Integrand size = 25, antiderivative size = 117 \[ \int (c+d x)^{5/2} \left (A+B x+C x^2+D x^3\right ) \, dx=\frac {2 \left (c^2 C d-B c d^2+A d^3-c^3 D\right ) (c+d x)^{7/2}}{7 d^4}-\frac {2 \left (2 c C d-B d^2-3 c^2 D\right ) (c+d x)^{9/2}}{9 d^4}+\frac {2 (C d-3 c D) (c+d x)^{11/2}}{11 d^4}+\frac {2 D (c+d x)^{13/2}}{13 d^4} \] Output:
2/7*(A*d^3-B*c*d^2+C*c^2*d-D*c^3)*(d*x+c)^(7/2)/d^4-2/9*(-B*d^2+2*C*c*d-3* D*c^2)*(d*x+c)^(9/2)/d^4+2/11*(C*d-3*D*c)*(d*x+c)^(11/2)/d^4+2/13*D*(d*x+c )^(13/2)/d^4
Time = 0.08 (sec) , antiderivative size = 83, normalized size of antiderivative = 0.71 \[ \int (c+d x)^{5/2} \left (A+B x+C x^2+D x^3\right ) \, dx=\frac {2 (c+d x)^{7/2} \left (-48 c^3 D+8 c^2 d (13 C+21 D x)-2 c d^2 (143 B+7 x (26 C+27 D x))+d^3 \left (1287 A+7 x \left (143 B+117 C x+99 D x^2\right )\right )\right )}{9009 d^4} \] Input:
Integrate[(c + d*x)^(5/2)*(A + B*x + C*x^2 + D*x^3),x]
Output:
(2*(c + d*x)^(7/2)*(-48*c^3*D + 8*c^2*d*(13*C + 21*D*x) - 2*c*d^2*(143*B + 7*x*(26*C + 27*D*x)) + d^3*(1287*A + 7*x*(143*B + 117*C*x + 99*D*x^2))))/ (9009*d^4)
Time = 0.30 (sec) , antiderivative size = 117, normalized size of antiderivative = 1.00, number of steps used = 2, number of rules used = 2, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.080, Rules used = {2389, 2009}
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.
\(\displaystyle \int (c+d x)^{5/2} \left (A+B x+C x^2+D x^3\right ) \, dx\) |
\(\Big \downarrow \) 2389 |
\(\displaystyle \int \left (\frac {(c+d x)^{5/2} \left (A d^3-B c d^2+c^3 (-D)+c^2 C d\right )}{d^3}+\frac {(c+d x)^{7/2} \left (B d^2+3 c^2 D-2 c C d\right )}{d^3}+\frac {(c+d x)^{9/2} (C d-3 c D)}{d^3}+\frac {D (c+d x)^{11/2}}{d^3}\right )dx\) |
\(\Big \downarrow \) 2009 |
\(\displaystyle \frac {2 (c+d x)^{7/2} \left (A d^3-B c d^2+c^3 (-D)+c^2 C d\right )}{7 d^4}-\frac {2 (c+d x)^{9/2} \left (-B d^2-3 c^2 D+2 c C d\right )}{9 d^4}+\frac {2 (c+d x)^{11/2} (C d-3 c D)}{11 d^4}+\frac {2 D (c+d x)^{13/2}}{13 d^4}\) |
Input:
Int[(c + d*x)^(5/2)*(A + B*x + C*x^2 + D*x^3),x]
Output:
(2*(c^2*C*d - B*c*d^2 + A*d^3 - c^3*D)*(c + d*x)^(7/2))/(7*d^4) - (2*(2*c* C*d - B*d^2 - 3*c^2*D)*(c + d*x)^(9/2))/(9*d^4) + (2*(C*d - 3*c*D)*(c + d* x)^(11/2))/(11*d^4) + (2*D*(c + d*x)^(13/2))/(13*d^4)
Int[(Pq_)*((a_) + (b_.)*(x_)^(n_.))^(p_.), x_Symbol] :> Int[ExpandIntegrand [Pq*(a + b*x^n)^p, x], x] /; FreeQ[{a, b, n}, x] && PolyQ[Pq, x] && (IGtQ[p , 0] || EqQ[n, 1])
Time = 0.39 (sec) , antiderivative size = 72, normalized size of antiderivative = 0.62
method | result | size |
pseudoelliptic | \(\frac {2 \left (\left (\frac {7}{13} D x^{3}+\frac {7}{11} C \,x^{2}+\frac {7}{9} B x +A \right ) d^{3}-\frac {2 \left (\frac {189}{143} D x^{2}+\frac {14}{11} C x +B \right ) c \,d^{2}}{9}+\frac {8 c^{2} \left (\frac {21 D x}{13}+C \right ) d}{99}-\frac {16 D c^{3}}{429}\right ) \left (x d +c \right )^{\frac {7}{2}}}{7 d^{4}}\) | \(72\) |
gosper | \(\frac {2 \left (x d +c \right )^{\frac {7}{2}} \left (693 D x^{3} d^{3}+819 C \,x^{2} d^{3}-378 D x^{2} c \,d^{2}+1001 B x \,d^{3}-364 C x c \,d^{2}+168 D x \,c^{2} d +1287 A \,d^{3}-286 B c \,d^{2}+104 C \,c^{2} d -48 D c^{3}\right )}{9009 d^{4}}\) | \(91\) |
orering | \(\frac {2 \left (x d +c \right )^{\frac {7}{2}} \left (693 D x^{3} d^{3}+819 C \,x^{2} d^{3}-378 D x^{2} c \,d^{2}+1001 B x \,d^{3}-364 C x c \,d^{2}+168 D x \,c^{2} d +1287 A \,d^{3}-286 B c \,d^{2}+104 C \,c^{2} d -48 D c^{3}\right )}{9009 d^{4}}\) | \(91\) |
derivativedivides | \(\frac {\frac {2 D \left (x d +c \right )^{\frac {13}{2}}}{13}+\frac {2 \left (C d -3 D c \right ) \left (x d +c \right )^{\frac {11}{2}}}{11}+\frac {2 \left (B \,d^{2}-2 C c d +3 D c^{2}\right ) \left (x d +c \right )^{\frac {9}{2}}}{9}+\frac {2 \left (A \,d^{3}-B c \,d^{2}+C \,c^{2} d -D c^{3}\right ) \left (x d +c \right )^{\frac {7}{2}}}{7}}{d^{4}}\) | \(94\) |
default | \(\frac {\frac {2 D \left (x d +c \right )^{\frac {13}{2}}}{13}+\frac {2 \left (C d -3 D c \right ) \left (x d +c \right )^{\frac {11}{2}}}{11}+\frac {2 \left (B \,d^{2}-2 C c d +3 D c^{2}\right ) \left (x d +c \right )^{\frac {9}{2}}}{9}+\frac {2 \left (A \,d^{3}-B c \,d^{2}+C \,c^{2} d -D c^{3}\right ) \left (x d +c \right )^{\frac {7}{2}}}{7}}{d^{4}}\) | \(94\) |
trager | \(\frac {2 \left (693 d^{6} D x^{6}+819 C \,d^{6} x^{5}+1701 D c \,d^{5} x^{5}+1001 B \,d^{6} x^{4}+2093 C c \,d^{5} x^{4}+1113 D c^{2} d^{4} x^{4}+1287 A \,d^{6} x^{3}+2717 B c \,d^{5} x^{3}+1469 C \,c^{2} d^{4} x^{3}+15 D c^{3} d^{3} x^{3}+3861 A c \,d^{5} x^{2}+2145 B \,c^{2} d^{4} x^{2}+39 C \,c^{3} d^{3} x^{2}-18 D c^{4} d^{2} x^{2}+3861 A \,c^{2} d^{4} x +143 B \,c^{3} d^{3} x -52 C \,c^{4} d^{2} x +24 D c^{5} d x +1287 A \,c^{3} d^{3}-286 B \,c^{4} d^{2}+104 C \,c^{5} d -48 D c^{6}\right ) \sqrt {x d +c}}{9009 d^{4}}\) | \(231\) |
Input:
int((d*x+c)^(5/2)*(D*x^3+C*x^2+B*x+A),x,method=_RETURNVERBOSE)
Output:
2/7*((7/13*D*x^3+7/11*C*x^2+7/9*B*x+A)*d^3-2/9*(189/143*D*x^2+14/11*C*x+B) *c*d^2+8/99*c^2*(21/13*D*x+C)*d-16/429*D*c^3)*(d*x+c)^(7/2)/d^4
Leaf count of result is larger than twice the leaf count of optimal. 213 vs. \(2 (101) = 202\).
Time = 0.07 (sec) , antiderivative size = 213, normalized size of antiderivative = 1.82 \[ \int (c+d x)^{5/2} \left (A+B x+C x^2+D x^3\right ) \, dx=\frac {2 \, {\left (693 \, D d^{6} x^{6} - 48 \, D c^{6} + 104 \, C c^{5} d - 286 \, B c^{4} d^{2} + 1287 \, A c^{3} d^{3} + 63 \, {\left (27 \, D c d^{5} + 13 \, C d^{6}\right )} x^{5} + 7 \, {\left (159 \, D c^{2} d^{4} + 299 \, C c d^{5} + 143 \, B d^{6}\right )} x^{4} + {\left (15 \, D c^{3} d^{3} + 1469 \, C c^{2} d^{4} + 2717 \, B c d^{5} + 1287 \, A d^{6}\right )} x^{3} - 3 \, {\left (6 \, D c^{4} d^{2} - 13 \, C c^{3} d^{3} - 715 \, B c^{2} d^{4} - 1287 \, A c d^{5}\right )} x^{2} + {\left (24 \, D c^{5} d - 52 \, C c^{4} d^{2} + 143 \, B c^{3} d^{3} + 3861 \, A c^{2} d^{4}\right )} x\right )} \sqrt {d x + c}}{9009 \, d^{4}} \] Input:
integrate((d*x+c)^(5/2)*(D*x^3+C*x^2+B*x+A),x, algorithm="fricas")
Output:
2/9009*(693*D*d^6*x^6 - 48*D*c^6 + 104*C*c^5*d - 286*B*c^4*d^2 + 1287*A*c^ 3*d^3 + 63*(27*D*c*d^5 + 13*C*d^6)*x^5 + 7*(159*D*c^2*d^4 + 299*C*c*d^5 + 143*B*d^6)*x^4 + (15*D*c^3*d^3 + 1469*C*c^2*d^4 + 2717*B*c*d^5 + 1287*A*d^ 6)*x^3 - 3*(6*D*c^4*d^2 - 13*C*c^3*d^3 - 715*B*c^2*d^4 - 1287*A*c*d^5)*x^2 + (24*D*c^5*d - 52*C*c^4*d^2 + 143*B*c^3*d^3 + 3861*A*c^2*d^4)*x)*sqrt(d* x + c)/d^4
Leaf count of result is larger than twice the leaf count of optimal. 483 vs. \(2 (128) = 256\).
Time = 0.45 (sec) , antiderivative size = 483, normalized size of antiderivative = 4.13 \[ \int (c+d x)^{5/2} \left (A+B x+C x^2+D x^3\right ) \, dx=\begin {cases} \frac {2 A c^{3} \sqrt {c + d x}}{7 d} + \frac {6 A c^{2} x \sqrt {c + d x}}{7} + \frac {6 A c d x^{2} \sqrt {c + d x}}{7} + \frac {2 A d^{2} x^{3} \sqrt {c + d x}}{7} - \frac {4 B c^{4} \sqrt {c + d x}}{63 d^{2}} + \frac {2 B c^{3} x \sqrt {c + d x}}{63 d} + \frac {10 B c^{2} x^{2} \sqrt {c + d x}}{21} + \frac {38 B c d x^{3} \sqrt {c + d x}}{63} + \frac {2 B d^{2} x^{4} \sqrt {c + d x}}{9} + \frac {16 C c^{5} \sqrt {c + d x}}{693 d^{3}} - \frac {8 C c^{4} x \sqrt {c + d x}}{693 d^{2}} + \frac {2 C c^{3} x^{2} \sqrt {c + d x}}{231 d} + \frac {226 C c^{2} x^{3} \sqrt {c + d x}}{693} + \frac {46 C c d x^{4} \sqrt {c + d x}}{99} + \frac {2 C d^{2} x^{5} \sqrt {c + d x}}{11} - \frac {32 D c^{6} \sqrt {c + d x}}{3003 d^{4}} + \frac {16 D c^{5} x \sqrt {c + d x}}{3003 d^{3}} - \frac {4 D c^{4} x^{2} \sqrt {c + d x}}{1001 d^{2}} + \frac {10 D c^{3} x^{3} \sqrt {c + d x}}{3003 d} + \frac {106 D c^{2} x^{4} \sqrt {c + d x}}{429} + \frac {54 D c d x^{5} \sqrt {c + d x}}{143} + \frac {2 D d^{2} x^{6} \sqrt {c + d x}}{13} & \text {for}\: d \neq 0 \\c^{\frac {5}{2}} \left (A x + \frac {B x^{2}}{2} + \frac {C x^{3}}{3} + \frac {D x^{4}}{4}\right ) & \text {otherwise} \end {cases} \] Input:
integrate((d*x+c)**(5/2)*(D*x**3+C*x**2+B*x+A),x)
Output:
Piecewise((2*A*c**3*sqrt(c + d*x)/(7*d) + 6*A*c**2*x*sqrt(c + d*x)/7 + 6*A *c*d*x**2*sqrt(c + d*x)/7 + 2*A*d**2*x**3*sqrt(c + d*x)/7 - 4*B*c**4*sqrt( c + d*x)/(63*d**2) + 2*B*c**3*x*sqrt(c + d*x)/(63*d) + 10*B*c**2*x**2*sqrt (c + d*x)/21 + 38*B*c*d*x**3*sqrt(c + d*x)/63 + 2*B*d**2*x**4*sqrt(c + d*x )/9 + 16*C*c**5*sqrt(c + d*x)/(693*d**3) - 8*C*c**4*x*sqrt(c + d*x)/(693*d **2) + 2*C*c**3*x**2*sqrt(c + d*x)/(231*d) + 226*C*c**2*x**3*sqrt(c + d*x) /693 + 46*C*c*d*x**4*sqrt(c + d*x)/99 + 2*C*d**2*x**5*sqrt(c + d*x)/11 - 3 2*D*c**6*sqrt(c + d*x)/(3003*d**4) + 16*D*c**5*x*sqrt(c + d*x)/(3003*d**3) - 4*D*c**4*x**2*sqrt(c + d*x)/(1001*d**2) + 10*D*c**3*x**3*sqrt(c + d*x)/ (3003*d) + 106*D*c**2*x**4*sqrt(c + d*x)/429 + 54*D*c*d*x**5*sqrt(c + d*x) /143 + 2*D*d**2*x**6*sqrt(c + d*x)/13, Ne(d, 0)), (c**(5/2)*(A*x + B*x**2/ 2 + C*x**3/3 + D*x**4/4), True))
Time = 0.04 (sec) , antiderivative size = 94, normalized size of antiderivative = 0.80 \[ \int (c+d x)^{5/2} \left (A+B x+C x^2+D x^3\right ) \, dx=\frac {2 \, {\left (693 \, {\left (d x + c\right )}^{\frac {13}{2}} D - 819 \, {\left (3 \, D c - C d\right )} {\left (d x + c\right )}^{\frac {11}{2}} + 1001 \, {\left (3 \, D c^{2} - 2 \, C c d + B d^{2}\right )} {\left (d x + c\right )}^{\frac {9}{2}} - 1287 \, {\left (D c^{3} - C c^{2} d + B c d^{2} - A d^{3}\right )} {\left (d x + c\right )}^{\frac {7}{2}}\right )}}{9009 \, d^{4}} \] Input:
integrate((d*x+c)^(5/2)*(D*x^3+C*x^2+B*x+A),x, algorithm="maxima")
Output:
2/9009*(693*(d*x + c)^(13/2)*D - 819*(3*D*c - C*d)*(d*x + c)^(11/2) + 1001 *(3*D*c^2 - 2*C*c*d + B*d^2)*(d*x + c)^(9/2) - 1287*(D*c^3 - C*c^2*d + B*c *d^2 - A*d^3)*(d*x + c)^(7/2))/d^4
Leaf count of result is larger than twice the leaf count of optimal. 817 vs. \(2 (101) = 202\).
Time = 0.13 (sec) , antiderivative size = 817, normalized size of antiderivative = 6.98 \[ \int (c+d x)^{5/2} \left (A+B x+C x^2+D x^3\right ) \, dx=\text {Too large to display} \] Input:
integrate((d*x+c)^(5/2)*(D*x^3+C*x^2+B*x+A),x, algorithm="giac")
Output:
2/45045*(45045*sqrt(d*x + c)*A*c^3 + 45045*((d*x + c)^(3/2) - 3*sqrt(d*x + c)*c)*A*c^2 + 15015*((d*x + c)^(3/2) - 3*sqrt(d*x + c)*c)*B*c^3/d + 9009* (3*(d*x + c)^(5/2) - 10*(d*x + c)^(3/2)*c + 15*sqrt(d*x + c)*c^2)*A*c + 30 03*(3*(d*x + c)^(5/2) - 10*(d*x + c)^(3/2)*c + 15*sqrt(d*x + c)*c^2)*C*c^3 /d^2 + 9009*(3*(d*x + c)^(5/2) - 10*(d*x + c)^(3/2)*c + 15*sqrt(d*x + c)*c ^2)*B*c^2/d + 1287*(5*(d*x + c)^(7/2) - 21*(d*x + c)^(5/2)*c + 35*(d*x + c )^(3/2)*c^2 - 35*sqrt(d*x + c)*c^3)*A + 1287*(5*(d*x + c)^(7/2) - 21*(d*x + c)^(5/2)*c + 35*(d*x + c)^(3/2)*c^2 - 35*sqrt(d*x + c)*c^3)*D*c^3/d^3 + 3861*(5*(d*x + c)^(7/2) - 21*(d*x + c)^(5/2)*c + 35*(d*x + c)^(3/2)*c^2 - 35*sqrt(d*x + c)*c^3)*C*c^2/d^2 + 3861*(5*(d*x + c)^(7/2) - 21*(d*x + c)^( 5/2)*c + 35*(d*x + c)^(3/2)*c^2 - 35*sqrt(d*x + c)*c^3)*B*c/d + 429*(35*(d *x + c)^(9/2) - 180*(d*x + c)^(7/2)*c + 378*(d*x + c)^(5/2)*c^2 - 420*(d*x + c)^(3/2)*c^3 + 315*sqrt(d*x + c)*c^4)*D*c^2/d^3 + 429*(35*(d*x + c)^(9/ 2) - 180*(d*x + c)^(7/2)*c + 378*(d*x + c)^(5/2)*c^2 - 420*(d*x + c)^(3/2) *c^3 + 315*sqrt(d*x + c)*c^4)*C*c/d^2 + 143*(35*(d*x + c)^(9/2) - 180*(d*x + c)^(7/2)*c + 378*(d*x + c)^(5/2)*c^2 - 420*(d*x + c)^(3/2)*c^3 + 315*sq rt(d*x + c)*c^4)*B/d + 195*(63*(d*x + c)^(11/2) - 385*(d*x + c)^(9/2)*c + 990*(d*x + c)^(7/2)*c^2 - 1386*(d*x + c)^(5/2)*c^3 + 1155*(d*x + c)^(3/2)* c^4 - 693*sqrt(d*x + c)*c^5)*D*c/d^3 + 65*(63*(d*x + c)^(11/2) - 385*(d*x + c)^(9/2)*c + 990*(d*x + c)^(7/2)*c^2 - 1386*(d*x + c)^(5/2)*c^3 + 115...
Timed out. \[ \int (c+d x)^{5/2} \left (A+B x+C x^2+D x^3\right ) \, dx=\int {\left (c+d\,x\right )}^{5/2}\,\left (A+B\,x+C\,x^2+x^3\,D\right ) \,d x \] Input:
int((c + d*x)^(5/2)*(A + B*x + C*x^2 + x^3*D),x)
Output:
int((c + d*x)^(5/2)*(A + B*x + C*x^2 + x^3*D), x)
Time = 0.14 (sec) , antiderivative size = 160, normalized size of antiderivative = 1.37 \[ \int (c+d x)^{5/2} \left (A+B x+C x^2+D x^3\right ) \, dx=\frac {2 \sqrt {d x +c}\, \left (693 d^{6} x^{6}+2520 c \,d^{5} x^{5}+1001 b \,d^{5} x^{4}+3206 c^{2} d^{4} x^{4}+1287 a \,d^{5} x^{3}+2717 b c \,d^{4} x^{3}+1484 c^{3} d^{3} x^{3}+3861 a c \,d^{4} x^{2}+2145 b \,c^{2} d^{3} x^{2}+21 c^{4} d^{2} x^{2}+3861 a \,c^{2} d^{3} x +143 b \,c^{3} d^{2} x -28 c^{5} d x +1287 a \,c^{3} d^{2}-286 b \,c^{4} d +56 c^{6}\right )}{9009 d^{3}} \] Input:
int((d*x+c)^(5/2)*(D*x^3+C*x^2+B*x+A),x)
Output:
(2*sqrt(c + d*x)*(1287*a*c**3*d**2 + 3861*a*c**2*d**3*x + 3861*a*c*d**4*x* *2 + 1287*a*d**5*x**3 - 286*b*c**4*d + 143*b*c**3*d**2*x + 2145*b*c**2*d** 3*x**2 + 2717*b*c*d**4*x**3 + 1001*b*d**5*x**4 + 56*c**6 - 28*c**5*d*x + 2 1*c**4*d**2*x**2 + 1484*c**3*d**3*x**3 + 3206*c**2*d**4*x**4 + 2520*c*d**5 *x**5 + 693*d**6*x**6))/(9009*d**3)