\(\int \frac {(c+d x)^{5/4}}{(a+b x)^{3/2}} \, dx\) [447]

Optimal result

Integrand size = 19, antiderivative size = 132 \[ \int \frac {(c+d x)^{5/4}}{(a+b x)^{3/2}} \, dx=\frac {10 d \sqrt {a+b x} \sqrt [4]{c+d x}}{3 b^2}-\frac {2 (c+d x)^{5/4}}{b \sqrt {a+b x}}+\frac {10 (b c-a d)^{5/4} \sqrt {-\frac {d (a+b x)}{b c-a d}} \operatorname {EllipticF}\left (\arcsin \left (\frac {\sqrt [4]{b} \sqrt [4]{c+d x}}{\sqrt [4]{b c-a d}}\right ),-1\right )}{3 b^{9/4} \sqrt {a+b x}} \] Output:

10/3*d*(b*x+a)^(1/2)*(d*x+c)^(1/4)/b^2-2*(d*x+c)^(5/4)/b/(b*x+a)^(1/2)+10/ 
3*(-a*d+b*c)^(5/4)*(-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)/b^(9/4)/(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 = 71, normalized size of antiderivative = 0.54 \[ \int \frac {(c+d x)^{5/4}}{(a+b x)^{3/2}} \, dx=-\frac {2 (c+d x)^{5/4} \operatorname {Hypergeometric2F1}\left (-\frac {5}{4},-\frac {1}{2},\frac {1}{2},\frac {d (a+b x)}{-b c+a d}\right )}{b \sqrt {a+b x} \left (\frac {b (c+d x)}{b c-a d}\right )^{5/4}} \] Input:

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

Output:

(-2*(c + d*x)^(5/4)*Hypergeometric2F1[-5/4, -1/2, 1/2, (d*(a + b*x))/(-(b* 
c) + a*d)])/(b*Sqrt[a + b*x]*((b*(c + d*x))/(b*c - a*d))^(5/4))
 

Rubi [A] (verified)

Time = 0.24 (sec) , antiderivative size = 159, normalized size of antiderivative = 1.20, number of steps used = 6, number of rules used = 5, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.263, Rules used = {57, 60, 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.

\(\displaystyle \int \frac {(c+d x)^{5/4}}{(a+b x)^{3/2}} \, dx\)

\(\Big \downarrow \) 57

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

\(\Big \downarrow \) 60

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

\(\Big \downarrow \) 73

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

\(\Big \downarrow \) 765

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

\(\Big \downarrow \) 762

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

Input:

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

Output:

(-2*(c + d*x)^(5/4))/(b*Sqrt[a + b*x]) + (5*d*((4*Sqrt[a + b*x]*(c + d*x)^ 
(1/4))/(3*b) + (4*(b*c - a*d)^(5/4)*Sqrt[1 - (b*(c + d*x))/(b*c - a*d)]*El 
lipticF[ArcSin[(b^(1/4)*(c + d*x)^(1/4))/(b*c - a*d)^(1/4)], -1])/(3*b^(5/ 
4)*d*Sqrt[a - (b*c)/d + (b*(c + d*x))/d])))/(2*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 60

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

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

Input:

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

Output:

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

Fricas [F]

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

integrate((d*x+c)^(5/4)/(b*x+a)^(3/2),x, algorithm="fricas")
 

Output:

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

Sympy [F]

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

integrate((d*x+c)**(5/4)/(b*x+a)**(3/2),x)
 

Output:

Integral((c + d*x)**(5/4)/(a + b*x)**(3/2), x)
 

Maxima [F]

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

integrate((d*x+c)^(5/4)/(b*x+a)^(3/2),x, algorithm="maxima")
 

Output:

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

Giac [F]

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

integrate((d*x+c)^(5/4)/(b*x+a)^(3/2),x, algorithm="giac")
 

Output:

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

Mupad [F(-1)]

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

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

Output:

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

Reduce [F]

\[ \int \frac {(c+d x)^{5/4}}{(a+b x)^{3/2}} \, dx =\text {Too large to display} \] Input:

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

Output:

( - 16*(c + d*x)**(1/4)*sqrt(a + b*x)*a*c*d + 4*(c + d*x)**(1/4)*sqrt(a + 
b*x)*a*d**2*x + 12*(c + d*x)**(1/4)*sqrt(a + b*x)*b*c**2 - 8*(c + d*x)**(1 
/4)*sqrt(a + b*x)*b*c*d*x - 5*int(((c + d*x)**(1/4)*sqrt(a + b*x)*x)/(a**3 
*c*d + a**3*d**2*x - 2*a**2*b*c**2 + 2*a**2*b*d**2*x**2 - 4*a*b**2*c**2*x 
- 3*a*b**2*c*d*x**2 + a*b**2*d**2*x**3 - 2*b**3*c**2*x**2 - 2*b**3*c*d*x** 
3),x)*a**4*d**4 + 20*int(((c + d*x)**(1/4)*sqrt(a + b*x)*x)/(a**3*c*d + a* 
*3*d**2*x - 2*a**2*b*c**2 + 2*a**2*b*d**2*x**2 - 4*a*b**2*c**2*x - 3*a*b** 
2*c*d*x**2 + a*b**2*d**2*x**3 - 2*b**3*c**2*x**2 - 2*b**3*c*d*x**3),x)*a** 
3*b*c*d**3 - 5*int(((c + d*x)**(1/4)*sqrt(a + b*x)*x)/(a**3*c*d + a**3*d** 
2*x - 2*a**2*b*c**2 + 2*a**2*b*d**2*x**2 - 4*a*b**2*c**2*x - 3*a*b**2*c*d* 
x**2 + a*b**2*d**2*x**3 - 2*b**3*c**2*x**2 - 2*b**3*c*d*x**3),x)*a**3*b*d* 
*4*x - 25*int(((c + d*x)**(1/4)*sqrt(a + b*x)*x)/(a**3*c*d + a**3*d**2*x - 
 2*a**2*b*c**2 + 2*a**2*b*d**2*x**2 - 4*a*b**2*c**2*x - 3*a*b**2*c*d*x**2 
+ a*b**2*d**2*x**3 - 2*b**3*c**2*x**2 - 2*b**3*c*d*x**3),x)*a**2*b**2*c**2 
*d**2 + 20*int(((c + d*x)**(1/4)*sqrt(a + b*x)*x)/(a**3*c*d + a**3*d**2*x 
- 2*a**2*b*c**2 + 2*a**2*b*d**2*x**2 - 4*a*b**2*c**2*x - 3*a*b**2*c*d*x**2 
 + a*b**2*d**2*x**3 - 2*b**3*c**2*x**2 - 2*b**3*c*d*x**3),x)*a**2*b**2*c*d 
**3*x + 10*int(((c + d*x)**(1/4)*sqrt(a + b*x)*x)/(a**3*c*d + a**3*d**2*x 
- 2*a**2*b*c**2 + 2*a**2*b*d**2*x**2 - 4*a*b**2*c**2*x - 3*a*b**2*c*d*x**2 
 + a*b**2*d**2*x**3 - 2*b**3*c**2*x**2 - 2*b**3*c*d*x**3),x)*a*b**3*c**...