\(\int \frac {c+d x}{x^3 (a+b x^2)^{7/2}} \, dx\) [1316]

Optimal result

Integrand size = 20, antiderivative size = 150 \[ \int \frac {c+d x}{x^3 \left (a+b x^2\right )^{7/2}} \, dx=-\frac {b (c+d x)}{5 a^2 \left (a+b x^2\right )^{5/2}}-\frac {b (10 c+9 d x)}{15 a^3 \left (a+b x^2\right )^{3/2}}-\frac {b (15 c+11 d x)}{5 a^4 \sqrt {a+b x^2}}-\frac {c \sqrt {a+b x^2}}{2 a^4 x^2}-\frac {d \sqrt {a+b x^2}}{a^4 x}+\frac {7 b c \text {arctanh}\left (\frac {\sqrt {a+b x^2}}{\sqrt {a}}\right )}{2 a^{9/2}} \] Output:

-1/5*b*(d*x+c)/a^2/(b*x^2+a)^(5/2)-1/15*b*(9*d*x+10*c)/a^3/(b*x^2+a)^(3/2) 
-1/5*b*(11*d*x+15*c)/a^4/(b*x^2+a)^(1/2)-1/2*c*(b*x^2+a)^(1/2)/a^4/x^2-d*( 
b*x^2+a)^(1/2)/a^4/x+7/2*b*c*arctanh((b*x^2+a)^(1/2)/a^(1/2))/a^(9/2)
 

Mathematica [A] (verified)

Time = 0.71 (sec) , antiderivative size = 120, normalized size of antiderivative = 0.80 \[ \int \frac {c+d x}{x^3 \left (a+b x^2\right )^{7/2}} \, dx=-\frac {15 a^3 (c+2 d x)+3 b^3 x^6 (35 c+32 d x)+5 a b^2 x^4 (49 c+48 d x)+a^2 b x^2 (161 c+180 d x)}{30 a^4 x^2 \left (a+b x^2\right )^{5/2}}-\frac {7 b c \text {arctanh}\left (\frac {\sqrt {b} x-\sqrt {a+b x^2}}{\sqrt {a}}\right )}{a^{9/2}} \] Input:

Integrate[(c + d*x)/(x^3*(a + b*x^2)^(7/2)),x]
 

Output:

-1/30*(15*a^3*(c + 2*d*x) + 3*b^3*x^6*(35*c + 32*d*x) + 5*a*b^2*x^4*(49*c 
+ 48*d*x) + a^2*b*x^2*(161*c + 180*d*x))/(a^4*x^2*(a + b*x^2)^(5/2)) - (7* 
b*c*ArcTanh[(Sqrt[b]*x - Sqrt[a + b*x^2])/Sqrt[a]])/a^(9/2)
 

Rubi [A] (verified)

Time = 1.05 (sec) , antiderivative size = 170, normalized size of antiderivative = 1.13, number of steps used = 13, number of rules used = 12, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.600, Rules used = {532, 25, 2336, 27, 2336, 27, 2338, 25, 534, 243, 73, 221}

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.

\(\displaystyle \int \frac {c+d x}{x^3 \left (a+b x^2\right )^{7/2}} \, dx\)

\(\Big \downarrow \) 532

\(\displaystyle -\frac {\int -\frac {-\frac {4 b d x^3}{a}-\frac {5 b c x^2}{a}+5 d x+5 c}{x^3 \left (b x^2+a\right )^{5/2}}dx}{5 a}-\frac {b (c+d x)}{5 a^2 \left (a+b x^2\right )^{5/2}}\)

\(\Big \downarrow \) 25

\(\displaystyle \frac {\int \frac {-\frac {4 b d x^3}{a}-\frac {5 b c x^2}{a}+5 d x+5 c}{x^3 \left (b x^2+a\right )^{5/2}}dx}{5 a}-\frac {b (c+d x)}{5 a^2 \left (a+b x^2\right )^{5/2}}\)

\(\Big \downarrow \) 2336

\(\displaystyle \frac {-\frac {\int -\frac {3 \left (-\frac {6 b d x^3}{a}-\frac {10 b c x^2}{a}+5 d x+5 c\right )}{x^3 \left (b x^2+a\right )^{3/2}}dx}{3 a}-\frac {b (10 c+9 d x)}{3 a^2 \left (a+b x^2\right )^{3/2}}}{5 a}-\frac {b (c+d x)}{5 a^2 \left (a+b x^2\right )^{5/2}}\)

\(\Big \downarrow \) 27

\(\displaystyle \frac {\frac {\int \frac {-\frac {6 b d x^3}{a}-\frac {10 b c x^2}{a}+5 d x+5 c}{x^3 \left (b x^2+a\right )^{3/2}}dx}{a}-\frac {b (10 c+9 d x)}{3 a^2 \left (a+b x^2\right )^{3/2}}}{5 a}-\frac {b (c+d x)}{5 a^2 \left (a+b x^2\right )^{5/2}}\)

\(\Big \downarrow \) 2336

\(\displaystyle \frac {\frac {-\frac {\int -\frac {5 \left (-\frac {3 b c x^2}{a}+d x+c\right )}{x^3 \sqrt {b x^2+a}}dx}{a}-\frac {b (15 c+11 d x)}{a^2 \sqrt {a+b x^2}}}{a}-\frac {b (10 c+9 d x)}{3 a^2 \left (a+b x^2\right )^{3/2}}}{5 a}-\frac {b (c+d x)}{5 a^2 \left (a+b x^2\right )^{5/2}}\)

\(\Big \downarrow \) 27

\(\displaystyle \frac {\frac {\frac {5 \int \frac {-\frac {3 b c x^2}{a}+d x+c}{x^3 \sqrt {b x^2+a}}dx}{a}-\frac {b (15 c+11 d x)}{a^2 \sqrt {a+b x^2}}}{a}-\frac {b (10 c+9 d x)}{3 a^2 \left (a+b x^2\right )^{3/2}}}{5 a}-\frac {b (c+d x)}{5 a^2 \left (a+b x^2\right )^{5/2}}\)

\(\Big \downarrow \) 2338

\(\displaystyle \frac {\frac {\frac {5 \left (-\frac {\int -\frac {2 a d-7 b c x}{x^2 \sqrt {b x^2+a}}dx}{2 a}-\frac {c \sqrt {a+b x^2}}{2 a x^2}\right )}{a}-\frac {b (15 c+11 d x)}{a^2 \sqrt {a+b x^2}}}{a}-\frac {b (10 c+9 d x)}{3 a^2 \left (a+b x^2\right )^{3/2}}}{5 a}-\frac {b (c+d x)}{5 a^2 \left (a+b x^2\right )^{5/2}}\)

\(\Big \downarrow \) 25

\(\displaystyle \frac {\frac {\frac {5 \left (\frac {\int \frac {2 a d-7 b c x}{x^2 \sqrt {b x^2+a}}dx}{2 a}-\frac {c \sqrt {a+b x^2}}{2 a x^2}\right )}{a}-\frac {b (15 c+11 d x)}{a^2 \sqrt {a+b x^2}}}{a}-\frac {b (10 c+9 d x)}{3 a^2 \left (a+b x^2\right )^{3/2}}}{5 a}-\frac {b (c+d x)}{5 a^2 \left (a+b x^2\right )^{5/2}}\)

\(\Big \downarrow \) 534

\(\displaystyle \frac {\frac {\frac {5 \left (\frac {-7 b c \int \frac {1}{x \sqrt {b x^2+a}}dx-\frac {2 d \sqrt {a+b x^2}}{x}}{2 a}-\frac {c \sqrt {a+b x^2}}{2 a x^2}\right )}{a}-\frac {b (15 c+11 d x)}{a^2 \sqrt {a+b x^2}}}{a}-\frac {b (10 c+9 d x)}{3 a^2 \left (a+b x^2\right )^{3/2}}}{5 a}-\frac {b (c+d x)}{5 a^2 \left (a+b x^2\right )^{5/2}}\)

\(\Big \downarrow \) 243

\(\displaystyle \frac {\frac {\frac {5 \left (\frac {-\frac {7}{2} b c \int \frac {1}{x^2 \sqrt {b x^2+a}}dx^2-\frac {2 d \sqrt {a+b x^2}}{x}}{2 a}-\frac {c \sqrt {a+b x^2}}{2 a x^2}\right )}{a}-\frac {b (15 c+11 d x)}{a^2 \sqrt {a+b x^2}}}{a}-\frac {b (10 c+9 d x)}{3 a^2 \left (a+b x^2\right )^{3/2}}}{5 a}-\frac {b (c+d x)}{5 a^2 \left (a+b x^2\right )^{5/2}}\)

\(\Big \downarrow \) 73

\(\displaystyle \frac {\frac {\frac {5 \left (\frac {-7 c \int \frac {1}{\frac {x^4}{b}-\frac {a}{b}}d\sqrt {b x^2+a}-\frac {2 d \sqrt {a+b x^2}}{x}}{2 a}-\frac {c \sqrt {a+b x^2}}{2 a x^2}\right )}{a}-\frac {b (15 c+11 d x)}{a^2 \sqrt {a+b x^2}}}{a}-\frac {b (10 c+9 d x)}{3 a^2 \left (a+b x^2\right )^{3/2}}}{5 a}-\frac {b (c+d x)}{5 a^2 \left (a+b x^2\right )^{5/2}}\)

\(\Big \downarrow \) 221

\(\displaystyle \frac {\frac {\frac {5 \left (\frac {\frac {7 b c \text {arctanh}\left (\frac {\sqrt {a+b x^2}}{\sqrt {a}}\right )}{\sqrt {a}}-\frac {2 d \sqrt {a+b x^2}}{x}}{2 a}-\frac {c \sqrt {a+b x^2}}{2 a x^2}\right )}{a}-\frac {b (15 c+11 d x)}{a^2 \sqrt {a+b x^2}}}{a}-\frac {b (10 c+9 d x)}{3 a^2 \left (a+b x^2\right )^{3/2}}}{5 a}-\frac {b (c+d x)}{5 a^2 \left (a+b x^2\right )^{5/2}}\)

Input:

Int[(c + d*x)/(x^3*(a + b*x^2)^(7/2)),x]
 

Output:

-1/5*(b*(c + d*x))/(a^2*(a + b*x^2)^(5/2)) + (-1/3*(b*(10*c + 9*d*x))/(a^2 
*(a + b*x^2)^(3/2)) + (-((b*(15*c + 11*d*x))/(a^2*Sqrt[a + b*x^2])) + (5*( 
-1/2*(c*Sqrt[a + b*x^2])/(a*x^2) + ((-2*d*Sqrt[a + b*x^2])/x + (7*b*c*ArcT 
anh[Sqrt[a + b*x^2]/Sqrt[a]])/Sqrt[a])/(2*a)))/a)/a)/(5*a)
 

Defintions of rubi rules used

rule 25

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

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 73

Int[((a_.) + (b_.)*(x_))^(m_)*((c_.) + (d_.)*(x_))^(n_), x_Symbol] :> With[ 
{p = Denominator[m]}, Simp[p/b   Subst[Int[x^(p*(m + 1) - 1)*(c - a*(d/b) + 
 d*(x^p/b))^n, x], x, (a + b*x)^(1/p)], x]] /; FreeQ[{a, b, c, d}, x] && Lt 
Q[-1, m, 0] && LeQ[-1, n, 0] && LeQ[Denominator[n], Denominator[m]] && IntL 
inearQ[a, b, c, d, m, n, x]
 

rule 221

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

rule 243

Int[(x_)^(m_.)*((a_) + (b_.)*(x_)^2)^(p_), x_Symbol] :> Simp[1/2   Subst[In 
t[x^((m - 1)/2)*(a + b*x)^p, x], x, x^2], x] /; FreeQ[{a, b, m, p}, x] && I 
ntegerQ[(m - 1)/2]
 

rule 532

Int[(x_)^(m_)*((c_) + (d_.)*(x_))^(n_.)*((a_) + (b_.)*(x_)^2)^(p_), x_Symbo 
l] :> With[{Qx = PolynomialQuotient[x^m*(c + d*x)^n, a + b*x^2, x], e = Coe 
ff[PolynomialRemainder[x^m*(c + d*x)^n, a + b*x^2, x], x, 0], f = Coeff[Pol 
ynomialRemainder[x^m*(c + d*x)^n, a + b*x^2, x], x, 1]}, Simp[(a*f - b*e*x) 
*((a + b*x^2)^(p + 1)/(2*a*b*(p + 1))), x] + Simp[1/(2*a*(p + 1))   Int[x^m 
*(a + b*x^2)^(p + 1)*ExpandToSum[2*a*(p + 1)*(Qx/x^m) + e*((2*p + 3)/x^m), 
x], x], x]] /; FreeQ[{a, b, c, d}, x] && IGtQ[n, 0] && ILtQ[m, 0] && LtQ[p, 
 -1] && IntegerQ[2*p]
 

rule 534

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

rule 2336

Int[(Pq_)*((c_.)*(x_))^(m_.)*((a_) + (b_.)*(x_)^2)^(p_), x_Symbol] :> With[ 
{Q = PolynomialQuotient[(c*x)^m*Pq, a + b*x^2, x], f = Coeff[PolynomialRema 
inder[(c*x)^m*Pq, a + b*x^2, x], x, 0], g = Coeff[PolynomialRemainder[(c*x) 
^m*Pq, a + b*x^2, x], x, 1]}, Simp[(a*g - b*f*x)*((a + b*x^2)^(p + 1)/(2*a* 
b*(p + 1))), x] + Simp[1/(2*a*(p + 1))   Int[(c*x)^m*(a + b*x^2)^(p + 1)*Ex 
pandToSum[(2*a*(p + 1)*Q)/(c*x)^m + (f*(2*p + 3))/(c*x)^m, x], x], x]] /; F 
reeQ[{a, b, c}, x] && PolyQ[Pq, x] && LtQ[p, -1] && ILtQ[m, 0]
 

rule 2338

Int[(Pq_)*((c_.)*(x_))^(m_)*((a_) + (b_.)*(x_)^2)^(p_), x_Symbol] :> With[{ 
Q = PolynomialQuotient[Pq, c*x, x], R = PolynomialRemainder[Pq, c*x, x]}, S 
imp[R*(c*x)^(m + 1)*((a + b*x^2)^(p + 1)/(a*c*(m + 1))), x] + Simp[1/(a*c*( 
m + 1))   Int[(c*x)^(m + 1)*(a + b*x^2)^p*ExpandToSum[a*c*(m + 1)*Q - b*R*( 
m + 2*p + 3)*x, x], x], x]] /; FreeQ[{a, b, c, p}, x] && PolyQ[Pq, x] && Lt 
Q[m, -1] && (IntegerQ[2*p] || NeQ[Expon[Pq, x], 1])
 

Maple [A] (verified)

Time = 0.34 (sec) , antiderivative size = 186, normalized size of antiderivative = 1.24

method	result	size
default	\(c \left (-\frac {1}{2 a \,x^{2} \left (b \,x^{2}+a \right )^{\frac {5}{2}}}-\frac {7 b \left (\frac {1}{5 a \left (b \,x^{2}+a \right )^{\frac {5}{2}}}+\frac {\frac {1}{3 a \left (b \,x^{2}+a \right )^{\frac {3}{2}}}+\frac {\frac {1}{a \sqrt {b \,x^{2}+a}}-\frac {\ln \left (\frac {2 a +2 \sqrt {a}\, \sqrt {b \,x^{2}+a}}{x}\right )}{a^{\frac {3}{2}}}}{a}}{a}\right )}{2 a}\right )+d \left (-\frac {1}{a x \left (b \,x^{2}+a \right )^{\frac {5}{2}}}-\frac {6 b \left (\frac {x}{5 a \left (b \,x^{2}+a \right )^{\frac {5}{2}}}+\frac {\frac {4 x}{15 a \left (b \,x^{2}+a \right )^{\frac {3}{2}}}+\frac {8 x}{15 a^{2} \sqrt {b \,x^{2}+a}}}{a}\right )}{a}\right )\)	\(186\)
risch	\(\text {Expression too large to display}\)	\(1097\)

Input:

int((d*x+c)/x^3/(b*x^2+a)^(7/2),x,method=_RETURNVERBOSE)
 

Output:

c*(-1/2/a/x^2/(b*x^2+a)^(5/2)-7/2*b/a*(1/5/a/(b*x^2+a)^(5/2)+1/a*(1/3/a/(b 
*x^2+a)^(3/2)+1/a*(1/a/(b*x^2+a)^(1/2)-1/a^(3/2)*ln((2*a+2*a^(1/2)*(b*x^2+ 
a)^(1/2))/x)))))+d*(-1/a/x/(b*x^2+a)^(5/2)-6*b/a*(1/5*x/a/(b*x^2+a)^(5/2)+ 
4/5/a*(1/3*x/a/(b*x^2+a)^(3/2)+2/3/a^2/(b*x^2+a)^(1/2)*x)))
 

Fricas [A] (verification not implemented)

Time = 0.13 (sec) , antiderivative size = 404, normalized size of antiderivative = 2.69 \[ \int \frac {c+d x}{x^3 \left (a+b x^2\right )^{7/2}} \, dx=\left [\frac {105 \, {\left (b^{4} c x^{8} + 3 \, a b^{3} c x^{6} + 3 \, a^{2} b^{2} c x^{4} + a^{3} b c x^{2}\right )} \sqrt {a} \log \left (-\frac {b x^{2} + 2 \, \sqrt {b x^{2} + a} \sqrt {a} + 2 \, a}{x^{2}}\right ) - 2 \, {\left (96 \, a b^{3} d x^{7} + 105 \, a b^{3} c x^{6} + 240 \, a^{2} b^{2} d x^{5} + 245 \, a^{2} b^{2} c x^{4} + 180 \, a^{3} b d x^{3} + 161 \, a^{3} b c x^{2} + 30 \, a^{4} d x + 15 \, a^{4} c\right )} \sqrt {b x^{2} + a}}{60 \, {\left (a^{5} b^{3} x^{8} + 3 \, a^{6} b^{2} x^{6} + 3 \, a^{7} b x^{4} + a^{8} x^{2}\right )}}, -\frac {105 \, {\left (b^{4} c x^{8} + 3 \, a b^{3} c x^{6} + 3 \, a^{2} b^{2} c x^{4} + a^{3} b c x^{2}\right )} \sqrt {-a} \arctan \left (\frac {\sqrt {b x^{2} + a} \sqrt {-a}}{a}\right ) + {\left (96 \, a b^{3} d x^{7} + 105 \, a b^{3} c x^{6} + 240 \, a^{2} b^{2} d x^{5} + 245 \, a^{2} b^{2} c x^{4} + 180 \, a^{3} b d x^{3} + 161 \, a^{3} b c x^{2} + 30 \, a^{4} d x + 15 \, a^{4} c\right )} \sqrt {b x^{2} + a}}{30 \, {\left (a^{5} b^{3} x^{8} + 3 \, a^{6} b^{2} x^{6} + 3 \, a^{7} b x^{4} + a^{8} x^{2}\right )}}\right ] \] Input:

integrate((d*x+c)/x^3/(b*x^2+a)^(7/2),x, algorithm="fricas")
 

Output:

[1/60*(105*(b^4*c*x^8 + 3*a*b^3*c*x^6 + 3*a^2*b^2*c*x^4 + a^3*b*c*x^2)*sqr 
t(a)*log(-(b*x^2 + 2*sqrt(b*x^2 + a)*sqrt(a) + 2*a)/x^2) - 2*(96*a*b^3*d*x 
^7 + 105*a*b^3*c*x^6 + 240*a^2*b^2*d*x^5 + 245*a^2*b^2*c*x^4 + 180*a^3*b*d 
*x^3 + 161*a^3*b*c*x^2 + 30*a^4*d*x + 15*a^4*c)*sqrt(b*x^2 + a))/(a^5*b^3* 
x^8 + 3*a^6*b^2*x^6 + 3*a^7*b*x^4 + a^8*x^2), -1/30*(105*(b^4*c*x^8 + 3*a* 
b^3*c*x^6 + 3*a^2*b^2*c*x^4 + a^3*b*c*x^2)*sqrt(-a)*arctan(sqrt(b*x^2 + a) 
*sqrt(-a)/a) + (96*a*b^3*d*x^7 + 105*a*b^3*c*x^6 + 240*a^2*b^2*d*x^5 + 245 
*a^2*b^2*c*x^4 + 180*a^3*b*d*x^3 + 161*a^3*b*c*x^2 + 30*a^4*d*x + 15*a^4*c 
)*sqrt(b*x^2 + a))/(a^5*b^3*x^8 + 3*a^6*b^2*x^6 + 3*a^7*b*x^4 + a^8*x^2)]
 

Sympy [B] (verification not implemented)

Leaf count of result is larger than twice the leaf count of optimal. 2654 vs. \(2 (138) = 276\).

Time = 17.51 (sec) , antiderivative size = 2654, normalized size of antiderivative = 17.69 \[ \int \frac {c+d x}{x^3 \left (a+b x^2\right )^{7/2}} \, dx=\text {Too large to display} \] Input:

integrate((d*x+c)/x**3/(b*x**2+a)**(7/2),x)
 

Output:

c*(-30*a**29*sqrt(1 + b*x**2/a)/(60*a**(65/2)*x**2 + 360*a**(63/2)*b*x**4 
+ 900*a**(61/2)*b**2*x**6 + 1200*a**(59/2)*b**3*x**8 + 900*a**(57/2)*b**4* 
x**10 + 360*a**(55/2)*b**5*x**12 + 60*a**(53/2)*b**6*x**14) - 412*a**28*b* 
x**2*sqrt(1 + b*x**2/a)/(60*a**(65/2)*x**2 + 360*a**(63/2)*b*x**4 + 900*a* 
*(61/2)*b**2*x**6 + 1200*a**(59/2)*b**3*x**8 + 900*a**(57/2)*b**4*x**10 + 
360*a**(55/2)*b**5*x**12 + 60*a**(53/2)*b**6*x**14) - 105*a**28*b*x**2*log 
(b*x**2/a)/(60*a**(65/2)*x**2 + 360*a**(63/2)*b*x**4 + 900*a**(61/2)*b**2* 
x**6 + 1200*a**(59/2)*b**3*x**8 + 900*a**(57/2)*b**4*x**10 + 360*a**(55/2) 
*b**5*x**12 + 60*a**(53/2)*b**6*x**14) + 210*a**28*b*x**2*log(sqrt(1 + b*x 
**2/a) + 1)/(60*a**(65/2)*x**2 + 360*a**(63/2)*b*x**4 + 900*a**(61/2)*b**2 
*x**6 + 1200*a**(59/2)*b**3*x**8 + 900*a**(57/2)*b**4*x**10 + 360*a**(55/2 
)*b**5*x**12 + 60*a**(53/2)*b**6*x**14) - 1546*a**27*b**2*x**4*sqrt(1 + b* 
x**2/a)/(60*a**(65/2)*x**2 + 360*a**(63/2)*b*x**4 + 900*a**(61/2)*b**2*x** 
6 + 1200*a**(59/2)*b**3*x**8 + 900*a**(57/2)*b**4*x**10 + 360*a**(55/2)*b* 
*5*x**12 + 60*a**(53/2)*b**6*x**14) - 630*a**27*b**2*x**4*log(b*x**2/a)/(6 
0*a**(65/2)*x**2 + 360*a**(63/2)*b*x**4 + 900*a**(61/2)*b**2*x**6 + 1200*a 
**(59/2)*b**3*x**8 + 900*a**(57/2)*b**4*x**10 + 360*a**(55/2)*b**5*x**12 + 
 60*a**(53/2)*b**6*x**14) + 1260*a**27*b**2*x**4*log(sqrt(1 + b*x**2/a) + 
1)/(60*a**(65/2)*x**2 + 360*a**(63/2)*b*x**4 + 900*a**(61/2)*b**2*x**6 + 1 
200*a**(59/2)*b**3*x**8 + 900*a**(57/2)*b**4*x**10 + 360*a**(55/2)*b**5...
                                                                                    
                                                                                    
 

Maxima [A] (verification not implemented)

Time = 0.04 (sec) , antiderivative size = 155, normalized size of antiderivative = 1.03 \[ \int \frac {c+d x}{x^3 \left (a+b x^2\right )^{7/2}} \, dx=-\frac {16 \, b d x}{5 \, \sqrt {b x^{2} + a} a^{4}} - \frac {8 \, b d x}{5 \, {\left (b x^{2} + a\right )}^{\frac {3}{2}} a^{3}} - \frac {6 \, b d x}{5 \, {\left (b x^{2} + a\right )}^{\frac {5}{2}} a^{2}} + \frac {7 \, b c \operatorname {arsinh}\left (\frac {a}{\sqrt {a b} {\left | x \right |}}\right )}{2 \, a^{\frac {9}{2}}} - \frac {7 \, b c}{2 \, \sqrt {b x^{2} + a} a^{4}} - \frac {7 \, b c}{6 \, {\left (b x^{2} + a\right )}^{\frac {3}{2}} a^{3}} - \frac {7 \, b c}{10 \, {\left (b x^{2} + a\right )}^{\frac {5}{2}} a^{2}} - \frac {d}{{\left (b x^{2} + a\right )}^{\frac {5}{2}} a x} - \frac {c}{2 \, {\left (b x^{2} + a\right )}^{\frac {5}{2}} a x^{2}} \] Input:

integrate((d*x+c)/x^3/(b*x^2+a)^(7/2),x, algorithm="maxima")
 

Output:

-16/5*b*d*x/(sqrt(b*x^2 + a)*a^4) - 8/5*b*d*x/((b*x^2 + a)^(3/2)*a^3) - 6/ 
5*b*d*x/((b*x^2 + a)^(5/2)*a^2) + 7/2*b*c*arcsinh(a/(sqrt(a*b)*abs(x)))/a^ 
(9/2) - 7/2*b*c/(sqrt(b*x^2 + a)*a^4) - 7/6*b*c/((b*x^2 + a)^(3/2)*a^3) - 
7/10*b*c/((b*x^2 + a)^(5/2)*a^2) - d/((b*x^2 + a)^(5/2)*a*x) - 1/2*c/((b*x 
^2 + a)^(5/2)*a*x^2)
 

Giac [A] (verification not implemented)

Time = 0.16 (sec) , antiderivative size = 222, normalized size of antiderivative = 1.48 \[ \int \frac {c+d x}{x^3 \left (a+b x^2\right )^{7/2}} \, dx=-\frac {{\left ({\left (3 \, {\left ({\left (\frac {11 \, b^{3} d x}{a^{4}} + \frac {15 \, b^{3} c}{a^{4}}\right )} x + \frac {25 \, b^{2} d}{a^{3}}\right )} x + \frac {100 \, b^{2} c}{a^{3}}\right )} x + \frac {45 \, b d}{a^{2}}\right )} x + \frac {58 \, b c}{a^{2}}}{15 \, {\left (b x^{2} + a\right )}^{\frac {5}{2}}} - \frac {7 \, b c \arctan \left (-\frac {\sqrt {b} x - \sqrt {b x^{2} + a}}{\sqrt {-a}}\right )}{\sqrt {-a} a^{4}} + \frac {{\left (\sqrt {b} x - \sqrt {b x^{2} + a}\right )}^{3} b c + 2 \, {\left (\sqrt {b} x - \sqrt {b x^{2} + a}\right )}^{2} a \sqrt {b} d + {\left (\sqrt {b} x - \sqrt {b x^{2} + a}\right )} a b c - 2 \, a^{2} \sqrt {b} d}{{\left ({\left (\sqrt {b} x - \sqrt {b x^{2} + a}\right )}^{2} - a\right )}^{2} a^{4}} \] Input:

integrate((d*x+c)/x^3/(b*x^2+a)^(7/2),x, algorithm="giac")
 

Output:

-1/15*(((3*((11*b^3*d*x/a^4 + 15*b^3*c/a^4)*x + 25*b^2*d/a^3)*x + 100*b^2* 
c/a^3)*x + 45*b*d/a^2)*x + 58*b*c/a^2)/(b*x^2 + a)^(5/2) - 7*b*c*arctan(-( 
sqrt(b)*x - sqrt(b*x^2 + a))/sqrt(-a))/(sqrt(-a)*a^4) + ((sqrt(b)*x - sqrt 
(b*x^2 + a))^3*b*c + 2*(sqrt(b)*x - sqrt(b*x^2 + a))^2*a*sqrt(b)*d + (sqrt 
(b)*x - sqrt(b*x^2 + a))*a*b*c - 2*a^2*sqrt(b)*d)/(((sqrt(b)*x - sqrt(b*x^ 
2 + a))^2 - a)^2*a^4)
 

Mupad [B] (verification not implemented)

Time = 9.27 (sec) , antiderivative size = 159, normalized size of antiderivative = 1.06 \[ \int \frac {c+d x}{x^3 \left (a+b x^2\right )^{7/2}} \, dx=\frac {7\,b\,c\,\mathrm {atanh}\left (\frac {\sqrt {b\,x^2+a}}{\sqrt {a}}\right )}{2\,a^{9/2}}-\frac {c}{2\,a\,x^2\,{\left (b\,x^2+a\right )}^{5/2}}-\frac {161\,b\,c}{30\,a^2\,{\left (b\,x^2+a\right )}^{5/2}}+\frac {a^3\,d-16\,d\,{\left (b\,x^2+a\right )}^3+8\,a\,d\,{\left (b\,x^2+a\right )}^2+2\,a^2\,d\,\left (b\,x^2+a\right )}{5\,a^4\,x\,{\left (b\,x^2+a\right )}^{5/2}}-\frac {49\,b^2\,c\,x^2}{6\,a^3\,{\left (b\,x^2+a\right )}^{5/2}}-\frac {7\,b^3\,c\,x^4}{2\,a^4\,{\left (b\,x^2+a\right )}^{5/2}} \] Input:

int((c + d*x)/(x^3*(a + b*x^2)^(7/2)),x)
 

Output:

(7*b*c*atanh((a + b*x^2)^(1/2)/a^(1/2)))/(2*a^(9/2)) - c/(2*a*x^2*(a + b*x 
^2)^(5/2)) - (161*b*c)/(30*a^2*(a + b*x^2)^(5/2)) + (a^3*d - 16*d*(a + b*x 
^2)^3 + 8*a*d*(a + b*x^2)^2 + 2*a^2*d*(a + b*x^2))/(5*a^4*x*(a + b*x^2)^(5 
/2)) - (49*b^2*c*x^2)/(6*a^3*(a + b*x^2)^(5/2)) - (7*b^3*c*x^4)/(2*a^4*(a 
+ b*x^2)^(5/2))
 

Reduce [B] (verification not implemented)

Time = 0.23 (sec) , antiderivative size = 504, normalized size of antiderivative = 3.36 \[ \int \frac {c+d x}{x^3 \left (a+b x^2\right )^{7/2}} \, dx=\frac {-15 \sqrt {b \,x^{2}+a}\, a^{4} c -30 \sqrt {b \,x^{2}+a}\, a^{4} d x -161 \sqrt {b \,x^{2}+a}\, a^{3} b c \,x^{2}-180 \sqrt {b \,x^{2}+a}\, a^{3} b d \,x^{3}-245 \sqrt {b \,x^{2}+a}\, a^{2} b^{2} c \,x^{4}-240 \sqrt {b \,x^{2}+a}\, a^{2} b^{2} d \,x^{5}-105 \sqrt {b \,x^{2}+a}\, a \,b^{3} c \,x^{6}-96 \sqrt {b \,x^{2}+a}\, a \,b^{3} d \,x^{7}-105 \sqrt {a}\, \mathrm {log}\left (\frac {\sqrt {b \,x^{2}+a}-\sqrt {a}+\sqrt {b}\, x}{\sqrt {a}}\right ) a^{3} b c \,x^{2}-315 \sqrt {a}\, \mathrm {log}\left (\frac {\sqrt {b \,x^{2}+a}-\sqrt {a}+\sqrt {b}\, x}{\sqrt {a}}\right ) a^{2} b^{2} c \,x^{4}-315 \sqrt {a}\, \mathrm {log}\left (\frac {\sqrt {b \,x^{2}+a}-\sqrt {a}+\sqrt {b}\, x}{\sqrt {a}}\right ) a \,b^{3} c \,x^{6}-105 \sqrt {a}\, \mathrm {log}\left (\frac {\sqrt {b \,x^{2}+a}-\sqrt {a}+\sqrt {b}\, x}{\sqrt {a}}\right ) b^{4} c \,x^{8}+105 \sqrt {a}\, \mathrm {log}\left (\frac {\sqrt {b \,x^{2}+a}+\sqrt {a}+\sqrt {b}\, x}{\sqrt {a}}\right ) a^{3} b c \,x^{2}+315 \sqrt {a}\, \mathrm {log}\left (\frac {\sqrt {b \,x^{2}+a}+\sqrt {a}+\sqrt {b}\, x}{\sqrt {a}}\right ) a^{2} b^{2} c \,x^{4}+315 \sqrt {a}\, \mathrm {log}\left (\frac {\sqrt {b \,x^{2}+a}+\sqrt {a}+\sqrt {b}\, x}{\sqrt {a}}\right ) a \,b^{3} c \,x^{6}+105 \sqrt {a}\, \mathrm {log}\left (\frac {\sqrt {b \,x^{2}+a}+\sqrt {a}+\sqrt {b}\, x}{\sqrt {a}}\right ) b^{4} c \,x^{8}+96 \sqrt {b}\, a^{4} d \,x^{2}+288 \sqrt {b}\, a^{3} b d \,x^{4}+288 \sqrt {b}\, a^{2} b^{2} d \,x^{6}+96 \sqrt {b}\, a \,b^{3} d \,x^{8}}{30 a^{5} x^{2} \left (b^{3} x^{6}+3 a \,b^{2} x^{4}+3 a^{2} b \,x^{2}+a^{3}\right )} \] Input:

int((d*x+c)/x^3/(b*x^2+a)^(7/2),x)
 

Output:

( - 15*sqrt(a + b*x**2)*a**4*c - 30*sqrt(a + b*x**2)*a**4*d*x - 161*sqrt(a 
 + b*x**2)*a**3*b*c*x**2 - 180*sqrt(a + b*x**2)*a**3*b*d*x**3 - 245*sqrt(a 
 + b*x**2)*a**2*b**2*c*x**4 - 240*sqrt(a + b*x**2)*a**2*b**2*d*x**5 - 105* 
sqrt(a + b*x**2)*a*b**3*c*x**6 - 96*sqrt(a + b*x**2)*a*b**3*d*x**7 - 105*s 
qrt(a)*log((sqrt(a + b*x**2) - sqrt(a) + sqrt(b)*x)/sqrt(a))*a**3*b*c*x**2 
 - 315*sqrt(a)*log((sqrt(a + b*x**2) - sqrt(a) + sqrt(b)*x)/sqrt(a))*a**2* 
b**2*c*x**4 - 315*sqrt(a)*log((sqrt(a + b*x**2) - sqrt(a) + sqrt(b)*x)/sqr 
t(a))*a*b**3*c*x**6 - 105*sqrt(a)*log((sqrt(a + b*x**2) - sqrt(a) + sqrt(b 
)*x)/sqrt(a))*b**4*c*x**8 + 105*sqrt(a)*log((sqrt(a + b*x**2) + sqrt(a) + 
sqrt(b)*x)/sqrt(a))*a**3*b*c*x**2 + 315*sqrt(a)*log((sqrt(a + b*x**2) + sq 
rt(a) + sqrt(b)*x)/sqrt(a))*a**2*b**2*c*x**4 + 315*sqrt(a)*log((sqrt(a + b 
*x**2) + sqrt(a) + sqrt(b)*x)/sqrt(a))*a*b**3*c*x**6 + 105*sqrt(a)*log((sq 
rt(a + b*x**2) + sqrt(a) + sqrt(b)*x)/sqrt(a))*b**4*c*x**8 + 96*sqrt(b)*a* 
*4*d*x**2 + 288*sqrt(b)*a**3*b*d*x**4 + 288*sqrt(b)*a**2*b**2*d*x**6 + 96* 
sqrt(b)*a*b**3*d*x**8)/(30*a**5*x**2*(a**3 + 3*a**2*b*x**2 + 3*a*b**2*x**4 
 + b**3*x**6))