\(\int \frac {a+b x^2}{(e x)^{13/2} \sqrt {c+d x}} \, dx\) [771]

Optimal result

Integrand size = 24, antiderivative size = 219 \[ \int \frac {a+b x^2}{(e x)^{13/2} \sqrt {c+d x}} \, dx=-\frac {2 a \sqrt {c+d x}}{11 c e (e x)^{11/2}}+\frac {20 a d \sqrt {c+d x}}{99 c^2 e^2 (e x)^{9/2}}-\frac {2 \left (99 b c^2+80 a d^2\right ) \sqrt {c+d x}}{693 c^3 e^3 (e x)^{7/2}}+\frac {4 d \left (99 b c^2+80 a d^2\right ) \sqrt {c+d x}}{1155 c^4 e^4 (e x)^{5/2}}-\frac {16 d^2 \left (99 b c^2+80 a d^2\right ) \sqrt {c+d x}}{3465 c^5 e^5 (e x)^{3/2}}+\frac {32 d^3 \left (99 b c^2+80 a d^2\right ) \sqrt {c+d x}}{3465 c^6 e^6 \sqrt {e x}} \] Output:

-2/11*a*(d*x+c)^(1/2)/c/e/(e*x)^(11/2)+20/99*a*d*(d*x+c)^(1/2)/c^2/e^2/(e* 
x)^(9/2)-2/693*(80*a*d^2+99*b*c^2)*(d*x+c)^(1/2)/c^3/e^3/(e*x)^(7/2)+4/115 
5*d*(80*a*d^2+99*b*c^2)*(d*x+c)^(1/2)/c^4/e^4/(e*x)^(5/2)-16/3465*d^2*(80* 
a*d^2+99*b*c^2)*(d*x+c)^(1/2)/c^5/e^5/(e*x)^(3/2)+32/3465*d^3*(80*a*d^2+99 
*b*c^2)*(d*x+c)^(1/2)/c^6/e^6/(e*x)^(1/2)
 

Mathematica [A] (verified)

Time = 0.22 (sec) , antiderivative size = 125, normalized size of antiderivative = 0.57 \[ \int \frac {a+b x^2}{(e x)^{13/2} \sqrt {c+d x}} \, dx=-\frac {2 x \sqrt {c+d x} \left (315 a c^5-350 a c^4 d x+495 b c^5 x^2+400 a c^3 d^2 x^2-594 b c^4 d x^3-480 a c^2 d^3 x^3+792 b c^3 d^2 x^4+640 a c d^4 x^4-1584 b c^2 d^3 x^5-1280 a d^5 x^5\right )}{3465 c^6 (e x)^{13/2}} \] Input:

Integrate[(a + b*x^2)/((e*x)^(13/2)*Sqrt[c + d*x]),x]
 

Output:

(-2*x*Sqrt[c + d*x]*(315*a*c^5 - 350*a*c^4*d*x + 495*b*c^5*x^2 + 400*a*c^3 
*d^2*x^2 - 594*b*c^4*d*x^3 - 480*a*c^2*d^3*x^3 + 792*b*c^3*d^2*x^4 + 640*a 
*c*d^4*x^4 - 1584*b*c^2*d^3*x^5 - 1280*a*d^5*x^5))/(3465*c^6*(e*x)^(13/2))
 

Rubi [A] (verified)

Time = 0.48 (sec) , antiderivative size = 220, normalized size of antiderivative = 1.00, number of steps used = 7, number of rules used = 7, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.292, Rules used = {520, 27, 87, 55, 55, 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.

\(\displaystyle \int \frac {a+b x^2}{(e x)^{13/2} \sqrt {c+d x}} \, dx\)

\(\Big \downarrow \) 520

\(\displaystyle -\frac {2 \int \frac {10 a d-11 b c x}{2 (e x)^{11/2} \sqrt {c+d x}}dx}{11 c e}-\frac {2 a \sqrt {c+d x}}{11 c e (e x)^{11/2}}\)

\(\Big \downarrow \) 27

\(\displaystyle -\frac {\int \frac {10 a d-11 b c x}{(e x)^{11/2} \sqrt {c+d x}}dx}{11 c e}-\frac {2 a \sqrt {c+d x}}{11 c e (e x)^{11/2}}\)

\(\Big \downarrow \) 87

\(\displaystyle -\frac {-\frac {\left (80 a d^2+99 b c^2\right ) \int \frac {1}{(e x)^{9/2} \sqrt {c+d x}}dx}{9 c e}-\frac {20 a d \sqrt {c+d x}}{9 c e (e x)^{9/2}}}{11 c e}-\frac {2 a \sqrt {c+d x}}{11 c e (e x)^{11/2}}\)

\(\Big \downarrow \) 55

\(\displaystyle -\frac {-\frac {\left (80 a d^2+99 b c^2\right ) \left (-\frac {6 d \int \frac {1}{(e x)^{7/2} \sqrt {c+d x}}dx}{7 c e}-\frac {2 \sqrt {c+d x}}{7 c e (e x)^{7/2}}\right )}{9 c e}-\frac {20 a d \sqrt {c+d x}}{9 c e (e x)^{9/2}}}{11 c e}-\frac {2 a \sqrt {c+d x}}{11 c e (e x)^{11/2}}\)

\(\Big \downarrow \) 55

\(\displaystyle -\frac {-\frac {\left (80 a d^2+99 b c^2\right ) \left (-\frac {6 d \left (-\frac {4 d \int \frac {1}{(e x)^{5/2} \sqrt {c+d x}}dx}{5 c e}-\frac {2 \sqrt {c+d x}}{5 c e (e x)^{5/2}}\right )}{7 c e}-\frac {2 \sqrt {c+d x}}{7 c e (e x)^{7/2}}\right )}{9 c e}-\frac {20 a d \sqrt {c+d x}}{9 c e (e x)^{9/2}}}{11 c e}-\frac {2 a \sqrt {c+d x}}{11 c e (e x)^{11/2}}\)

\(\Big \downarrow \) 55

\(\displaystyle -\frac {-\frac {\left (80 a d^2+99 b c^2\right ) \left (-\frac {6 d \left (-\frac {4 d \left (-\frac {2 d \int \frac {1}{(e x)^{3/2} \sqrt {c+d x}}dx}{3 c e}-\frac {2 \sqrt {c+d x}}{3 c e (e x)^{3/2}}\right )}{5 c e}-\frac {2 \sqrt {c+d x}}{5 c e (e x)^{5/2}}\right )}{7 c e}-\frac {2 \sqrt {c+d x}}{7 c e (e x)^{7/2}}\right )}{9 c e}-\frac {20 a d \sqrt {c+d x}}{9 c e (e x)^{9/2}}}{11 c e}-\frac {2 a \sqrt {c+d x}}{11 c e (e x)^{11/2}}\)

\(\Big \downarrow \) 48

\(\displaystyle -\frac {-\frac {\left (80 a d^2+99 b c^2\right ) \left (-\frac {6 d \left (-\frac {4 d \left (\frac {4 d \sqrt {c+d x}}{3 c^2 e^2 \sqrt {e x}}-\frac {2 \sqrt {c+d x}}{3 c e (e x)^{3/2}}\right )}{5 c e}-\frac {2 \sqrt {c+d x}}{5 c e (e x)^{5/2}}\right )}{7 c e}-\frac {2 \sqrt {c+d x}}{7 c e (e x)^{7/2}}\right )}{9 c e}-\frac {20 a d \sqrt {c+d x}}{9 c e (e x)^{9/2}}}{11 c e}-\frac {2 a \sqrt {c+d x}}{11 c e (e x)^{11/2}}\)

Input:

Int[(a + b*x^2)/((e*x)^(13/2)*Sqrt[c + d*x]),x]
 

Output:

(-2*a*Sqrt[c + d*x])/(11*c*e*(e*x)^(11/2)) - ((-20*a*d*Sqrt[c + d*x])/(9*c 
*e*(e*x)^(9/2)) - ((99*b*c^2 + 80*a*d^2)*((-2*Sqrt[c + d*x])/(7*c*e*(e*x)^ 
(7/2)) - (6*d*((-2*Sqrt[c + d*x])/(5*c*e*(e*x)^(5/2)) - (4*d*((-2*Sqrt[c + 
 d*x])/(3*c*e*(e*x)^(3/2)) + (4*d*Sqrt[c + d*x])/(3*c^2*e^2*Sqrt[e*x])))/( 
5*c*e)))/(7*c*e)))/(9*c*e))/(11*c*e)
 

Defintions of rubi rules used

rule 27

Int[(a_)*(Fx_), x_Symbol] :> Simp[a   Int[Fx, x], x] /; FreeQ[a, x] &&  !Ma 
tchQ[Fx, (b_)*(Gx_) /; FreeQ[b, x]]
 

rule 48

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]
 

rule 55

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])
 

rule 87

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]))))
 

rule 520

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]
 

Maple [A] (verified)

Time = 0.25 (sec) , antiderivative size = 120, normalized size of antiderivative = 0.55

method	result	size
gosper	\(-\frac {2 x \sqrt {d x +c}\, \left (-1280 a \,d^{5} x^{5}-1584 b \,c^{2} d^{3} x^{5}+640 a c \,d^{4} x^{4}+792 b \,c^{3} d^{2} x^{4}-480 a \,c^{2} d^{3} x^{3}-594 b \,c^{4} d \,x^{3}+400 a \,c^{3} d^{2} x^{2}+495 b \,c^{5} x^{2}-350 a \,c^{4} d x +315 a \,c^{5}\right )}{3465 c^{6} \left (e x \right )^{\frac {13}{2}}}\)	\(120\)
orering	\(-\frac {2 x \sqrt {d x +c}\, \left (-1280 a \,d^{5} x^{5}-1584 b \,c^{2} d^{3} x^{5}+640 a c \,d^{4} x^{4}+792 b \,c^{3} d^{2} x^{4}-480 a \,c^{2} d^{3} x^{3}-594 b \,c^{4} d \,x^{3}+400 a \,c^{3} d^{2} x^{2}+495 b \,c^{5} x^{2}-350 a \,c^{4} d x +315 a \,c^{5}\right )}{3465 c^{6} \left (e x \right )^{\frac {13}{2}}}\)	\(120\)
default	\(-\frac {2 \sqrt {d x +c}\, \left (-1280 a \,d^{5} x^{5}-1584 b \,c^{2} d^{3} x^{5}+640 a c \,d^{4} x^{4}+792 b \,c^{3} d^{2} x^{4}-480 a \,c^{2} d^{3} x^{3}-594 b \,c^{4} d \,x^{3}+400 a \,c^{3} d^{2} x^{2}+495 b \,c^{5} x^{2}-350 a \,c^{4} d x +315 a \,c^{5}\right )}{3465 x^{5} c^{6} e^{6} \sqrt {e x}}\)	\(125\)
risch	\(-\frac {2 \sqrt {d x +c}\, \left (-1280 a \,d^{5} x^{5}-1584 b \,c^{2} d^{3} x^{5}+640 a c \,d^{4} x^{4}+792 b \,c^{3} d^{2} x^{4}-480 a \,c^{2} d^{3} x^{3}-594 b \,c^{4} d \,x^{3}+400 a \,c^{3} d^{2} x^{2}+495 b \,c^{5} x^{2}-350 a \,c^{4} d x +315 a \,c^{5}\right )}{3465 x^{5} c^{6} e^{6} \sqrt {e x}}\)	\(125\)

Input:

int((b*x^2+a)/(e*x)^(13/2)/(d*x+c)^(1/2),x,method=_RETURNVERBOSE)
 

Output:

-2/3465*x*(d*x+c)^(1/2)*(-1280*a*d^5*x^5-1584*b*c^2*d^3*x^5+640*a*c*d^4*x^ 
4+792*b*c^3*d^2*x^4-480*a*c^2*d^3*x^3-594*b*c^4*d*x^3+400*a*c^3*d^2*x^2+49 
5*b*c^5*x^2-350*a*c^4*d*x+315*a*c^5)/c^6/(e*x)^(13/2)
 

Fricas [A] (verification not implemented)

Time = 0.09 (sec) , antiderivative size = 124, normalized size of antiderivative = 0.57 \[ \int \frac {a+b x^2}{(e x)^{13/2} \sqrt {c+d x}} \, dx=\frac {2 \, {\left (350 \, a c^{4} d x - 315 \, a c^{5} + 16 \, {\left (99 \, b c^{2} d^{3} + 80 \, a d^{5}\right )} x^{5} - 8 \, {\left (99 \, b c^{3} d^{2} + 80 \, a c d^{4}\right )} x^{4} + 6 \, {\left (99 \, b c^{4} d + 80 \, a c^{2} d^{3}\right )} x^{3} - 5 \, {\left (99 \, b c^{5} + 80 \, a c^{3} d^{2}\right )} x^{2}\right )} \sqrt {d x + c} \sqrt {e x}}{3465 \, c^{6} e^{7} x^{6}} \] Input:

integrate((b*x^2+a)/(e*x)^(13/2)/(d*x+c)^(1/2),x, algorithm="fricas")
 

Output:

2/3465*(350*a*c^4*d*x - 315*a*c^5 + 16*(99*b*c^2*d^3 + 80*a*d^5)*x^5 - 8*( 
99*b*c^3*d^2 + 80*a*c*d^4)*x^4 + 6*(99*b*c^4*d + 80*a*c^2*d^3)*x^3 - 5*(99 
*b*c^5 + 80*a*c^3*d^2)*x^2)*sqrt(d*x + c)*sqrt(e*x)/(c^6*e^7*x^6)
                                                                                    
                                                                                    
 

Sympy [F(-1)]

Timed out. \[ \int \frac {a+b x^2}{(e x)^{13/2} \sqrt {c+d x}} \, dx=\text {Timed out} \] Input:

integrate((b*x**2+a)/(e*x)**(13/2)/(d*x+c)**(1/2),x)
 

Output:

Timed out
 

Maxima [F(-2)]

Exception generated. \[ \int \frac {a+b x^2}{(e x)^{13/2} \sqrt {c+d x}} \, dx=\text {Exception raised: ValueError} \] Input:

integrate((b*x^2+a)/(e*x)^(13/2)/(d*x+c)^(1/2),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
 

Giac [A] (verification not implemented)

Time = 0.15 (sec) , antiderivative size = 244, normalized size of antiderivative = 1.11 \[ \int \frac {a+b x^2}{(e x)^{13/2} \sqrt {c+d x}} \, dx=\frac {2 \, {\left ({\left ({\left (2 \, {\left (d x + c\right )} {\left (4 \, {\left (d x + c\right )} {\left (\frac {2 \, {\left (99 \, b c^{2} d^{3} e^{5} + 80 \, a d^{5} e^{5}\right )} {\left (d x + c\right )}}{c^{6}} - \frac {11 \, {\left (99 \, b c^{3} d^{3} e^{5} + 80 \, a c d^{5} e^{5}\right )}}{c^{6}}\right )} + \frac {99 \, {\left (99 \, b c^{4} d^{3} e^{5} + 80 \, a c^{2} d^{5} e^{5}\right )}}{c^{6}}\right )} - \frac {231 \, {\left (99 \, b c^{5} d^{3} e^{5} + 80 \, a c^{3} d^{5} e^{5}\right )}}{c^{6}}\right )} {\left (d x + c\right )} + \frac {2310 \, {\left (6 \, b c^{6} d^{3} e^{5} + 5 \, a c^{4} d^{5} e^{5}\right )}}{c^{6}}\right )} {\left (d x + c\right )} - \frac {3465 \, {\left (b c^{7} d^{3} e^{5} + a c^{5} d^{5} e^{5}\right )}}{c^{6}}\right )} \sqrt {d x + c} d^{7}}{3465 \, {\left ({\left (d x + c\right )} d e - c d e\right )}^{\frac {11}{2}} e^{6} {\left | d \right |}} \] Input:

integrate((b*x^2+a)/(e*x)^(13/2)/(d*x+c)^(1/2),x, algorithm="giac")
 

Output:

2/3465*(((2*(d*x + c)*(4*(d*x + c)*(2*(99*b*c^2*d^3*e^5 + 80*a*d^5*e^5)*(d 
*x + c)/c^6 - 11*(99*b*c^3*d^3*e^5 + 80*a*c*d^5*e^5)/c^6) + 99*(99*b*c^4*d 
^3*e^5 + 80*a*c^2*d^5*e^5)/c^6) - 231*(99*b*c^5*d^3*e^5 + 80*a*c^3*d^5*e^5 
)/c^6)*(d*x + c) + 2310*(6*b*c^6*d^3*e^5 + 5*a*c^4*d^5*e^5)/c^6)*(d*x + c) 
 - 3465*(b*c^7*d^3*e^5 + a*c^5*d^5*e^5)/c^6)*sqrt(d*x + c)*d^7/(((d*x + c) 
*d*e - c*d*e)^(11/2)*e^6*abs(d))
 

Mupad [B] (verification not implemented)

Time = 8.63 (sec) , antiderivative size = 148, normalized size of antiderivative = 0.68 \[ \int \frac {a+b x^2}{(e x)^{13/2} \sqrt {c+d x}} \, dx=-\frac {\sqrt {c+d\,x}\,\left (\frac {2\,a}{11\,c\,e^6}-\frac {x^3\,\left (1188\,b\,c^4\,d+960\,a\,c^2\,d^3\right )}{3465\,c^6\,e^6}+\frac {x^4\,\left (1584\,b\,c^3\,d^2+1280\,a\,c\,d^4\right )}{3465\,c^6\,e^6}+\frac {x^2\,\left (990\,b\,c^5+800\,a\,c^3\,d^2\right )}{3465\,c^6\,e^6}-\frac {x^5\,\left (3168\,b\,c^2\,d^3+2560\,a\,d^5\right )}{3465\,c^6\,e^6}-\frac {20\,a\,d\,x}{99\,c^2\,e^6}\right )}{x^5\,\sqrt {e\,x}} \] Input:

int((a + b*x^2)/((e*x)^(13/2)*(c + d*x)^(1/2)),x)
 

Output:

-((c + d*x)^(1/2)*((2*a)/(11*c*e^6) - (x^3*(960*a*c^2*d^3 + 1188*b*c^4*d)) 
/(3465*c^6*e^6) + (x^4*(1584*b*c^3*d^2 + 1280*a*c*d^4))/(3465*c^6*e^6) + ( 
x^2*(990*b*c^5 + 800*a*c^3*d^2))/(3465*c^6*e^6) - (x^5*(2560*a*d^5 + 3168* 
b*c^2*d^3))/(3465*c^6*e^6) - (20*a*d*x)/(99*c^2*e^6)))/(x^5*(e*x)^(1/2))
 

Reduce [B] (verification not implemented)

Time = 0.23 (sec) , antiderivative size = 219, normalized size of antiderivative = 1.00 \[ \int \frac {a+b x^2}{(e x)^{13/2} \sqrt {c+d x}} \, dx=\frac {2 \sqrt {e}\, \left (-315 \sqrt {x}\, \sqrt {d x +c}\, a \,c^{5}+350 \sqrt {x}\, \sqrt {d x +c}\, a \,c^{4} d x -400 \sqrt {x}\, \sqrt {d x +c}\, a \,c^{3} d^{2} x^{2}+480 \sqrt {x}\, \sqrt {d x +c}\, a \,c^{2} d^{3} x^{3}-640 \sqrt {x}\, \sqrt {d x +c}\, a c \,d^{4} x^{4}+1280 \sqrt {x}\, \sqrt {d x +c}\, a \,d^{5} x^{5}-495 \sqrt {x}\, \sqrt {d x +c}\, b \,c^{5} x^{2}+594 \sqrt {x}\, \sqrt {d x +c}\, b \,c^{4} d \,x^{3}-792 \sqrt {x}\, \sqrt {d x +c}\, b \,c^{3} d^{2} x^{4}+1584 \sqrt {x}\, \sqrt {d x +c}\, b \,c^{2} d^{3} x^{5}-1280 \sqrt {d}\, a \,d^{5} x^{6}-1584 \sqrt {d}\, b \,c^{2} d^{3} x^{6}\right )}{3465 c^{6} e^{7} x^{6}} \] Input:

int((b*x^2+a)/(e*x)^(13/2)/(d*x+c)^(1/2),x)
 

Output:

(2*sqrt(e)*( - 315*sqrt(x)*sqrt(c + d*x)*a*c**5 + 350*sqrt(x)*sqrt(c + d*x 
)*a*c**4*d*x - 400*sqrt(x)*sqrt(c + d*x)*a*c**3*d**2*x**2 + 480*sqrt(x)*sq 
rt(c + d*x)*a*c**2*d**3*x**3 - 640*sqrt(x)*sqrt(c + d*x)*a*c*d**4*x**4 + 1 
280*sqrt(x)*sqrt(c + d*x)*a*d**5*x**5 - 495*sqrt(x)*sqrt(c + d*x)*b*c**5*x 
**2 + 594*sqrt(x)*sqrt(c + d*x)*b*c**4*d*x**3 - 792*sqrt(x)*sqrt(c + d*x)* 
b*c**3*d**2*x**4 + 1584*sqrt(x)*sqrt(c + d*x)*b*c**2*d**3*x**5 - 1280*sqrt 
(d)*a*d**5*x**6 - 1584*sqrt(d)*b*c**2*d**3*x**6))/(3465*c**6*e**7*x**6)