\(\int \frac {(A+B x) (c+d x)^5}{(a+b x^2)^{5/2}} \, dx\) [186]

Optimal result

Integrand size = 24, antiderivative size = 299 \[ \int \frac {(A+B x) (c+d x)^5}{\left (a+b x^2\right )^{5/2}} \, dx=-\frac {(c+d x)^4 (a (B c+A d)-(A b c-a B d) x)}{3 a b \left (a+b x^2\right )^{3/2}}-\frac {(c+d x)^2 \left (2 a^2 d^2 (5 B c+2 A d)-\left (5 a B d \left (b c^2-a d^2\right )+2 A b c \left (b c^2+3 a d^2\right )\right ) x\right )}{3 a^2 b^2 \sqrt {a+b x^2}}-\frac {d \left (4 \left (5 a B c d \left (b c^2-4 a d^2\right )+2 A \left (b^2 c^4+4 a b c^2 d^2-2 a^2 d^4\right )\right )+d \left (5 a B d \left (2 b c^2-3 a d^2\right )+2 A b c \left (2 b c^2+7 a d^2\right )\right ) x\right ) \sqrt {a+b x^2}}{6 a^2 b^3}-\frac {5 d^3 \left (a B d^2-2 b c (2 B c+A d)\right ) \text {arctanh}\left (\frac {\sqrt {b} x}{\sqrt {a+b x^2}}\right )}{2 b^{7/2}} \] Output:

-1/3*(d*x+c)^4*(a*(A*d+B*c)-(A*b*c-B*a*d)*x)/a/b/(b*x^2+a)^(3/2)-1/3*(d*x+ 
c)^2*(2*a^2*d^2*(2*A*d+5*B*c)-(5*a*B*d*(-a*d^2+b*c^2)+2*A*b*c*(3*a*d^2+b*c 
^2))*x)/a^2/b^2/(b*x^2+a)^(1/2)-1/6*d*(20*a*B*c*d*(-4*a*d^2+b*c^2)+8*A*(-2 
*a^2*d^4+4*a*b*c^2*d^2+b^2*c^4)+d*(5*a*B*d*(-3*a*d^2+2*b*c^2)+2*A*b*c*(7*a 
*d^2+2*b*c^2))*x)*(b*x^2+a)^(1/2)/a^2/b^3-5/2*d^3*(a*B*d^2-2*b*c*(A*d+2*B* 
c))*arctanh(b^(1/2)*x/(b*x^2+a)^(1/2))/b^(7/2)
 

Mathematica [A] (verified)

Time = 2.38 (sec) , antiderivative size = 301, normalized size of antiderivative = 1.01 \[ \int \frac {(A+B x) (c+d x)^5}{\left (a+b x^2\right )^{5/2}} \, dx=\frac {4 A b^4 c^5 x^3+a^4 d^4 (80 B c+16 A d+15 B d x)+2 a b^3 c^3 x \left (3 A c^2+5 B c d x^2+10 A d^2 x^2\right )-2 a^3 b d^2 \left (A d \left (20 c^2+15 c d x-12 d^2 x^2\right )+10 B \left (2 c^3+3 c^2 d x-6 c d^2 x^2-d^3 x^3\right )\right )-a^2 b^2 \left (2 A d \left (5 c^4+30 c^2 d^2 x^2+20 c d^3 x^3-3 d^4 x^4\right )+B \left (2 c^5+60 c^3 d^2 x^2+80 c^2 d^3 x^3-30 c d^4 x^4-3 d^5 x^5\right )\right )}{6 a^2 b^3 \left (a+b x^2\right )^{3/2}}+\frac {5 d^3 \left (a B d^2-2 b c (2 B c+A d)\right ) \log \left (-\sqrt {b} x+\sqrt {a+b x^2}\right )}{2 b^{7/2}} \] Input:

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

Output:

(4*A*b^4*c^5*x^3 + a^4*d^4*(80*B*c + 16*A*d + 15*B*d*x) + 2*a*b^3*c^3*x*(3 
*A*c^2 + 5*B*c*d*x^2 + 10*A*d^2*x^2) - 2*a^3*b*d^2*(A*d*(20*c^2 + 15*c*d*x 
 - 12*d^2*x^2) + 10*B*(2*c^3 + 3*c^2*d*x - 6*c*d^2*x^2 - d^3*x^3)) - a^2*b 
^2*(2*A*d*(5*c^4 + 30*c^2*d^2*x^2 + 20*c*d^3*x^3 - 3*d^4*x^4) + B*(2*c^5 + 
 60*c^3*d^2*x^2 + 80*c^2*d^3*x^3 - 30*c*d^4*x^4 - 3*d^5*x^5)))/(6*a^2*b^3* 
(a + b*x^2)^(3/2)) + (5*d^3*(a*B*d^2 - 2*b*c*(2*B*c + A*d))*Log[-(Sqrt[b]* 
x) + Sqrt[a + b*x^2]])/(2*b^(7/2))
 

Rubi [A] (verified)

Time = 0.56 (sec) , antiderivative size = 329, 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 = {684, 684, 25, 27, 676, 224, 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.

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

\(\Big \downarrow \) 684

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

\(\Big \downarrow \) 684

\(\displaystyle \frac {\frac {\int -\frac {d (c+d x) \left (a d \left (2 A b c^2-25 a B d c-8 a A d^2\right )+\left (5 a B d \left (2 b c^2-3 a d^2\right )+2 A b c \left (2 b c^2+7 a d^2\right )\right ) x\right )}{\sqrt {b x^2+a}}dx}{a b}-\frac {(c+d x)^2 \left (2 a^2 d^2 (2 A d+5 B c)-x \left (2 A b c \left (3 a d^2+b c^2\right )+5 a B d \left (b c^2-a d^2\right )\right )\right )}{a b \sqrt {a+b x^2}}}{3 a b}-\frac {(c+d x)^4 (a (A d+B c)-x (A b c-a B d))}{3 a b \left (a+b x^2\right )^{3/2}}\)

\(\Big \downarrow \) 25

\(\displaystyle \frac {-\frac {\int \frac {d (c+d x) \left (a d \left (2 A b c^2-25 a B d c-8 a A d^2\right )+\left (5 a B d \left (2 b c^2-3 a d^2\right )+2 A b c \left (2 b c^2+7 a d^2\right )\right ) x\right )}{\sqrt {b x^2+a}}dx}{a b}-\frac {(c+d x)^2 \left (2 a^2 d^2 (2 A d+5 B c)-x \left (2 A b c \left (3 a d^2+b c^2\right )+5 a B d \left (b c^2-a d^2\right )\right )\right )}{a b \sqrt {a+b x^2}}}{3 a b}-\frac {(c+d x)^4 (a (A d+B c)-x (A b c-a B d))}{3 a b \left (a+b x^2\right )^{3/2}}\)

\(\Big \downarrow \) 27

\(\displaystyle \frac {-\frac {d \int \frac {(c+d x) \left (a d \left (2 A b c^2-25 a B d c-8 a A d^2\right )+\left (5 a B d \left (2 b c^2-3 a d^2\right )+2 A b c \left (2 b c^2+7 a d^2\right )\right ) x\right )}{\sqrt {b x^2+a}}dx}{a b}-\frac {(c+d x)^2 \left (2 a^2 d^2 (2 A d+5 B c)-x \left (2 A b c \left (3 a d^2+b c^2\right )+5 a B d \left (b c^2-a d^2\right )\right )\right )}{a b \sqrt {a+b x^2}}}{3 a b}-\frac {(c+d x)^4 (a (A d+B c)-x (A b c-a B d))}{3 a b \left (a+b x^2\right )^{3/2}}\)

\(\Big \downarrow \) 676

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

\(\Big \downarrow \) 224

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

\(\Big \downarrow \) 219

\(\displaystyle \frac {-\frac {d \left (\frac {15 a^2 d^2 \text {arctanh}\left (\frac {\sqrt {b} x}{\sqrt {a+b x^2}}\right ) \left (a B d^2-2 b c (A d+2 B c)\right )}{2 b^{3/2}}+\frac {2 \sqrt {a+b x^2} \left (2 A \left (-2 a^2 d^4+4 a b c^2 d^2+b^2 c^4\right )+5 a B c d \left (b c^2-4 a d^2\right )\right )}{b}+\frac {d x \sqrt {a+b x^2} \left (2 A b c \left (7 a d^2+2 b c^2\right )+5 a B d \left (2 b c^2-3 a d^2\right )\right )}{2 b}\right )}{a b}-\frac {(c+d x)^2 \left (2 a^2 d^2 (2 A d+5 B c)-x \left (2 A b c \left (3 a d^2+b c^2\right )+5 a B d \left (b c^2-a d^2\right )\right )\right )}{a b \sqrt {a+b x^2}}}{3 a b}-\frac {(c+d x)^4 (a (A d+B c)-x (A b c-a B d))}{3 a b \left (a+b x^2\right )^{3/2}}\)

Input:

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

Output:

-1/3*((c + d*x)^4*(a*(B*c + A*d) - (A*b*c - a*B*d)*x))/(a*b*(a + b*x^2)^(3 
/2)) + (-(((c + d*x)^2*(2*a^2*d^2*(5*B*c + 2*A*d) - (5*a*B*d*(b*c^2 - a*d^ 
2) + 2*A*b*c*(b*c^2 + 3*a*d^2))*x))/(a*b*Sqrt[a + b*x^2])) - (d*((2*(5*a*B 
*c*d*(b*c^2 - 4*a*d^2) + 2*A*(b^2*c^4 + 4*a*b*c^2*d^2 - 2*a^2*d^4))*Sqrt[a 
 + b*x^2])/b + (d*(5*a*B*d*(2*b*c^2 - 3*a*d^2) + 2*A*b*c*(2*b*c^2 + 7*a*d^ 
2))*x*Sqrt[a + b*x^2])/(2*b) + (15*a^2*d^2*(a*B*d^2 - 2*b*c*(2*B*c + A*d)) 
*ArcTanh[(Sqrt[b]*x)/Sqrt[a + b*x^2]])/(2*b^(3/2))))/(a*b))/(3*a*b)
 

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 224

Int[1/Sqrt[(a_) + (b_.)*(x_)^2], x_Symbol] :> Subst[Int[1/(1 - b*x^2), x], 
x, x/Sqrt[a + b*x^2]] /; FreeQ[{a, b}, x] &&  !GtQ[a, 0]
 

rule 676

Int[((d_.) + (e_.)*(x_))*((f_.) + (g_.)*(x_))*((a_) + (c_.)*(x_)^2)^(p_), x 
_Symbol] :> Simp[(e*f + d*g)*((a + c*x^2)^(p + 1)/(2*c*(p + 1))), x] + (Sim 
p[e*g*x*((a + c*x^2)^(p + 1)/(c*(2*p + 3))), x] - Simp[(a*e*g - c*d*f*(2*p 
+ 3))/(c*(2*p + 3))   Int[(a + c*x^2)^p, x], x]) /; FreeQ[{a, c, d, e, f, g 
, p}, x] &&  !LeQ[p, -1]
 

rule 684

Int[((d_.) + (e_.)*(x_))^(m_)*((f_.) + (g_.)*(x_))*((a_) + (c_.)*(x_)^2)^(p 
_.), x_Symbol] :> Simp[(d + e*x)^(m - 1)*(a + c*x^2)^(p + 1)*((a*(e*f + d*g 
) - (c*d*f - a*e*g)*x)/(2*a*c*(p + 1))), x] - Simp[1/(2*a*c*(p + 1))   Int[ 
(d + e*x)^(m - 2)*(a + c*x^2)^(p + 1)*Simp[a*e*(e*f*(m - 1) + d*g*m) - c*d^ 
2*f*(2*p + 3) + e*(a*e*g*m - c*d*f*(m + 2*p + 2))*x, x], x], x] /; FreeQ[{a 
, c, d, e, f, g}, x] && LtQ[p, -1] && GtQ[m, 1] && (EqQ[d, 0] || (EqQ[m, 2] 
 && EqQ[p, -3] && RationalQ[a, c, d, e, f, g]) ||  !ILtQ[m + 2*p + 3, 0])
 

Maple [A] (verified)

Time = 1.60 (sec) , antiderivative size = 405, normalized size of antiderivative = 1.35

method	result	size
default	\(A \,c^{5} \left (\frac {x}{3 a \left (b \,x^{2}+a \right )^{\frac {3}{2}}}+\frac {2 x}{3 a^{2} \sqrt {b \,x^{2}+a}}\right )+d^{4} \left (A d +5 B c \right ) \left (\frac {x^{4}}{b \left (b \,x^{2}+a \right )^{\frac {3}{2}}}-\frac {4 a \left (-\frac {x^{2}}{b \left (b \,x^{2}+a \right )^{\frac {3}{2}}}-\frac {2 a}{3 b^{2} \left (b \,x^{2}+a \right )^{\frac {3}{2}}}\right )}{b}\right )+5 c \,d^{3} \left (A d +2 B c \right ) \left (-\frac {x^{3}}{3 b \left (b \,x^{2}+a \right )^{\frac {3}{2}}}+\frac {-\frac {x}{b \sqrt {b \,x^{2}+a}}+\frac {\ln \left (\sqrt {b}\, x +\sqrt {b \,x^{2}+a}\right )}{b^{\frac {3}{2}}}}{b}\right )+10 c^{2} d^{2} \left (A d +B c \right ) \left (-\frac {x^{2}}{b \left (b \,x^{2}+a \right )^{\frac {3}{2}}}-\frac {2 a}{3 b^{2} \left (b \,x^{2}+a \right )^{\frac {3}{2}}}\right )+5 d \,c^{3} \left (2 A d +B c \right ) \left (-\frac {x}{2 b \left (b \,x^{2}+a \right )^{\frac {3}{2}}}+\frac {a \left (\frac {x}{3 a \left (b \,x^{2}+a \right )^{\frac {3}{2}}}+\frac {2 x}{3 a^{2} \sqrt {b \,x^{2}+a}}\right )}{2 b}\right )-\frac {c^{4} \left (5 A d +B c \right )}{3 b \left (b \,x^{2}+a \right )^{\frac {3}{2}}}+B \,d^{5} \left (\frac {x^{5}}{2 b \left (b \,x^{2}+a \right )^{\frac {3}{2}}}-\frac {5 a \left (-\frac {x^{3}}{3 b \left (b \,x^{2}+a \right )^{\frac {3}{2}}}+\frac {-\frac {x}{b \sqrt {b \,x^{2}+a}}+\frac {\ln \left (\sqrt {b}\, x +\sqrt {b \,x^{2}+a}\right )}{b^{\frac {3}{2}}}}{b}\right )}{2 b}\right )\)	\(405\)
risch	\(\text {Expression too large to display}\)	\(1053\)

Input:

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

Output:

A*c^5*(1/3*x/a/(b*x^2+a)^(3/2)+2/3/a^2/(b*x^2+a)^(1/2)*x)+d^4*(A*d+5*B*c)* 
(x^4/b/(b*x^2+a)^(3/2)-4*a/b*(-x^2/b/(b*x^2+a)^(3/2)-2/3*a/b^2/(b*x^2+a)^( 
3/2)))+5*c*d^3*(A*d+2*B*c)*(-1/3*x^3/b/(b*x^2+a)^(3/2)+1/b*(-x/b/(b*x^2+a) 
^(1/2)+1/b^(3/2)*ln(b^(1/2)*x+(b*x^2+a)^(1/2))))+10*c^2*d^2*(A*d+B*c)*(-x^ 
2/b/(b*x^2+a)^(3/2)-2/3*a/b^2/(b*x^2+a)^(3/2))+5*d*c^3*(2*A*d+B*c)*(-1/2*x 
/b/(b*x^2+a)^(3/2)+1/2*a/b*(1/3*x/a/(b*x^2+a)^(3/2)+2/3/a^2/(b*x^2+a)^(1/2 
)*x))-1/3*c^4*(5*A*d+B*c)/b/(b*x^2+a)^(3/2)+B*d^5*(1/2*x^5/b/(b*x^2+a)^(3/ 
2)-5/2*a/b*(-1/3*x^3/b/(b*x^2+a)^(3/2)+1/b*(-x/b/(b*x^2+a)^(1/2)+1/b^(3/2) 
*ln(b^(1/2)*x+(b*x^2+a)^(1/2)))))
 

Fricas [A] (verification not implemented)

Time = 0.15 (sec) , antiderivative size = 1020, normalized size of antiderivative = 3.41 \[ \int \frac {(A+B x) (c+d x)^5}{\left (a+b x^2\right )^{5/2}} \, dx =\text {Too large to display} \] Input:

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

Output:

[-1/12*(15*(4*B*a^4*b*c^2*d^3 + 2*A*a^4*b*c*d^4 - B*a^5*d^5 + (4*B*a^2*b^3 
*c^2*d^3 + 2*A*a^2*b^3*c*d^4 - B*a^3*b^2*d^5)*x^4 + 2*(4*B*a^3*b^2*c^2*d^3 
 + 2*A*a^3*b^2*c*d^4 - B*a^4*b*d^5)*x^2)*sqrt(b)*log(-2*b*x^2 + 2*sqrt(b*x 
^2 + a)*sqrt(b)*x - a) - 2*(3*B*a^2*b^3*d^5*x^5 - 2*B*a^2*b^3*c^5 - 10*A*a 
^2*b^3*c^4*d - 40*B*a^3*b^2*c^3*d^2 - 40*A*a^3*b^2*c^2*d^3 + 80*B*a^4*b*c* 
d^4 + 16*A*a^4*b*d^5 + 6*(5*B*a^2*b^3*c*d^4 + A*a^2*b^3*d^5)*x^4 + 2*(2*A* 
b^5*c^5 + 5*B*a*b^4*c^4*d + 10*A*a*b^4*c^3*d^2 - 40*B*a^2*b^3*c^2*d^3 - 20 
*A*a^2*b^3*c*d^4 + 10*B*a^3*b^2*d^5)*x^3 - 12*(5*B*a^2*b^3*c^3*d^2 + 5*A*a 
^2*b^3*c^2*d^3 - 10*B*a^3*b^2*c*d^4 - 2*A*a^3*b^2*d^5)*x^2 + 3*(2*A*a*b^4* 
c^5 - 20*B*a^3*b^2*c^2*d^3 - 10*A*a^3*b^2*c*d^4 + 5*B*a^4*b*d^5)*x)*sqrt(b 
*x^2 + a))/(a^2*b^6*x^4 + 2*a^3*b^5*x^2 + a^4*b^4), -1/6*(15*(4*B*a^4*b*c^ 
2*d^3 + 2*A*a^4*b*c*d^4 - B*a^5*d^5 + (4*B*a^2*b^3*c^2*d^3 + 2*A*a^2*b^3*c 
*d^4 - B*a^3*b^2*d^5)*x^4 + 2*(4*B*a^3*b^2*c^2*d^3 + 2*A*a^3*b^2*c*d^4 - B 
*a^4*b*d^5)*x^2)*sqrt(-b)*arctan(sqrt(-b)*x/sqrt(b*x^2 + a)) - (3*B*a^2*b^ 
3*d^5*x^5 - 2*B*a^2*b^3*c^5 - 10*A*a^2*b^3*c^4*d - 40*B*a^3*b^2*c^3*d^2 - 
40*A*a^3*b^2*c^2*d^3 + 80*B*a^4*b*c*d^4 + 16*A*a^4*b*d^5 + 6*(5*B*a^2*b^3* 
c*d^4 + A*a^2*b^3*d^5)*x^4 + 2*(2*A*b^5*c^5 + 5*B*a*b^4*c^4*d + 10*A*a*b^4 
*c^3*d^2 - 40*B*a^2*b^3*c^2*d^3 - 20*A*a^2*b^3*c*d^4 + 10*B*a^3*b^2*d^5)*x 
^3 - 12*(5*B*a^2*b^3*c^3*d^2 + 5*A*a^2*b^3*c^2*d^3 - 10*B*a^3*b^2*c*d^4 - 
2*A*a^3*b^2*d^5)*x^2 + 3*(2*A*a*b^4*c^5 - 20*B*a^3*b^2*c^2*d^3 - 10*A*a...
 

Sympy [F]

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

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

Output:

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

Maxima [A] (verification not implemented)

Time = 0.06 (sec) , antiderivative size = 514, normalized size of antiderivative = 1.72 \[ \int \frac {(A+B x) (c+d x)^5}{\left (a+b x^2\right )^{5/2}} \, dx=\frac {B d^{5} x^{5}}{2 \, {\left (b x^{2} + a\right )}^{\frac {3}{2}} b} + \frac {5 \, B a d^{5} x {\left (\frac {3 \, x^{2}}{{\left (b x^{2} + a\right )}^{\frac {3}{2}} b} + \frac {2 \, a}{{\left (b x^{2} + a\right )}^{\frac {3}{2}} b^{2}}\right )}}{6 \, b} + \frac {2 \, A c^{5} x}{3 \, \sqrt {b x^{2} + a} a^{2}} + \frac {A c^{5} x}{3 \, {\left (b x^{2} + a\right )}^{\frac {3}{2}} a} + \frac {5 \, B a d^{5} x}{6 \, \sqrt {b x^{2} + a} b^{3}} - \frac {5 \, B a d^{5} \operatorname {arsinh}\left (\frac {b x}{\sqrt {a b}}\right )}{2 \, b^{\frac {7}{2}}} - \frac {B c^{5}}{3 \, {\left (b x^{2} + a\right )}^{\frac {3}{2}} b} - \frac {5 \, A c^{4} d}{3 \, {\left (b x^{2} + a\right )}^{\frac {3}{2}} b} - \frac {5}{3} \, {\left (2 \, B c^{2} d^{3} + A c d^{4}\right )} x {\left (\frac {3 \, x^{2}}{{\left (b x^{2} + a\right )}^{\frac {3}{2}} b} + \frac {2 \, a}{{\left (b x^{2} + a\right )}^{\frac {3}{2}} b^{2}}\right )} + \frac {{\left (5 \, B c d^{4} + A d^{5}\right )} x^{4}}{{\left (b x^{2} + a\right )}^{\frac {3}{2}} b} + \frac {4 \, {\left (5 \, B c d^{4} + A d^{5}\right )} a x^{2}}{{\left (b x^{2} + a\right )}^{\frac {3}{2}} b^{2}} - \frac {10 \, {\left (B c^{3} d^{2} + A c^{2} d^{3}\right )} x^{2}}{{\left (b x^{2} + a\right )}^{\frac {3}{2}} b} - \frac {5 \, {\left (2 \, B c^{2} d^{3} + A c d^{4}\right )} x}{3 \, \sqrt {b x^{2} + a} b^{2}} - \frac {5 \, {\left (B c^{4} d + 2 \, A c^{3} d^{2}\right )} x}{3 \, {\left (b x^{2} + a\right )}^{\frac {3}{2}} b} + \frac {5 \, {\left (B c^{4} d + 2 \, A c^{3} d^{2}\right )} x}{3 \, \sqrt {b x^{2} + a} a b} + \frac {5 \, {\left (2 \, B c^{2} d^{3} + A c d^{4}\right )} \operatorname {arsinh}\left (\frac {b x}{\sqrt {a b}}\right )}{b^{\frac {5}{2}}} + \frac {8 \, {\left (5 \, B c d^{4} + A d^{5}\right )} a^{2}}{3 \, {\left (b x^{2} + a\right )}^{\frac {3}{2}} b^{3}} - \frac {20 \, {\left (B c^{3} d^{2} + A c^{2} d^{3}\right )} a}{3 \, {\left (b x^{2} + a\right )}^{\frac {3}{2}} b^{2}} \] Input:

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

Output:

1/2*B*d^5*x^5/((b*x^2 + a)^(3/2)*b) + 5/6*B*a*d^5*x*(3*x^2/((b*x^2 + a)^(3 
/2)*b) + 2*a/((b*x^2 + a)^(3/2)*b^2))/b + 2/3*A*c^5*x/(sqrt(b*x^2 + a)*a^2 
) + 1/3*A*c^5*x/((b*x^2 + a)^(3/2)*a) + 5/6*B*a*d^5*x/(sqrt(b*x^2 + a)*b^3 
) - 5/2*B*a*d^5*arcsinh(b*x/sqrt(a*b))/b^(7/2) - 1/3*B*c^5/((b*x^2 + a)^(3 
/2)*b) - 5/3*A*c^4*d/((b*x^2 + a)^(3/2)*b) - 5/3*(2*B*c^2*d^3 + A*c*d^4)*x 
*(3*x^2/((b*x^2 + a)^(3/2)*b) + 2*a/((b*x^2 + a)^(3/2)*b^2)) + (5*B*c*d^4 
+ A*d^5)*x^4/((b*x^2 + a)^(3/2)*b) + 4*(5*B*c*d^4 + A*d^5)*a*x^2/((b*x^2 + 
 a)^(3/2)*b^2) - 10*(B*c^3*d^2 + A*c^2*d^3)*x^2/((b*x^2 + a)^(3/2)*b) - 5/ 
3*(2*B*c^2*d^3 + A*c*d^4)*x/(sqrt(b*x^2 + a)*b^2) - 5/3*(B*c^4*d + 2*A*c^3 
*d^2)*x/((b*x^2 + a)^(3/2)*b) + 5/3*(B*c^4*d + 2*A*c^3*d^2)*x/(sqrt(b*x^2 
+ a)*a*b) + 5*(2*B*c^2*d^3 + A*c*d^4)*arcsinh(b*x/sqrt(a*b))/b^(5/2) + 8/3 
*(5*B*c*d^4 + A*d^5)*a^2/((b*x^2 + a)^(3/2)*b^3) - 20/3*(B*c^3*d^2 + A*c^2 
*d^3)*a/((b*x^2 + a)^(3/2)*b^2)
 

Giac [A] (verification not implemented)

Time = 0.13 (sec) , antiderivative size = 410, normalized size of antiderivative = 1.37 \[ \int \frac {(A+B x) (c+d x)^5}{\left (a+b x^2\right )^{5/2}} \, dx=\frac {{\left ({\left ({\left (3 \, {\left (\frac {B d^{5} x}{b} + \frac {2 \, {\left (5 \, B a^{2} b^{5} c d^{4} + A a^{2} b^{5} d^{5}\right )}}{a^{2} b^{6}}\right )} x + \frac {2 \, {\left (2 \, A b^{7} c^{5} + 5 \, B a b^{6} c^{4} d + 10 \, A a b^{6} c^{3} d^{2} - 40 \, B a^{2} b^{5} c^{2} d^{3} - 20 \, A a^{2} b^{5} c d^{4} + 10 \, B a^{3} b^{4} d^{5}\right )}}{a^{2} b^{6}}\right )} x - \frac {12 \, {\left (5 \, B a^{2} b^{5} c^{3} d^{2} + 5 \, A a^{2} b^{5} c^{2} d^{3} - 10 \, B a^{3} b^{4} c d^{4} - 2 \, A a^{3} b^{4} d^{5}\right )}}{a^{2} b^{6}}\right )} x + \frac {3 \, {\left (2 \, A a b^{6} c^{5} - 20 \, B a^{3} b^{4} c^{2} d^{3} - 10 \, A a^{3} b^{4} c d^{4} + 5 \, B a^{4} b^{3} d^{5}\right )}}{a^{2} b^{6}}\right )} x - \frac {2 \, {\left (B a^{2} b^{5} c^{5} + 5 \, A a^{2} b^{5} c^{4} d + 20 \, B a^{3} b^{4} c^{3} d^{2} + 20 \, A a^{3} b^{4} c^{2} d^{3} - 40 \, B a^{4} b^{3} c d^{4} - 8 \, A a^{4} b^{3} d^{5}\right )}}{a^{2} b^{6}}}{6 \, {\left (b x^{2} + a\right )}^{\frac {3}{2}}} - \frac {5 \, {\left (4 \, B b c^{2} d^{3} + 2 \, A b c d^{4} - B a d^{5}\right )} \log \left ({\left | -\sqrt {b} x + \sqrt {b x^{2} + a} \right |}\right )}{2 \, b^{\frac {7}{2}}} \] Input:

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

Output:

1/6*((((3*(B*d^5*x/b + 2*(5*B*a^2*b^5*c*d^4 + A*a^2*b^5*d^5)/(a^2*b^6))*x 
+ 2*(2*A*b^7*c^5 + 5*B*a*b^6*c^4*d + 10*A*a*b^6*c^3*d^2 - 40*B*a^2*b^5*c^2 
*d^3 - 20*A*a^2*b^5*c*d^4 + 10*B*a^3*b^4*d^5)/(a^2*b^6))*x - 12*(5*B*a^2*b 
^5*c^3*d^2 + 5*A*a^2*b^5*c^2*d^3 - 10*B*a^3*b^4*c*d^4 - 2*A*a^3*b^4*d^5)/( 
a^2*b^6))*x + 3*(2*A*a*b^6*c^5 - 20*B*a^3*b^4*c^2*d^3 - 10*A*a^3*b^4*c*d^4 
 + 5*B*a^4*b^3*d^5)/(a^2*b^6))*x - 2*(B*a^2*b^5*c^5 + 5*A*a^2*b^5*c^4*d + 
20*B*a^3*b^4*c^3*d^2 + 20*A*a^3*b^4*c^2*d^3 - 40*B*a^4*b^3*c*d^4 - 8*A*a^4 
*b^3*d^5)/(a^2*b^6))/(b*x^2 + a)^(3/2) - 5/2*(4*B*b*c^2*d^3 + 2*A*b*c*d^4 
- B*a*d^5)*log(abs(-sqrt(b)*x + sqrt(b*x^2 + a)))/b^(7/2)
 

Mupad [F(-1)]

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

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

Output:

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

Reduce [F]

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

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

Output:

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