\(\int \frac {\sqrt {a+b x}}{(c+d x)^{7/4}} \, dx\) [471]

Optimal result

Integrand size = 19, antiderivative size = 111 \[ \int \frac {\sqrt {a+b x}}{(c+d x)^{7/4}} \, dx=-\frac {4 \sqrt {a+b x}}{3 d (c+d x)^{3/4}}+\frac {8 b^{3/4} \sqrt [4]{b c-a d} \sqrt {-\frac {d (a+b x)}{b c-a d}} \operatorname {EllipticF}\left (\arcsin \left (\frac {\sqrt [4]{b} \sqrt [4]{c+d x}}{\sqrt [4]{b c-a d}}\right ),-1\right )}{3 d^2 \sqrt {a+b x}} \] Output:

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

Mathematica [C] (verified)

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

Time = 0.03 (sec) , antiderivative size = 73, normalized size of antiderivative = 0.66 \[ \int \frac {\sqrt {a+b x}}{(c+d x)^{7/4}} \, dx=\frac {2 (a+b x)^{3/2} \left (\frac {b (c+d x)}{b c-a d}\right )^{7/4} \operatorname {Hypergeometric2F1}\left (\frac {3}{2},\frac {7}{4},\frac {5}{2},\frac {d (a+b x)}{-b c+a d}\right )}{3 b (c+d x)^{7/4}} \] Input:

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

Output:

(2*(a + b*x)^(3/2)*((b*(c + d*x))/(b*c - a*d))^(7/4)*Hypergeometric2F1[3/2 
, 7/4, 5/2, (d*(a + b*x))/(-(b*c) + a*d)])/(3*b*(c + d*x)^(7/4))
 

Rubi [A] (verified)

Time = 0.21 (sec) , antiderivative size = 127, normalized size of antiderivative = 1.14, number of steps used = 5, number of rules used = 4, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.211, Rules used = {57, 73, 765, 762}

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.

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

\(\Big \downarrow \) 57

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

\(\Big \downarrow \) 73

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

\(\Big \downarrow \) 765

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

\(\Big \downarrow \) 762

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

Input:

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

Output:

(-4*Sqrt[a + b*x])/(3*d*(c + d*x)^(3/4)) + (8*b^(3/4)*(b*c - a*d)^(1/4)*Sq 
rt[1 - (b*(c + d*x))/(b*c - a*d)]*EllipticF[ArcSin[(b^(1/4)*(c + d*x)^(1/4 
))/(b*c - a*d)^(1/4)], -1])/(3*d^2*Sqrt[a - (b*c)/d + (b*(c + d*x))/d])
 

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 762

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

rule 765

Int[1/Sqrt[(a_) + (b_.)*(x_)^4], x_Symbol] :> Simp[Sqrt[1 + b*(x^4/a)]/Sqrt 
[a + b*x^4]   Int[1/Sqrt[1 + b*(x^4/a)], x], x] /; FreeQ[{a, b}, x] && NegQ 
[b/a] &&  !GtQ[a, 0]
 

Maple [F]

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

Input:

int((b*x+a)^(1/2)/(d*x+c)^(7/4),x)
 

Output:

int((b*x+a)^(1/2)/(d*x+c)^(7/4),x)
 

Fricas [F]

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

integrate((b*x+a)^(1/2)/(d*x+c)^(7/4),x, algorithm="fricas")
 

Output:

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

Sympy [F]

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

integrate((b*x+a)**(1/2)/(d*x+c)**(7/4),x)
 

Output:

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

Maxima [F]

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

integrate((b*x+a)^(1/2)/(d*x+c)^(7/4),x, algorithm="maxima")
 

Output:

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

Giac [F]

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

integrate((b*x+a)^(1/2)/(d*x+c)^(7/4),x, algorithm="giac")
 

Output:

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

Mupad [F(-1)]

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

int((a + b*x)^(1/2)/(c + d*x)^(7/4),x)
 

Output:

int((a + b*x)^(1/2)/(c + d*x)^(7/4), x)
 

Reduce [F]

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

int((b*x+a)^(1/2)/(d*x+c)^(7/4),x)
 

Output:

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