Integrand size = 24, antiderivative size = 86 \[ \int \frac {(c+d x)^3}{\left (c^2-d^2 x^2\right )^{7/2}} \, dx=\frac {2 (c+d x)^2}{5 d \left (c^2-d^2 x^2\right )^{5/2}}+\frac {c+d x}{15 c d \left (c^2-d^2 x^2\right )^{3/2}}+\frac {2 x}{15 c^3 \sqrt {c^2-d^2 x^2}} \] Output:
2/5*(d*x+c)^2/d/(-d^2*x^2+c^2)^(5/2)+1/15*(d*x+c)/c/d/(-d^2*x^2+c^2)^(3/2) +2/15*x/c^3/(-d^2*x^2+c^2)^(1/2)
Time = 0.37 (sec) , antiderivative size = 53, normalized size of antiderivative = 0.62 \[ \int \frac {(c+d x)^3}{\left (c^2-d^2 x^2\right )^{7/2}} \, dx=\frac {\sqrt {c^2-d^2 x^2} \left (7 c^2-6 c d x+2 d^2 x^2\right )}{15 c^3 d (c-d x)^3} \] Input:
Integrate[(c + d*x)^3/(c^2 - d^2*x^2)^(7/2),x]
Output:
(Sqrt[c^2 - d^2*x^2]*(7*c^2 - 6*c*d*x + 2*d^2*x^2))/(15*c^3*d*(c - d*x)^3)
Time = 0.38 (sec) , antiderivative size = 111, normalized size of antiderivative = 1.29, number of steps used = 4, number of rules used = 4, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.167, Rules used = {464, 461, 461, 460}
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.
\(\displaystyle \int \frac {(c+d x)^3}{\left (c^2-d^2 x^2\right )^{7/2}} \, dx\) |
\(\Big \downarrow \) 464 |
\(\displaystyle \int \frac {1}{(c-d x)^3 \sqrt {c^2-d^2 x^2}}dx\) |
\(\Big \downarrow \) 461 |
\(\displaystyle \frac {2 \int \frac {1}{(c-d x)^2 \sqrt {c^2-d^2 x^2}}dx}{5 c}+\frac {\sqrt {c^2-d^2 x^2}}{5 c d (c-d x)^3}\) |
\(\Big \downarrow \) 461 |
\(\displaystyle \frac {2 \left (\frac {\int \frac {1}{(c-d x) \sqrt {c^2-d^2 x^2}}dx}{3 c}+\frac {\sqrt {c^2-d^2 x^2}}{3 c d (c-d x)^2}\right )}{5 c}+\frac {\sqrt {c^2-d^2 x^2}}{5 c d (c-d x)^3}\) |
\(\Big \downarrow \) 460 |
\(\displaystyle \frac {2 \left (\frac {\sqrt {c^2-d^2 x^2}}{3 c^2 d (c-d x)}+\frac {\sqrt {c^2-d^2 x^2}}{3 c d (c-d x)^2}\right )}{5 c}+\frac {\sqrt {c^2-d^2 x^2}}{5 c d (c-d x)^3}\) |
Input:
Int[(c + d*x)^3/(c^2 - d^2*x^2)^(7/2),x]
Output:
Sqrt[c^2 - d^2*x^2]/(5*c*d*(c - d*x)^3) + (2*(Sqrt[c^2 - d^2*x^2]/(3*c*d*( c - d*x)^2) + Sqrt[c^2 - d^2*x^2]/(3*c^2*d*(c - d*x))))/(5*c)
Int[((c_) + (d_.)*(x_))^(n_)*((a_) + (b_.)*(x_)^2)^(p_), x_Symbol] :> Simp[ (-d)*(c + d*x)^n*((a + b*x^2)^(p + 1)/(b*c*n)), x] /; FreeQ[{a, b, c, d, n, p}, x] && EqQ[b*c^2 + a*d^2, 0] && EqQ[n + 2*p + 2, 0]
Int[((c_) + (d_.)*(x_))^(n_)*((a_) + (b_.)*(x_)^2)^(p_), x_Symbol] :> Simp[ (-d)*(c + d*x)^n*((a + b*x^2)^(p + 1)/(2*b*c*(n + p + 1))), x] + Simp[Simpl ify[n + 2*p + 2]/(2*c*(n + p + 1)) Int[(c + d*x)^(n + 1)*(a + b*x^2)^p, x ], x] /; FreeQ[{a, b, c, d, n, p}, x] && EqQ[b*c^2 + a*d^2, 0] && ILtQ[Simp lify[n + 2*p + 2], 0] && (LtQ[n, -1] || GtQ[n + p, 0])
Int[((c_) + (d_.)*(x_))^(n_)*((a_) + (b_.)*(x_)^2)^(p_), x_Symbol] :> Int[( a + b*x^2)^(n + p)/(a/c + b*(x/d))^n, x] /; FreeQ[{a, b, c, d}, x] && EqQ[b *c^2 + a*d^2, 0] && IntegerQ[n] && RationalQ[p] && (LtQ[0, -n, p] || LtQ[p, -n, 0]) && NeQ[n, 2] && NeQ[n, -1]
Time = 0.26 (sec) , antiderivative size = 50, normalized size of antiderivative = 0.58
method | result | size |
trager | \(\frac {\left (2 d^{2} x^{2}-6 c d x +7 c^{2}\right ) \sqrt {-d^{2} x^{2}+c^{2}}}{15 c^{3} \left (-d x +c \right )^{3} d}\) | \(50\) |
gosper | \(\frac {\left (d x +c \right )^{4} \left (-d x +c \right ) \left (2 d^{2} x^{2}-6 c d x +7 c^{2}\right )}{15 d \,c^{3} \left (-d^{2} x^{2}+c^{2}\right )^{\frac {7}{2}}}\) | \(55\) |
orering | \(\frac {\left (d x +c \right )^{4} \left (-d x +c \right ) \left (2 d^{2} x^{2}-6 c d x +7 c^{2}\right )}{15 d \,c^{3} \left (-d^{2} x^{2}+c^{2}\right )^{\frac {7}{2}}}\) | \(55\) |
default | \(c^{3} \left (\frac {x}{5 c^{2} \left (-d^{2} x^{2}+c^{2}\right )^{\frac {5}{2}}}+\frac {\frac {4 x}{15 c^{2} \left (-d^{2} x^{2}+c^{2}\right )^{\frac {3}{2}}}+\frac {8 x}{15 c^{4} \sqrt {-d^{2} x^{2}+c^{2}}}}{c^{2}}\right )+d^{3} \left (\frac {x^{2}}{3 d^{2} \left (-d^{2} x^{2}+c^{2}\right )^{\frac {5}{2}}}-\frac {2 c^{2}}{15 d^{4} \left (-d^{2} x^{2}+c^{2}\right )^{\frac {5}{2}}}\right )+3 c \,d^{2} \left (\frac {x}{4 d^{2} \left (-d^{2} x^{2}+c^{2}\right )^{\frac {5}{2}}}-\frac {c^{2} \left (\frac {x}{5 c^{2} \left (-d^{2} x^{2}+c^{2}\right )^{\frac {5}{2}}}+\frac {\frac {4 x}{15 c^{2} \left (-d^{2} x^{2}+c^{2}\right )^{\frac {3}{2}}}+\frac {8 x}{15 c^{4} \sqrt {-d^{2} x^{2}+c^{2}}}}{c^{2}}\right )}{4 d^{2}}\right )+\frac {3 c^{2}}{5 d \left (-d^{2} x^{2}+c^{2}\right )^{\frac {5}{2}}}\) | \(246\) |
Input:
int((d*x+c)^3/(-d^2*x^2+c^2)^(7/2),x,method=_RETURNVERBOSE)
Output:
1/15*(2*d^2*x^2-6*c*d*x+7*c^2)/c^3/(-d*x+c)^3/d*(-d^2*x^2+c^2)^(1/2)
Time = 0.12 (sec) , antiderivative size = 106, normalized size of antiderivative = 1.23 \[ \int \frac {(c+d x)^3}{\left (c^2-d^2 x^2\right )^{7/2}} \, dx=\frac {7 \, d^{3} x^{3} - 21 \, c d^{2} x^{2} + 21 \, c^{2} d x - 7 \, c^{3} - {\left (2 \, d^{2} x^{2} - 6 \, c d x + 7 \, c^{2}\right )} \sqrt {-d^{2} x^{2} + c^{2}}}{15 \, {\left (c^{3} d^{4} x^{3} - 3 \, c^{4} d^{3} x^{2} + 3 \, c^{5} d^{2} x - c^{6} d\right )}} \] Input:
integrate((d*x+c)^3/(-d^2*x^2+c^2)^(7/2),x, algorithm="fricas")
Output:
1/15*(7*d^3*x^3 - 21*c*d^2*x^2 + 21*c^2*d*x - 7*c^3 - (2*d^2*x^2 - 6*c*d*x + 7*c^2)*sqrt(-d^2*x^2 + c^2))/(c^3*d^4*x^3 - 3*c^4*d^3*x^2 + 3*c^5*d^2*x - c^6*d)
\[ \int \frac {(c+d x)^3}{\left (c^2-d^2 x^2\right )^{7/2}} \, dx=\int \frac {\left (c + d x\right )^{3}}{\left (- \left (- c + d x\right ) \left (c + d x\right )\right )^{\frac {7}{2}}}\, dx \] Input:
integrate((d*x+c)**3/(-d**2*x**2+c**2)**(7/2),x)
Output:
Integral((c + d*x)**3/(-(-c + d*x)*(c + d*x))**(7/2), x)
Time = 0.04 (sec) , antiderivative size = 101, normalized size of antiderivative = 1.17 \[ \int \frac {(c+d x)^3}{\left (c^2-d^2 x^2\right )^{7/2}} \, dx=\frac {d x^{2}}{3 \, {\left (-d^{2} x^{2} + c^{2}\right )}^{\frac {5}{2}}} + \frac {4 \, c x}{5 \, {\left (-d^{2} x^{2} + c^{2}\right )}^{\frac {5}{2}}} + \frac {7 \, c^{2}}{15 \, {\left (-d^{2} x^{2} + c^{2}\right )}^{\frac {5}{2}} d} + \frac {x}{15 \, {\left (-d^{2} x^{2} + c^{2}\right )}^{\frac {3}{2}} c} + \frac {2 \, x}{15 \, \sqrt {-d^{2} x^{2} + c^{2}} c^{3}} \] Input:
integrate((d*x+c)^3/(-d^2*x^2+c^2)^(7/2),x, algorithm="maxima")
Output:
1/3*d*x^2/(-d^2*x^2 + c^2)^(5/2) + 4/5*c*x/(-d^2*x^2 + c^2)^(5/2) + 7/15*c ^2/((-d^2*x^2 + c^2)^(5/2)*d) + 1/15*x/((-d^2*x^2 + c^2)^(3/2)*c) + 2/15*x /(sqrt(-d^2*x^2 + c^2)*c^3)
Leaf count of result is larger than twice the leaf count of optimal. 165 vs. \(2 (74) = 148\).
Time = 0.14 (sec) , antiderivative size = 165, normalized size of antiderivative = 1.92 \[ \int \frac {(c+d x)^3}{\left (c^2-d^2 x^2\right )^{7/2}} \, dx=-\frac {2 \, {\left (\frac {20 \, {\left (c d + \sqrt {-d^{2} x^{2} + c^{2}} {\left | d \right |}\right )}}{d^{2} x} - \frac {40 \, {\left (c d + \sqrt {-d^{2} x^{2} + c^{2}} {\left | d \right |}\right )}^{2}}{d^{4} x^{2}} + \frac {30 \, {\left (c d + \sqrt {-d^{2} x^{2} + c^{2}} {\left | d \right |}\right )}^{3}}{d^{6} x^{3}} - \frac {15 \, {\left (c d + \sqrt {-d^{2} x^{2} + c^{2}} {\left | d \right |}\right )}^{4}}{d^{8} x^{4}} - 7\right )}}{15 \, c^{3} {\left (\frac {c d + \sqrt {-d^{2} x^{2} + c^{2}} {\left | d \right |}}{d^{2} x} - 1\right )}^{5} {\left | d \right |}} \] Input:
integrate((d*x+c)^3/(-d^2*x^2+c^2)^(7/2),x, algorithm="giac")
Output:
-2/15*(20*(c*d + sqrt(-d^2*x^2 + c^2)*abs(d))/(d^2*x) - 40*(c*d + sqrt(-d^ 2*x^2 + c^2)*abs(d))^2/(d^4*x^2) + 30*(c*d + sqrt(-d^2*x^2 + c^2)*abs(d))^ 3/(d^6*x^3) - 15*(c*d + sqrt(-d^2*x^2 + c^2)*abs(d))^4/(d^8*x^4) - 7)/(c^3 *((c*d + sqrt(-d^2*x^2 + c^2)*abs(d))/(d^2*x) - 1)^5*abs(d))
Time = 7.34 (sec) , antiderivative size = 49, normalized size of antiderivative = 0.57 \[ \int \frac {(c+d x)^3}{\left (c^2-d^2 x^2\right )^{7/2}} \, dx=\frac {\sqrt {c^2-d^2\,x^2}\,\left (7\,c^2-6\,c\,d\,x+2\,d^2\,x^2\right )}{15\,c^3\,d\,{\left (c-d\,x\right )}^3} \] Input:
int((c + d*x)^3/(c^2 - d^2*x^2)^(7/2),x)
Output:
((c^2 - d^2*x^2)^(1/2)*(7*c^2 + 2*d^2*x^2 - 6*c*d*x))/(15*c^3*d*(c - d*x)^ 3)
Time = 0.20 (sec) , antiderivative size = 120, normalized size of antiderivative = 1.40 \[ \int \frac {(c+d x)^3}{\left (c^2-d^2 x^2\right )^{7/2}} \, dx=\frac {-\frac {2 \tan \left (\frac {\mathit {asin} \left (\frac {d x}{c}\right )}{2}\right )^{5}}{5}-\frac {4 \tan \left (\frac {\mathit {asin} \left (\frac {d x}{c}\right )}{2}\right )^{2}}{3}+\frac {2 \tan \left (\frac {\mathit {asin} \left (\frac {d x}{c}\right )}{2}\right )}{3}-\frac {8}{15}}{c^{3} d \left (\tan \left (\frac {\mathit {asin} \left (\frac {d x}{c}\right )}{2}\right )^{5}-5 \tan \left (\frac {\mathit {asin} \left (\frac {d x}{c}\right )}{2}\right )^{4}+10 \tan \left (\frac {\mathit {asin} \left (\frac {d x}{c}\right )}{2}\right )^{3}-10 \tan \left (\frac {\mathit {asin} \left (\frac {d x}{c}\right )}{2}\right )^{2}+5 \tan \left (\frac {\mathit {asin} \left (\frac {d x}{c}\right )}{2}\right )-1\right )} \] Input:
int((d*x+c)^3/(-d^2*x^2+c^2)^(7/2),x)
Output:
(2*( - 3*tan(asin((d*x)/c)/2)**5 - 10*tan(asin((d*x)/c)/2)**2 + 5*tan(asin ((d*x)/c)/2) - 4))/(15*c**3*d*(tan(asin((d*x)/c)/2)**5 - 5*tan(asin((d*x)/ c)/2)**4 + 10*tan(asin((d*x)/c)/2)**3 - 10*tan(asin((d*x)/c)/2)**2 + 5*tan (asin((d*x)/c)/2) - 1))