\(\int \frac {c+d x}{(c^2-d^2 x^2)^{3/8}} \, dx\) [333]

Optimal result

Integrand size = 22, antiderivative size = 67 \[ \int \frac {c+d x}{\left (c^2-d^2 x^2\right )^{3/8}} \, dx=\frac {4\ 2^{5/8} \left (c^2-d^2 x^2\right )^{13/8} \operatorname {Hypergeometric2F1}\left (\frac {3}{8},\frac {13}{8},\frac {21}{8},\frac {c+d x}{2 c}\right )}{13 c^2 d \left (\frac {c-d x}{c}\right )^{13/8}} \] Output:

4/13*2^(5/8)*(-d^2*x^2+c^2)^(13/8)*hypergeom([3/8, 13/8],[21/8],1/2*(d*x+c 
)/c)/c^2/d/((-d*x+c)/c)^(13/8)
 

Mathematica [A] (verified)

Time = 10.03 (sec) , antiderivative size = 80, normalized size of antiderivative = 1.19 \[ \int \frac {c+d x}{\left (c^2-d^2 x^2\right )^{3/8}} \, dx=-\frac {4 \left (c^2-d^2 x^2\right )^{5/8}}{5 d}+\frac {c x \left (1-\frac {d^2 x^2}{c^2}\right )^{3/8} \operatorname {Hypergeometric2F1}\left (\frac {3}{8},\frac {1}{2},\frac {3}{2},\frac {d^2 x^2}{c^2}\right )}{\left (c^2-d^2 x^2\right )^{3/8}} \] Input:

Integrate[(c + d*x)/(c^2 - d^2*x^2)^(3/8),x]
 

Output:

(-4*(c^2 - d^2*x^2)^(5/8))/(5*d) + (c*x*(1 - (d^2*x^2)/c^2)^(3/8)*Hypergeo 
metric2F1[3/8, 1/2, 3/2, (d^2*x^2)/c^2])/(c^2 - d^2*x^2)^(3/8)
 

Rubi [A] (verified)

Time = 0.30 (sec) , antiderivative size = 80, normalized size of antiderivative = 1.19, number of steps used = 3, number of rules used = 3, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.136, Rules used = {455, 238, 237}

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.

\(\displaystyle \int \frac {c+d x}{\left (c^2-d^2 x^2\right )^{3/8}} \, dx\)

\(\Big \downarrow \) 455

\(\displaystyle c \int \frac {1}{\left (c^2-d^2 x^2\right )^{3/8}}dx-\frac {4 \left (c^2-d^2 x^2\right )^{5/8}}{5 d}\)

\(\Big \downarrow \) 238

\(\displaystyle \frac {c \left (1-\frac {d^2 x^2}{c^2}\right )^{3/8} \int \frac {1}{\left (1-\frac {d^2 x^2}{c^2}\right )^{3/8}}dx}{\left (c^2-d^2 x^2\right )^{3/8}}-\frac {4 \left (c^2-d^2 x^2\right )^{5/8}}{5 d}\)

\(\Big \downarrow \) 237

\(\displaystyle \frac {c x \left (1-\frac {d^2 x^2}{c^2}\right )^{3/8} \operatorname {Hypergeometric2F1}\left (\frac {3}{8},\frac {1}{2},\frac {3}{2},\frac {d^2 x^2}{c^2}\right )}{\left (c^2-d^2 x^2\right )^{3/8}}-\frac {4 \left (c^2-d^2 x^2\right )^{5/8}}{5 d}\)

Input:

Int[(c + d*x)/(c^2 - d^2*x^2)^(3/8),x]
 

Output:

(-4*(c^2 - d^2*x^2)^(5/8))/(5*d) + (c*x*(1 - (d^2*x^2)/c^2)^(3/8)*Hypergeo 
metric2F1[3/8, 1/2, 3/2, (d^2*x^2)/c^2])/(c^2 - d^2*x^2)^(3/8)
 

Defintions of rubi rules used

rule 237

Int[((a_) + (b_.)*(x_)^2)^(p_), x_Symbol] :> Simp[a^p*x*Hypergeometric2F1[- 
p, 1/2, 1/2 + 1, (-b)*(x^2/a)], x] /; FreeQ[{a, b, p}, x] &&  !IntegerQ[2*p 
] && GtQ[a, 0]
 

rule 238

Int[((a_) + (b_.)*(x_)^2)^(p_), x_Symbol] :> Simp[a^IntPart[p]*((a + b*x^2) 
^FracPart[p]/(1 + b*(x^2/a))^FracPart[p])   Int[(1 + b*(x^2/a))^p, x], x] / 
; FreeQ[{a, b, p}, x] &&  !IntegerQ[2*p] &&  !GtQ[a, 0]
 

rule 455

Int[((c_) + (d_.)*(x_))*((a_) + (b_.)*(x_)^2)^(p_.), x_Symbol] :> Simp[d*(( 
a + b*x^2)^(p + 1)/(2*b*(p + 1))), x] + Simp[c   Int[(a + b*x^2)^p, x], x] 
/; FreeQ[{a, b, c, d, p}, x] &&  !LeQ[p, -1]
 

Maple [F]

\[\int \frac {d x +c}{\left (-d^{2} x^{2}+c^{2}\right )^{\frac {3}{8}}}d x\]

Input:

int((d*x+c)/(-d^2*x^2+c^2)^(3/8),x)
 

Output:

int((d*x+c)/(-d^2*x^2+c^2)^(3/8),x)
 

Fricas [F]

\[ \int \frac {c+d x}{\left (c^2-d^2 x^2\right )^{3/8}} \, dx=\int { \frac {d x + c}{{\left (-d^{2} x^{2} + c^{2}\right )}^{\frac {3}{8}}} \,d x } \] Input:

integrate((d*x+c)/(-d^2*x^2+c^2)^(3/8),x, algorithm="fricas")
 

Output:

integral(-(-d^2*x^2 + c^2)^(5/8)/(d*x - c), x)
 

Sympy [A] (verification not implemented)

Time = 1.08 (sec) , antiderivative size = 66, normalized size of antiderivative = 0.99 \[ \int \frac {c+d x}{\left (c^2-d^2 x^2\right )^{3/8}} \, dx=\sqrt [4]{c} x {{}_{2}F_{1}\left (\begin {matrix} \frac {3}{8}, \frac {1}{2} \\ \frac {3}{2} \end {matrix}\middle | {\frac {d^{2} x^{2} e^{2 i \pi }}{c^{2}}} \right )} + d \left (\begin {cases} \frac {x^{2}}{2 \left (c^{2}\right )^{\frac {3}{8}}} & \text {for}\: d^{2} = 0 \\- \frac {4 \left (c^{2} - d^{2} x^{2}\right )^{\frac {5}{8}}}{5 d^{2}} & \text {otherwise} \end {cases}\right ) \] Input:

integrate((d*x+c)/(-d**2*x**2+c**2)**(3/8),x)
 

Output:

c**(1/4)*x*hyper((3/8, 1/2), (3/2,), d**2*x**2*exp_polar(2*I*pi)/c**2) + d 
*Piecewise((x**2/(2*(c**2)**(3/8)), Eq(d**2, 0)), (-4*(c**2 - d**2*x**2)** 
(5/8)/(5*d**2), True))
 

Maxima [F]

\[ \int \frac {c+d x}{\left (c^2-d^2 x^2\right )^{3/8}} \, dx=\int { \frac {d x + c}{{\left (-d^{2} x^{2} + c^{2}\right )}^{\frac {3}{8}}} \,d x } \] Input:

integrate((d*x+c)/(-d^2*x^2+c^2)^(3/8),x, algorithm="maxima")
 

Output:

integrate((d*x + c)/(-d^2*x^2 + c^2)^(3/8), x)
 

Giac [F]

\[ \int \frac {c+d x}{\left (c^2-d^2 x^2\right )^{3/8}} \, dx=\int { \frac {d x + c}{{\left (-d^{2} x^{2} + c^{2}\right )}^{\frac {3}{8}}} \,d x } \] Input:

integrate((d*x+c)/(-d^2*x^2+c^2)^(3/8),x, algorithm="giac")
 

Output:

integrate((d*x + c)/(-d^2*x^2 + c^2)^(3/8), x)
 

Mupad [B] (verification not implemented)

Time = 6.70 (sec) , antiderivative size = 67, normalized size of antiderivative = 1.00 \[ \int \frac {c+d x}{\left (c^2-d^2 x^2\right )^{3/8}} \, dx=\frac {c\,x\,{\left (1-\frac {d^2\,x^2}{c^2}\right )}^{3/8}\,{{}}_2{\mathrm {F}}_1\left (\frac {3}{8},\frac {1}{2};\ \frac {3}{2};\ \frac {d^2\,x^2}{c^2}\right )}{{\left (c^2-d^2\,x^2\right )}^{3/8}}-\frac {4\,{\left (c^2-d^2\,x^2\right )}^{5/8}}{5\,d} \] Input:

int((c + d*x)/(c^2 - d^2*x^2)^(3/8),x)
 

Output:

(c*x*(1 - (d^2*x^2)/c^2)^(3/8)*hypergeom([3/8, 1/2], 3/2, (d^2*x^2)/c^2))/ 
(c^2 - d^2*x^2)^(3/8) - (4*(c^2 - d^2*x^2)^(5/8))/(5*d)
 

Reduce [F]

\[ \int \frac {c+d x}{\left (c^2-d^2 x^2\right )^{3/8}} \, dx=\frac {4 \left (-d^{2} x^{2}+c^{2}\right )^{\frac {3}{8}} c^{2}-4 \left (-d^{2} x^{2}+c^{2}\right )^{\frac {3}{8}} d^{2} x^{2}+5 \left (-d^{2} x^{2}+c^{2}\right )^{\frac {3}{4}} \left (\int \frac {1}{\left (-d^{2} x^{2}+c^{2}\right )^{\frac {3}{8}}}d x \right ) c d}{5 \left (-d^{2} x^{2}+c^{2}\right )^{\frac {3}{4}} d} \] Input:

int((d*x+c)/(-d^2*x^2+c^2)^(3/8),x)
 

Output:

(4*(c**2 - d**2*x**2)**(3/8)*c**2 - 4*(c**2 - d**2*x**2)**(3/8)*d**2*x**2 
+ 5*(c**2 - d**2*x**2)**(3/4)*int((c**2 - d**2*x**2)**(1/4)/(c**2 - d**2*x 
**2)**(5/8),x)*c*d)/(5*(c**2 - d**2*x**2)**(3/4)*d)