{5/4}}{b \sqrt [4]{a+b x}}\)

\(\Big \downarrow \) 221

\(\displaystyle \frac {5 d \left (\frac {(b c-a d) \left (\frac {\text {arctanh}\left (\frac {\sqrt [4]{d} \sqrt [4]{a+b x}}{\sqrt [4]{b} \sqrt [4]{\frac {d (a+b x)}{b}-\frac {a d}{b}+c}}\right )}{2 \sqrt [4]{b} d^{3/4}}-\frac {\arctan \left (\frac {\sqrt [4]{d} \sqrt [4]{a+b x}}{\sqrt [4]{b} \sqrt [4]{\frac {d (a+b x)}{b}-\frac {a d}{b}+c}}\right )}{2 \sqrt [4]{b} d^{3/4}}\right )}{b}+\frac {(a+b x)^{3/4} \sqrt [4]{c+d x}}{b}\right )}{b}-\frac {4 (c+d x)^{5/4}}{b \sqrt [4]{a+b x}}\)

Input:

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

Output:

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

Defintions of rubi rules used

rule 27

Int[(a_)*(Fx_), x_Symbol] :> Simp[a   Int[Fx, x], x] /; FreeQ[a, x] &&  !Ma 
tchQ[Fx, (b_)*(Gx_) /; FreeQ[b, x]]
 

rule 57

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] - Simp[d*(n/(b*(m + 1))) 
Int[(a + b*x)^(m + 1)*(c + d*x)^(n - 1), x], x] /; FreeQ[{a, b, c, d}, x] & 
& 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 60

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] + Simp[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] && GtQ[n, 0] && NeQ[m + n + 1, 0] &&  !(IGtQ[m, 0] && ( !Integer 
Q[n] || (GtQ[m, 0] && LtQ[m - n, 0]))) &&  !ILtQ[m + n + 2, 0] && IntLinear 
Q[a, b, c, d, m, n, x]
 

rule 73

Int[((a_.) + (b_.)*(x_))^(m_)*((c_.) + (d_.)*(x_))^(n_), x_Symbol] :> With[ 
{p = Denominator[m]}, Simp[p/b   Subst[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] && Lt 
Q[-1, m, 0] && LeQ[-1, n, 0] && LeQ[Denominator[n], Denominator[m]] && IntL 
inearQ[a, b, c, d, m, n, x]
 

rule 218

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

rule 221

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 827

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

rule 854

Int[(x_)^(m_.)*((a_) + (b_.)*(x_)^(n_))^(p_), x_Symbol] :> Simp[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]
 

Maple [F]

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

Input:

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

Output:

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

Fricas [C] (verification not implemented)

Result contains complex when optimal does not.

Time = 0.09 (sec) , antiderivative size = 754, normalized size of antiderivative = 4.96 \[ \int \frac {(c+d x)^{5/4}}{(a+b x)^{5/4}} \, dx =\text {Too large to display} \] Input:

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

Output:

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

Sympy [F]

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

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

Output:

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

Maxima [F]

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

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

Output:

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

Giac [F]

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

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

Output:

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

Mupad [F(-1)]

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

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

Output:

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

Reduce [F]

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

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

Output:

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