\(\int \frac {1}{x^{15/4} (a+b x)^{3/4}} \, dx\) [726]

Optimal result

Integrand size = 15, antiderivative size = 130 \[ \int \frac {1}{x^{15/4} (a+b x)^{3/4}} \, dx=-\frac {4 \sqrt [4]{a+b x}}{11 a x^{11/4}}+\frac {40 b \sqrt [4]{a+b x}}{77 a^2 x^{7/4}}-\frac {80 b^2 \sqrt [4]{a+b x}}{77 a^3 x^{3/4}}+\frac {160 b^2 \left (\frac {b x}{a+b x}\right )^{3/4} (a+b x)^{3/4} \operatorname {EllipticF}\left (\frac {1}{2} \arcsin \left (\frac {\sqrt {a}}{\sqrt {a+b x}}\right ),2\right )}{77 a^{7/2} x^{3/4}} \] Output:

-4/11*(b*x+a)^(1/4)/a/x^(11/4)+40/77*b*(b*x+a)^(1/4)/a^2/x^(7/4)-80/77*b^2 
*(b*x+a)^(1/4)/a^3/x^(3/4)+160/77*b^2*(b*x/(b*x+a))^(3/4)*(b*x+a)^(3/4)*In 
verseJacobiAM(1/2*arcsin(a^(1/2)/(b*x+a)^(1/2)),2^(1/2))/a^(7/2)/x^(3/4)
 

Mathematica [C] (verified)

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

Time = 10.01 (sec) , antiderivative size = 47, normalized size of antiderivative = 0.36 \[ \int \frac {1}{x^{15/4} (a+b x)^{3/4}} \, dx=-\frac {4 \left (1+\frac {b x}{a}\right )^{3/4} \operatorname {Hypergeometric2F1}\left (-\frac {11}{4},\frac {3}{4},-\frac {7}{4},-\frac {b x}{a}\right )}{11 x^{11/4} (a+b x)^{3/4}} \] Input:

Integrate[1/(x^(15/4)*(a + b*x)^(3/4)),x]
 

Output:

(-4*(1 + (b*x)/a)^(3/4)*Hypergeometric2F1[-11/4, 3/4, -7/4, -((b*x)/a)])/( 
11*x^(11/4)*(a + b*x)^(3/4))
 

Rubi [A] (warning: unable to verify)

Time = 0.25 (sec) , antiderivative size = 147, normalized size of antiderivative = 1.13, number of steps used = 9, number of rules used = 8, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.533, Rules used = {61, 61, 61, 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.

\(\displaystyle \int \frac {1}{x^{15/4} (a+b x)^{3/4}} \, dx\)

\(\Big \downarrow \) 61

\(\displaystyle -\frac {10 b \int \frac {1}{x^{11/4} (a+b x)^{3/4}}dx}{11 a}-\frac {4 \sqrt [4]{a+b x}}{11 a x^{11/4}}\)

\(\Big \downarrow \) 61

\(\displaystyle -\frac {10 b \left (-\frac {6 b \int \frac {1}{x^{7/4} (a+b x)^{3/4}}dx}{7 a}-\frac {4 \sqrt [4]{a+b x}}{7 a x^{7/4}}\right )}{11 a}-\frac {4 \sqrt [4]{a+b x}}{11 a x^{11/4}}\)

\(\Big \downarrow \) 61

\(\displaystyle -\frac {10 b \left (-\frac {6 b \left (-\frac {2 b \int \frac {1}{x^{3/4} (a+b x)^{3/4}}dx}{3 a}-\frac {4 \sqrt [4]{a+b x}}{3 a x^{3/4}}\right )}{7 a}-\frac {4 \sqrt [4]{a+b x}}{7 a x^{7/4}}\right )}{11 a}-\frac {4 \sqrt [4]{a+b x}}{11 a x^{11/4}}\)

\(\Big \downarrow \) 73

\(\displaystyle -\frac {10 b \left (-\frac {6 b \left (-\frac {8 b \int \frac {1}{(a+b x)^{3/4}}d\sqrt [4]{x}}{3 a}-\frac {4 \sqrt [4]{a+b x}}{3 a x^{3/4}}\right )}{7 a}-\frac {4 \sqrt [4]{a+b x}}{7 a x^{7/4}}\right )}{11 a}-\frac {4 \sqrt [4]{a+b x}}{11 a x^{11/4}}\)

\(\Big \downarrow \) 768

\(\displaystyle -\frac {10 b \left (-\frac {6 b \left (-\frac {8 b x^{3/4} \left (\frac {a}{b x}+1\right )^{3/4} \int \frac {1}{\left (\frac {a}{b x}+1\right )^{3/4} x^{3/4}}d\sqrt [4]{x}}{3 a (a+b x)^{3/4}}-\frac {4 \sqrt [4]{a+b x}}{3 a x^{3/4}}\right )}{7 a}-\frac {4 \sqrt [4]{a+b x}}{7 a x^{7/4}}\right )}{11 a}-\frac {4 \sqrt [4]{a+b x}}{11 a x^{11/4}}\)

\(\Big \downarrow \) 858

\(\displaystyle -\frac {10 b \left (-\frac {6 b \left (\frac {8 b x^{3/4} \left (\frac {a}{b x}+1\right )^{3/4} \int \frac {1}{\sqrt [4]{x} \left (\frac {a x}{b}+1\right )^{3/4}}d\frac {1}{\sqrt [4]{x}}}{3 a (a+b x)^{3/4}}-\frac {4 \sqrt [4]{a+b x}}{3 a x^{3/4}}\right )}{7 a}-\frac {4 \sqrt [4]{a+b x}}{7 a x^{7/4}}\right )}{11 a}-\frac {4 \sqrt [4]{a+b x}}{11 a x^{11/4}}\)

\(\Big \downarrow \) 807

\(\displaystyle -\frac {10 b \left (-\frac {6 b \left (\frac {4 b x^{3/4} \left (\frac {a}{b x}+1\right )^{3/4} \int \frac {1}{\left (\frac {\sqrt {x} a}{b}+1\right )^{3/4}}d\sqrt {x}}{3 a (a+b x)^{3/4}}-\frac {4 \sqrt [4]{a+b x}}{3 a x^{3/4}}\right )}{7 a}-\frac {4 \sqrt [4]{a+b x}}{7 a x^{7/4}}\right )}{11 a}-\frac {4 \sqrt [4]{a+b x}}{11 a x^{11/4}}\)

\(\Big \downarrow \) 229

\(\displaystyle -\frac {10 b \left (-\frac {6 b \left (\frac {8 b^{3/2} x^{3/4} \left (\frac {a}{b x}+1\right )^{3/4} \operatorname {EllipticF}\left (\frac {1}{2} \arctan \left (\frac {\sqrt {a} \sqrt {x}}{\sqrt {b}}\right ),2\right )}{3 a^{3/2} (a+b x)^{3/4}}-\frac {4 \sqrt [4]{a+b x}}{3 a x^{3/4}}\right )}{7 a}-\frac {4 \sqrt [4]{a+b x}}{7 a x^{7/4}}\right )}{11 a}-\frac {4 \sqrt [4]{a+b x}}{11 a x^{11/4}}\)

Input:

Int[1/(x^(15/4)*(a + b*x)^(3/4)),x]
 

Output:

(-4*(a + b*x)^(1/4))/(11*a*x^(11/4)) - (10*b*((-4*(a + b*x)^(1/4))/(7*a*x^ 
(7/4)) - (6*b*((-4*(a + b*x)^(1/4))/(3*a*x^(3/4)) + (8*b^(3/2)*(1 + a/(b*x 
))^(3/4)*x^(3/4)*EllipticF[ArcTan[(Sqrt[a]*Sqrt[x])/Sqrt[b]]/2, 2])/(3*a^( 
3/2)*(a + b*x)^(3/4))))/(7*a)))/(11*a)
 

Defintions of rubi rules used

rule 61

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

\[\int \frac {1}{x^{\frac {15}{4}} \left (b x +a \right )^{\frac {3}{4}}}d x\]

Input:

int(1/x^(15/4)/(b*x+a)^(3/4),x)
 

Output:

int(1/x^(15/4)/(b*x+a)^(3/4),x)
 

Fricas [F]

\[ \int \frac {1}{x^{15/4} (a+b x)^{3/4}} \, dx=\int { \frac {1}{{\left (b x + a\right )}^{\frac {3}{4}} x^{\frac {15}{4}}} \,d x } \] Input:

integrate(1/x^(15/4)/(b*x+a)^(3/4),x, algorithm="fricas")
 

Output:

integral((b*x + a)^(1/4)*x^(1/4)/(b*x^5 + a*x^4), x)
 

Sympy [F(-1)]

Timed out. \[ \int \frac {1}{x^{15/4} (a+b x)^{3/4}} \, dx=\text {Timed out} \] Input:

integrate(1/x**(15/4)/(b*x+a)**(3/4),x)
 

Output:

Timed out
 

Maxima [F]

\[ \int \frac {1}{x^{15/4} (a+b x)^{3/4}} \, dx=\int { \frac {1}{{\left (b x + a\right )}^{\frac {3}{4}} x^{\frac {15}{4}}} \,d x } \] Input:

integrate(1/x^(15/4)/(b*x+a)^(3/4),x, algorithm="maxima")
 

Output:

integrate(1/((b*x + a)^(3/4)*x^(15/4)), x)
 

Giac [F]

\[ \int \frac {1}{x^{15/4} (a+b x)^{3/4}} \, dx=\int { \frac {1}{{\left (b x + a\right )}^{\frac {3}{4}} x^{\frac {15}{4}}} \,d x } \] Input:

integrate(1/x^(15/4)/(b*x+a)^(3/4),x, algorithm="giac")
 

Output:

integrate(1/((b*x + a)^(3/4)*x^(15/4)), x)
 

Mupad [F(-1)]

Timed out. \[ \int \frac {1}{x^{15/4} (a+b x)^{3/4}} \, dx=\int \frac {1}{x^{15/4}\,{\left (a+b\,x\right )}^{3/4}} \,d x \] Input:

int(1/(x^(15/4)*(a + b*x)^(3/4)),x)
 

Output:

int(1/(x^(15/4)*(a + b*x)^(3/4)), x)
 

Reduce [B] (verification not implemented)

Time = 0.16 (sec) , antiderivative size = 38, normalized size of antiderivative = 0.29 \[ \int \frac {1}{x^{15/4} (a+b x)^{3/4}} \, dx=\frac {4 \left (b x +a \right )^{\frac {1}{4}} \left (-32 b^{2} x^{2}+8 a b x -5 a^{2}\right )}{45 x^{\frac {9}{4}} \sqrt {x}\, a^{3}} \] Input:

int(1/x^(15/4)/(b*x+a)^(3/4),x)
 

Output:

(4*x**(3/4)*(a + b*x)**(1/4)*( - 5*a**2 + 8*a*b*x - 32*b**2*x**2))/(45*sqr 
t(x)*a**3*x**3)