\(\int \frac {(c+d x)^n}{(c^2-d^2 x^2)^{5/2}} \, dx\) [346]

Optimal result

Integrand size = 24, antiderivative size = 83 \[ \int \frac {(c+d x)^n}{\left (c^2-d^2 x^2\right )^{5/2}} \, dx=\frac {2^{-\frac {3}{2}+n} (c+d x)^n \left (1+\frac {d x}{c}\right )^{\frac {3}{2}-n} \operatorname {Hypergeometric2F1}\left (-\frac {3}{2},\frac {5}{2}-n,-\frac {1}{2},\frac {c-d x}{2 c}\right )}{3 c d \left (c^2-d^2 x^2\right )^{3/2}} \] Output:

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

Mathematica [A] (verified)

Time = 0.59 (sec) , antiderivative size = 92, normalized size of antiderivative = 1.11 \[ \int \frac {(c+d x)^n}{\left (c^2-d^2 x^2\right )^{5/2}} \, dx=\frac {2^{-\frac {3}{2}+n} (c+d x)^n \left (1+\frac {d x}{c}\right )^{\frac {1}{2}-n} \operatorname {Hypergeometric2F1}\left (-\frac {3}{2},\frac {5}{2}-n,-\frac {1}{2},\frac {c-d x}{2 c}\right )}{\left (3 c^3 d-3 c^2 d^2 x\right ) \sqrt {c^2-d^2 x^2}} \] Input:

Integrate[(c + d*x)^n/(c^2 - d^2*x^2)^(5/2),x]
 

Output:

(2^(-3/2 + n)*(c + d*x)^n*(1 + (d*x)/c)^(1/2 - n)*Hypergeometric2F1[-3/2, 
5/2 - n, -1/2, (c - d*x)/(2*c)])/((3*c^3*d - 3*c^2*d^2*x)*Sqrt[c^2 - d^2*x 
^2])
 

Rubi [A] (verified)

Time = 0.35 (sec) , antiderivative size = 83, normalized size of antiderivative = 1.00, number of steps used = 3, number of rules used = 3, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.125, Rules used = {474, 473, 79}

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.

\(\displaystyle \int \frac {(c+d x)^n}{\left (c^2-d^2 x^2\right )^{5/2}} \, dx\)

\(\Big \downarrow \) 474

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

\(\Big \downarrow \) 473

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

\(\Big \downarrow \) 79

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

Input:

Int[(c + d*x)^n/(c^2 - d^2*x^2)^(5/2),x]
 

Output:

(2^(-3/2 + n)*(c + d*x)^n*(1 + (d*x)/c)^(3/2 - n)*Hypergeometric2F1[-3/2, 
5/2 - n, -1/2, (c - d*x)/(2*c)])/(3*c*d*(c^2 - d^2*x^2)^(3/2))
 

Defintions of rubi rules used

rule 79

Int[((a_) + (b_.)*(x_))^(m_)*((c_) + (d_.)*(x_))^(n_), x_Symbol] :> Simp[(( 
a + b*x)^(m + 1)/(b*(m + 1)*(b/(b*c - a*d))^n))*Hypergeometric2F1[-n, m + 1 
, m + 2, (-d)*((a + b*x)/(b*c - a*d))], x] /; FreeQ[{a, b, c, d, m, n}, x] 
&&  !IntegerQ[m] &&  !IntegerQ[n] && GtQ[b/(b*c - a*d), 0] && (RationalQ[m] 
 ||  !(RationalQ[n] && GtQ[-d/(b*c - a*d), 0]))
 

rule 473

Int[((c_) + (d_.)*(x_))^(n_)*((a_) + (b_.)*(x_)^2)^(p_), x_Symbol] :> Simp[ 
c^(n - 1)*((a + b*x^2)^(p + 1)/((1 + d*(x/c))^(p + 1)*(a/c + (b*x)/d)^(p + 
1)))   Int[(1 + d*(x/c))^(n + p)*(a/c + (b/d)*x)^p, x], x] /; FreeQ[{a, b, 
c, d, n}, x] && EqQ[b*c^2 + a*d^2, 0] && (IntegerQ[n] || GtQ[c, 0]) &&  !Gt 
Q[a, 0] &&  !(IntegerQ[n] && (IntegerQ[3*p] || IntegerQ[4*p]))
 

rule 474

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

Maple [F]

\[\int \frac {\left (d x +c \right )^{n}}{\left (-d^{2} x^{2}+c^{2}\right )^{\frac {5}{2}}}d x\]

Input:

int((d*x+c)^n/(-d^2*x^2+c^2)^(5/2),x)
 

Output:

int((d*x+c)^n/(-d^2*x^2+c^2)^(5/2),x)
 

Fricas [F]

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

integrate((d*x+c)^n/(-d^2*x^2+c^2)^(5/2),x, algorithm="fricas")
 

Output:

integral(-sqrt(-d^2*x^2 + c^2)*(d*x + c)^n/(d^6*x^6 - 3*c^2*d^4*x^4 + 3*c^ 
4*d^2*x^2 - c^6), x)
 

Sympy [F]

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

integrate((d*x+c)**n/(-d**2*x**2+c**2)**(5/2),x)
 

Output:

Integral((c + d*x)**n/(-(-c + d*x)*(c + d*x))**(5/2), x)
 

Maxima [F]

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

integrate((d*x+c)^n/(-d^2*x^2+c^2)^(5/2),x, algorithm="maxima")
 

Output:

integrate((d*x + c)^n/(-d^2*x^2 + c^2)^(5/2), x)
 

Giac [F]

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

integrate((d*x+c)^n/(-d^2*x^2+c^2)^(5/2),x, algorithm="giac")
 

Output:

integrate((d*x + c)^n/(-d^2*x^2 + c^2)^(5/2), x)
 

Mupad [F(-1)]

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

int((c + d*x)^n/(c^2 - d^2*x^2)^(5/2),x)
 

Output:

int((c + d*x)^n/(c^2 - d^2*x^2)^(5/2), x)
 

Reduce [F]

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

int((d*x+c)^n/(-d^2*x^2+c^2)^(5/2),x)
 

Output:

int((c + d*x)**n/(sqrt(c**2 - d**2*x**2)*c**4 - 2*sqrt(c**2 - d**2*x**2)*c 
**2*d**2*x**2 + sqrt(c**2 - d**2*x**2)*d**4*x**4),x)