Integrand size = 24, antiderivative size = 94 \[ \int \frac {(c+d x) \left (a x^2+b x^3\right )^{5/2}}{x^4} \, dx=-\frac {2 a (b c-a d) \left (a x^2+b x^3\right )^{7/2}}{7 b^3 x^7}+\frac {2 (b c-2 a d) \left (a x^2+b x^3\right )^{9/2}}{9 b^3 x^9}+\frac {2 d \left (a x^2+b x^3\right )^{11/2}}{11 b^3 x^{11}} \] Output:
-2/7*a*(-a*d+b*c)*(b*x^3+a*x^2)^(7/2)/b^3/x^7+2/9*(-2*a*d+b*c)*(b*x^3+a*x^ 2)^(9/2)/b^3/x^9+2/11*d*(b*x^3+a*x^2)^(11/2)/b^3/x^11
Time = 0.05 (sec) , antiderivative size = 61, normalized size of antiderivative = 0.65 \[ \int \frac {(c+d x) \left (a x^2+b x^3\right )^{5/2}}{x^4} \, dx=\frac {2 x (a+b x)^4 \left (8 a^2 d+7 b^2 x (11 c+9 d x)-2 a b (11 c+14 d x)\right )}{693 b^3 \sqrt {x^2 (a+b x)}} \] Input:
Integrate[((c + d*x)*(a*x^2 + b*x^3)^(5/2))/x^4,x]
Output:
(2*x*(a + b*x)^4*(8*a^2*d + 7*b^2*x*(11*c + 9*d*x) - 2*a*b*(11*c + 14*d*x) ))/(693*b^3*Sqrt[x^2*(a + b*x)])
Time = 0.44 (sec) , antiderivative size = 95, normalized size of antiderivative = 1.01, number of steps used = 3, number of rules used = 3, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.125, Rules used = {1945, 1922, 1920}
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.
\(\displaystyle \int \frac {\left (a x^2+b x^3\right )^{5/2} (c+d x)}{x^4} \, dx\) |
\(\Big \downarrow \) 1945 |
\(\displaystyle \frac {(11 b c-4 a d) \int \frac {\left (b x^3+a x^2\right )^{5/2}}{x^4}dx}{11 b}+\frac {2 d \left (a x^2+b x^3\right )^{7/2}}{11 b x^5}\) |
\(\Big \downarrow \) 1922 |
\(\displaystyle \frac {(11 b c-4 a d) \left (\frac {2 \left (a x^2+b x^3\right )^{7/2}}{9 b x^6}-\frac {2 a \int \frac {\left (b x^3+a x^2\right )^{5/2}}{x^5}dx}{9 b}\right )}{11 b}+\frac {2 d \left (a x^2+b x^3\right )^{7/2}}{11 b x^5}\) |
\(\Big \downarrow \) 1920 |
\(\displaystyle \frac {\left (\frac {2 \left (a x^2+b x^3\right )^{7/2}}{9 b x^6}-\frac {4 a \left (a x^2+b x^3\right )^{7/2}}{63 b^2 x^7}\right ) (11 b c-4 a d)}{11 b}+\frac {2 d \left (a x^2+b x^3\right )^{7/2}}{11 b x^5}\) |
Input:
Int[((c + d*x)*(a*x^2 + b*x^3)^(5/2))/x^4,x]
Output:
(2*d*(a*x^2 + b*x^3)^(7/2))/(11*b*x^5) + ((11*b*c - 4*a*d)*((-4*a*(a*x^2 + b*x^3)^(7/2))/(63*b^2*x^7) + (2*(a*x^2 + b*x^3)^(7/2))/(9*b*x^6)))/(11*b)
Int[((c_.)*(x_))^(m_.)*((a_.)*(x_)^(j_.) + (b_.)*(x_)^(n_.))^(p_), x_Symbol ] :> Simp[(-c^(j - 1))*(c*x)^(m - j + 1)*((a*x^j + b*x^n)^(p + 1)/(a*(n - j )*(p + 1))), x] /; FreeQ[{a, b, c, j, m, n, p}, x] && !IntegerQ[p] && NeQ[ n, j] && EqQ[m + n*p + n - j + 1, 0] && (IntegerQ[j] || GtQ[c, 0])
Int[((c_.)*(x_))^(m_.)*((a_.)*(x_)^(j_.) + (b_.)*(x_)^(n_.))^(p_), x_Symbol ] :> Simp[c^(j - 1)*(c*x)^(m - j + 1)*((a*x^j + b*x^n)^(p + 1)/(a*(m + j*p + 1))), x] - Simp[b*((m + n*p + n - j + 1)/(a*c^(n - j)*(m + j*p + 1))) I nt[(c*x)^(m + n - j)*(a*x^j + b*x^n)^p, x], x] /; FreeQ[{a, b, c, j, m, n, p}, x] && !IntegerQ[p] && NeQ[n, j] && ILtQ[Simplify[(m + n*p + n - j + 1) /(n - j)], 0] && NeQ[m + j*p + 1, 0] && (IntegersQ[j, n] || GtQ[c, 0])
Int[((e_.)*(x_))^(m_.)*((a_.)*(x_)^(j_.) + (b_.)*(x_)^(jn_.))^(p_)*((c_) + (d_.)*(x_)^(n_.)), x_Symbol] :> Simp[d*e^(j - 1)*(e*x)^(m - j + 1)*((a*x^j + b*x^(j + n))^(p + 1)/(b*(m + n + p*(j + n) + 1))), x] - Simp[(a*d*(m + j* p + 1) - b*c*(m + n + p*(j + n) + 1))/(b*(m + n + p*(j + n) + 1)) Int[(e* x)^m*(a*x^j + b*x^(j + n))^p, x], x] /; FreeQ[{a, b, c, d, e, j, m, n, p}, x] && EqQ[jn, j + n] && !IntegerQ[p] && NeQ[b*c - a*d, 0] && NeQ[m + n + p *(j + n) + 1, 0] && (GtQ[e, 0] || IntegerQ[j])
Time = 0.57 (sec) , antiderivative size = 61, normalized size of antiderivative = 0.65
method | result | size |
gosper | \(\frac {2 \left (b x +a \right ) \left (63 b^{2} d \,x^{2}-28 a b d x +77 b^{2} c x +8 a^{2} d -22 a b c \right ) \left (b \,x^{3}+a \,x^{2}\right )^{\frac {5}{2}}}{693 b^{3} x^{5}}\) | \(61\) |
default | \(\frac {2 \left (b x +a \right ) \left (63 b^{2} d \,x^{2}-28 a b d x +77 b^{2} c x +8 a^{2} d -22 a b c \right ) \left (b \,x^{3}+a \,x^{2}\right )^{\frac {5}{2}}}{693 b^{3} x^{5}}\) | \(61\) |
orering | \(\frac {2 \left (b x +a \right ) \left (63 b^{2} d \,x^{2}-28 a b d x +77 b^{2} c x +8 a^{2} d -22 a b c \right ) \left (b \,x^{3}+a \,x^{2}\right )^{\frac {5}{2}}}{693 b^{3} x^{5}}\) | \(61\) |
pseudoelliptic | \(-\frac {15 \left (b^{2} x^{3} \left (a d +\frac {b c}{6}\right ) \operatorname {arctanh}\left (\frac {\sqrt {b x +a}}{\sqrt {a}}\right )+\frac {11 \left (\frac {26 x b \left (\frac {27 d x}{13}+c \right ) a^{\frac {3}{2}}}{33}+\frac {4 \left (d x +\frac {2 c}{3}\right ) a^{\frac {5}{2}}}{11}+b^{2} x^{2} \sqrt {a}\, \left (-\frac {16 d x}{11}+c \right )\right ) \sqrt {b x +a}}{30}\right )}{4 \sqrt {a}\, x^{3}}\) | \(88\) |
risch | \(\frac {2 \sqrt {x^{2} \left (b x +a \right )}\, \left (63 d \,x^{5} b^{5}+161 a \,b^{4} d \,x^{4}+77 b^{5} c \,x^{4}+113 a^{2} b^{3} d \,x^{3}+209 a \,b^{4} c \,x^{3}+3 a^{3} b^{2} d \,x^{2}+165 a^{2} b^{3} c \,x^{2}-4 a^{4} b d x +11 a^{3} b^{2} c x +8 a^{5} d -22 a^{4} b c \right )}{693 x \,b^{3}}\) | \(126\) |
trager | \(\frac {2 \left (63 d \,x^{5} b^{5}+161 a \,b^{4} d \,x^{4}+77 b^{5} c \,x^{4}+113 a^{2} b^{3} d \,x^{3}+209 a \,b^{4} c \,x^{3}+3 a^{3} b^{2} d \,x^{2}+165 a^{2} b^{3} c \,x^{2}-4 a^{4} b d x +11 a^{3} b^{2} c x +8 a^{5} d -22 a^{4} b c \right ) \sqrt {b \,x^{3}+a \,x^{2}}}{693 b^{3} x}\) | \(128\) |
Input:
int((d*x+c)*(b*x^3+a*x^2)^(5/2)/x^4,x,method=_RETURNVERBOSE)
Output:
2/693*(b*x+a)*(63*b^2*d*x^2-28*a*b*d*x+77*b^2*c*x+8*a^2*d-22*a*b*c)*(b*x^3 +a*x^2)^(5/2)/b^3/x^5
Time = 0.11 (sec) , antiderivative size = 126, normalized size of antiderivative = 1.34 \[ \int \frac {(c+d x) \left (a x^2+b x^3\right )^{5/2}}{x^4} \, dx=\frac {2 \, {\left (63 \, b^{5} d x^{5} - 22 \, a^{4} b c + 8 \, a^{5} d + 7 \, {\left (11 \, b^{5} c + 23 \, a b^{4} d\right )} x^{4} + {\left (209 \, a b^{4} c + 113 \, a^{2} b^{3} d\right )} x^{3} + 3 \, {\left (55 \, a^{2} b^{3} c + a^{3} b^{2} d\right )} x^{2} + {\left (11 \, a^{3} b^{2} c - 4 \, a^{4} b d\right )} x\right )} \sqrt {b x^{3} + a x^{2}}}{693 \, b^{3} x} \] Input:
integrate((d*x+c)*(b*x^3+a*x^2)^(5/2)/x^4,x, algorithm="fricas")
Output:
2/693*(63*b^5*d*x^5 - 22*a^4*b*c + 8*a^5*d + 7*(11*b^5*c + 23*a*b^4*d)*x^4 + (209*a*b^4*c + 113*a^2*b^3*d)*x^3 + 3*(55*a^2*b^3*c + a^3*b^2*d)*x^2 + (11*a^3*b^2*c - 4*a^4*b*d)*x)*sqrt(b*x^3 + a*x^2)/(b^3*x)
\[ \int \frac {(c+d x) \left (a x^2+b x^3\right )^{5/2}}{x^4} \, dx=\int \frac {\left (x^{2} \left (a + b x\right )\right )^{\frac {5}{2}} \left (c + d x\right )}{x^{4}}\, dx \] Input:
integrate((d*x+c)*(b*x**3+a*x**2)**(5/2)/x**4,x)
Output:
Integral((x**2*(a + b*x))**(5/2)*(c + d*x)/x**4, x)
Time = 0.04 (sec) , antiderivative size = 119, normalized size of antiderivative = 1.27 \[ \int \frac {(c+d x) \left (a x^2+b x^3\right )^{5/2}}{x^4} \, dx=\frac {2 \, {\left (7 \, b^{4} x^{4} + 19 \, a b^{3} x^{3} + 15 \, a^{2} b^{2} x^{2} + a^{3} b x - 2 \, a^{4}\right )} \sqrt {b x + a} c}{63 \, b^{2}} + \frac {2 \, {\left (63 \, b^{5} x^{5} + 161 \, a b^{4} x^{4} + 113 \, a^{2} b^{3} x^{3} + 3 \, a^{3} b^{2} x^{2} - 4 \, a^{4} b x + 8 \, a^{5}\right )} \sqrt {b x + a} d}{693 \, b^{3}} \] Input:
integrate((d*x+c)*(b*x^3+a*x^2)^(5/2)/x^4,x, algorithm="maxima")
Output:
2/63*(7*b^4*x^4 + 19*a*b^3*x^3 + 15*a^2*b^2*x^2 + a^3*b*x - 2*a^4)*sqrt(b* x + a)*c/b^2 + 2/693*(63*b^5*x^5 + 161*a*b^4*x^4 + 113*a^2*b^3*x^3 + 3*a^3 *b^2*x^2 - 4*a^4*b*x + 8*a^5)*sqrt(b*x + a)*d/b^3
Leaf count of result is larger than twice the leaf count of optimal. 456 vs. \(2 (82) = 164\).
Time = 0.17 (sec) , antiderivative size = 456, normalized size of antiderivative = 4.85 \[ \int \frac {(c+d x) \left (a x^2+b x^3\right )^{5/2}}{x^4} \, dx =\text {Too large to display} \] Input:
integrate((d*x+c)*(b*x^3+a*x^2)^(5/2)/x^4,x, algorithm="giac")
Output:
2/3465*(1155*((b*x + a)^(3/2) - 3*sqrt(b*x + a)*a)*a^3*c*sgn(x)/b + 693*(3 *(b*x + a)^(5/2) - 10*(b*x + a)^(3/2)*a + 15*sqrt(b*x + a)*a^2)*a^2*c*sgn( x)/b + 231*(3*(b*x + a)^(5/2) - 10*(b*x + a)^(3/2)*a + 15*sqrt(b*x + a)*a^ 2)*a^3*d*sgn(x)/b^2 + 297*(5*(b*x + a)^(7/2) - 21*(b*x + a)^(5/2)*a + 35*( b*x + a)^(3/2)*a^2 - 35*sqrt(b*x + a)*a^3)*a*c*sgn(x)/b + 297*(5*(b*x + a) ^(7/2) - 21*(b*x + a)^(5/2)*a + 35*(b*x + a)^(3/2)*a^2 - 35*sqrt(b*x + a)* a^3)*a^2*d*sgn(x)/b^2 + 11*(35*(b*x + a)^(9/2) - 180*(b*x + a)^(7/2)*a + 3 78*(b*x + a)^(5/2)*a^2 - 420*(b*x + a)^(3/2)*a^3 + 315*sqrt(b*x + a)*a^4)* c*sgn(x)/b + 33*(35*(b*x + a)^(9/2) - 180*(b*x + a)^(7/2)*a + 378*(b*x + a )^(5/2)*a^2 - 420*(b*x + a)^(3/2)*a^3 + 315*sqrt(b*x + a)*a^4)*a*d*sgn(x)/ b^2 + 5*(63*(b*x + a)^(11/2) - 385*(b*x + a)^(9/2)*a + 990*(b*x + a)^(7/2) *a^2 - 1386*(b*x + a)^(5/2)*a^3 + 1155*(b*x + a)^(3/2)*a^4 - 693*sqrt(b*x + a)*a^5)*d*sgn(x)/b^2)/b + 4/693*(11*a^(9/2)*b*c - 4*a^(11/2)*d)*sgn(x)/b ^3
Time = 8.87 (sec) , antiderivative size = 120, normalized size of antiderivative = 1.28 \[ \int \frac {(c+d x) \left (a x^2+b x^3\right )^{5/2}}{x^4} \, dx=\frac {\sqrt {b\,x^3+a\,x^2}\,\left (\frac {16\,a^5\,d-44\,a^4\,b\,c}{693\,b^3}+\frac {2\,a\,x^3\,\left (113\,a\,d+209\,b\,c\right )}{693}+\frac {2\,b^2\,d\,x^5}{11}+\frac {x^4\,\left (154\,c\,b^5+322\,a\,d\,b^4\right )}{693\,b^3}-\frac {2\,a^3\,x\,\left (4\,a\,d-11\,b\,c\right )}{693\,b^2}+\frac {2\,a^2\,x^2\,\left (a\,d+55\,b\,c\right )}{231\,b}\right )}{x} \] Input:
int(((a*x^2 + b*x^3)^(5/2)*(c + d*x))/x^4,x)
Output:
((a*x^2 + b*x^3)^(1/2)*((16*a^5*d - 44*a^4*b*c)/(693*b^3) + (2*a*x^3*(113* a*d + 209*b*c))/693 + (2*b^2*d*x^5)/11 + (x^4*(154*b^5*c + 322*a*b^4*d))/( 693*b^3) - (2*a^3*x*(4*a*d - 11*b*c))/(693*b^2) + (2*a^2*x^2*(a*d + 55*b*c ))/(231*b)))/x
Time = 0.20 (sec) , antiderivative size = 117, normalized size of antiderivative = 1.24 \[ \int \frac {(c+d x) \left (a x^2+b x^3\right )^{5/2}}{x^4} \, dx=\frac {2 \sqrt {b x +a}\, \left (63 b^{5} d \,x^{5}+161 a \,b^{4} d \,x^{4}+77 b^{5} c \,x^{4}+113 a^{2} b^{3} d \,x^{3}+209 a \,b^{4} c \,x^{3}+3 a^{3} b^{2} d \,x^{2}+165 a^{2} b^{3} c \,x^{2}-4 a^{4} b d x +11 a^{3} b^{2} c x +8 a^{5} d -22 a^{4} b c \right )}{693 b^{3}} \] Input:
int((d*x+c)*(b*x^3+a*x^2)^(5/2)/x^4,x)
Output:
(2*sqrt(a + b*x)*(8*a**5*d - 22*a**4*b*c - 4*a**4*b*d*x + 11*a**3*b**2*c*x + 3*a**3*b**2*d*x**2 + 165*a**2*b**3*c*x**2 + 113*a**2*b**3*d*x**3 + 209* a*b**4*c*x**3 + 161*a*b**4*d*x**4 + 77*b**5*c*x**4 + 63*b**5*d*x**5))/(693 *b**3)