\(\int \frac {A+B x}{(c+d x) (a+b x^2)^{5/2}} \, dx\) [192]

Optimal result

Integrand size = 24, antiderivative size = 204 \[ \int \frac {A+B x}{(c+d x) \left (a+b x^2\right )^{5/2}} \, dx=-\frac {a (B c-A d)-(A b c+a B d) x}{3 a \left (b c^2+a d^2\right ) \left (a+b x^2\right )^{3/2}}-\frac {3 a^2 d^2 (B c-A d)+\left (a B d \left (b c^2-2 a d^2\right )-A b c \left (2 b c^2+5 a d^2\right )\right ) x}{3 a^2 \left (b c^2+a d^2\right )^2 \sqrt {a+b x^2}}+\frac {d^3 (B c-A d) \text {arctanh}\left (\frac {a d-b c x}{\sqrt {b c^2+a d^2} \sqrt {a+b x^2}}\right )}{\left (b c^2+a d^2\right )^{5/2}} \] Output:

-1/3*(a*(-A*d+B*c)-(A*b*c+B*a*d)*x)/a/(a*d^2+b*c^2)/(b*x^2+a)^(3/2)-1/3*(3 
*a^2*d^2*(-A*d+B*c)+(a*B*d*(-2*a*d^2+b*c^2)-A*b*c*(5*a*d^2+2*b*c^2))*x)/a^ 
2/(a*d^2+b*c^2)^2/(b*x^2+a)^(1/2)+d^3*(-A*d+B*c)*arctanh((-b*c*x+a*d)/(a*d 
^2+b*c^2)^(1/2)/(b*x^2+a)^(1/2))/(a*d^2+b*c^2)^(5/2)
 

Mathematica [A] (verified)

Time = 1.60 (sec) , antiderivative size = 222, normalized size of antiderivative = 1.09 \[ \int \frac {A+B x}{(c+d x) \left (a+b x^2\right )^{5/2}} \, dx=\frac {2 A b^3 c^3 x^3+a^3 d^2 (-4 B c+4 A d+3 B d x)+a b^2 c x \left (3 A c^2-B c d x^2+5 A d^2 x^2\right )+a^2 b \left (A d \left (c^2+6 c d x+3 d^2 x^2\right )-B \left (c^3+3 c d^2 x^2-2 d^3 x^3\right )\right )}{3 a^2 \left (b c^2+a d^2\right )^2 \left (a+b x^2\right )^{3/2}}+\frac {2 d^3 (B c-A d) \arctan \left (\frac {\sqrt {b} (c+d x)-d \sqrt {a+b x^2}}{\sqrt {-b c^2-a d^2}}\right )}{\left (-b c^2-a d^2\right )^{5/2}} \] Input:

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

Output:

(2*A*b^3*c^3*x^3 + a^3*d^2*(-4*B*c + 4*A*d + 3*B*d*x) + a*b^2*c*x*(3*A*c^2 
 - B*c*d*x^2 + 5*A*d^2*x^2) + a^2*b*(A*d*(c^2 + 6*c*d*x + 3*d^2*x^2) - B*( 
c^3 + 3*c*d^2*x^2 - 2*d^3*x^3)))/(3*a^2*(b*c^2 + a*d^2)^2*(a + b*x^2)^(3/2 
)) + (2*d^3*(B*c - A*d)*ArcTan[(Sqrt[b]*(c + d*x) - d*Sqrt[a + b*x^2])/Sqr 
t[-(b*c^2) - a*d^2]])/(-(b*c^2) - a*d^2)^(5/2)
 

Rubi [A] (verified)

Time = 0.40 (sec) , antiderivative size = 225, normalized size of antiderivative = 1.10, number of steps used = 8, number of rules used = 7, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.292, Rules used = {686, 25, 27, 686, 27, 488, 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.

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

\(\Big \downarrow \) 686

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

\(\Big \downarrow \) 25

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

\(\Big \downarrow \) 27

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

\(\Big \downarrow \) 686

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

\(\Big \downarrow \) 27

\(\displaystyle \frac {-\frac {3 a d^3 (B c-A d) \int \frac {1}{(c+d x) \sqrt {b x^2+a}}dx}{a d^2+b c^2}-\frac {3 a^2 d^2 (B c-A d)+x \left (a B d \left (b c^2-2 a d^2\right )-A b c \left (5 a d^2+2 b c^2\right )\right )}{a \sqrt {a+b x^2} \left (a d^2+b c^2\right )}}{3 a \left (a d^2+b c^2\right )}-\frac {a (B c-A d)-x (a B d+A b c)}{3 a \left (a+b x^2\right )^{3/2} \left (a d^2+b c^2\right )}\)

\(\Big \downarrow \) 488

\(\displaystyle \frac {\frac {3 a d^3 (B c-A d) \int \frac {1}{b c^2+a d^2-\frac {(a d-b c x)^2}{b x^2+a}}d\frac {a d-b c x}{\sqrt {b x^2+a}}}{a d^2+b c^2}-\frac {3 a^2 d^2 (B c-A d)+x \left (a B d \left (b c^2-2 a d^2\right )-A b c \left (5 a d^2+2 b c^2\right )\right )}{a \sqrt {a+b x^2} \left (a d^2+b c^2\right )}}{3 a \left (a d^2+b c^2\right )}-\frac {a (B c-A d)-x (a B d+A b c)}{3 a \left (a+b x^2\right )^{3/2} \left (a d^2+b c^2\right )}\)

\(\Big \downarrow \) 219

\(\displaystyle \frac {\frac {3 a d^3 (B c-A d) \text {arctanh}\left (\frac {a d-b c x}{\sqrt {a+b x^2} \sqrt {a d^2+b c^2}}\right )}{\left (a d^2+b c^2\right )^{3/2}}-\frac {3 a^2 d^2 (B c-A d)+x \left (a B d \left (b c^2-2 a d^2\right )-A b c \left (5 a d^2+2 b c^2\right )\right )}{a \sqrt {a+b x^2} \left (a d^2+b c^2\right )}}{3 a \left (a d^2+b c^2\right )}-\frac {a (B c-A d)-x (a B d+A b c)}{3 a \left (a+b x^2\right )^{3/2} \left (a d^2+b c^2\right )}\)

Input:

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

Output:

-1/3*(a*(B*c - A*d) - (A*b*c + a*B*d)*x)/(a*(b*c^2 + a*d^2)*(a + b*x^2)^(3 
/2)) + (-((3*a^2*d^2*(B*c - A*d) + (a*B*d*(b*c^2 - 2*a*d^2) - A*b*c*(2*b*c 
^2 + 5*a*d^2))*x)/(a*(b*c^2 + a*d^2)*Sqrt[a + b*x^2])) + (3*a*d^3*(B*c - A 
*d)*ArcTanh[(a*d - b*c*x)/(Sqrt[b*c^2 + a*d^2]*Sqrt[a + b*x^2])])/(b*c^2 + 
 a*d^2)^(3/2))/(3*a*(b*c^2 + a*d^2))
 

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 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 488

Int[1/(((c_) + (d_.)*(x_))*Sqrt[(a_) + (b_.)*(x_)^2]), x_Symbol] :> -Subst[ 
Int[1/(b*c^2 + a*d^2 - x^2), x], x, (a*d - b*c*x)/Sqrt[a + b*x^2]] /; FreeQ 
[{a, b, c, d}, x]
 

rule 686

Int[((d_.) + (e_.)*(x_))^(m_)*((f_.) + (g_.)*(x_))*((a_) + (c_.)*(x_)^2)^(p 
_), x_Symbol] :> Simp[(-(d + e*x)^(m + 1))*(f*a*c*e - a*g*c*d + c*(c*d*f + 
a*e*g)*x)*((a + c*x^2)^(p + 1)/(2*a*c*(p + 1)*(c*d^2 + a*e^2))), x] + Simp[ 
1/(2*a*c*(p + 1)*(c*d^2 + a*e^2))   Int[(d + e*x)^m*(a + c*x^2)^(p + 1)*Sim 
p[f*(c^2*d^2*(2*p + 3) + a*c*e^2*(m + 2*p + 3)) - a*c*d*e*g*m + c*e*(c*d*f 
+ a*e*g)*(m + 2*p + 4)*x, x], x], x] /; FreeQ[{a, c, d, e, f, g}, x] && LtQ 
[p, -1] && (IntegerQ[m] || IntegerQ[p] || IntegersQ[2*m, 2*p])
 

Maple [B] (verified)

Leaf count of result is larger than twice the leaf count of optimal. \(645\) vs. \(2(190)=380\).

Time = 1.32 (sec) , antiderivative size = 646, normalized size of antiderivative = 3.17

method	result	size
default	\(\frac {B \left (\frac {x}{3 a \left (b \,x^{2}+a \right )^{\frac {3}{2}}}+\frac {2 x}{3 a^{2} \sqrt {b \,x^{2}+a}}\right )}{d}+\frac {\left (A d -B c \right ) \left (\frac {d^{2}}{3 \left (a \,d^{2}+b \,c^{2}\right ) \left (b \left (x +\frac {c}{d}\right )^{2}-\frac {2 b c \left (x +\frac {c}{d}\right )}{d}+\frac {a \,d^{2}+b \,c^{2}}{d^{2}}\right )^{\frac {3}{2}}}+\frac {b c d \left (\frac {\frac {4 b \left (x +\frac {c}{d}\right )}{3}-\frac {4 b c}{3 d}}{\left (\frac {4 b \left (a \,d^{2}+b \,c^{2}\right )}{d^{2}}-\frac {4 b^{2} c^{2}}{d^{2}}\right ) \left (b \left (x +\frac {c}{d}\right )^{2}-\frac {2 b c \left (x +\frac {c}{d}\right )}{d}+\frac {a \,d^{2}+b \,c^{2}}{d^{2}}\right )^{\frac {3}{2}}}+\frac {16 b \left (2 b \left (x +\frac {c}{d}\right )-\frac {2 b c}{d}\right )}{3 {\left (\frac {4 b \left (a \,d^{2}+b \,c^{2}\right )}{d^{2}}-\frac {4 b^{2} c^{2}}{d^{2}}\right )}^{2} \sqrt {b \left (x +\frac {c}{d}\right )^{2}-\frac {2 b c \left (x +\frac {c}{d}\right )}{d}+\frac {a \,d^{2}+b \,c^{2}}{d^{2}}}}\right )}{a \,d^{2}+b \,c^{2}}+\frac {d^{2} \left (\frac {d^{2}}{\left (a \,d^{2}+b \,c^{2}\right ) \sqrt {b \left (x +\frac {c}{d}\right )^{2}-\frac {2 b c \left (x +\frac {c}{d}\right )}{d}+\frac {a \,d^{2}+b \,c^{2}}{d^{2}}}}+\frac {2 b c d \left (2 b \left (x +\frac {c}{d}\right )-\frac {2 b c}{d}\right )}{\left (a \,d^{2}+b \,c^{2}\right ) \left (\frac {4 b \left (a \,d^{2}+b \,c^{2}\right )}{d^{2}}-\frac {4 b^{2} c^{2}}{d^{2}}\right ) \sqrt {b \left (x +\frac {c}{d}\right )^{2}-\frac {2 b c \left (x +\frac {c}{d}\right )}{d}+\frac {a \,d^{2}+b \,c^{2}}{d^{2}}}}-\frac {d^{2} \ln \left (\frac {\frac {2 a \,d^{2}+2 b \,c^{2}}{d^{2}}-\frac {2 b c \left (x +\frac {c}{d}\right )}{d}+2 \sqrt {\frac {a \,d^{2}+b \,c^{2}}{d^{2}}}\, \sqrt {b \left (x +\frac {c}{d}\right )^{2}-\frac {2 b c \left (x +\frac {c}{d}\right )}{d}+\frac {a \,d^{2}+b \,c^{2}}{d^{2}}}}{x +\frac {c}{d}}\right )}{\left (a \,d^{2}+b \,c^{2}\right ) \sqrt {\frac {a \,d^{2}+b \,c^{2}}{d^{2}}}}\right )}{a \,d^{2}+b \,c^{2}}\right )}{d^{2}}\)	\(646\)

Input:

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

Output:

B/d*(1/3*x/a/(b*x^2+a)^(3/2)+2/3/a^2/(b*x^2+a)^(1/2)*x)+(A*d-B*c)/d^2*(1/3 
/(a*d^2+b*c^2)*d^2/(b*(x+c/d)^2-2*b*c/d*(x+c/d)+(a*d^2+b*c^2)/d^2)^(3/2)+b 
*c*d/(a*d^2+b*c^2)*(2/3*(2*b*(x+c/d)-2*b*c/d)/(4*b*(a*d^2+b*c^2)/d^2-4*b^2 
*c^2/d^2)/(b*(x+c/d)^2-2*b*c/d*(x+c/d)+(a*d^2+b*c^2)/d^2)^(3/2)+16/3*b/(4* 
b*(a*d^2+b*c^2)/d^2-4*b^2*c^2/d^2)^2*(2*b*(x+c/d)-2*b*c/d)/(b*(x+c/d)^2-2* 
b*c/d*(x+c/d)+(a*d^2+b*c^2)/d^2)^(1/2))+1/(a*d^2+b*c^2)*d^2*(1/(a*d^2+b*c^ 
2)*d^2/(b*(x+c/d)^2-2*b*c/d*(x+c/d)+(a*d^2+b*c^2)/d^2)^(1/2)+2*b*c*d/(a*d^ 
2+b*c^2)*(2*b*(x+c/d)-2*b*c/d)/(4*b*(a*d^2+b*c^2)/d^2-4*b^2*c^2/d^2)/(b*(x 
+c/d)^2-2*b*c/d*(x+c/d)+(a*d^2+b*c^2)/d^2)^(1/2)-1/(a*d^2+b*c^2)*d^2/((a*d 
^2+b*c^2)/d^2)^(1/2)*ln((2*(a*d^2+b*c^2)/d^2-2*b*c/d*(x+c/d)+2*((a*d^2+b*c 
^2)/d^2)^(1/2)*(b*(x+c/d)^2-2*b*c/d*(x+c/d)+(a*d^2+b*c^2)/d^2)^(1/2))/(x+c 
/d))))
 

Fricas [B] (verification not implemented)

Leaf count of result is larger than twice the leaf count of optimal. 585 vs. \(2 (192) = 384\).

Time = 0.50 (sec) , antiderivative size = 1196, normalized size of antiderivative = 5.86 \[ \int \frac {A+B x}{(c+d x) \left (a+b x^2\right )^{5/2}} \, dx =\text {Too large to display} \] Input:

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

Output:

[-1/6*(3*(B*a^4*c*d^3 - A*a^4*d^4 + (B*a^2*b^2*c*d^3 - A*a^2*b^2*d^4)*x^4 
+ 2*(B*a^3*b*c*d^3 - A*a^3*b*d^4)*x^2)*sqrt(b*c^2 + a*d^2)*log((2*a*b*c*d* 
x - a*b*c^2 - 2*a^2*d^2 - (2*b^2*c^2 + a*b*d^2)*x^2 - 2*sqrt(b*c^2 + a*d^2 
)*(b*c*x - a*d)*sqrt(b*x^2 + a))/(d^2*x^2 + 2*c*d*x + c^2)) + 2*(B*a^2*b^2 
*c^5 - A*a^2*b^2*c^4*d + 5*B*a^3*b*c^3*d^2 - 5*A*a^3*b*c^2*d^3 + 4*B*a^4*c 
*d^4 - 4*A*a^4*d^5 - (2*A*b^4*c^5 - B*a*b^3*c^4*d + 7*A*a*b^3*c^3*d^2 + B* 
a^2*b^2*c^2*d^3 + 5*A*a^2*b^2*c*d^4 + 2*B*a^3*b*d^5)*x^3 + 3*(B*a^2*b^2*c^ 
3*d^2 - A*a^2*b^2*c^2*d^3 + B*a^3*b*c*d^4 - A*a^3*b*d^5)*x^2 - 3*(A*a*b^3* 
c^5 + 3*A*a^2*b^2*c^3*d^2 + B*a^3*b*c^2*d^3 + 2*A*a^3*b*c*d^4 + B*a^4*d^5) 
*x)*sqrt(b*x^2 + a))/(a^4*b^3*c^6 + 3*a^5*b^2*c^4*d^2 + 3*a^6*b*c^2*d^4 + 
a^7*d^6 + (a^2*b^5*c^6 + 3*a^3*b^4*c^4*d^2 + 3*a^4*b^3*c^2*d^4 + a^5*b^2*d 
^6)*x^4 + 2*(a^3*b^4*c^6 + 3*a^4*b^3*c^4*d^2 + 3*a^5*b^2*c^2*d^4 + a^6*b*d 
^6)*x^2), 1/3*(3*(B*a^4*c*d^3 - A*a^4*d^4 + (B*a^2*b^2*c*d^3 - A*a^2*b^2*d 
^4)*x^4 + 2*(B*a^3*b*c*d^3 - A*a^3*b*d^4)*x^2)*sqrt(-b*c^2 - a*d^2)*arctan 
(sqrt(-b*c^2 - a*d^2)*(b*c*x - a*d)*sqrt(b*x^2 + a)/(a*b*c^2 + a^2*d^2 + ( 
b^2*c^2 + a*b*d^2)*x^2)) - (B*a^2*b^2*c^5 - A*a^2*b^2*c^4*d + 5*B*a^3*b*c^ 
3*d^2 - 5*A*a^3*b*c^2*d^3 + 4*B*a^4*c*d^4 - 4*A*a^4*d^5 - (2*A*b^4*c^5 - B 
*a*b^3*c^4*d + 7*A*a*b^3*c^3*d^2 + B*a^2*b^2*c^2*d^3 + 5*A*a^2*b^2*c*d^4 + 
 2*B*a^3*b*d^5)*x^3 + 3*(B*a^2*b^2*c^3*d^2 - A*a^2*b^2*c^2*d^3 + B*a^3*b*c 
*d^4 - A*a^3*b*d^5)*x^2 - 3*(A*a*b^3*c^5 + 3*A*a^2*b^2*c^3*d^2 + B*a^3*...
 

Sympy [F]

\[ \int \frac {A+B x}{(c+d x) \left (a+b x^2\right )^{5/2}} \, dx=\int \frac {A + B x}{\left (a + b x^{2}\right )^{\frac {5}{2}} \left (c + d x\right )}\, dx \] Input:

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

Output:

Integral((A + B*x)/((a + b*x**2)**(5/2)*(c + d*x)), x)
 

Maxima [B] (verification not implemented)

Leaf count of result is larger than twice the leaf count of optimal. 630 vs. \(2 (192) = 384\).

Time = 0.11 (sec) , antiderivative size = 630, normalized size of antiderivative = 3.09 \[ \int \frac {A+B x}{(c+d x) \left (a+b x^2\right )^{5/2}} \, dx=-\frac {B b c^{2} x}{\frac {\sqrt {b x^{2} + a} a b^{2} c^{4}}{d} + 2 \, \sqrt {b x^{2} + a} a^{2} b c^{2} d + \sqrt {b x^{2} + a} a^{3} d^{3}} - \frac {B b c^{2} x}{3 \, {\left ({\left (b x^{2} + a\right )}^{\frac {3}{2}} a b c^{2} d + {\left (b x^{2} + a\right )}^{\frac {3}{2}} a^{2} d^{3}\right )}} - \frac {2 \, B b c^{2} x}{3 \, {\left (\sqrt {b x^{2} + a} a^{2} b c^{2} d + \sqrt {b x^{2} + a} a^{3} d^{3}\right )}} + \frac {A b c x}{3 \, {\left ({\left (b x^{2} + a\right )}^{\frac {3}{2}} a b c^{2} + {\left (b x^{2} + a\right )}^{\frac {3}{2}} a^{2} d^{2}\right )}} + \frac {A b c x}{2 \, \sqrt {b x^{2} + a} a^{2} b c^{2} + \frac {\sqrt {b x^{2} + a} a b^{2} c^{4}}{d^{2}} + \sqrt {b x^{2} + a} a^{3} d^{2}} + \frac {2 \, A b c x}{3 \, {\left (\sqrt {b x^{2} + a} a^{2} b c^{2} + \sqrt {b x^{2} + a} a^{3} d^{2}\right )}} - \frac {B c}{3 \, {\left ({\left (b x^{2} + a\right )}^{\frac {3}{2}} b c^{2} + {\left (b x^{2} + a\right )}^{\frac {3}{2}} a d^{2}\right )}} - \frac {B c}{2 \, \sqrt {b x^{2} + a} a b c^{2} + \frac {\sqrt {b x^{2} + a} b^{2} c^{4}}{d^{2}} + \sqrt {b x^{2} + a} a^{2} d^{2}} + \frac {A}{\frac {\sqrt {b x^{2} + a} b^{2} c^{4}}{d^{3}} + \frac {2 \, \sqrt {b x^{2} + a} a b c^{2}}{d} + \sqrt {b x^{2} + a} a^{2} d} + \frac {A}{3 \, {\left (\frac {{\left (b x^{2} + a\right )}^{\frac {3}{2}} b c^{2}}{d} + {\left (b x^{2} + a\right )}^{\frac {3}{2}} a d\right )}} + \frac {2 \, B x}{3 \, \sqrt {b x^{2} + a} a^{2} d} + \frac {B x}{3 \, {\left (b x^{2} + a\right )}^{\frac {3}{2}} a d} - \frac {B c \operatorname {arsinh}\left (\frac {b c x}{\sqrt {a b} {\left | d x + c \right |}} - \frac {a d}{\sqrt {a b} {\left | d x + c \right |}}\right )}{{\left (a + \frac {b c^{2}}{d^{2}}\right )}^{\frac {5}{2}} d^{2}} + \frac {A \operatorname {arsinh}\left (\frac {b c x}{\sqrt {a b} {\left | d x + c \right |}} - \frac {a d}{\sqrt {a b} {\left | d x + c \right |}}\right )}{{\left (a + \frac {b c^{2}}{d^{2}}\right )}^{\frac {5}{2}} d} \] Input:

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

Output:

-B*b*c^2*x/(sqrt(b*x^2 + a)*a*b^2*c^4/d + 2*sqrt(b*x^2 + a)*a^2*b*c^2*d + 
sqrt(b*x^2 + a)*a^3*d^3) - 1/3*B*b*c^2*x/((b*x^2 + a)^(3/2)*a*b*c^2*d + (b 
*x^2 + a)^(3/2)*a^2*d^3) - 2/3*B*b*c^2*x/(sqrt(b*x^2 + a)*a^2*b*c^2*d + sq 
rt(b*x^2 + a)*a^3*d^3) + 1/3*A*b*c*x/((b*x^2 + a)^(3/2)*a*b*c^2 + (b*x^2 + 
 a)^(3/2)*a^2*d^2) + A*b*c*x/(2*sqrt(b*x^2 + a)*a^2*b*c^2 + sqrt(b*x^2 + a 
)*a*b^2*c^4/d^2 + sqrt(b*x^2 + a)*a^3*d^2) + 2/3*A*b*c*x/(sqrt(b*x^2 + a)* 
a^2*b*c^2 + sqrt(b*x^2 + a)*a^3*d^2) - 1/3*B*c/((b*x^2 + a)^(3/2)*b*c^2 + 
(b*x^2 + a)^(3/2)*a*d^2) - B*c/(2*sqrt(b*x^2 + a)*a*b*c^2 + sqrt(b*x^2 + a 
)*b^2*c^4/d^2 + sqrt(b*x^2 + a)*a^2*d^2) + A/(sqrt(b*x^2 + a)*b^2*c^4/d^3 
+ 2*sqrt(b*x^2 + a)*a*b*c^2/d + sqrt(b*x^2 + a)*a^2*d) + 1/3*A/((b*x^2 + a 
)^(3/2)*b*c^2/d + (b*x^2 + a)^(3/2)*a*d) + 2/3*B*x/(sqrt(b*x^2 + a)*a^2*d) 
 + 1/3*B*x/((b*x^2 + a)^(3/2)*a*d) - B*c*arcsinh(b*c*x/(sqrt(a*b)*abs(d*x 
+ c)) - a*d/(sqrt(a*b)*abs(d*x + c)))/((a + b*c^2/d^2)^(5/2)*d^2) + A*arcs 
inh(b*c*x/(sqrt(a*b)*abs(d*x + c)) - a*d/(sqrt(a*b)*abs(d*x + c)))/((a + b 
*c^2/d^2)^(5/2)*d)
                                                                                    
                                                                                    
 

Giac [B] (verification not implemented)

Leaf count of result is larger than twice the leaf count of optimal. 1467 vs. \(2 (192) = 384\).

Time = 0.13 (sec) , antiderivative size = 1467, normalized size of antiderivative = 7.19 \[ \int \frac {A+B x}{(c+d x) \left (a+b x^2\right )^{5/2}} \, dx=\text {Too large to display} \] Input:

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

Output:

2*(B*c*d^3 - A*d^4)*arctan(((sqrt(b)*x - sqrt(b*x^2 + a))*d + sqrt(b)*c)/s 
qrt(-b*c^2 - a*d^2))/((b^2*c^4 + 2*a*b*c^2*d^2 + a^2*d^4)*sqrt(-b*c^2 - a* 
d^2)) + 1/3*((((2*A*b^10*c^15 - B*a*b^9*c^14*d + 17*A*a*b^9*c^13*d^2 - 4*B 
*a^2*b^8*c^12*d^3 + 60*A*a^2*b^8*c^11*d^4 - 3*B*a^3*b^7*c^10*d^5 + 115*A*a 
^3*b^7*c^9*d^6 + 10*B*a^4*b^6*c^8*d^7 + 130*A*a^4*b^6*c^7*d^8 + 25*B*a^5*b 
^5*c^6*d^9 + 87*A*a^5*b^5*c^5*d^10 + 24*B*a^6*b^4*c^4*d^11 + 32*A*a^6*b^4* 
c^3*d^12 + 11*B*a^7*b^3*c^2*d^13 + 5*A*a^7*b^3*c*d^14 + 2*B*a^8*b^2*d^15)* 
x/(a^2*b^9*c^16 + 8*a^3*b^8*c^14*d^2 + 28*a^4*b^7*c^12*d^4 + 56*a^5*b^6*c^ 
10*d^6 + 70*a^6*b^5*c^8*d^8 + 56*a^7*b^4*c^6*d^10 + 28*a^8*b^3*c^4*d^12 + 
8*a^9*b^2*c^2*d^14 + a^10*b*d^16) - 3*(B*a^2*b^8*c^13*d^2 - A*a^2*b^8*c^12 
*d^3 + 6*B*a^3*b^7*c^11*d^4 - 6*A*a^3*b^7*c^10*d^5 + 15*B*a^4*b^6*c^9*d^6 
- 15*A*a^4*b^6*c^8*d^7 + 20*B*a^5*b^5*c^7*d^8 - 20*A*a^5*b^5*c^6*d^9 + 15* 
B*a^6*b^4*c^5*d^10 - 15*A*a^6*b^4*c^4*d^11 + 6*B*a^7*b^3*c^3*d^12 - 6*A*a^ 
7*b^3*c^2*d^13 + B*a^8*b^2*c*d^14 - A*a^8*b^2*d^15)/(a^2*b^9*c^16 + 8*a^3* 
b^8*c^14*d^2 + 28*a^4*b^7*c^12*d^4 + 56*a^5*b^6*c^10*d^6 + 70*a^6*b^5*c^8* 
d^8 + 56*a^7*b^4*c^6*d^10 + 28*a^8*b^3*c^4*d^12 + 8*a^9*b^2*c^2*d^14 + a^1 
0*b*d^16))*x + 3*(A*a*b^9*c^15 + 8*A*a^2*b^8*c^13*d^2 + B*a^3*b^7*c^12*d^3 
 + 27*A*a^3*b^7*c^11*d^4 + 6*B*a^4*b^6*c^10*d^5 + 50*A*a^4*b^6*c^9*d^6 + 1 
5*B*a^5*b^5*c^8*d^7 + 55*A*a^5*b^5*c^7*d^8 + 20*B*a^6*b^4*c^6*d^9 + 36*A*a 
^6*b^4*c^5*d^10 + 15*B*a^7*b^3*c^4*d^11 + 13*A*a^7*b^3*c^3*d^12 + 6*B*a...
 

Mupad [F(-1)]

Timed out. \[ \int \frac {A+B x}{(c+d x) \left (a+b x^2\right )^{5/2}} \, dx=\int \frac {A+B\,x}{{\left (b\,x^2+a\right )}^{5/2}\,\left (c+d\,x\right )} \,d x \] Input:

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

Output:

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

Reduce [B] (verification not implemented)

Time = 0.64 (sec) , antiderivative size = 7379, normalized size of antiderivative = 36.17 \[ \int \frac {A+B x}{(c+d x) \left (a+b x^2\right )^{5/2}} \, dx =\text {Too large to display} \] Input:

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

Output:

( - 6*sqrt(b)*sqrt(2*sqrt(b)*sqrt(a*d**2 + b*c**2)*c - a*d**2 - 2*b*c**2)* 
sqrt(a*d**2 + b*c**2)*atan((sqrt(a + b*x**2)*d + sqrt(b)*d*x)/sqrt(2*sqrt( 
b)*sqrt(a*d**2 + b*c**2)*c - a*d**2 - 2*b*c**2))*a**3*c*d**2 + 6*sqrt(b)*s 
qrt(2*sqrt(b)*sqrt(a*d**2 + b*c**2)*c - a*d**2 - 2*b*c**2)*sqrt(a*d**2 + b 
*c**2)*atan((sqrt(a + b*x**2)*d + sqrt(b)*d*x)/sqrt(2*sqrt(b)*sqrt(a*d**2 
+ b*c**2)*c - a*d**2 - 2*b*c**2))*a**2*b*c**2*d - 12*sqrt(b)*sqrt(2*sqrt(b 
)*sqrt(a*d**2 + b*c**2)*c - a*d**2 - 2*b*c**2)*sqrt(a*d**2 + b*c**2)*atan( 
(sqrt(a + b*x**2)*d + sqrt(b)*d*x)/sqrt(2*sqrt(b)*sqrt(a*d**2 + b*c**2)*c 
- a*d**2 - 2*b*c**2))*a**2*b*c*d**2*x**2 + 12*sqrt(b)*sqrt(2*sqrt(b)*sqrt( 
a*d**2 + b*c**2)*c - a*d**2 - 2*b*c**2)*sqrt(a*d**2 + b*c**2)*atan((sqrt(a 
 + b*x**2)*d + sqrt(b)*d*x)/sqrt(2*sqrt(b)*sqrt(a*d**2 + b*c**2)*c - a*d** 
2 - 2*b*c**2))*a*b**2*c**2*d*x**2 - 6*sqrt(b)*sqrt(2*sqrt(b)*sqrt(a*d**2 + 
 b*c**2)*c - a*d**2 - 2*b*c**2)*sqrt(a*d**2 + b*c**2)*atan((sqrt(a + b*x** 
2)*d + sqrt(b)*d*x)/sqrt(2*sqrt(b)*sqrt(a*d**2 + b*c**2)*c - a*d**2 - 2*b* 
c**2))*a*b**2*c*d**2*x**4 + 6*sqrt(b)*sqrt(2*sqrt(b)*sqrt(a*d**2 + b*c**2) 
*c - a*d**2 - 2*b*c**2)*sqrt(a*d**2 + b*c**2)*atan((sqrt(a + b*x**2)*d + s 
qrt(b)*d*x)/sqrt(2*sqrt(b)*sqrt(a*d**2 + b*c**2)*c - a*d**2 - 2*b*c**2))*b 
**3*c**2*d*x**4 - 6*sqrt(2*sqrt(b)*sqrt(a*d**2 + b*c**2)*c - a*d**2 - 2*b* 
c**2)*atan((sqrt(a + b*x**2)*d + sqrt(b)*d*x)/sqrt(2*sqrt(b)*sqrt(a*d**2 + 
 b*c**2)*c - a*d**2 - 2*b*c**2))*a**4*d**4 - 6*sqrt(2*sqrt(b)*sqrt(a*d*...