\(\int \frac {a+b x^2}{x^5 \sqrt {c+d x}} \, dx\) [497]

Optimal result

Integrand size = 20, antiderivative size = 148 \[ \int \frac {a+b x^2}{x^5 \sqrt {c+d x}} \, dx=-\frac {a \sqrt {c+d x}}{4 c x^4}+\frac {7 a d \sqrt {c+d x}}{24 c^2 x^3}-\frac {\left (48 b c^2+35 a d^2\right ) \sqrt {c+d x}}{96 c^3 x^2}+\frac {d \left (48 b c^2+35 a d^2\right ) \sqrt {c+d x}}{64 c^4 x}-\frac {d^2 \left (48 b c^2+35 a d^2\right ) \text {arctanh}\left (\frac {\sqrt {c+d x}}{\sqrt {c}}\right )}{64 c^{9/2}} \] Output:

-1/4*a*(d*x+c)^(1/2)/c/x^4+7/24*a*d*(d*x+c)^(1/2)/c^2/x^3-1/96*(35*a*d^2+4 
8*b*c^2)*(d*x+c)^(1/2)/c^3/x^2+1/64*d*(35*a*d^2+48*b*c^2)*(d*x+c)^(1/2)/c^ 
4/x-1/64*d^2*(35*a*d^2+48*b*c^2)*arctanh((d*x+c)^(1/2)/c^(1/2))/c^(9/2)
 

Mathematica [A] (verified)

Time = 0.25 (sec) , antiderivative size = 112, normalized size of antiderivative = 0.76 \[ \int \frac {a+b x^2}{x^5 \sqrt {c+d x}} \, dx=\frac {\frac {\sqrt {c} \sqrt {c+d x} \left (48 b c^2 x^2 (-2 c+3 d x)+a \left (-48 c^3+56 c^2 d x-70 c d^2 x^2+105 d^3 x^3\right )\right )}{x^4}-3 d^2 \left (48 b c^2+35 a d^2\right ) \text {arctanh}\left (\frac {\sqrt {c+d x}}{\sqrt {c}}\right )}{192 c^{9/2}} \] Input:

Integrate[(a + b*x^2)/(x^5*Sqrt[c + d*x]),x]
 

Output:

((Sqrt[c]*Sqrt[c + d*x]*(48*b*c^2*x^2*(-2*c + 3*d*x) + a*(-48*c^3 + 56*c^2 
*d*x - 70*c*d^2*x^2 + 105*d^3*x^3)))/x^4 - 3*d^2*(48*b*c^2 + 35*a*d^2)*Arc 
Tanh[Sqrt[c + d*x]/Sqrt[c]])/(192*c^(9/2))
 

Rubi [A] (verified)

Time = 0.46 (sec) , antiderivative size = 158, normalized size of antiderivative = 1.07, number of steps used = 9, number of rules used = 8, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.400, Rules used = {517, 25, 1471, 25, 298, 215, 215, 219}

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.

\(\displaystyle \int \frac {a+b x^2}{x^5 \sqrt {c+d x}} \, dx\)

\(\Big \downarrow \) 517

\(\displaystyle 2 d^2 \int \frac {b c^2-2 b (c+d x) c+a d^2+b (c+d x)^2}{d^5 x^5}d\sqrt {c+d x}\)

\(\Big \downarrow \) 25

\(\displaystyle -2 d^2 \int -\frac {b c^2-2 b (c+d x) c+a d^2+b (c+d x)^2}{d^5 x^5}d\sqrt {c+d x}\)

\(\Big \downarrow \) 1471

\(\displaystyle 2 d^2 \left (\frac {\int -\frac {8 b c^2-8 b (c+d x) c+7 a d^2}{d^4 x^4}d\sqrt {c+d x}}{8 c}-\frac {a \sqrt {c+d x}}{8 c d^2 x^4}\right )\)

\(\Big \downarrow \) 25

\(\displaystyle 2 d^2 \left (-\frac {\int \frac {8 b c^2-8 b (c+d x) c+7 a d^2}{d^4 x^4}d\sqrt {c+d x}}{8 c}-\frac {a \sqrt {c+d x}}{8 c d^2 x^4}\right )\)

\(\Big \downarrow \) 298

\(\displaystyle 2 d^2 \left (-\frac {\frac {\left (35 a d^2+48 b c^2\right ) \int -\frac {1}{d^3 x^3}d\sqrt {c+d x}}{6 c}-\frac {7 a \sqrt {c+d x}}{6 c d x^3}}{8 c}-\frac {a \sqrt {c+d x}}{8 c d^2 x^4}\right )\)

\(\Big \downarrow \) 215

\(\displaystyle 2 d^2 \left (-\frac {\frac {\left (35 a d^2+48 b c^2\right ) \left (\frac {3 \int \frac {1}{d^2 x^2}d\sqrt {c+d x}}{4 c}+\frac {\sqrt {c+d x}}{4 c d^2 x^2}\right )}{6 c}-\frac {7 a \sqrt {c+d x}}{6 c d x^3}}{8 c}-\frac {a \sqrt {c+d x}}{8 c d^2 x^4}\right )\)

\(\Big \downarrow \) 215

\(\displaystyle 2 d^2 \left (-\frac {\frac {\left (35 a d^2+48 b c^2\right ) \left (\frac {3 \left (\frac {\int -\frac {1}{d x}d\sqrt {c+d x}}{2 c}-\frac {\sqrt {c+d x}}{2 c d x}\right )}{4 c}+\frac {\sqrt {c+d x}}{4 c d^2 x^2}\right )}{6 c}-\frac {7 a \sqrt {c+d x}}{6 c d x^3}}{8 c}-\frac {a \sqrt {c+d x}}{8 c d^2 x^4}\right )\)

\(\Big \downarrow \) 219

\(\displaystyle 2 d^2 \left (-\frac {\frac {\left (35 a d^2+48 b c^2\right ) \left (\frac {3 \left (\frac {\text {arctanh}\left (\frac {\sqrt {c+d x}}{\sqrt {c}}\right )}{2 c^{3/2}}-\frac {\sqrt {c+d x}}{2 c d x}\right )}{4 c}+\frac {\sqrt {c+d x}}{4 c d^2 x^2}\right )}{6 c}-\frac {7 a \sqrt {c+d x}}{6 c d x^3}}{8 c}-\frac {a \sqrt {c+d x}}{8 c d^2 x^4}\right )\)

Input:

Int[(a + b*x^2)/(x^5*Sqrt[c + d*x]),x]
 

Output:

2*d^2*(-1/8*(a*Sqrt[c + d*x])/(c*d^2*x^4) - ((-7*a*Sqrt[c + d*x])/(6*c*d*x 
^3) + ((48*b*c^2 + 35*a*d^2)*(Sqrt[c + d*x]/(4*c*d^2*x^2) + (3*(-1/2*Sqrt[ 
c + d*x]/(c*d*x) + ArcTanh[Sqrt[c + d*x]/Sqrt[c]]/(2*c^(3/2))))/(4*c)))/(6 
*c))/(8*c))
 

Defintions of rubi rules used

rule 25

Int[-(Fx_), x_Symbol] :> Simp[Identity[-1]   Int[Fx, x], x]
 

rule 215

Int[((a_) + (b_.)*(x_)^2)^(p_), x_Symbol] :> Simp[(-x)*((a + b*x^2)^(p + 1) 
/(2*a*(p + 1))), x] + Simp[(2*p + 3)/(2*a*(p + 1))   Int[(a + b*x^2)^(p + 1 
), x], x] /; FreeQ[{a, b}, x] && LtQ[p, -1] && (IntegerQ[4*p] || IntegerQ[6 
*p])
 

rule 219

Int[((a_) + (b_.)*(x_)^2)^(-1), x_Symbol] :> Simp[(1/(Rt[a, 2]*Rt[-b, 2]))* 
ArcTanh[Rt[-b, 2]*(x/Rt[a, 2])], x] /; FreeQ[{a, b}, x] && NegQ[a/b] && (Gt 
Q[a, 0] || LtQ[b, 0])
 

rule 298

Int[((a_) + (b_.)*(x_)^2)^(p_)*((c_) + (d_.)*(x_)^2), x_Symbol] :> Simp[(-( 
b*c - a*d))*x*((a + b*x^2)^(p + 1)/(2*a*b*(p + 1))), x] - Simp[(a*d - b*c*( 
2*p + 3))/(2*a*b*(p + 1))   Int[(a + b*x^2)^(p + 1), x], x] /; FreeQ[{a, b, 
 c, d, p}, x] && NeQ[b*c - a*d, 0] && (LtQ[p, -1] || ILtQ[1/2 + p, 0])
 

rule 517

Int[((e_.)*(x_))^(m_)*((c_) + (d_.)*(x_))^(n_)*((a_) + (b_.)*(x_)^2)^(p_.), 
 x_Symbol] :> Simp[2*(e^m/d^(m + 2*p + 1))   Subst[Int[x^(2*n + 1)*(-c + x^ 
2)^m*(b*c^2 + a*d^2 - 2*b*c*x^2 + b*x^4)^p, x], x, Sqrt[c + d*x]], x] /; Fr 
eeQ[{a, b, c, d, e}, x] && IGtQ[p, 0] && ILtQ[m, 0] && IntegerQ[n + 1/2]
 

rule 1471

Int[((d_) + (e_.)*(x_)^2)^(q_)*((a_) + (b_.)*(x_)^2 + (c_.)*(x_)^4)^(p_.), 
x_Symbol] :> With[{Qx = PolynomialQuotient[(a + b*x^2 + c*x^4)^p, d + e*x^2 
, x], R = Coeff[PolynomialRemainder[(a + b*x^2 + c*x^4)^p, d + e*x^2, x], x 
, 0]}, Simp[(-R)*x*((d + e*x^2)^(q + 1)/(2*d*(q + 1))), x] + Simp[1/(2*d*(q 
 + 1))   Int[(d + e*x^2)^(q + 1)*ExpandToSum[2*d*(q + 1)*Qx + R*(2*q + 3), 
x], x], x]] /; FreeQ[{a, b, c, d, e}, x] && NeQ[b^2 - 4*a*c, 0] && NeQ[c*d^ 
2 - b*d*e + a*e^2, 0] && IGtQ[p, 0] && LtQ[q, -1]
 

Maple [A] (verified)

Time = 0.29 (sec) , antiderivative size = 101, normalized size of antiderivative = 0.68

method	result	size
pseudoelliptic	\(\frac {-\frac {35 d^{2} \left (a \,d^{2}+\frac {48 b \,c^{2}}{35}\right ) x^{4} \operatorname {arctanh}\left (\frac {\sqrt {d x +c}}{\sqrt {c}}\right )}{64}+\frac {7 \left (d x \left (\frac {18 b \,x^{2}}{7}+a \right ) c^{\frac {5}{2}}+\frac {6 \left (-2 b \,x^{2}-a \right ) c^{\frac {7}{2}}}{7}+\frac {15 d^{2} x^{2} \left (d x \sqrt {c}-\frac {2 c^{\frac {3}{2}}}{3}\right ) a}{8}\right ) \sqrt {d x +c}}{24}}{c^{\frac {9}{2}} x^{4}}\)	\(101\)
risch	\(-\frac {\sqrt {d x +c}\, \left (-105 a \,x^{3} d^{3}-144 b \,c^{2} d \,x^{3}+70 a \,d^{2} x^{2} c +96 b \,c^{3} x^{2}-56 a d x \,c^{2}+48 c^{3} a \right )}{192 c^{4} x^{4}}-\frac {d^{2} \left (35 a \,d^{2}+48 b \,c^{2}\right ) \operatorname {arctanh}\left (\frac {\sqrt {d x +c}}{\sqrt {c}}\right )}{64 c^{\frac {9}{2}}}\)	\(103\)
derivativedivides	\(2 d^{2} \left (-\frac {-\frac {\left (35 a \,d^{2}+48 b \,c^{2}\right ) \left (d x +c \right )^{\frac {7}{2}}}{128 c^{4}}+\frac {11 \left (35 a \,d^{2}+48 b \,c^{2}\right ) \left (d x +c \right )^{\frac {5}{2}}}{384 c^{3}}-\frac {\left (511 a \,d^{2}+624 b \,c^{2}\right ) \left (d x +c \right )^{\frac {3}{2}}}{384 c^{2}}+\frac {\left (93 a \,d^{2}+80 b \,c^{2}\right ) \sqrt {d x +c}}{128 c}}{d^{4} x^{4}}-\frac {\left (35 a \,d^{2}+48 b \,c^{2}\right ) \operatorname {arctanh}\left (\frac {\sqrt {d x +c}}{\sqrt {c}}\right )}{128 c^{\frac {9}{2}}}\right )\)	\(146\)
default	\(2 d^{2} \left (-\frac {-\frac {\left (35 a \,d^{2}+48 b \,c^{2}\right ) \left (d x +c \right )^{\frac {7}{2}}}{128 c^{4}}+\frac {11 \left (35 a \,d^{2}+48 b \,c^{2}\right ) \left (d x +c \right )^{\frac {5}{2}}}{384 c^{3}}-\frac {\left (511 a \,d^{2}+624 b \,c^{2}\right ) \left (d x +c \right )^{\frac {3}{2}}}{384 c^{2}}+\frac {\left (93 a \,d^{2}+80 b \,c^{2}\right ) \sqrt {d x +c}}{128 c}}{d^{4} x^{4}}-\frac {\left (35 a \,d^{2}+48 b \,c^{2}\right ) \operatorname {arctanh}\left (\frac {\sqrt {d x +c}}{\sqrt {c}}\right )}{128 c^{\frac {9}{2}}}\right )\)	\(146\)

Input:

int((b*x^2+a)/x^5/(d*x+c)^(1/2),x,method=_RETURNVERBOSE)
 

Output:

7/24/c^(9/2)*(-15/8*d^2*(a*d^2+48/35*b*c^2)*x^4*arctanh((d*x+c)^(1/2)/c^(1 
/2))+(d*x*(18/7*b*x^2+a)*c^(5/2)+6/7*(-2*b*x^2-a)*c^(7/2)+15/8*d^2*x^2*(d* 
x*c^(1/2)-2/3*c^(3/2))*a)*(d*x+c)^(1/2))/x^4
 

Fricas [A] (verification not implemented)

Time = 0.09 (sec) , antiderivative size = 236, normalized size of antiderivative = 1.59 \[ \int \frac {a+b x^2}{x^5 \sqrt {c+d x}} \, dx=\left [\frac {3 \, {\left (48 \, b c^{2} d^{2} + 35 \, a d^{4}\right )} \sqrt {c} x^{4} \log \left (\frac {d x - 2 \, \sqrt {d x + c} \sqrt {c} + 2 \, c}{x}\right ) + 2 \, {\left (56 \, a c^{3} d x - 48 \, a c^{4} + 3 \, {\left (48 \, b c^{3} d + 35 \, a c d^{3}\right )} x^{3} - 2 \, {\left (48 \, b c^{4} + 35 \, a c^{2} d^{2}\right )} x^{2}\right )} \sqrt {d x + c}}{384 \, c^{5} x^{4}}, \frac {3 \, {\left (48 \, b c^{2} d^{2} + 35 \, a d^{4}\right )} \sqrt {-c} x^{4} \arctan \left (\frac {\sqrt {-c}}{\sqrt {d x + c}}\right ) + {\left (56 \, a c^{3} d x - 48 \, a c^{4} + 3 \, {\left (48 \, b c^{3} d + 35 \, a c d^{3}\right )} x^{3} - 2 \, {\left (48 \, b c^{4} + 35 \, a c^{2} d^{2}\right )} x^{2}\right )} \sqrt {d x + c}}{192 \, c^{5} x^{4}}\right ] \] Input:

integrate((b*x^2+a)/x^5/(d*x+c)^(1/2),x, algorithm="fricas")
 

Output:

[1/384*(3*(48*b*c^2*d^2 + 35*a*d^4)*sqrt(c)*x^4*log((d*x - 2*sqrt(d*x + c) 
*sqrt(c) + 2*c)/x) + 2*(56*a*c^3*d*x - 48*a*c^4 + 3*(48*b*c^3*d + 35*a*c*d 
^3)*x^3 - 2*(48*b*c^4 + 35*a*c^2*d^2)*x^2)*sqrt(d*x + c))/(c^5*x^4), 1/192 
*(3*(48*b*c^2*d^2 + 35*a*d^4)*sqrt(-c)*x^4*arctan(sqrt(-c)/sqrt(d*x + c)) 
+ (56*a*c^3*d*x - 48*a*c^4 + 3*(48*b*c^3*d + 35*a*c*d^3)*x^3 - 2*(48*b*c^4 
 + 35*a*c^2*d^2)*x^2)*sqrt(d*x + c))/(c^5*x^4)]
 

Sympy [A] (verification not implemented)

Time = 72.54 (sec) , antiderivative size = 274, normalized size of antiderivative = 1.85 \[ \int \frac {a+b x^2}{x^5 \sqrt {c+d x}} \, dx=- \frac {a}{4 \sqrt {d} x^{\frac {9}{2}} \sqrt {\frac {c}{d x} + 1}} + \frac {a \sqrt {d}}{24 c x^{\frac {7}{2}} \sqrt {\frac {c}{d x} + 1}} - \frac {7 a d^{\frac {3}{2}}}{96 c^{2} x^{\frac {5}{2}} \sqrt {\frac {c}{d x} + 1}} + \frac {35 a d^{\frac {5}{2}}}{192 c^{3} x^{\frac {3}{2}} \sqrt {\frac {c}{d x} + 1}} + \frac {35 a d^{\frac {7}{2}}}{64 c^{4} \sqrt {x} \sqrt {\frac {c}{d x} + 1}} - \frac {35 a d^{4} \operatorname {asinh}{\left (\frac {\sqrt {c}}{\sqrt {d} \sqrt {x}} \right )}}{64 c^{\frac {9}{2}}} - \frac {b}{2 \sqrt {d} x^{\frac {5}{2}} \sqrt {\frac {c}{d x} + 1}} + \frac {b \sqrt {d}}{4 c x^{\frac {3}{2}} \sqrt {\frac {c}{d x} + 1}} + \frac {3 b d^{\frac {3}{2}}}{4 c^{2} \sqrt {x} \sqrt {\frac {c}{d x} + 1}} - \frac {3 b d^{2} \operatorname {asinh}{\left (\frac {\sqrt {c}}{\sqrt {d} \sqrt {x}} \right )}}{4 c^{\frac {5}{2}}} \] Input:

integrate((b*x**2+a)/x**5/(d*x+c)**(1/2),x)
 

Output:

-a/(4*sqrt(d)*x**(9/2)*sqrt(c/(d*x) + 1)) + a*sqrt(d)/(24*c*x**(7/2)*sqrt( 
c/(d*x) + 1)) - 7*a*d**(3/2)/(96*c**2*x**(5/2)*sqrt(c/(d*x) + 1)) + 35*a*d 
**(5/2)/(192*c**3*x**(3/2)*sqrt(c/(d*x) + 1)) + 35*a*d**(7/2)/(64*c**4*sqr 
t(x)*sqrt(c/(d*x) + 1)) - 35*a*d**4*asinh(sqrt(c)/(sqrt(d)*sqrt(x)))/(64*c 
**(9/2)) - b/(2*sqrt(d)*x**(5/2)*sqrt(c/(d*x) + 1)) + b*sqrt(d)/(4*c*x**(3 
/2)*sqrt(c/(d*x) + 1)) + 3*b*d**(3/2)/(4*c**2*sqrt(x)*sqrt(c/(d*x) + 1)) - 
 3*b*d**2*asinh(sqrt(c)/(sqrt(d)*sqrt(x)))/(4*c**(5/2))
 

Maxima [A] (verification not implemented)

Time = 0.12 (sec) , antiderivative size = 219, normalized size of antiderivative = 1.48 \[ \int \frac {a+b x^2}{x^5 \sqrt {c+d x}} \, dx=\frac {1}{384} \, d^{4} {\left (\frac {2 \, {\left (3 \, {\left (48 \, b c^{2} + 35 \, a d^{2}\right )} {\left (d x + c\right )}^{\frac {7}{2}} - 11 \, {\left (48 \, b c^{3} + 35 \, a c d^{2}\right )} {\left (d x + c\right )}^{\frac {5}{2}} + {\left (624 \, b c^{4} + 511 \, a c^{2} d^{2}\right )} {\left (d x + c\right )}^{\frac {3}{2}} - 3 \, {\left (80 \, b c^{5} + 93 \, a c^{3} d^{2}\right )} \sqrt {d x + c}\right )}}{{\left (d x + c\right )}^{4} c^{4} d^{2} - 4 \, {\left (d x + c\right )}^{3} c^{5} d^{2} + 6 \, {\left (d x + c\right )}^{2} c^{6} d^{2} - 4 \, {\left (d x + c\right )} c^{7} d^{2} + c^{8} d^{2}} + \frac {3 \, {\left (48 \, b c^{2} + 35 \, a d^{2}\right )} \log \left (\frac {\sqrt {d x + c} - \sqrt {c}}{\sqrt {d x + c} + \sqrt {c}}\right )}{c^{\frac {9}{2}} d^{2}}\right )} \] Input:

integrate((b*x^2+a)/x^5/(d*x+c)^(1/2),x, algorithm="maxima")
                                                                                    
                                                                                    
 

Output:

1/384*d^4*(2*(3*(48*b*c^2 + 35*a*d^2)*(d*x + c)^(7/2) - 11*(48*b*c^3 + 35* 
a*c*d^2)*(d*x + c)^(5/2) + (624*b*c^4 + 511*a*c^2*d^2)*(d*x + c)^(3/2) - 3 
*(80*b*c^5 + 93*a*c^3*d^2)*sqrt(d*x + c))/((d*x + c)^4*c^4*d^2 - 4*(d*x + 
c)^3*c^5*d^2 + 6*(d*x + c)^2*c^6*d^2 - 4*(d*x + c)*c^7*d^2 + c^8*d^2) + 3* 
(48*b*c^2 + 35*a*d^2)*log((sqrt(d*x + c) - sqrt(c))/(sqrt(d*x + c) + sqrt( 
c)))/(c^(9/2)*d^2))
 

Giac [A] (verification not implemented)

Time = 0.13 (sec) , antiderivative size = 180, normalized size of antiderivative = 1.22 \[ \int \frac {a+b x^2}{x^5 \sqrt {c+d x}} \, dx=\frac {\frac {3 \, {\left (48 \, b c^{2} d^{3} + 35 \, a d^{5}\right )} \arctan \left (\frac {\sqrt {d x + c}}{\sqrt {-c}}\right )}{\sqrt {-c} c^{4}} + \frac {144 \, {\left (d x + c\right )}^{\frac {7}{2}} b c^{2} d^{3} - 528 \, {\left (d x + c\right )}^{\frac {5}{2}} b c^{3} d^{3} + 624 \, {\left (d x + c\right )}^{\frac {3}{2}} b c^{4} d^{3} - 240 \, \sqrt {d x + c} b c^{5} d^{3} + 105 \, {\left (d x + c\right )}^{\frac {7}{2}} a d^{5} - 385 \, {\left (d x + c\right )}^{\frac {5}{2}} a c d^{5} + 511 \, {\left (d x + c\right )}^{\frac {3}{2}} a c^{2} d^{5} - 279 \, \sqrt {d x + c} a c^{3} d^{5}}{c^{4} d^{4} x^{4}}}{192 \, d} \] Input:

integrate((b*x^2+a)/x^5/(d*x+c)^(1/2),x, algorithm="giac")
 

Output:

1/192*(3*(48*b*c^2*d^3 + 35*a*d^5)*arctan(sqrt(d*x + c)/sqrt(-c))/(sqrt(-c 
)*c^4) + (144*(d*x + c)^(7/2)*b*c^2*d^3 - 528*(d*x + c)^(5/2)*b*c^3*d^3 + 
624*(d*x + c)^(3/2)*b*c^4*d^3 - 240*sqrt(d*x + c)*b*c^5*d^3 + 105*(d*x + c 
)^(7/2)*a*d^5 - 385*(d*x + c)^(5/2)*a*c*d^5 + 511*(d*x + c)^(3/2)*a*c^2*d^ 
5 - 279*sqrt(d*x + c)*a*c^3*d^5)/(c^4*d^4*x^4))/d
 

Mupad [B] (verification not implemented)

Time = 0.14 (sec) , antiderivative size = 194, normalized size of antiderivative = 1.31 \[ \int \frac {a+b x^2}{x^5 \sqrt {c+d x}} \, dx=-\frac {\frac {11\,\left (48\,b\,c^2\,d^2+35\,a\,d^4\right )\,{\left (c+d\,x\right )}^{5/2}}{192\,c^3}-\frac {\left (48\,b\,c^2\,d^2+35\,a\,d^4\right )\,{\left (c+d\,x\right )}^{7/2}}{64\,c^4}+\frac {\left (80\,b\,c^2\,d^2+93\,a\,d^4\right )\,\sqrt {c+d\,x}}{64\,c}-\frac {\left (624\,b\,c^2\,d^2+511\,a\,d^4\right )\,{\left (c+d\,x\right )}^{3/2}}{192\,c^2}}{{\left (c+d\,x\right )}^4-4\,c^3\,\left (c+d\,x\right )-4\,c\,{\left (c+d\,x\right )}^3+6\,c^2\,{\left (c+d\,x\right )}^2+c^4}-\frac {d^2\,\mathrm {atanh}\left (\frac {\sqrt {c+d\,x}}{\sqrt {c}}\right )\,\left (48\,b\,c^2+35\,a\,d^2\right )}{64\,c^{9/2}} \] Input:

int((a + b*x^2)/(x^5*(c + d*x)^(1/2)),x)
 

Output:

- ((11*(35*a*d^4 + 48*b*c^2*d^2)*(c + d*x)^(5/2))/(192*c^3) - ((35*a*d^4 + 
 48*b*c^2*d^2)*(c + d*x)^(7/2))/(64*c^4) + ((93*a*d^4 + 80*b*c^2*d^2)*(c + 
 d*x)^(1/2))/(64*c) - ((511*a*d^4 + 624*b*c^2*d^2)*(c + d*x)^(3/2))/(192*c 
^2))/((c + d*x)^4 - 4*c^3*(c + d*x) - 4*c*(c + d*x)^3 + 6*c^2*(c + d*x)^2 
+ c^4) - (d^2*atanh((c + d*x)^(1/2)/c^(1/2))*(35*a*d^2 + 48*b*c^2))/(64*c^ 
(9/2))
 

Reduce [B] (verification not implemented)

Time = 0.20 (sec) , antiderivative size = 194, normalized size of antiderivative = 1.31 \[ \int \frac {a+b x^2}{x^5 \sqrt {c+d x}} \, dx=\frac {-96 \sqrt {d x +c}\, a \,c^{4}+112 \sqrt {d x +c}\, a \,c^{3} d x -140 \sqrt {d x +c}\, a \,c^{2} d^{2} x^{2}+210 \sqrt {d x +c}\, a c \,d^{3} x^{3}-192 \sqrt {d x +c}\, b \,c^{4} x^{2}+288 \sqrt {d x +c}\, b \,c^{3} d \,x^{3}+105 \sqrt {c}\, \mathrm {log}\left (\sqrt {d x +c}-\sqrt {c}\right ) a \,d^{4} x^{4}+144 \sqrt {c}\, \mathrm {log}\left (\sqrt {d x +c}-\sqrt {c}\right ) b \,c^{2} d^{2} x^{4}-105 \sqrt {c}\, \mathrm {log}\left (\sqrt {d x +c}+\sqrt {c}\right ) a \,d^{4} x^{4}-144 \sqrt {c}\, \mathrm {log}\left (\sqrt {d x +c}+\sqrt {c}\right ) b \,c^{2} d^{2} x^{4}}{384 c^{5} x^{4}} \] Input:

int((b*x^2+a)/x^5/(d*x+c)^(1/2),x)
 

Output:

( - 96*sqrt(c + d*x)*a*c**4 + 112*sqrt(c + d*x)*a*c**3*d*x - 140*sqrt(c + 
d*x)*a*c**2*d**2*x**2 + 210*sqrt(c + d*x)*a*c*d**3*x**3 - 192*sqrt(c + d*x 
)*b*c**4*x**2 + 288*sqrt(c + d*x)*b*c**3*d*x**3 + 105*sqrt(c)*log(sqrt(c + 
 d*x) - sqrt(c))*a*d**4*x**4 + 144*sqrt(c)*log(sqrt(c + d*x) - sqrt(c))*b* 
c**2*d**2*x**4 - 105*sqrt(c)*log(sqrt(c + d*x) + sqrt(c))*a*d**4*x**4 - 14 
4*sqrt(c)*log(sqrt(c + d*x) + sqrt(c))*b*c**2*d**2*x**4)/(384*c**5*x**4)