Integrand size = 25, antiderivative size = 117 \[ \int (c+d x)^{3/2} \left (A+B x+C x^2+D x^3\right ) \, dx=\frac {2 \left (c^2 C d-B c d^2+A d^3-c^3 D\right ) (c+d x)^{5/2}}{5 d^4}-\frac {2 \left (2 c C d-B d^2-3 c^2 D\right ) (c+d x)^{7/2}}{7 d^4}+\frac {2 (C d-3 c D) (c+d x)^{9/2}}{9 d^4}+\frac {2 D (c+d x)^{11/2}}{11 d^4} \] Output:
2/5*(A*d^3-B*c*d^2+C*c^2*d-D*c^3)*(d*x+c)^(5/2)/d^4-2/7*(-B*d^2+2*C*c*d-3* D*c^2)*(d*x+c)^(7/2)/d^4+2/9*(C*d-3*D*c)*(d*x+c)^(9/2)/d^4+2/11*D*(d*x+c)^ (11/2)/d^4
Time = 0.07 (sec) , antiderivative size = 83, normalized size of antiderivative = 0.71 \[ \int (c+d x)^{3/2} \left (A+B x+C x^2+D x^3\right ) \, dx=\frac {2 (c+d x)^{5/2} \left (-48 c^3 D+8 c^2 d (11 C+15 D x)-2 c d^2 (99 B+5 x (22 C+21 D x))+d^3 \left (693 A+5 x \left (99 B+77 C x+63 D x^2\right )\right )\right )}{3465 d^4} \] Input:
Integrate[(c + d*x)^(3/2)*(A + B*x + C*x^2 + D*x^3),x]
Output:
(2*(c + d*x)^(5/2)*(-48*c^3*D + 8*c^2*d*(11*C + 15*D*x) - 2*c*d^2*(99*B + 5*x*(22*C + 21*D*x)) + d^3*(693*A + 5*x*(99*B + 77*C*x + 63*D*x^2))))/(346 5*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.
\(\displaystyle \int (c+d x)^{3/2} \left (A+B x+C x^2+D x^3\right ) \, dx\) |
\(\Big \downarrow \) 2389 |
\(\displaystyle \int \left (\frac {(c+d x)^{3/2} \left (A d^3-B c d^2+c^3 (-D)+c^2 C d\right )}{d^3}+\frac {(c+d x)^{5/2} \left (B d^2+3 c^2 D-2 c C d\right )}{d^3}+\frac {(c+d x)^{7/2} (C d-3 c D)}{d^3}+\frac {D (c+d x)^{9/2}}{d^3}\right )dx\) |
\(\Big \downarrow \) 2009 |
\(\displaystyle \frac {2 (c+d x)^{5/2} \left (A d^3-B c d^2+c^3 (-D)+c^2 C d\right )}{5 d^4}-\frac {2 (c+d x)^{7/2} \left (-B d^2-3 c^2 D+2 c C d\right )}{7 d^4}+\frac {2 (c+d x)^{9/2} (C d-3 c D)}{9 d^4}+\frac {2 D (c+d x)^{11/2}}{11 d^4}\) |
Input:
Int[(c + d*x)^(3/2)*(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)^(5/2))/(5*d^4) - (2*(2*c* C*d - B*d^2 - 3*c^2*D)*(c + d*x)^(7/2))/(7*d^4) + (2*(C*d - 3*c*D)*(c + d* x)^(9/2))/(9*d^4) + (2*D*(c + d*x)^(11/2))/(11*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 = 0.38 (sec) , antiderivative size = 72, normalized size of antiderivative = 0.62
method | result | size |
pseudoelliptic | \(\frac {2 \left (\left (\frac {5}{11} D x^{3}+\frac {5}{9} C \,x^{2}+\frac {5}{7} B x +A \right ) d^{3}-\frac {2 \left (\frac {35}{33} D x^{2}+\frac {10}{9} C x +B \right ) c \,d^{2}}{7}+\frac {8 \left (\frac {15 D x}{11}+C \right ) c^{2} d}{63}-\frac {16 D c^{3}}{231}\right ) \left (x d +c \right )^{\frac {5}{2}}}{5 d^{4}}\) | \(72\) |
gosper | \(\frac {2 \left (x d +c \right )^{\frac {5}{2}} \left (315 D x^{3} d^{3}+385 C \,x^{2} d^{3}-210 D x^{2} c \,d^{2}+495 B x \,d^{3}-220 C x c \,d^{2}+120 D x \,c^{2} d +693 A \,d^{3}-198 B c \,d^{2}+88 C \,c^{2} d -48 D c^{3}\right )}{3465 d^{4}}\) | \(91\) |
orering | \(\frac {2 \left (x d +c \right )^{\frac {5}{2}} \left (315 D x^{3} d^{3}+385 C \,x^{2} d^{3}-210 D x^{2} c \,d^{2}+495 B x \,d^{3}-220 C x c \,d^{2}+120 D x \,c^{2} d +693 A \,d^{3}-198 B c \,d^{2}+88 C \,c^{2} d -48 D c^{3}\right )}{3465 d^{4}}\) | \(91\) |
derivativedivides | \(\frac {\frac {2 D \left (x d +c \right )^{\frac {11}{2}}}{11}+\frac {2 \left (C d -3 D c \right ) \left (x d +c \right )^{\frac {9}{2}}}{9}+\frac {2 \left (B \,d^{2}-2 C c d +3 D c^{2}\right ) \left (x d +c \right )^{\frac {7}{2}}}{7}+\frac {2 \left (A \,d^{3}-B c \,d^{2}+C \,c^{2} d -D c^{3}\right ) \left (x d +c \right )^{\frac {5}{2}}}{5}}{d^{4}}\) | \(94\) |
default | \(\frac {\frac {2 D \left (x d +c \right )^{\frac {11}{2}}}{11}+\frac {2 \left (C d -3 D c \right ) \left (x d +c \right )^{\frac {9}{2}}}{9}+\frac {2 \left (B \,d^{2}-2 C c d +3 D c^{2}\right ) \left (x d +c \right )^{\frac {7}{2}}}{7}+\frac {2 \left (A \,d^{3}-B c \,d^{2}+C \,c^{2} d -D c^{3}\right ) \left (x d +c \right )^{\frac {5}{2}}}{5}}{d^{4}}\) | \(94\) |
trager | \(\frac {2 \left (315 d^{5} D x^{5}+385 C \,d^{5} x^{4}+420 D c \,d^{4} x^{4}+495 B \,d^{5} x^{3}+550 C c \,d^{4} x^{3}+15 D c^{2} d^{3} x^{3}+693 A \,d^{5} x^{2}+792 B c \,d^{4} x^{2}+33 C \,c^{2} d^{3} x^{2}-18 D c^{3} d^{2} x^{2}+1386 A c \,d^{4} x +99 B \,c^{2} d^{3} x -44 C \,c^{3} d^{2} x +24 D c^{4} d x +693 A \,c^{2} d^{3}-198 B \,c^{3} d^{2}+88 C \,c^{4} d -48 D c^{5}\right ) \sqrt {x d +c}}{3465 d^{4}}\) | \(183\) |
Input:
int((d*x+c)^(3/2)*(D*x^3+C*x^2+B*x+A),x,method=_RETURNVERBOSE)
Output:
2/5*((5/11*D*x^3+5/9*C*x^2+5/7*B*x+A)*d^3-2/7*(35/33*D*x^2+10/9*C*x+B)*c*d ^2+8/63*(15/11*D*x+C)*c^2*d-16/231*D*c^3)*(d*x+c)^(5/2)/d^4
Time = 0.07 (sec) , antiderivative size = 172, normalized size of antiderivative = 1.47 \[ \int (c+d x)^{3/2} \left (A+B x+C x^2+D x^3\right ) \, dx=\frac {2 \, {\left (315 \, D d^{5} x^{5} - 48 \, D c^{5} + 88 \, C c^{4} d - 198 \, B c^{3} d^{2} + 693 \, A c^{2} d^{3} + 35 \, {\left (12 \, D c d^{4} + 11 \, C d^{5}\right )} x^{4} + 5 \, {\left (3 \, D c^{2} d^{3} + 110 \, C c d^{4} + 99 \, B d^{5}\right )} x^{3} - 3 \, {\left (6 \, D c^{3} d^{2} - 11 \, C c^{2} d^{3} - 264 \, B c d^{4} - 231 \, A d^{5}\right )} x^{2} + {\left (24 \, D c^{4} d - 44 \, C c^{3} d^{2} + 99 \, B c^{2} d^{3} + 1386 \, A c d^{4}\right )} x\right )} \sqrt {d x + c}}{3465 \, d^{4}} \] Input:
integrate((d*x+c)^(3/2)*(D*x^3+C*x^2+B*x+A),x, algorithm="fricas")
Output:
2/3465*(315*D*d^5*x^5 - 48*D*c^5 + 88*C*c^4*d - 198*B*c^3*d^2 + 693*A*c^2* d^3 + 35*(12*D*c*d^4 + 11*C*d^5)*x^4 + 5*(3*D*c^2*d^3 + 110*C*c*d^4 + 99*B *d^5)*x^3 - 3*(6*D*c^3*d^2 - 11*C*c^2*d^3 - 264*B*c*d^4 - 231*A*d^5)*x^2 + (24*D*c^4*d - 44*C*c^3*d^2 + 99*B*c^2*d^3 + 1386*A*c*d^4)*x)*sqrt(d*x + c )/d^4
Time = 0.93 (sec) , antiderivative size = 143, normalized size of antiderivative = 1.22 \[ \int (c+d x)^{3/2} \left (A+B x+C x^2+D x^3\right ) \, dx=\begin {cases} \frac {2 \left (\frac {D \left (c + d x\right )^{\frac {11}{2}}}{11 d^{3}} + \frac {\left (c + d x\right )^{\frac {9}{2}} \left (C d - 3 D c\right )}{9 d^{3}} + \frac {\left (c + d x\right )^{\frac {7}{2}} \left (B d^{2} - 2 C c d + 3 D c^{2}\right )}{7 d^{3}} + \frac {\left (c + d x\right )^{\frac {5}{2}} \left (A d^{3} - B c d^{2} + C c^{2} d - D c^{3}\right )}{5 d^{3}}\right )}{d} & \text {for}\: d \neq 0 \\c^{\frac {3}{2}} \left (A x + \frac {B x^{2}}{2} + \frac {C x^{3}}{3} + \frac {D x^{4}}{4}\right ) & \text {otherwise} \end {cases} \] Input:
integrate((d*x+c)**(3/2)*(D*x**3+C*x**2+B*x+A),x)
Output:
Piecewise((2*(D*(c + d*x)**(11/2)/(11*d**3) + (c + d*x)**(9/2)*(C*d - 3*D* c)/(9*d**3) + (c + d*x)**(7/2)*(B*d**2 - 2*C*c*d + 3*D*c**2)/(7*d**3) + (c + d*x)**(5/2)*(A*d**3 - B*c*d**2 + C*c**2*d - D*c**3)/(5*d**3))/d, Ne(d, 0)), (c**(3/2)*(A*x + B*x**2/2 + C*x**3/3 + D*x**4/4), True))
Time = 0.03 (sec) , antiderivative size = 94, normalized size of antiderivative = 0.80 \[ \int (c+d x)^{3/2} \left (A+B x+C x^2+D x^3\right ) \, dx=\frac {2 \, {\left (315 \, {\left (d x + c\right )}^{\frac {11}{2}} D - 385 \, {\left (3 \, D c - C d\right )} {\left (d x + c\right )}^{\frac {9}{2}} + 495 \, {\left (3 \, D c^{2} - 2 \, C c d + B d^{2}\right )} {\left (d x + c\right )}^{\frac {7}{2}} - 693 \, {\left (D c^{3} - C c^{2} d + B c d^{2} - A d^{3}\right )} {\left (d x + c\right )}^{\frac {5}{2}}\right )}}{3465 \, d^{4}} \] Input:
integrate((d*x+c)^(3/2)*(D*x^3+C*x^2+B*x+A),x, algorithm="maxima")
Output:
2/3465*(315*(d*x + c)^(11/2)*D - 385*(3*D*c - C*d)*(d*x + c)^(9/2) + 495*( 3*D*c^2 - 2*C*c*d + B*d^2)*(d*x + c)^(7/2) - 693*(D*c^3 - C*c^2*d + B*c*d^ 2 - A*d^3)*(d*x + c)^(5/2))/d^4
Leaf count of result is larger than twice the leaf count of optimal. 536 vs. \(2 (101) = 202\).
Time = 0.13 (sec) , antiderivative size = 536, normalized size of antiderivative = 4.58 \[ \int (c+d x)^{3/2} \left (A+B x+C x^2+D x^3\right ) \, dx =\text {Too large to display} \] Input:
integrate((d*x+c)^(3/2)*(D*x^3+C*x^2+B*x+A),x, algorithm="giac")
Output:
2/3465*(3465*sqrt(d*x + c)*A*c^2 + 2310*((d*x + c)^(3/2) - 3*sqrt(d*x + c) *c)*A*c + 1155*((d*x + c)^(3/2) - 3*sqrt(d*x + c)*c)*B*c^2/d + 231*(3*(d*x + c)^(5/2) - 10*(d*x + c)^(3/2)*c + 15*sqrt(d*x + c)*c^2)*A + 231*(3*(d*x + c)^(5/2) - 10*(d*x + c)^(3/2)*c + 15*sqrt(d*x + c)*c^2)*C*c^2/d^2 + 462 *(3*(d*x + c)^(5/2) - 10*(d*x + c)^(3/2)*c + 15*sqrt(d*x + c)*c^2)*B*c/d + 99*(5*(d*x + c)^(7/2) - 21*(d*x + c)^(5/2)*c + 35*(d*x + c)^(3/2)*c^2 - 3 5*sqrt(d*x + c)*c^3)*D*c^2/d^3 + 198*(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/d^2 + 99*(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/d + 22*(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/d^3 + 11*(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/d^2 + 5*(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/d^3)/d
Timed out. \[ \int (c+d x)^{3/2} \left (A+B x+C x^2+D x^3\right ) \, dx=\int {\left (c+d\,x\right )}^{3/2}\,\left (A+B\,x+C\,x^2+x^3\,D\right ) \,d x \] Input:
int((c + d*x)^(3/2)*(A + B*x + C*x^2 + x^3*D),x)
Output:
int((c + d*x)^(3/2)*(A + B*x + C*x^2 + x^3*D), x)
Time = 0.14 (sec) , antiderivative size = 125, normalized size of antiderivative = 1.07 \[ \int (c+d x)^{3/2} \left (A+B x+C x^2+D x^3\right ) \, dx=\frac {2 \sqrt {d x +c}\, \left (315 d^{5} x^{5}+805 c \,d^{4} x^{4}+495 b \,d^{4} x^{3}+565 c^{2} d^{3} x^{3}+693 a \,d^{4} x^{2}+792 b c \,d^{3} x^{2}+15 c^{3} d^{2} x^{2}+1386 a c \,d^{3} x +99 b \,c^{2} d^{2} x -20 c^{4} d x +693 a \,c^{2} d^{2}-198 b \,c^{3} d +40 c^{5}\right )}{3465 d^{3}} \] Input:
int((d*x+c)^(3/2)*(D*x^3+C*x^2+B*x+A),x)
Output:
(2*sqrt(c + d*x)*(693*a*c**2*d**2 + 1386*a*c*d**3*x + 693*a*d**4*x**2 - 19 8*b*c**3*d + 99*b*c**2*d**2*x + 792*b*c*d**3*x**2 + 495*b*d**4*x**3 + 40*c **5 - 20*c**4*d*x + 15*c**3*d**2*x**2 + 565*c**2*d**3*x**3 + 805*c*d**4*x* *4 + 315*d**5*x**5))/(3465*d**3)