3.18.4 \(\int \frac {1}{(a+b x)^{9/4} \sqrt [4]{c+d x}} \, dx\) [1704]

3.18.4.1 Optimal result

Integrand size = 19, antiderivative size = 760 \[ \int \frac {1}{(a+b x)^{9/4} \sqrt [4]{c+d x}} \, dx=-\frac {4 (c+d x)^{3/4}}{5 (b c-a d) (a+b x)^{5/4}}+\frac {8 d (c+d x)^{3/4}}{5 (b c-a d)^2 \sqrt [4]{a+b x}}-\frac {8 d^{3/2} \sqrt {(a+b x) (c+d x)} \sqrt {(b c+a d+2 b d x)^2} \sqrt {(a d+b (c+2 d x))^2}}{5 \sqrt {b} (b c-a d)^3 \sqrt [4]{a+b x} \sqrt [4]{c+d x} (b c+a d+2 b d x) \left (1+\frac {2 \sqrt {b} \sqrt {d} \sqrt {(a+b x) (c+d x)}}{b c-a d}\right )}+\frac {4 \sqrt {2} d^{5/4} \sqrt [4]{(a+b x) (c+d x)} \sqrt {(b c+a d+2 b d x)^2} \left (1+\frac {2 \sqrt {b} \sqrt {d} \sqrt {(a+b x) (c+d x)}}{b c-a d}\right ) \sqrt {\frac {(a d+b (c+2 d x))^2}{(b c-a d)^2 \left (1+\frac {2 \sqrt {b} \sqrt {d} \sqrt {(a+b x) (c+d x)}}{b c-a d}\right )^2}} E\left (2 \arctan \left (\frac {\sqrt {2} \sqrt [4]{b} \sqrt [4]{d} \sqrt [4]{(a+b x) (c+d x)}}{\sqrt {b c-a d}}\right )|\frac {1}{2}\right )}{5 b^{3/4} \sqrt {b c-a d} \sqrt [4]{a+b x} \sqrt [4]{c+d x} (b c+a d+2 b d x) \sqrt {(a d+b (c+2 d x))^2}}-\frac {2 \sqrt {2} d^{5/4} \sqrt [4]{(a+b x) (c+d x)} \sqrt {(b c+a d+2 b d x)^2} \left (1+\frac {2 \sqrt {b} \sqrt {d} \sqrt {(a+b x) (c+d x)}}{b c-a d}\right ) \sqrt {\frac {(a d+b (c+2 d x))^2}{(b c-a d)^2 \left (1+\frac {2 \sqrt {b} \sqrt {d} \sqrt {(a+b x) (c+d x)}}{b c-a d}\right )^2}} \operatorname {EllipticF}\left (2 \arctan \left (\frac {\sqrt {2} \sqrt [4]{b} \sqrt [4]{d} \sqrt [4]{(a+b x) (c+d x)}}{\sqrt {b c-a d}}\right ),\frac {1}{2}\right )}{5 b^{3/4} \sqrt {b c-a d} \sqrt [4]{a+b x} \sqrt [4]{c+d x} (b c+a d+2 b d x) \sqrt {(a d+b (c+2 d x))^2}} \]

output

-4/5*(d*x+c)^(3/4)/(-a*d+b*c)/(b*x+a)^(5/4)+8/5*d*(d*x+c)^(3/4)/(-a*d+b*c) 
^2/(b*x+a)^(1/4)-8/5*d^(3/2)*((b*x+a)*(d*x+c))^(1/2)*((2*b*d*x+a*d+b*c)^2) 
^(1/2)*((a*d+b*(2*d*x+c))^2)^(1/2)/(-a*d+b*c)^3/(b*x+a)^(1/4)/(d*x+c)^(1/4 
)/(2*b*d*x+a*d+b*c)/b^(1/2)/(1+2*b^(1/2)*d^(1/2)*((b*x+a)*(d*x+c))^(1/2)/( 
-a*d+b*c))+4/5*d^(5/4)*((b*x+a)*(d*x+c))^(1/4)*(cos(2*arctan(b^(1/4)*d^(1/ 
4)*((b*x+a)*(d*x+c))^(1/4)*2^(1/2)/(-a*d+b*c)^(1/2)))^2)^(1/2)/cos(2*arcta 
n(b^(1/4)*d^(1/4)*((b*x+a)*(d*x+c))^(1/4)*2^(1/2)/(-a*d+b*c)^(1/2)))*Ellip 
ticE(sin(2*arctan(b^(1/4)*d^(1/4)*((b*x+a)*(d*x+c))^(1/4)*2^(1/2)/(-a*d+b* 
c)^(1/2))),1/2*2^(1/2))*2^(1/2)*(1+2*b^(1/2)*d^(1/2)*((b*x+a)*(d*x+c))^(1/ 
2)/(-a*d+b*c))*((2*b*d*x+a*d+b*c)^2)^(1/2)*((a*d+b*(2*d*x+c))^2/(-a*d+b*c) 
^2/(1+2*b^(1/2)*d^(1/2)*((b*x+a)*(d*x+c))^(1/2)/(-a*d+b*c))^2)^(1/2)/b^(3/ 
4)/(b*x+a)^(1/4)/(d*x+c)^(1/4)/(2*b*d*x+a*d+b*c)/(-a*d+b*c)^(1/2)/((a*d+b* 
(2*d*x+c))^2)^(1/2)-2/5*d^(5/4)*((b*x+a)*(d*x+c))^(1/4)*(cos(2*arctan(b^(1 
/4)*d^(1/4)*((b*x+a)*(d*x+c))^(1/4)*2^(1/2)/(-a*d+b*c)^(1/2)))^2)^(1/2)/co 
s(2*arctan(b^(1/4)*d^(1/4)*((b*x+a)*(d*x+c))^(1/4)*2^(1/2)/(-a*d+b*c)^(1/2 
)))*EllipticF(sin(2*arctan(b^(1/4)*d^(1/4)*((b*x+a)*(d*x+c))^(1/4)*2^(1/2) 
/(-a*d+b*c)^(1/2))),1/2*2^(1/2))*2^(1/2)*(1+2*b^(1/2)*d^(1/2)*((b*x+a)*(d* 
x+c))^(1/2)/(-a*d+b*c))*((2*b*d*x+a*d+b*c)^2)^(1/2)*((a*d+b*(2*d*x+c))^2/( 
-a*d+b*c)^2/(1+2*b^(1/2)*d^(1/2)*((b*x+a)*(d*x+c))^(1/2)/(-a*d+b*c))^2)^(1 
/2)/b^(3/4)/(b*x+a)^(1/4)/(d*x+c)^(1/4)/(2*b*d*x+a*d+b*c)/(-a*d+b*c)^(1...
 

3.18.4.2 Mathematica [C] (verified)

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

Time = 0.02 (sec) , antiderivative size = 73, normalized size of antiderivative = 0.10 \[ \int \frac {1}{(a+b x)^{9/4} \sqrt [4]{c+d x}} \, dx=-\frac {4 \sqrt [4]{\frac {b (c+d x)}{b c-a d}} \operatorname {Hypergeometric2F1}\left (-\frac {5}{4},\frac {1}{4},-\frac {1}{4},\frac {d (a+b x)}{-b c+a d}\right )}{5 b (a+b x)^{5/4} \sqrt [4]{c+d x}} \]

input

Integrate[1/((a + b*x)^(9/4)*(c + d*x)^(1/4)),x]
 

output

(-4*((b*(c + d*x))/(b*c - a*d))^(1/4)*Hypergeometric2F1[-5/4, 1/4, -1/4, ( 
d*(a + b*x))/(-(b*c) + a*d)])/(5*b*(a + b*x)^(5/4)*(c + d*x)^(1/4))
 

3.18.4.3 Rubi [A] (warning: unable to verify)

Time = 0.38 (sec) , antiderivative size = 254, normalized size of antiderivative = 0.33, number of steps used = 9, number of rules used = 8, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.421, Rules used = {61, 61, 73, 839, 813, 858, 807, 212}

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.

\(\displaystyle \int \frac {1}{(a+b x)^{9/4} \sqrt [4]{c+d x}} \, dx\)

\(\Big \downarrow \) 61

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

\(\Big \downarrow \) 61

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

\(\Big \downarrow \) 73

\(\displaystyle -\frac {2 d \left (\frac {8 d \int \frac {\sqrt {a+b x}}{\sqrt [4]{c-\frac {a d}{b}+\frac {d (a+b x)}{b}}}d\sqrt [4]{a+b x}}{b (b c-a d)}-\frac {4 (c+d x)^{3/4}}{\sqrt [4]{a+b x} (b c-a d)}\right )}{5 (b c-a d)}-\frac {4 (c+d x)^{3/4}}{5 (a+b x)^{5/4} (b c-a d)}\)

\(\Big \downarrow \) 839

\(\displaystyle -\frac {2 d \left (\frac {8 d \left (\frac {(a+b x)^{3/4}}{2 \sqrt [4]{\frac {d (a+b x)}{b}-\frac {a d}{b}+c}}-\frac {1}{2} \left (c-\frac {a d}{b}\right ) \int \frac {\sqrt {a+b x}}{\left (c-\frac {a d}{b}+\frac {d (a+b x)}{b}\right )^{5/4}}d\sqrt [4]{a+b x}\right )}{b (b c-a d)}-\frac {4 (c+d x)^{3/4}}{\sqrt [4]{a+b x} (b c-a d)}\right )}{5 (b c-a d)}-\frac {4 (c+d x)^{3/4}}{5 (a+b x)^{5/4} (b c-a d)}\)

\(\Big \downarrow \) 813

\(\displaystyle -\frac {2 d \left (\frac {8 d \left (\frac {(a+b x)^{3/4}}{2 \sqrt [4]{\frac {d (a+b x)}{b}-\frac {a d}{b}+c}}-\frac {b \sqrt [4]{a+b x} \left (c-\frac {a d}{b}\right ) \sqrt [4]{\frac {b c-a d}{d (a+b x)}+1} \int \frac {1}{(a+b x)^{3/4} \left (\frac {b c-a d}{d (a+b x)}+1\right )^{5/4}}d\sqrt [4]{a+b x}}{2 d \sqrt [4]{\frac {d (a+b x)}{b}-\frac {a d}{b}+c}}\right )}{b (b c-a d)}-\frac {4 (c+d x)^{3/4}}{\sqrt [4]{a+b x} (b c-a d)}\right )}{5 (b c-a d)}-\frac {4 (c+d x)^{3/4}}{5 (a+b x)^{5/4} (b c-a d)}\)

\(\Big \downarrow \) 858

\(\displaystyle -\frac {2 d \left (\frac {8 d \left (\frac {b \sqrt [4]{a+b x} \left (c-\frac {a d}{b}\right ) \sqrt [4]{\frac {b c-a d}{d (a+b x)}+1} \int \frac {1}{\sqrt [4]{a+b x} \left (\frac {(b c-a d) (a+b x)}{d}+1\right )^{5/4}}d\frac {1}{\sqrt [4]{a+b x}}}{2 d \sqrt [4]{\frac {d (a+b x)}{b}-\frac {a d}{b}+c}}+\frac {(a+b x)^{3/4}}{2 \sqrt [4]{\frac {d (a+b x)}{b}-\frac {a d}{b}+c}}\right )}{b (b c-a d)}-\frac {4 (c+d x)^{3/4}}{\sqrt [4]{a+b x} (b c-a d)}\right )}{5 (b c-a d)}-\frac {4 (c+d x)^{3/4}}{5 (a+b x)^{5/4} (b c-a d)}\)

\(\Big \downarrow \) 807

\(\displaystyle -\frac {2 d \left (\frac {8 d \left (\frac {b \sqrt [4]{a+b x} \left (c-\frac {a d}{b}\right ) \sqrt [4]{\frac {b c-a d}{d (a+b x)}+1} \int \frac {1}{\left (\frac {\sqrt {a+b x} (b c-a d)}{d}+1\right )^{5/4}}d\sqrt {a+b x}}{4 d \sqrt [4]{\frac {d (a+b x)}{b}-\frac {a d}{b}+c}}+\frac {(a+b x)^{3/4}}{2 \sqrt [4]{\frac {d (a+b x)}{b}-\frac {a d}{b}+c}}\right )}{b (b c-a d)}-\frac {4 (c+d x)^{3/4}}{\sqrt [4]{a+b x} (b c-a d)}\right )}{5 (b c-a d)}-\frac {4 (c+d x)^{3/4}}{5 (a+b x)^{5/4} (b c-a d)}\)

\(\Big \downarrow \) 212

\(\displaystyle -\frac {2 d \left (\frac {8 d \left (\frac {b \sqrt [4]{a+b x} \left (c-\frac {a d}{b}\right ) \sqrt [4]{\frac {b c-a d}{d (a+b x)}+1} E\left (\left .\frac {1}{2} \arctan \left (\frac {\sqrt {b c-a d} \sqrt {a+b x}}{\sqrt {d}}\right )\right |2\right )}{2 \sqrt {d} \sqrt {b c-a d} \sqrt [4]{\frac {d (a+b x)}{b}-\frac {a d}{b}+c}}+\frac {(a+b x)^{3/4}}{2 \sqrt [4]{\frac {d (a+b x)}{b}-\frac {a d}{b}+c}}\right )}{b (b c-a d)}-\frac {4 (c+d x)^{3/4}}{\sqrt [4]{a+b x} (b c-a d)}\right )}{5 (b c-a d)}-\frac {4 (c+d x)^{3/4}}{5 (a+b x)^{5/4} (b c-a d)}\)

input

Int[1/((a + b*x)^(9/4)*(c + d*x)^(1/4)),x]
 

output

(-4*(c + d*x)^(3/4))/(5*(b*c - a*d)*(a + b*x)^(5/4)) - (2*d*((-4*(c + d*x) 
^(3/4))/((b*c - a*d)*(a + b*x)^(1/4)) + (8*d*((a + b*x)^(3/4)/(2*(c - (a*d 
)/b + (d*(a + b*x))/b)^(1/4)) + (b*(c - (a*d)/b)*(a + b*x)^(1/4)*(1 + (b*c 
 - a*d)/(d*(a + b*x)))^(1/4)*EllipticE[ArcTan[(Sqrt[b*c - a*d]*Sqrt[a + b* 
x])/Sqrt[d]]/2, 2])/(2*Sqrt[d]*Sqrt[b*c - a*d]*(c - (a*d)/b + (d*(a + b*x) 
)/b)^(1/4))))/(b*(b*c - a*d))))/(5*(b*c - a*d))
 

3.18.4.3.1 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 212

Int[((a_) + (b_.)*(x_)^2)^(-5/4), x_Symbol] :> Simp[(2/(a^(5/4)*Rt[b/a, 2]) 
)*EllipticE[(1/2)*ArcTan[Rt[b/a, 2]*x], 2], x] /; FreeQ[{a, b}, x] && GtQ[a 
, 0] && PosQ[b/a]
 

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 813

Int[(x_)^2/((a_) + (b_.)*(x_)^4)^(5/4), x_Symbol] :> Simp[x*((1 + a/(b*x^4) 
)^(1/4)/(b*(a + b*x^4)^(1/4)))   Int[1/(x^3*(1 + a/(b*x^4))^(5/4)), x], x] 
/; FreeQ[{a, b}, x] && PosQ[b/a]
 

rule 839

Int[(x_)^2/((a_) + (b_.)*(x_)^4)^(1/4), x_Symbol] :> Simp[x^3/(2*(a + b*x^4 
)^(1/4)), x] - Simp[a/2   Int[x^2/(a + b*x^4)^(5/4), x], x] /; FreeQ[{a, b} 
, x] && PosQ[b/a]
 

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]
 

3.18.4.4 Maple [F]

\[\int \frac {1}{\left (b x +a \right )^{\frac {9}{4}} \left (d x +c \right )^{\frac {1}{4}}}d x\]

input

int(1/(b*x+a)^(9/4)/(d*x+c)^(1/4),x)
 

output

int(1/(b*x+a)^(9/4)/(d*x+c)^(1/4),x)
 

3.18.4.5 Fricas [F]

\[ \int \frac {1}{(a+b x)^{9/4} \sqrt [4]{c+d x}} \, dx=\int { \frac {1}{{\left (b x + a\right )}^{\frac {9}{4}} {\left (d x + c\right )}^{\frac {1}{4}}} \,d x } \]

input

integrate(1/(b*x+a)^(9/4)/(d*x+c)^(1/4),x, algorithm="fricas")
 

output

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

3.18.4.6 Sympy [F]

\[ \int \frac {1}{(a+b x)^{9/4} \sqrt [4]{c+d x}} \, dx=\int \frac {1}{\left (a + b x\right )^{\frac {9}{4}} \sqrt [4]{c + d x}}\, dx \]

input

integrate(1/(b*x+a)**(9/4)/(d*x+c)**(1/4),x)
 

output

Integral(1/((a + b*x)**(9/4)*(c + d*x)**(1/4)), x)
 

3.18.4.7 Maxima [F]

\[ \int \frac {1}{(a+b x)^{9/4} \sqrt [4]{c+d x}} \, dx=\int { \frac {1}{{\left (b x + a\right )}^{\frac {9}{4}} {\left (d x + c\right )}^{\frac {1}{4}}} \,d x } \]

input

integrate(1/(b*x+a)^(9/4)/(d*x+c)^(1/4),x, algorithm="maxima")
 

output

integrate(1/((b*x + a)^(9/4)*(d*x + c)^(1/4)), x)
 

3.18.4.8 Giac [F]

\[ \int \frac {1}{(a+b x)^{9/4} \sqrt [4]{c+d x}} \, dx=\int { \frac {1}{{\left (b x + a\right )}^{\frac {9}{4}} {\left (d x + c\right )}^{\frac {1}{4}}} \,d x } \]

input

integrate(1/(b*x+a)^(9/4)/(d*x+c)^(1/4),x, algorithm="giac")
 

output

integrate(1/((b*x + a)^(9/4)*(d*x + c)^(1/4)), x)
 

3.18.4.9 Mupad [F(-1)]

Timed out. \[ \int \frac {1}{(a+b x)^{9/4} \sqrt [4]{c+d x}} \, dx=\int \frac {1}{{\left (a+b\,x\right )}^{9/4}\,{\left (c+d\,x\right )}^{1/4}} \,d x \]

input

int(1/((a + b*x)^(9/4)*(c + d*x)^(1/4)),x)
 

output

int(1/((a + b*x)^(9/4)*(c + d*x)^(1/4)), x)