\(\int \frac {1}{(a+b x)^{7/4} (c+d x)^{3/4}} \, dx\) [611]

Optimal result
Mathematica [C] (verified)
Rubi [A] (warning: unable to verify)
Maple [F]
Fricas [F]
Sympy [F]
Maxima [F]
Giac [F]
Mupad [F(-1)]
Reduce [B] (verification not implemented)

Optimal result

Integrand size = 19, antiderivative size = 130 \[ \int \frac {1}{(a+b x)^{7/4} (c+d x)^{3/4}} \, dx=-\frac {4 \sqrt [4]{c+d x}}{3 (b c-a d) (a+b x)^{3/4}}+\frac {8 d^{3/2} (a+b x)^{3/4} \left (\frac {b (c+d x)}{d (a+b x)}\right )^{3/4} \operatorname {EllipticF}\left (\frac {1}{2} \arctan \left (\frac {\sqrt {b c-a d}}{\sqrt {d} \sqrt {a+b x}}\right ),2\right )}{3 b (b c-a d)^{3/2} (c+d x)^{3/4}} \] Output:

-4/3*(d*x+c)^(1/4)/(-a*d+b*c)/(b*x+a)^(3/4)+8/3*d^(3/2)*(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/(-a*d+b*c)^(3/2)/(d*x+c)^(3/4)
 

Mathematica [C] (verified)

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

Time = 0.02 (sec) , antiderivative size = 73, normalized size of antiderivative = 0.56 \[ \int \frac {1}{(a+b x)^{7/4} (c+d x)^{3/4}} \, dx=-\frac {4 \left (\frac {b (c+d x)}{b c-a d}\right )^{3/4} \operatorname {Hypergeometric2F1}\left (-\frac {3}{4},\frac {3}{4},\frac {1}{4},\frac {d (a+b x)}{-b c+a d}\right )}{3 b (a+b x)^{3/4} (c+d x)^{3/4}} \] Input:

Integrate[1/((a + b*x)^(7/4)*(c + d*x)^(3/4)),x]
 

Output:

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

Rubi [A] (warning: unable to verify)

Time = 0.27 (sec) , antiderivative size = 148, normalized size of antiderivative = 1.14, number of steps used = 7, number of rules used = 6, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.316, Rules used = {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.

\(\displaystyle \int \frac {1}{(a+b x)^{7/4} (c+d x)^{3/4}} \, dx\)

\(\Big \downarrow \) 61

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

\(\Big \downarrow \) 73

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

\(\Big \downarrow \) 768

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

\(\Big \downarrow \) 858

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

\(\Big \downarrow \) 807

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

\(\Big \downarrow \) 229

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

Input:

Int[1/((a + b*x)^(7/4)*(c + d*x)^(3/4)),x]
 

Output:

(-4*(c + d*x)^(1/4))/(3*(b*c - a*d)*(a + b*x)^(3/4)) + (8*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*(b*c - a*d)^(3/2)*(c - (a*d)/b + 
 (d*(a + b*x))/b)^(3/4))
 

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]

\[\int \frac {1}{\left (b x +a \right )^{\frac {7}{4}} \left (x d +c \right )^{\frac {3}{4}}}d x\]

Input:

int(1/(b*x+a)^(7/4)/(d*x+c)^(3/4),x)
 

Output:

int(1/(b*x+a)^(7/4)/(d*x+c)^(3/4),x)
 

Fricas [F]

\[ \int \frac {1}{(a+b x)^{7/4} (c+d x)^{3/4}} \, dx=\int { \frac {1}{{\left (b x + a\right )}^{\frac {7}{4}} {\left (d x + c\right )}^{\frac {3}{4}}} \,d x } \] Input:

integrate(1/(b*x+a)^(7/4)/(d*x+c)^(3/4),x, algorithm="fricas")
 

Output:

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

Sympy [F]

\[ \int \frac {1}{(a+b x)^{7/4} (c+d x)^{3/4}} \, dx=\int \frac {1}{\left (a + b x\right )^{\frac {7}{4}} \left (c + d x\right )^{\frac {3}{4}}}\, dx \] Input:

integrate(1/(b*x+a)**(7/4)/(d*x+c)**(3/4),x)
 

Output:

Integral(1/((a + b*x)**(7/4)*(c + d*x)**(3/4)), x)
 

Maxima [F]

\[ \int \frac {1}{(a+b x)^{7/4} (c+d x)^{3/4}} \, dx=\int { \frac {1}{{\left (b x + a\right )}^{\frac {7}{4}} {\left (d x + c\right )}^{\frac {3}{4}}} \,d x } \] Input:

integrate(1/(b*x+a)^(7/4)/(d*x+c)^(3/4),x, algorithm="maxima")
 

Output:

integrate(1/((b*x + a)^(7/4)*(d*x + c)^(3/4)), x)
 

Giac [F]

\[ \int \frac {1}{(a+b x)^{7/4} (c+d x)^{3/4}} \, dx=\int { \frac {1}{{\left (b x + a\right )}^{\frac {7}{4}} {\left (d x + c\right )}^{\frac {3}{4}}} \,d x } \] Input:

integrate(1/(b*x+a)^(7/4)/(d*x+c)^(3/4),x, algorithm="giac")
 

Output:

integrate(1/((b*x + a)^(7/4)*(d*x + c)^(3/4)), x)
 

Mupad [F(-1)]

Timed out. \[ \int \frac {1}{(a+b x)^{7/4} (c+d x)^{3/4}} \, dx=\int \frac {1}{{\left (a+b\,x\right )}^{7/4}\,{\left (c+d\,x\right )}^{3/4}} \,d x \] Input:

int(1/((a + b*x)^(7/4)*(c + d*x)^(3/4)),x)
 

Output:

int(1/((a + b*x)^(7/4)*(c + d*x)^(3/4)), x)
 

Reduce [B] (verification not implemented)

Time = 0.18 (sec) , antiderivative size = 49, normalized size of antiderivative = 0.38 \[ \int \frac {1}{(a+b x)^{7/4} (c+d x)^{3/4}} \, dx=\frac {4 \left (d x +c \right )^{\frac {3}{4}} \left (b x +a \right )^{\frac {1}{4}}}{3 \sqrt {d x +c}\, \left (a b d x -b^{2} c x +a^{2} d -a b c \right )} \] Input:

int(1/(b*x+a)^(7/4)/(d*x+c)^(3/4),x)
 

Output:

(4*(c + d*x)**(3/4)*(a + b*x)**(1/4))/(3*sqrt(c + d*x)*(a**2*d - a*b*c + a 
*b*d*x - b**2*c*x))