\(\int \frac {a+b x^2}{x^4 (c+d x)^{5/2}} \, dx\) [514]

Optimal result

Integrand size = 20, antiderivative size = 168 \[ \int \frac {a+b x^2}{x^4 (c+d x)^{5/2}} \, dx=-\frac {2 d \left (b c^2+a d^2\right )}{3 c^4 (c+d x)^{3/2}}-\frac {4 d \left (b c^2+2 a d^2\right )}{c^5 \sqrt {c+d x}}-\frac {a \sqrt {c+d x}}{3 c^3 x^3}+\frac {17 a d \sqrt {c+d x}}{12 c^4 x^2}-\frac {\left (8 b c^2+41 a d^2\right ) \sqrt {c+d x}}{8 c^5 x}+\frac {5 d \left (8 b c^2+21 a d^2\right ) \text {arctanh}\left (\frac {\sqrt {c+d x}}{\sqrt {c}}\right )}{8 c^{11/2}} \] Output:

-2/3*d*(a*d^2+b*c^2)/c^4/(d*x+c)^(3/2)-4*d*(2*a*d^2+b*c^2)/c^5/(d*x+c)^(1/ 
2)-1/3*a*(d*x+c)^(1/2)/c^3/x^3+17/12*a*d*(d*x+c)^(1/2)/c^4/x^2-1/8*(41*a*d 
^2+8*b*c^2)*(d*x+c)^(1/2)/c^5/x+5/8*d*(21*a*d^2+8*b*c^2)*arctanh((d*x+c)^( 
1/2)/c^(1/2))/c^(11/2)
 

Mathematica [A] (verified)

Time = 0.28 (sec) , antiderivative size = 131, normalized size of antiderivative = 0.78 \[ \int \frac {a+b x^2}{x^4 (c+d x)^{5/2}} \, dx=-\frac {8 b c^2 x^2 \left (3 c^2+20 c d x+15 d^2 x^2\right )+a \left (8 c^4-18 c^3 d x+63 c^2 d^2 x^2+420 c d^3 x^3+315 d^4 x^4\right )}{24 c^5 x^3 (c+d x)^{3/2}}+\frac {5 d \left (8 b c^2+21 a d^2\right ) \text {arctanh}\left (\frac {\sqrt {c+d x}}{\sqrt {c}}\right )}{8 c^{11/2}} \] Input:

Integrate[(a + b*x^2)/(x^4*(c + d*x)^(5/2)),x]
 

Output:

-1/24*(8*b*c^2*x^2*(3*c^2 + 20*c*d*x + 15*d^2*x^2) + a*(8*c^4 - 18*c^3*d*x 
 + 63*c^2*d^2*x^2 + 420*c*d^3*x^3 + 315*d^4*x^4))/(c^5*x^3*(c + d*x)^(3/2) 
) + (5*d*(8*b*c^2 + 21*a*d^2)*ArcTanh[Sqrt[c + d*x]/Sqrt[c]])/(8*c^(11/2))
 

Rubi [A] (verified)

Time = 0.85 (sec) , antiderivative size = 193, normalized size of antiderivative = 1.15, number of steps used = 8, number of rules used = 7, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.350, Rules used = {517, 1582, 1582, 27, 1582, 1584, 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.

\(\displaystyle \int \frac {a+b x^2}{x^4 (c+d x)^{5/2}} \, dx\)

\(\Big \downarrow \) 517

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

\(\Big \downarrow \) 1582

\(\displaystyle 2 d \left (\frac {\int -\frac {6 \left (b c^2+a d^2\right ) c^2-6 \left (b c^2-a d^2\right ) (c+d x) c+5 a d^2 (c+d x)^2}{d^3 x^3 (c+d x)^2}d\sqrt {c+d x}}{6 c^3}-\frac {a \sqrt {c+d x}}{6 c^3 d x^3}\right )\)

\(\Big \downarrow \) 1582

\(\displaystyle 2 d \left (\frac {\frac {\int \frac {3 \left (8 \left (b c^2+a d^2\right ) c^4+16 a d^2 (c+d x) c^3+17 a d^2 (c+d x)^2 c^2\right )}{d^2 x^2 (c+d x)^2}d\sqrt {c+d x}}{4 c^3}+\frac {17 a \sqrt {c+d x}}{4 c x^2}}{6 c^3}-\frac {a \sqrt {c+d x}}{6 c^3 d x^3}\right )\)

\(\Big \downarrow \) 27

\(\displaystyle 2 d \left (\frac {\frac {3 \int \frac {8 \left (b c^2+a d^2\right ) c^4+16 a d^2 (c+d x) c^3+17 a d^2 (c+d x)^2 c^2}{d^2 x^2 (c+d x)^2}d\sqrt {c+d x}}{4 c^3}+\frac {17 a \sqrt {c+d x}}{4 c x^2}}{6 c^3}-\frac {a \sqrt {c+d x}}{6 c^3 d x^3}\right )\)

\(\Big \downarrow \) 1582

\(\displaystyle 2 d \left (\frac {\frac {3 \left (\frac {\int -\frac {16 \left (b c^2+a d^2\right ) c^6+16 \left (b c^2+3 a d^2\right ) (c+d x) c^5+\left (8 b c^2+41 a d^2\right ) (c+d x)^2 c^4}{d x (c+d x)^2}d\sqrt {c+d x}}{2 c^3}-\frac {c \sqrt {c+d x} \left (41 a d^2+8 b c^2\right )}{2 d x}\right )}{4 c^3}+\frac {17 a \sqrt {c+d x}}{4 c x^2}}{6 c^3}-\frac {a \sqrt {c+d x}}{6 c^3 d x^3}\right )\)

\(\Big \downarrow \) 1584

\(\displaystyle 2 d \left (\frac {\frac {3 \left (\frac {\int \left (\frac {32 \left (b c^6+2 a d^2 c^4\right )}{c+d x}-\frac {5 \left (8 b c^6+21 a d^2 c^4\right )}{d x}+\frac {16 \left (b c^7+a d^2 c^5\right )}{(c+d x)^2}\right )d\sqrt {c+d x}}{2 c^3}-\frac {c \sqrt {c+d x} \left (41 a d^2+8 b c^2\right )}{2 d x}\right )}{4 c^3}+\frac {17 a \sqrt {c+d x}}{4 c x^2}}{6 c^3}-\frac {a \sqrt {c+d x}}{6 c^3 d x^3}\right )\)

\(\Big \downarrow \) 2009

\(\displaystyle 2 d \left (\frac {\frac {3 \left (\frac {5 c^{7/2} \left (21 a d^2+8 b c^2\right ) \text {arctanh}\left (\frac {\sqrt {c+d x}}{\sqrt {c}}\right )-\frac {16 c^5 \left (a d^2+b c^2\right )}{3 (c+d x)^{3/2}}-\frac {32 c^4 \left (2 a d^2+b c^2\right )}{\sqrt {c+d x}}}{2 c^3}-\frac {c \sqrt {c+d x} \left (41 a d^2+8 b c^2\right )}{2 d x}\right )}{4 c^3}+\frac {17 a \sqrt {c+d x}}{4 c x^2}}{6 c^3}-\frac {a \sqrt {c+d x}}{6 c^3 d x^3}\right )\)

Input:

Int[(a + b*x^2)/(x^4*(c + d*x)^(5/2)),x]
 

Output:

2*d*(-1/6*(a*Sqrt[c + d*x])/(c^3*d*x^3) + ((17*a*Sqrt[c + d*x])/(4*c*x^2) 
+ (3*(-1/2*(c*(8*b*c^2 + 41*a*d^2)*Sqrt[c + d*x])/(d*x) + ((-16*c^5*(b*c^2 
 + a*d^2))/(3*(c + d*x)^(3/2)) - (32*c^4*(b*c^2 + 2*a*d^2))/Sqrt[c + d*x] 
+ 5*c^(7/2)*(8*b*c^2 + 21*a*d^2)*ArcTanh[Sqrt[c + d*x]/Sqrt[c]])/(2*c^3))) 
/(4*c^3))/(6*c^3))
 

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 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 1582

Int[(x_)^(m_)*((d_) + (e_.)*(x_)^2)^(q_)*((a_) + (b_.)*(x_)^2 + (c_.)*(x_)^ 
4)^(p_.), x_Symbol] :> Simp[(-d)^(m/2 - 1)*(c*d^2 - b*d*e + a*e^2)^p*x*((d 
+ e*x^2)^(q + 1)/(2*e^(2*p + m/2)*(q + 1))), x] + Simp[(-d)^(m/2 - 1)/(2*e^ 
(2*p)*(q + 1))   Int[x^m*(d + e*x^2)^(q + 1)*ExpandToSum[Together[(1/(d + e 
*x^2))*(2*(-d)^(-m/2 + 1)*e^(2*p)*(q + 1)*(a + b*x^2 + c*x^4)^p - ((c*d^2 - 
 b*d*e + a*e^2)^p/(e^(m/2)*x^m))*(d + e*(2*q + 3)*x^2))], x], x], x] /; Fre 
eQ[{a, b, c, d, e}, x] && NeQ[b^2 - 4*a*c, 0] && IGtQ[p, 0] && ILtQ[q, -1] 
&& ILtQ[m/2, 0]
 

rule 1584

Int[((f_.)*(x_))^(m_.)*((d_) + (e_.)*(x_)^2)^(q_.)*((a_) + (b_.)*(x_)^2 + ( 
c_.)*(x_)^4)^(p_.), x_Symbol] :> Int[ExpandIntegrand[(f*x)^m*(d + e*x^2)^q* 
(a + b*x^2 + c*x^4)^p, x], x] /; FreeQ[{a, b, c, d, e, f, m, q}, x] && NeQ[ 
b^2 - 4*a*c, 0] && IGtQ[p, 0] && IGtQ[q, -2]
 

rule 2009

Int[u_, x_Symbol] :> Simp[IntSum[u, x], x] /; SumQ[u]
 

Maple [A] (verified)

Time = 0.35 (sec) , antiderivative size = 122, normalized size of antiderivative = 0.73

method	result	size
pseudoelliptic	\(\frac {\frac {105 \left (d x +c \right )^{\frac {3}{2}} d \,x^{3} \left (a \,d^{2}+\frac {8 b \,c^{2}}{21}\right ) \operatorname {arctanh}\left (\frac {\sqrt {d x +c}}{\sqrt {c}}\right )}{8}-\frac {21 d^{2} \left (\frac {40 b \,x^{2}}{21}+a \right ) x^{2} c^{\frac {5}{2}}}{8}+\frac {3 d x \left (-\frac {80 b \,x^{2}}{9}+a \right ) c^{\frac {7}{2}}}{4}+\frac {3 \left (-\frac {4 b \,x^{2}}{3}-\frac {4 a}{9}\right ) c^{\frac {9}{2}}}{4}-\frac {105 d^{3} x^{3} \left (d x \sqrt {c}+\frac {4 c^{\frac {3}{2}}}{3}\right ) a}{8}}{c^{\frac {11}{2}} \left (d x +c \right )^{\frac {3}{2}} x^{3}}\)	\(122\)
risch	\(-\frac {\sqrt {d x +c}\, \left (123 a \,d^{2} x^{2}+24 b \,c^{2} x^{2}-34 a d x c +8 a \,c^{2}\right )}{24 c^{5} x^{3}}-\frac {d \left (-\frac {2 \left (-64 a \,d^{2}-32 b \,c^{2}\right )}{\sqrt {d x +c}}+\frac {32 c \left (a \,d^{2}+b \,c^{2}\right )}{3 \left (d x +c \right )^{\frac {3}{2}}}-\frac {2 \left (105 a \,d^{2}+40 b \,c^{2}\right ) \operatorname {arctanh}\left (\frac {\sqrt {d x +c}}{\sqrt {c}}\right )}{\sqrt {c}}\right )}{16 c^{5}}\)	\(128\)
derivativedivides	\(2 d \left (-\frac {a \,d^{2}+b \,c^{2}}{3 c^{4} \left (d x +c \right )^{\frac {3}{2}}}-\frac {4 a \,d^{2}+2 b \,c^{2}}{c^{5} \sqrt {d x +c}}+\frac {-\frac {\left (\frac {41 a \,d^{2}}{16}+\frac {b \,c^{2}}{2}\right ) \left (d x +c \right )^{\frac {5}{2}}+\left (-\frac {35}{6} a \,d^{2} c -b \,c^{3}\right ) \left (d x +c \right )^{\frac {3}{2}}+\left (\frac {55}{16} a \,c^{2} d^{2}+\frac {1}{2} b \,c^{4}\right ) \sqrt {d x +c}}{d^{3} x^{3}}+\frac {5 \left (21 a \,d^{2}+8 b \,c^{2}\right ) \operatorname {arctanh}\left (\frac {\sqrt {d x +c}}{\sqrt {c}}\right )}{16 \sqrt {c}}}{c^{5}}\right )\)	\(164\)
default	\(2 d \left (-\frac {a \,d^{2}+b \,c^{2}}{3 c^{4} \left (d x +c \right )^{\frac {3}{2}}}-\frac {4 a \,d^{2}+2 b \,c^{2}}{c^{5} \sqrt {d x +c}}+\frac {-\frac {\left (\frac {41 a \,d^{2}}{16}+\frac {b \,c^{2}}{2}\right ) \left (d x +c \right )^{\frac {5}{2}}+\left (-\frac {35}{6} a \,d^{2} c -b \,c^{3}\right ) \left (d x +c \right )^{\frac {3}{2}}+\left (\frac {55}{16} a \,c^{2} d^{2}+\frac {1}{2} b \,c^{4}\right ) \sqrt {d x +c}}{d^{3} x^{3}}+\frac {5 \left (21 a \,d^{2}+8 b \,c^{2}\right ) \operatorname {arctanh}\left (\frac {\sqrt {d x +c}}{\sqrt {c}}\right )}{16 \sqrt {c}}}{c^{5}}\right )\)	\(164\)

Input:

int((b*x^2+a)/x^4/(d*x+c)^(5/2),x,method=_RETURNVERBOSE)
 

Output:

3/4/c^(11/2)*(35/2*(d*x+c)^(3/2)*d*x^3*(a*d^2+8/21*b*c^2)*arctanh((d*x+c)^ 
(1/2)/c^(1/2))-7/2*d^2*(40/21*b*x^2+a)*x^2*c^(5/2)+d*x*(-80/9*b*x^2+a)*c^( 
7/2)+(-4/3*b*x^2-4/9*a)*c^(9/2)-35/2*d^3*x^3*(d*x*c^(1/2)+4/3*c^(3/2))*a)/ 
(d*x+c)^(3/2)/x^3
 

Fricas [A] (verification not implemented)

Time = 0.10 (sec) , antiderivative size = 421, normalized size of antiderivative = 2.51 \[ \int \frac {a+b x^2}{x^4 (c+d x)^{5/2}} \, dx=\left [\frac {15 \, {\left ({\left (8 \, b c^{2} d^{3} + 21 \, a d^{5}\right )} x^{5} + 2 \, {\left (8 \, b c^{3} d^{2} + 21 \, a c d^{4}\right )} x^{4} + {\left (8 \, b c^{4} d + 21 \, a c^{2} d^{3}\right )} x^{3}\right )} \sqrt {c} \log \left (\frac {d x + 2 \, \sqrt {d x + c} \sqrt {c} + 2 \, c}{x}\right ) + 2 \, {\left (18 \, a c^{4} d x - 8 \, a c^{5} - 15 \, {\left (8 \, b c^{3} d^{2} + 21 \, a c d^{4}\right )} x^{4} - 20 \, {\left (8 \, b c^{4} d + 21 \, a c^{2} d^{3}\right )} x^{3} - 3 \, {\left (8 \, b c^{5} + 21 \, a c^{3} d^{2}\right )} x^{2}\right )} \sqrt {d x + c}}{48 \, {\left (c^{6} d^{2} x^{5} + 2 \, c^{7} d x^{4} + c^{8} x^{3}\right )}}, -\frac {15 \, {\left ({\left (8 \, b c^{2} d^{3} + 21 \, a d^{5}\right )} x^{5} + 2 \, {\left (8 \, b c^{3} d^{2} + 21 \, a c d^{4}\right )} x^{4} + {\left (8 \, b c^{4} d + 21 \, a c^{2} d^{3}\right )} x^{3}\right )} \sqrt {-c} \arctan \left (\frac {\sqrt {-c}}{\sqrt {d x + c}}\right ) - {\left (18 \, a c^{4} d x - 8 \, a c^{5} - 15 \, {\left (8 \, b c^{3} d^{2} + 21 \, a c d^{4}\right )} x^{4} - 20 \, {\left (8 \, b c^{4} d + 21 \, a c^{2} d^{3}\right )} x^{3} - 3 \, {\left (8 \, b c^{5} + 21 \, a c^{3} d^{2}\right )} x^{2}\right )} \sqrt {d x + c}}{24 \, {\left (c^{6} d^{2} x^{5} + 2 \, c^{7} d x^{4} + c^{8} x^{3}\right )}}\right ] \] Input:

integrate((b*x^2+a)/x^4/(d*x+c)^(5/2),x, algorithm="fricas")
 

Output:

[1/48*(15*((8*b*c^2*d^3 + 21*a*d^5)*x^5 + 2*(8*b*c^3*d^2 + 21*a*c*d^4)*x^4 
 + (8*b*c^4*d + 21*a*c^2*d^3)*x^3)*sqrt(c)*log((d*x + 2*sqrt(d*x + c)*sqrt 
(c) + 2*c)/x) + 2*(18*a*c^4*d*x - 8*a*c^5 - 15*(8*b*c^3*d^2 + 21*a*c*d^4)* 
x^4 - 20*(8*b*c^4*d + 21*a*c^2*d^3)*x^3 - 3*(8*b*c^5 + 21*a*c^3*d^2)*x^2)* 
sqrt(d*x + c))/(c^6*d^2*x^5 + 2*c^7*d*x^4 + c^8*x^3), -1/24*(15*((8*b*c^2* 
d^3 + 21*a*d^5)*x^5 + 2*(8*b*c^3*d^2 + 21*a*c*d^4)*x^4 + (8*b*c^4*d + 21*a 
*c^2*d^3)*x^3)*sqrt(-c)*arctan(sqrt(-c)/sqrt(d*x + c)) - (18*a*c^4*d*x - 8 
*a*c^5 - 15*(8*b*c^3*d^2 + 21*a*c*d^4)*x^4 - 20*(8*b*c^4*d + 21*a*c^2*d^3) 
*x^3 - 3*(8*b*c^5 + 21*a*c^3*d^2)*x^2)*sqrt(d*x + c))/(c^6*d^2*x^5 + 2*c^7 
*d*x^4 + c^8*x^3)]
                                                                                    
                                                                                    
 

Sympy [B] (verification not implemented)

Leaf count of result is larger than twice the leaf count of optimal. 1355 vs. \(2 (160) = 320\).

Time = 133.71 (sec) , antiderivative size = 1355, normalized size of antiderivative = 8.07 \[ \int \frac {a+b x^2}{x^4 (c+d x)^{5/2}} \, dx=\text {Too large to display} \] Input:

integrate((b*x**2+a)/x**4/(d*x+c)**(5/2),x)
 

Output:

a*(-8*c**(133/2)*d**128*x**128/(24*c**(137/2)*d**(257/2)*x**(263/2)*sqrt(c 
/(d*x) + 1) + 24*c**(135/2)*d**(259/2)*x**(265/2)*sqrt(c/(d*x) + 1)) + 18* 
c**(131/2)*d**129*x**129/(24*c**(137/2)*d**(257/2)*x**(263/2)*sqrt(c/(d*x) 
 + 1) + 24*c**(135/2)*d**(259/2)*x**(265/2)*sqrt(c/(d*x) + 1)) - 63*c**(12 
9/2)*d**130*x**130/(24*c**(137/2)*d**(257/2)*x**(263/2)*sqrt(c/(d*x) + 1) 
+ 24*c**(135/2)*d**(259/2)*x**(265/2)*sqrt(c/(d*x) + 1)) - 420*c**(127/2)* 
d**131*x**131/(24*c**(137/2)*d**(257/2)*x**(263/2)*sqrt(c/(d*x) + 1) + 24* 
c**(135/2)*d**(259/2)*x**(265/2)*sqrt(c/(d*x) + 1)) - 315*c**(125/2)*d**13 
2*x**132/(24*c**(137/2)*d**(257/2)*x**(263/2)*sqrt(c/(d*x) + 1) + 24*c**(1 
35/2)*d**(259/2)*x**(265/2)*sqrt(c/(d*x) + 1)) + 315*c**63*d**(263/2)*x**( 
263/2)*sqrt(c/(d*x) + 1)*asinh(sqrt(c)/(sqrt(d)*sqrt(x)))/(24*c**(137/2)*d 
**(257/2)*x**(263/2)*sqrt(c/(d*x) + 1) + 24*c**(135/2)*d**(259/2)*x**(265/ 
2)*sqrt(c/(d*x) + 1)) + 315*c**62*d**(265/2)*x**(265/2)*sqrt(c/(d*x) + 1)* 
asinh(sqrt(c)/(sqrt(d)*sqrt(x)))/(24*c**(137/2)*d**(257/2)*x**(263/2)*sqrt 
(c/(d*x) + 1) + 24*c**(135/2)*d**(259/2)*x**(265/2)*sqrt(c/(d*x) + 1))) + 
b*(-6*c**17*sqrt(1 + d*x/c)/(6*c**(39/2)*x + 18*c**(37/2)*d*x**2 + 18*c**( 
35/2)*d**2*x**3 + 6*c**(33/2)*d**3*x**4) - 46*c**16*d*x*sqrt(1 + d*x/c)/(6 
*c**(39/2)*x + 18*c**(37/2)*d*x**2 + 18*c**(35/2)*d**2*x**3 + 6*c**(33/2)* 
d**3*x**4) - 15*c**16*d*x*log(d*x/c)/(6*c**(39/2)*x + 18*c**(37/2)*d*x**2 
+ 18*c**(35/2)*d**2*x**3 + 6*c**(33/2)*d**3*x**4) + 30*c**16*d*x*log(sq...
 

Maxima [A] (verification not implemented)

Time = 0.11 (sec) , antiderivative size = 227, normalized size of antiderivative = 1.35 \[ \int \frac {a+b x^2}{x^4 (c+d x)^{5/2}} \, dx=\frac {1}{48} \, d^{3} {\left (\frac {2 \, {\left (16 \, b c^{6} + 16 \, a c^{4} d^{2} - 15 \, {\left (8 \, b c^{2} + 21 \, a d^{2}\right )} {\left (d x + c\right )}^{4} + 40 \, {\left (8 \, b c^{3} + 21 \, a c d^{2}\right )} {\left (d x + c\right )}^{3} - 33 \, {\left (8 \, b c^{4} + 21 \, a c^{2} d^{2}\right )} {\left (d x + c\right )}^{2} + 48 \, {\left (b c^{5} + 3 \, a c^{3} d^{2}\right )} {\left (d x + c\right )}\right )}}{{\left (d x + c\right )}^{\frac {9}{2}} c^{5} d^{2} - 3 \, {\left (d x + c\right )}^{\frac {7}{2}} c^{6} d^{2} + 3 \, {\left (d x + c\right )}^{\frac {5}{2}} c^{7} d^{2} - {\left (d x + c\right )}^{\frac {3}{2}} c^{8} d^{2}} - \frac {15 \, {\left (8 \, b c^{2} + 21 \, a d^{2}\right )} \log \left (\frac {\sqrt {d x + c} - \sqrt {c}}{\sqrt {d x + c} + \sqrt {c}}\right )}{c^{\frac {11}{2}} d^{2}}\right )} \] Input:

integrate((b*x^2+a)/x^4/(d*x+c)^(5/2),x, algorithm="maxima")
 

Output:

1/48*d^3*(2*(16*b*c^6 + 16*a*c^4*d^2 - 15*(8*b*c^2 + 21*a*d^2)*(d*x + c)^4 
 + 40*(8*b*c^3 + 21*a*c*d^2)*(d*x + c)^3 - 33*(8*b*c^4 + 21*a*c^2*d^2)*(d* 
x + c)^2 + 48*(b*c^5 + 3*a*c^3*d^2)*(d*x + c))/((d*x + c)^(9/2)*c^5*d^2 - 
3*(d*x + c)^(7/2)*c^6*d^2 + 3*(d*x + c)^(5/2)*c^7*d^2 - (d*x + c)^(3/2)*c^ 
8*d^2) - 15*(8*b*c^2 + 21*a*d^2)*log((sqrt(d*x + c) - sqrt(c))/(sqrt(d*x + 
 c) + sqrt(c)))/(c^(11/2)*d^2))
 

Giac [A] (verification not implemented)

Time = 0.13 (sec) , antiderivative size = 192, normalized size of antiderivative = 1.14 \[ \int \frac {a+b x^2}{x^4 (c+d x)^{5/2}} \, dx=-\frac {5 \, {\left (8 \, b c^{2} d + 21 \, a d^{3}\right )} \arctan \left (\frac {\sqrt {d x + c}}{\sqrt {-c}}\right )}{8 \, \sqrt {-c} c^{5}} - \frac {120 \, {\left (d x + c\right )}^{4} b c^{2} d - 320 \, {\left (d x + c\right )}^{3} b c^{3} d + 264 \, {\left (d x + c\right )}^{2} b c^{4} d - 48 \, {\left (d x + c\right )} b c^{5} d - 16 \, b c^{6} d + 315 \, {\left (d x + c\right )}^{4} a d^{3} - 840 \, {\left (d x + c\right )}^{3} a c d^{3} + 693 \, {\left (d x + c\right )}^{2} a c^{2} d^{3} - 144 \, {\left (d x + c\right )} a c^{3} d^{3} - 16 \, a c^{4} d^{3}}{24 \, {\left ({\left (d x + c\right )}^{\frac {3}{2}} - \sqrt {d x + c} c\right )}^{3} c^{5}} \] Input:

integrate((b*x^2+a)/x^4/(d*x+c)^(5/2),x, algorithm="giac")
 

Output:

-5/8*(8*b*c^2*d + 21*a*d^3)*arctan(sqrt(d*x + c)/sqrt(-c))/(sqrt(-c)*c^5) 
- 1/24*(120*(d*x + c)^4*b*c^2*d - 320*(d*x + c)^3*b*c^3*d + 264*(d*x + c)^ 
2*b*c^4*d - 48*(d*x + c)*b*c^5*d - 16*b*c^6*d + 315*(d*x + c)^4*a*d^3 - 84 
0*(d*x + c)^3*a*c*d^3 + 693*(d*x + c)^2*a*c^2*d^3 - 144*(d*x + c)*a*c^3*d^ 
3 - 16*a*c^4*d^3)/(((d*x + c)^(3/2) - sqrt(d*x + c)*c)^3*c^5)
 

Mupad [B] (verification not implemented)

Time = 8.12 (sec) , antiderivative size = 198, normalized size of antiderivative = 1.18 \[ \int \frac {a+b x^2}{x^4 (c+d x)^{5/2}} \, dx=\frac {5\,d\,\mathrm {atanh}\left (\frac {\sqrt {c+d\,x}}{\sqrt {c}}\right )\,\left (8\,b\,c^2+21\,a\,d^2\right )}{8\,c^{11/2}}-\frac {\frac {2\,\left (b\,c^2\,d+a\,d^3\right )}{3\,c}-\frac {11\,\left (8\,b\,c^2\,d+21\,a\,d^3\right )\,{\left (c+d\,x\right )}^2}{8\,c^3}+\frac {5\,\left (8\,b\,c^2\,d+21\,a\,d^3\right )\,{\left (c+d\,x\right )}^3}{3\,c^4}-\frac {5\,\left (8\,b\,c^2\,d+21\,a\,d^3\right )\,{\left (c+d\,x\right )}^4}{8\,c^5}+\frac {2\,\left (b\,c^2\,d+3\,a\,d^3\right )\,\left (c+d\,x\right )}{c^2}}{3\,c\,{\left (c+d\,x\right )}^{7/2}-{\left (c+d\,x\right )}^{9/2}+c^3\,{\left (c+d\,x\right )}^{3/2}-3\,c^2\,{\left (c+d\,x\right )}^{5/2}} \] Input:

int((a + b*x^2)/(x^4*(c + d*x)^(5/2)),x)
 

Output:

(5*d*atanh((c + d*x)^(1/2)/c^(1/2))*(21*a*d^2 + 8*b*c^2))/(8*c^(11/2)) - ( 
(2*(a*d^3 + b*c^2*d))/(3*c) - (11*(21*a*d^3 + 8*b*c^2*d)*(c + d*x)^2)/(8*c 
^3) + (5*(21*a*d^3 + 8*b*c^2*d)*(c + d*x)^3)/(3*c^4) - (5*(21*a*d^3 + 8*b* 
c^2*d)*(c + d*x)^4)/(8*c^5) + (2*(3*a*d^3 + b*c^2*d)*(c + d*x))/c^2)/(3*c* 
(c + d*x)^(7/2) - (c + d*x)^(9/2) + c^3*(c + d*x)^(3/2) - 3*c^2*(c + d*x)^ 
(5/2))
 

Reduce [B] (verification not implemented)

Time = 0.19 (sec) , antiderivative size = 337, normalized size of antiderivative = 2.01 \[ \int \frac {a+b x^2}{x^4 (c+d x)^{5/2}} \, dx=\frac {-315 \sqrt {c}\, \sqrt {d x +c}\, \mathrm {log}\left (\sqrt {d x +c}-\sqrt {c}\right ) a c \,d^{3} x^{3}-315 \sqrt {c}\, \sqrt {d x +c}\, \mathrm {log}\left (\sqrt {d x +c}-\sqrt {c}\right ) a \,d^{4} x^{4}-120 \sqrt {c}\, \sqrt {d x +c}\, \mathrm {log}\left (\sqrt {d x +c}-\sqrt {c}\right ) b \,c^{3} d \,x^{3}-120 \sqrt {c}\, \sqrt {d x +c}\, \mathrm {log}\left (\sqrt {d x +c}-\sqrt {c}\right ) b \,c^{2} d^{2} x^{4}+315 \sqrt {c}\, \sqrt {d x +c}\, \mathrm {log}\left (\sqrt {d x +c}+\sqrt {c}\right ) a c \,d^{3} x^{3}+315 \sqrt {c}\, \sqrt {d x +c}\, \mathrm {log}\left (\sqrt {d x +c}+\sqrt {c}\right ) a \,d^{4} x^{4}+120 \sqrt {c}\, \sqrt {d x +c}\, \mathrm {log}\left (\sqrt {d x +c}+\sqrt {c}\right ) b \,c^{3} d \,x^{3}+120 \sqrt {c}\, \sqrt {d x +c}\, \mathrm {log}\left (\sqrt {d x +c}+\sqrt {c}\right ) b \,c^{2} d^{2} x^{4}-16 a \,c^{5}+36 a \,c^{4} d x -126 a \,c^{3} d^{2} x^{2}-840 a \,c^{2} d^{3} x^{3}-630 a c \,d^{4} x^{4}-48 b \,c^{5} x^{2}-320 b \,c^{4} d \,x^{3}-240 b \,c^{3} d^{2} x^{4}}{48 \sqrt {d x +c}\, c^{6} x^{3} \left (d x +c \right )} \] Input:

int((b*x^2+a)/x^4/(d*x+c)^(5/2),x)
 

Output:

( - 315*sqrt(c)*sqrt(c + d*x)*log(sqrt(c + d*x) - sqrt(c))*a*c*d**3*x**3 - 
 315*sqrt(c)*sqrt(c + d*x)*log(sqrt(c + d*x) - sqrt(c))*a*d**4*x**4 - 120* 
sqrt(c)*sqrt(c + d*x)*log(sqrt(c + d*x) - sqrt(c))*b*c**3*d*x**3 - 120*sqr 
t(c)*sqrt(c + d*x)*log(sqrt(c + d*x) - sqrt(c))*b*c**2*d**2*x**4 + 315*sqr 
t(c)*sqrt(c + d*x)*log(sqrt(c + d*x) + sqrt(c))*a*c*d**3*x**3 + 315*sqrt(c 
)*sqrt(c + d*x)*log(sqrt(c + d*x) + sqrt(c))*a*d**4*x**4 + 120*sqrt(c)*sqr 
t(c + d*x)*log(sqrt(c + d*x) + sqrt(c))*b*c**3*d*x**3 + 120*sqrt(c)*sqrt(c 
 + d*x)*log(sqrt(c + d*x) + sqrt(c))*b*c**2*d**2*x**4 - 16*a*c**5 + 36*a*c 
**4*d*x - 126*a*c**3*d**2*x**2 - 840*a*c**2*d**3*x**3 - 630*a*c*d**4*x**4 
- 48*b*c**5*x**2 - 320*b*c**4*d*x**3 - 240*b*c**3*d**2*x**4)/(48*sqrt(c + 
d*x)*c**6*x**3*(c + d*x))