\(\int \frac {(c+d x)^{5/4}}{(a+b x)^{11/4}} \, dx\) [585]

Optimal result

Integrand size = 19, antiderivative size = 149 \[ \int \frac {(c+d x)^{5/4}}{(a+b x)^{11/4}} \, dx=-\frac {20 d \sqrt [4]{c+d x}}{21 b^2 (a+b x)^{3/4}}-\frac {4 (c+d x)^{5/4}}{7 b (a+b x)^{7/4}}-\frac {20 d^{5/2} (a+b x)^{3/4} \left (\frac {b (c+d x)}{d (a+b x)}\right )^{3/4} \operatorname {EllipticF}\left (\frac {1}{2} \arctan \left (\frac {\sqrt {b c-a d}}{\sqrt {d} \sqrt {a+b x}}\right ),2\right )}{21 b^3 \sqrt {b c-a d} (c+d x)^{3/4}} \] Output:

-20/21*d*(d*x+c)^(1/4)/b^2/(b*x+a)^(3/4)-4/7*(d*x+c)^(5/4)/b/(b*x+a)^(7/4) 
-20/21*d^(5/2)*(b*x+a)^(3/4)*(b*(d*x+c)/d/(b*x+a))^(3/4)*InverseJacobiAM(1 
/2*arctan(1/d^(1/2)/(b*x+a)^(1/2)*(-a*d+b*c)^(1/2)),2^(1/2))/b^3/(-a*d+b*c 
)^(1/2)/(d*x+c)^(3/4)
 

Mathematica [C] (verified)

Result contains higher order function than in optimal. Order 5 vs. order 4 in optimal.

Time = 0.04 (sec) , antiderivative size = 73, normalized size of antiderivative = 0.49 \[ \int \frac {(c+d x)^{5/4}}{(a+b x)^{11/4}} \, dx=-\frac {4 (c+d x)^{5/4} \operatorname {Hypergeometric2F1}\left (-\frac {7}{4},-\frac {5}{4},-\frac {3}{4},\frac {d (a+b x)}{-b c+a d}\right )}{7 b (a+b x)^{7/4} \left (\frac {b (c+d x)}{b c-a d}\right )^{5/4}} \] Input:

Integrate[(c + d*x)^(5/4)/(a + b*x)^(11/4),x]
 

Output:

(-4*(c + d*x)^(5/4)*Hypergeometric2F1[-7/4, -5/4, -3/4, (d*(a + b*x))/(-(b 
*c) + a*d)])/(7*b*(a + b*x)^(7/4)*((b*(c + d*x))/(b*c - a*d))^(5/4))
 

Rubi [A] (warning: unable to verify)

Time = 0.29 (sec) , antiderivative size = 175, normalized size of antiderivative = 1.17, number of steps used = 8, number of rules used = 7, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.368, Rules used = {57, 57, 73, 768, 858, 807, 229}

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.

\(\displaystyle \int \frac {(c+d x)^{5/4}}{(a+b x)^{11/4}} \, dx\)

\(\Big \downarrow \) 57

\(\displaystyle \frac {5 d \int \frac {\sqrt [4]{c+d x}}{(a+b x)^{7/4}}dx}{7 b}-\frac {4 (c+d x)^{5/4}}{7 b (a+b x)^{7/4}}\)

\(\Big \downarrow \) 57

\(\displaystyle \frac {5 d \left (\frac {d \int \frac {1}{(a+b x)^{3/4} (c+d x)^{3/4}}dx}{3 b}-\frac {4 \sqrt [4]{c+d x}}{3 b (a+b x)^{3/4}}\right )}{7 b}-\frac {4 (c+d x)^{5/4}}{7 b (a+b x)^{7/4}}\)

\(\Big \downarrow \) 73

\(\displaystyle \frac {5 d \left (\frac {4 d \int \frac {1}{\left (c-\frac {a d}{b}+\frac {d (a+b x)}{b}\right )^{3/4}}d\sqrt [4]{a+b x}}{3 b^2}-\frac {4 \sqrt [4]{c+d x}}{3 b (a+b x)^{3/4}}\right )}{7 b}-\frac {4 (c+d x)^{5/4}}{7 b (a+b x)^{7/4}}\)

\(\Big \downarrow \) 768

\(\displaystyle \frac {5 d \left (\frac {4 d (a+b x)^{3/4} \left (\frac {b c-a d}{d (a+b x)}+1\right )^{3/4} \int \frac {1}{(a+b x)^{3/4} \left (\frac {b c-a d}{d (a+b x)}+1\right )^{3/4}}d\sqrt [4]{a+b x}}{3 b^2 \left (\frac {d (a+b x)}{b}-\frac {a d}{b}+c\right )^{3/4}}-\frac {4 \sqrt [4]{c+d x}}{3 b (a+b x)^{3/4}}\right )}{7 b}-\frac {4 (c+d x)^{5/4}}{7 b (a+b x)^{7/4}}\)

\(\Big \downarrow \) 858

\(\displaystyle \frac {5 d \left (-\frac {4 d (a+b x)^{3/4} \left (\frac {b c-a d}{d (a+b x)}+1\right )^{3/4} \int \frac {1}{\sqrt [4]{a+b x} \left (\frac {(b c-a d) (a+b x)}{d}+1\right )^{3/4}}d\frac {1}{\sqrt [4]{a+b x}}}{3 b^2 \left (\frac {d (a+b x)}{b}-\frac {a d}{b}+c\right )^{3/4}}-\frac {4 \sqrt [4]{c+d x}}{3 b (a+b x)^{3/4}}\right )}{7 b}-\frac {4 (c+d x)^{5/4}}{7 b (a+b x)^{7/4}}\)

\(\Big \downarrow \) 807

\(\displaystyle \frac {5 d \left (-\frac {2 d (a+b x)^{3/4} \left (\frac {b c-a d}{d (a+b x)}+1\right )^{3/4} \int \frac {1}{\left (\frac {\sqrt {a+b x} (b c-a d)}{d}+1\right )^{3/4}}d\sqrt {a+b x}}{3 b^2 \left (\frac {d (a+b x)}{b}-\frac {a d}{b}+c\right )^{3/4}}-\frac {4 \sqrt [4]{c+d x}}{3 b (a+b x)^{3/4}}\right )}{7 b}-\frac {4 (c+d x)^{5/4}}{7 b (a+b x)^{7/4}}\)

\(\Big \downarrow \) 229

\(\displaystyle \frac {5 d \left (-\frac {4 d^{3/2} (a+b x)^{3/4} \left (\frac {b c-a d}{d (a+b x)}+1\right )^{3/4} \operatorname {EllipticF}\left (\frac {1}{2} \arctan \left (\frac {\sqrt {b c-a d} \sqrt {a+b x}}{\sqrt {d}}\right ),2\right )}{3 b^2 \sqrt {b c-a d} \left (\frac {d (a+b x)}{b}-\frac {a d}{b}+c\right )^{3/4}}-\frac {4 \sqrt [4]{c+d x}}{3 b (a+b x)^{3/4}}\right )}{7 b}-\frac {4 (c+d x)^{5/4}}{7 b (a+b x)^{7/4}}\)

Input:

Int[(c + d*x)^(5/4)/(a + b*x)^(11/4),x]
 

Output:

(-4*(c + d*x)^(5/4))/(7*b*(a + b*x)^(7/4)) + (5*d*((-4*(c + d*x)^(1/4))/(3 
*b*(a + b*x)^(3/4)) - (4*d^(3/2)*(a + b*x)^(3/4)*(1 + (b*c - a*d)/(d*(a + 
b*x)))^(3/4)*EllipticF[ArcTan[(Sqrt[b*c - a*d]*Sqrt[a + b*x])/Sqrt[d]]/2, 
2])/(3*b^2*Sqrt[b*c - a*d]*(c - (a*d)/b + (d*(a + b*x))/b)^(3/4))))/(7*b)
 

Defintions of rubi rules used

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 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 229

Int[((a_) + (b_.)*(x_)^2)^(-3/4), x_Symbol] :> Simp[(2/(a^(3/4)*Rt[b/a, 2]) 
)*EllipticF[(1/2)*ArcTan[Rt[b/a, 2]*x], 2], x] /; FreeQ[{a, b}, x] && GtQ[a 
, 0] && PosQ[b/a]
 

rule 768

Int[((a_) + (b_.)*(x_)^4)^(-3/4), x_Symbol] :> Simp[x^3*((1 + a/(b*x^4))^(3 
/4)/(a + b*x^4)^(3/4))   Int[1/(x^3*(1 + a/(b*x^4))^(3/4)), x], x] /; FreeQ 
[{a, b}, x]
 

rule 807

Int[(x_)^(m_.)*((a_) + (b_.)*(x_)^(n_))^(p_), x_Symbol] :> With[{k = GCD[m 
+ 1, n]}, Simp[1/k   Subst[Int[x^((m + 1)/k - 1)*(a + b*x^(n/k))^p, x], x, 
x^k], x] /; k != 1] /; FreeQ[{a, b, p}, x] && IGtQ[n, 0] && IntegerQ[m]
 

rule 858

Int[(x_)^(m_.)*((a_) + (b_.)*(x_)^(n_))^(p_), x_Symbol] :> -Subst[Int[(a + 
b/x^n)^p/x^(m + 2), x], x, 1/x] /; FreeQ[{a, b, p}, x] && ILtQ[n, 0] && Int 
egerQ[m]
 

Maple [F]

\[\int \frac {\left (x d +c \right )^{\frac {5}{4}}}{\left (b x +a \right )^{\frac {11}{4}}}d x\]

Input:

int((d*x+c)^(5/4)/(b*x+a)^(11/4),x)
 

Output:

int((d*x+c)^(5/4)/(b*x+a)^(11/4),x)
 

Fricas [F]

\[ \int \frac {(c+d x)^{5/4}}{(a+b x)^{11/4}} \, dx=\int { \frac {{\left (d x + c\right )}^{\frac {5}{4}}}{{\left (b x + a\right )}^{\frac {11}{4}}} \,d x } \] Input:

integrate((d*x+c)^(5/4)/(b*x+a)^(11/4),x, algorithm="fricas")
 

Output:

integral((b*x + a)^(1/4)*(d*x + c)^(5/4)/(b^3*x^3 + 3*a*b^2*x^2 + 3*a^2*b* 
x + a^3), x)
 

Sympy [F]

\[ \int \frac {(c+d x)^{5/4}}{(a+b x)^{11/4}} \, dx=\int \frac {\left (c + d x\right )^{\frac {5}{4}}}{\left (a + b x\right )^{\frac {11}{4}}}\, dx \] Input:

integrate((d*x+c)**(5/4)/(b*x+a)**(11/4),x)
 

Output:

Integral((c + d*x)**(5/4)/(a + b*x)**(11/4), x)
 

Maxima [F]

\[ \int \frac {(c+d x)^{5/4}}{(a+b x)^{11/4}} \, dx=\int { \frac {{\left (d x + c\right )}^{\frac {5}{4}}}{{\left (b x + a\right )}^{\frac {11}{4}}} \,d x } \] Input:

integrate((d*x+c)^(5/4)/(b*x+a)^(11/4),x, algorithm="maxima")
 

Output:

integrate((d*x + c)^(5/4)/(b*x + a)^(11/4), x)
 

Giac [F]

\[ \int \frac {(c+d x)^{5/4}}{(a+b x)^{11/4}} \, dx=\int { \frac {{\left (d x + c\right )}^{\frac {5}{4}}}{{\left (b x + a\right )}^{\frac {11}{4}}} \,d x } \] Input:

integrate((d*x+c)^(5/4)/(b*x+a)^(11/4),x, algorithm="giac")
 

Output:

integrate((d*x + c)^(5/4)/(b*x + a)^(11/4), x)
 

Mupad [F(-1)]

Timed out. \[ \int \frac {(c+d x)^{5/4}}{(a+b x)^{11/4}} \, dx=\int \frac {{\left (c+d\,x\right )}^{5/4}}{{\left (a+b\,x\right )}^{11/4}} \,d x \] Input:

int((c + d*x)^(5/4)/(a + b*x)^(11/4),x)
 

Output:

int((c + d*x)^(5/4)/(a + b*x)^(11/4), x)
 

Reduce [F]

\[ \int \frac {(c+d x)^{5/4}}{(a+b x)^{11/4}} \, dx=\left (\int \frac {\left (d x +c \right )^{\frac {1}{4}}}{\left (b x +a \right )^{\frac {3}{4}} a^{2}+2 \left (b x +a \right )^{\frac {3}{4}} a b x +\left (b x +a \right )^{\frac {3}{4}} b^{2} x^{2}}d x \right ) c +\left (\int \frac {\left (d x +c \right )^{\frac {1}{4}} x}{\left (b x +a \right )^{\frac {3}{4}} a^{2}+2 \left (b x +a \right )^{\frac {3}{4}} a b x +\left (b x +a \right )^{\frac {3}{4}} b^{2} x^{2}}d x \right ) d \] Input:

int((d*x+c)^(5/4)/(b*x+a)^(11/4),x)
                                                                                    
                                                                                    
 

Output:

int((c + d*x)**(1/4)/((a + b*x)**(3/4)*a**2 + 2*(a + b*x)**(3/4)*a*b*x + ( 
a + b*x)**(3/4)*b**2*x**2),x)*c + int(((c + d*x)**(1/4)*x)/((a + b*x)**(3/ 
4)*a**2 + 2*(a + b*x)**(3/4)*a*b*x + (a + b*x)**(3/4)*b**2*x**2),x)*d