Integrand size = 24, antiderivative size = 135 \[ \int \frac {\sqrt {c+d x} \left (a+b x^2\right )}{(e x)^{11/2}} \, dx=-\frac {2 a (c+d x)^{3/2}}{9 c e (e x)^{9/2}}+\frac {4 a d (c+d x)^{3/2}}{21 c^2 e^2 (e x)^{7/2}}-\frac {2 \left (21 b c^2+8 a d^2\right ) (c+d x)^{3/2}}{105 c^3 e^3 (e x)^{5/2}}+\frac {4 d \left (21 b c^2+8 a d^2\right ) (c+d x)^{3/2}}{315 c^4 e^4 (e x)^{3/2}} \] Output:
-2/9*a*(d*x+c)^(3/2)/c/e/(e*x)^(9/2)+4/21*a*d*(d*x+c)^(3/2)/c^2/e^2/(e*x)^ (7/2)-2/105*(8*a*d^2+21*b*c^2)*(d*x+c)^(3/2)/c^3/e^3/(e*x)^(5/2)+4/315*d*( 8*a*d^2+21*b*c^2)*(d*x+c)^(3/2)/c^4/e^4/(e*x)^(3/2)
Time = 0.17 (sec) , antiderivative size = 74, normalized size of antiderivative = 0.55 \[ \int \frac {\sqrt {c+d x} \left (a+b x^2\right )}{(e x)^{11/2}} \, dx=-\frac {2 x (c+d x)^{3/2} \left (21 b c^2 x^2 (3 c-2 d x)+a \left (35 c^3-30 c^2 d x+24 c d^2 x^2-16 d^3 x^3\right )\right )}{315 c^4 (e x)^{11/2}} \] Input:
Integrate[(Sqrt[c + d*x]*(a + b*x^2))/(e*x)^(11/2),x]
Output:
(-2*x*(c + d*x)^(3/2)*(21*b*c^2*x^2*(3*c - 2*d*x) + a*(35*c^3 - 30*c^2*d*x + 24*c*d^2*x^2 - 16*d^3*x^3)))/(315*c^4*(e*x)^(11/2))
Time = 0.38 (sec) , antiderivative size = 144, normalized size of antiderivative = 1.07, number of steps used = 5, number of rules used = 5, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.208, Rules used = {520, 27, 87, 55, 48}
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.
\(\displaystyle \int \frac {\left (a+b x^2\right ) \sqrt {c+d x}}{(e x)^{11/2}} \, dx\) |
\(\Big \downarrow \) 520 |
\(\displaystyle -\frac {2 \int \frac {3 (2 a d-3 b c x) \sqrt {c+d x}}{2 (e x)^{9/2}}dx}{9 c e}-\frac {2 a (c+d x)^{3/2}}{9 c e (e x)^{9/2}}\) |
\(\Big \downarrow \) 27 |
\(\displaystyle -\frac {\int \frac {(2 a d-3 b c x) \sqrt {c+d x}}{(e x)^{9/2}}dx}{3 c e}-\frac {2 a (c+d x)^{3/2}}{9 c e (e x)^{9/2}}\) |
\(\Big \downarrow \) 87 |
\(\displaystyle -\frac {-\frac {\left (8 a d^2+21 b c^2\right ) \int \frac {\sqrt {c+d x}}{(e x)^{7/2}}dx}{7 c e}-\frac {4 a d (c+d x)^{3/2}}{7 c e (e x)^{7/2}}}{3 c e}-\frac {2 a (c+d x)^{3/2}}{9 c e (e x)^{9/2}}\) |
\(\Big \downarrow \) 55 |
\(\displaystyle -\frac {-\frac {\left (8 a d^2+21 b c^2\right ) \left (-\frac {2 d \int \frac {\sqrt {c+d x}}{(e x)^{5/2}}dx}{5 c e}-\frac {2 (c+d x)^{3/2}}{5 c e (e x)^{5/2}}\right )}{7 c e}-\frac {4 a d (c+d x)^{3/2}}{7 c e (e x)^{7/2}}}{3 c e}-\frac {2 a (c+d x)^{3/2}}{9 c e (e x)^{9/2}}\) |
\(\Big \downarrow \) 48 |
\(\displaystyle -\frac {-\frac {\left (8 a d^2+21 b c^2\right ) \left (\frac {4 d (c+d x)^{3/2}}{15 c^2 e^2 (e x)^{3/2}}-\frac {2 (c+d x)^{3/2}}{5 c e (e x)^{5/2}}\right )}{7 c e}-\frac {4 a d (c+d x)^{3/2}}{7 c e (e x)^{7/2}}}{3 c e}-\frac {2 a (c+d x)^{3/2}}{9 c e (e x)^{9/2}}\) |
Input:
Int[(Sqrt[c + d*x]*(a + b*x^2))/(e*x)^(11/2),x]
Output:
(-2*a*(c + d*x)^(3/2))/(9*c*e*(e*x)^(9/2)) - ((-4*a*d*(c + d*x)^(3/2))/(7* c*e*(e*x)^(7/2)) - ((21*b*c^2 + 8*a*d^2)*((-2*(c + d*x)^(3/2))/(5*c*e*(e*x )^(5/2)) + (4*d*(c + d*x)^(3/2))/(15*c^2*e^2*(e*x)^(3/2))))/(7*c*e))/(3*c* e)
Int[(a_)*(Fx_), x_Symbol] :> Simp[a Int[Fx, x], x] /; FreeQ[a, x] && !Ma tchQ[Fx, (b_)*(Gx_) /; FreeQ[b, x]]
Int[((a_.) + (b_.)*(x_))^(m_.)*((c_.) + (d_.)*(x_))^(n_), x_Symbol] :> Simp [(a + b*x)^(m + 1)*((c + d*x)^(n + 1)/((b*c - a*d)*(m + 1))), x] /; FreeQ[{ a, b, c, d, m, n}, x] && EqQ[m + n + 2, 0] && NeQ[m, -1]
Int[((a_.) + (b_.)*(x_))^(m_)*((c_.) + (d_.)*(x_))^(n_), x_Symbol] :> Simp[ (a + b*x)^(m + 1)*((c + d*x)^(n + 1)/((b*c - a*d)*(m + 1))), x] - Simp[d*(S implify[m + n + 2]/((b*c - a*d)*(m + 1))) Int[(a + b*x)^Simplify[m + 1]*( c + d*x)^n, x], x] /; FreeQ[{a, b, c, d, m, n}, x] && ILtQ[Simplify[m + n + 2], 0] && NeQ[m, -1] && !(LtQ[m, -1] && LtQ[n, -1] && (EqQ[a, 0] || (NeQ[ c, 0] && LtQ[m - n, 0] && IntegerQ[n]))) && (SumSimplerQ[m, 1] || !SumSimp lerQ[n, 1])
Int[((a_.) + (b_.)*(x_))*((c_.) + (d_.)*(x_))^(n_.)*((e_.) + (f_.)*(x_))^(p _.), x_] :> Simp[(-(b*e - a*f))*(c + d*x)^(n + 1)*((e + f*x)^(p + 1)/(f*(p + 1)*(c*f - d*e))), x] - Simp[(a*d*f*(n + p + 2) - b*(d*e*(n + 1) + c*f*(p + 1)))/(f*(p + 1)*(c*f - d*e)) Int[(c + d*x)^n*(e + f*x)^(p + 1), x], x] /; FreeQ[{a, b, c, d, e, f, n}, x] && LtQ[p, -1] && ( !LtQ[n, -1] || Intege rQ[p] || !(IntegerQ[n] || !(EqQ[e, 0] || !(EqQ[c, 0] || LtQ[p, n]))))
Int[((e_.)*(x_))^(m_)*((c_) + (d_.)*(x_))^(n_)*((a_) + (b_.)*(x_)^2)^(p_.), x_Symbol] :> With[{Qx = PolynomialQuotient[(a + b*x^2)^p, e*x, x], R = Pol ynomialRemainder[(a + b*x^2)^p, e*x, x]}, Simp[R*(e*x)^(m + 1)*((c + d*x)^( n + 1)/((m + 1)*(e*c))), x] + Simp[1/((m + 1)*(e*c)) Int[(e*x)^(m + 1)*(c + d*x)^n*ExpandToSum[(m + 1)*(e*c)*Qx - d*R*(m + n + 2), x], x], x]] /; Fr eeQ[{a, b, c, d, e, n}, x] && IGtQ[p, 0] && LtQ[m, -1] && !IntegerQ[n]
Time = 0.24 (sec) , antiderivative size = 72, normalized size of antiderivative = 0.53
method | result | size |
gosper | \(-\frac {2 x \left (d x +c \right )^{\frac {3}{2}} \left (-16 a \,x^{3} d^{3}-42 b \,c^{2} d \,x^{3}+24 a \,d^{2} x^{2} c +63 b \,c^{3} x^{2}-30 a d x \,c^{2}+35 c^{3} a \right )}{315 c^{4} \left (e x \right )^{\frac {11}{2}}}\) | \(72\) |
orering | \(-\frac {2 x \left (d x +c \right )^{\frac {3}{2}} \left (-16 a \,x^{3} d^{3}-42 b \,c^{2} d \,x^{3}+24 a \,d^{2} x^{2} c +63 b \,c^{3} x^{2}-30 a d x \,c^{2}+35 c^{3} a \right )}{315 c^{4} \left (e x \right )^{\frac {11}{2}}}\) | \(72\) |
default | \(-\frac {2 \left (d x +c \right )^{\frac {3}{2}} \left (-16 a \,x^{3} d^{3}-42 b \,c^{2} d \,x^{3}+24 a \,d^{2} x^{2} c +63 b \,c^{3} x^{2}-30 a d x \,c^{2}+35 c^{3} a \right )}{315 x^{4} c^{4} e^{5} \sqrt {e x}}\) | \(77\) |
risch | \(-\frac {2 \sqrt {d x +c}\, \left (-16 a \,d^{4} x^{4}-42 b \,c^{2} d^{2} x^{4}+8 a c \,d^{3} x^{3}+21 b \,c^{3} d \,x^{3}-6 a \,c^{2} d^{2} x^{2}+63 b \,c^{4} x^{2}+5 a \,c^{3} d x +35 a \,c^{4}\right )}{315 e^{5} \sqrt {e x}\, x^{4} c^{4}}\) | \(101\) |
Input:
int((d*x+c)^(1/2)*(b*x^2+a)/(e*x)^(11/2),x,method=_RETURNVERBOSE)
Output:
-2/315*x*(d*x+c)^(3/2)*(-16*a*d^3*x^3-42*b*c^2*d*x^3+24*a*c*d^2*x^2+63*b*c ^3*x^2-30*a*c^2*d*x+35*a*c^3)/c^4/(e*x)^(11/2)
Time = 0.10 (sec) , antiderivative size = 99, normalized size of antiderivative = 0.73 \[ \int \frac {\sqrt {c+d x} \left (a+b x^2\right )}{(e x)^{11/2}} \, dx=-\frac {2 \, {\left (5 \, a c^{3} d x + 35 \, a c^{4} - 2 \, {\left (21 \, b c^{2} d^{2} + 8 \, a d^{4}\right )} x^{4} + {\left (21 \, b c^{3} d + 8 \, a c d^{3}\right )} x^{3} + 3 \, {\left (21 \, b c^{4} - 2 \, a c^{2} d^{2}\right )} x^{2}\right )} \sqrt {d x + c} \sqrt {e x}}{315 \, c^{4} e^{6} x^{5}} \] Input:
integrate((d*x+c)^(1/2)*(b*x^2+a)/(e*x)^(11/2),x, algorithm="fricas")
Output:
-2/315*(5*a*c^3*d*x + 35*a*c^4 - 2*(21*b*c^2*d^2 + 8*a*d^4)*x^4 + (21*b*c^ 3*d + 8*a*c*d^3)*x^3 + 3*(21*b*c^4 - 2*a*c^2*d^2)*x^2)*sqrt(d*x + c)*sqrt( e*x)/(c^4*e^6*x^5)
Leaf count of result is larger than twice the leaf count of optimal. 823 vs. \(2 (131) = 262\).
Time = 115.57 (sec) , antiderivative size = 823, normalized size of antiderivative = 6.10 \[ \int \frac {\sqrt {c+d x} \left (a+b x^2\right )}{(e x)^{11/2}} \, dx =\text {Too large to display} \] Input:
integrate((d*x+c)**(1/2)*(b*x**2+a)/(e*x)**(11/2),x)
Output:
-70*a*c**7*d**(19/2)*sqrt(c/(d*x) + 1)/(315*c**7*d**9*e**(11/2)*x**4 + 945 *c**6*d**10*e**(11/2)*x**5 + 945*c**5*d**11*e**(11/2)*x**6 + 315*c**4*d**1 2*e**(11/2)*x**7) - 220*a*c**6*d**(21/2)*x*sqrt(c/(d*x) + 1)/(315*c**7*d** 9*e**(11/2)*x**4 + 945*c**6*d**10*e**(11/2)*x**5 + 945*c**5*d**11*e**(11/2 )*x**6 + 315*c**4*d**12*e**(11/2)*x**7) - 228*a*c**5*d**(23/2)*x**2*sqrt(c /(d*x) + 1)/(315*c**7*d**9*e**(11/2)*x**4 + 945*c**6*d**10*e**(11/2)*x**5 + 945*c**5*d**11*e**(11/2)*x**6 + 315*c**4*d**12*e**(11/2)*x**7) - 80*a*c* *4*d**(25/2)*x**3*sqrt(c/(d*x) + 1)/(315*c**7*d**9*e**(11/2)*x**4 + 945*c* *6*d**10*e**(11/2)*x**5 + 945*c**5*d**11*e**(11/2)*x**6 + 315*c**4*d**12*e **(11/2)*x**7) + 10*a*c**3*d**(27/2)*x**4*sqrt(c/(d*x) + 1)/(315*c**7*d**9 *e**(11/2)*x**4 + 945*c**6*d**10*e**(11/2)*x**5 + 945*c**5*d**11*e**(11/2) *x**6 + 315*c**4*d**12*e**(11/2)*x**7) + 60*a*c**2*d**(29/2)*x**5*sqrt(c/( d*x) + 1)/(315*c**7*d**9*e**(11/2)*x**4 + 945*c**6*d**10*e**(11/2)*x**5 + 945*c**5*d**11*e**(11/2)*x**6 + 315*c**4*d**12*e**(11/2)*x**7) + 80*a*c*d* *(31/2)*x**6*sqrt(c/(d*x) + 1)/(315*c**7*d**9*e**(11/2)*x**4 + 945*c**6*d* *10*e**(11/2)*x**5 + 945*c**5*d**11*e**(11/2)*x**6 + 315*c**4*d**12*e**(11 /2)*x**7) + 32*a*d**(33/2)*x**7*sqrt(c/(d*x) + 1)/(315*c**7*d**9*e**(11/2) *x**4 + 945*c**6*d**10*e**(11/2)*x**5 + 945*c**5*d**11*e**(11/2)*x**6 + 31 5*c**4*d**12*e**(11/2)*x**7) - 2*b*sqrt(d)*sqrt(c/(d*x) + 1)/(5*e**(11/2)* x**2) - 2*b*d**(3/2)*sqrt(c/(d*x) + 1)/(15*c*e**(11/2)*x) + 4*b*d**(5/2...
Exception generated. \[ \int \frac {\sqrt {c+d x} \left (a+b x^2\right )}{(e x)^{11/2}} \, dx=\text {Exception raised: ValueError} \] Input:
integrate((d*x+c)^(1/2)*(b*x^2+a)/(e*x)^(11/2),x, algorithm="maxima")
Output:
Exception raised: ValueError >> Computation failed since Maxima requested additional constraints; using the 'assume' command before evaluation *may* help (example of legal syntax is 'assume(e>0)', see `assume?` for more de tails)Is e
Time = 0.15 (sec) , antiderivative size = 165, normalized size of antiderivative = 1.22 \[ \int \frac {\sqrt {c+d x} \left (a+b x^2\right )}{(e x)^{11/2}} \, dx=\frac {2 \, {\left ({\left (d x + c\right )} {\left ({\left (d x + c\right )} {\left (\frac {2 \, {\left (21 \, b c^{2} d^{7} e^{4} + 8 \, a d^{9} e^{4}\right )} {\left (d x + c\right )}}{c^{4}} - \frac {9 \, {\left (21 \, b c^{3} d^{7} e^{4} + 8 \, a c d^{9} e^{4}\right )}}{c^{4}}\right )} + \frac {126 \, {\left (2 \, b c^{4} d^{7} e^{4} + a c^{2} d^{9} e^{4}\right )}}{c^{4}}\right )} - \frac {105 \, {\left (b c^{5} d^{7} e^{4} + a c^{3} d^{9} e^{4}\right )}}{c^{4}}\right )} {\left (d x + c\right )}^{\frac {3}{2}} d}{315 \, {\left ({\left (d x + c\right )} d e - c d e\right )}^{\frac {9}{2}} e^{5} {\left | d \right |}} \] Input:
integrate((d*x+c)^(1/2)*(b*x^2+a)/(e*x)^(11/2),x, algorithm="giac")
Output:
2/315*((d*x + c)*((d*x + c)*(2*(21*b*c^2*d^7*e^4 + 8*a*d^9*e^4)*(d*x + c)/ c^4 - 9*(21*b*c^3*d^7*e^4 + 8*a*c*d^9*e^4)/c^4) + 126*(2*b*c^4*d^7*e^4 + a *c^2*d^9*e^4)/c^4) - 105*(b*c^5*d^7*e^4 + a*c^3*d^9*e^4)/c^4)*(d*x + c)^(3 /2)*d/(((d*x + c)*d*e - c*d*e)^(9/2)*e^5*abs(d))
Time = 8.73 (sec) , antiderivative size = 115, normalized size of antiderivative = 0.85 \[ \int \frac {\sqrt {c+d x} \left (a+b x^2\right )}{(e x)^{11/2}} \, dx=-\frac {\sqrt {c+d\,x}\,\left (\frac {2\,a}{9\,e^5}+\frac {x^2\,\left (126\,b\,c^4-12\,a\,c^2\,d^2\right )}{315\,c^4\,e^5}-\frac {x^4\,\left (84\,b\,c^2\,d^2+32\,a\,d^4\right )}{315\,c^4\,e^5}+\frac {x^3\,\left (42\,b\,c^3\,d+16\,a\,c\,d^3\right )}{315\,c^4\,e^5}+\frac {2\,a\,d\,x}{63\,c\,e^5}\right )}{x^4\,\sqrt {e\,x}} \] Input:
int(((a + b*x^2)*(c + d*x)^(1/2))/(e*x)^(11/2),x)
Output:
-((c + d*x)^(1/2)*((2*a)/(9*e^5) + (x^2*(126*b*c^4 - 12*a*c^2*d^2))/(315*c ^4*e^5) - (x^4*(32*a*d^4 + 84*b*c^2*d^2))/(315*c^4*e^5) + (x^3*(16*a*c*d^3 + 42*b*c^3*d))/(315*c^4*e^5) + (2*a*d*x)/(63*c*e^5)))/(x^4*(e*x)^(1/2))
Time = 0.19 (sec) , antiderivative size = 179, normalized size of antiderivative = 1.33 \[ \int \frac {\sqrt {c+d x} \left (a+b x^2\right )}{(e x)^{11/2}} \, dx=\frac {2 \sqrt {e}\, \left (-35 \sqrt {x}\, \sqrt {d x +c}\, a \,c^{4}-5 \sqrt {x}\, \sqrt {d x +c}\, a \,c^{3} d x +6 \sqrt {x}\, \sqrt {d x +c}\, a \,c^{2} d^{2} x^{2}-8 \sqrt {x}\, \sqrt {d x +c}\, a c \,d^{3} x^{3}+16 \sqrt {x}\, \sqrt {d x +c}\, a \,d^{4} x^{4}-63 \sqrt {x}\, \sqrt {d x +c}\, b \,c^{4} x^{2}-21 \sqrt {x}\, \sqrt {d x +c}\, b \,c^{3} d \,x^{3}+42 \sqrt {x}\, \sqrt {d x +c}\, b \,c^{2} d^{2} x^{4}-16 \sqrt {d}\, a \,d^{4} x^{5}-42 \sqrt {d}\, b \,c^{2} d^{2} x^{5}\right )}{315 c^{4} e^{6} x^{5}} \] Input:
int((d*x+c)^(1/2)*(b*x^2+a)/(e*x)^(11/2),x)
Output:
(2*sqrt(e)*( - 35*sqrt(x)*sqrt(c + d*x)*a*c**4 - 5*sqrt(x)*sqrt(c + d*x)*a *c**3*d*x + 6*sqrt(x)*sqrt(c + d*x)*a*c**2*d**2*x**2 - 8*sqrt(x)*sqrt(c + d*x)*a*c*d**3*x**3 + 16*sqrt(x)*sqrt(c + d*x)*a*d**4*x**4 - 63*sqrt(x)*sqr t(c + d*x)*b*c**4*x**2 - 21*sqrt(x)*sqrt(c + d*x)*b*c**3*d*x**3 + 42*sqrt( x)*sqrt(c + d*x)*b*c**2*d**2*x**4 - 16*sqrt(d)*a*d**4*x**5 - 42*sqrt(d)*b* c**2*d**2*x**5))/(315*c**4*e**6*x**5)