\(\int \frac {(A+B x) (c+d x)^4}{(a+b x^2)^{3/2}} \, dx\) [178]

Optimal result

Integrand size = 24, antiderivative size = 235 \[ \int \frac {(A+B x) (c+d x)^4}{\left (a+b x^2\right )^{3/2}} \, dx=-\frac {(c+d x)^3 (a (B c+A d)-(A b c-a B d) x)}{a b \sqrt {a+b x^2}}-\frac {d (3 A b c-4 a B d) (c+d x)^2 \sqrt {a+b x^2}}{3 a b^2}-\frac {d \left (4 \left (3 A b c \left (b c^2-4 a d^2\right )-4 a B d \left (4 b c^2-a d^2\right )\right )+b d \left (6 A b c^2-20 a B c d-9 a A d^2\right ) x\right ) \sqrt {a+b x^2}}{6 a b^3}-\frac {d \left (3 a d^2 (4 B c+A d)-4 b c^2 (2 B c+3 A d)\right ) \text {arctanh}\left (\frac {\sqrt {b} x}{\sqrt {a+b x^2}}\right )}{2 b^{5/2}} \] Output:

-(d*x+c)^3*(a*(A*d+B*c)-(A*b*c-B*a*d)*x)/a/b/(b*x^2+a)^(1/2)-1/3*d*(3*A*b* 
c-4*B*a*d)*(d*x+c)^2*(b*x^2+a)^(1/2)/a/b^2-1/6*d*(12*A*b*c*(-4*a*d^2+b*c^2 
)-16*a*B*d*(-a*d^2+4*b*c^2)+b*d*(-9*A*a*d^2+6*A*b*c^2-20*B*a*c*d)*x)*(b*x^ 
2+a)^(1/2)/a/b^3-1/2*d*(3*a*d^2*(A*d+4*B*c)-4*b*c^2*(3*A*d+2*B*c))*arctanh 
(b^(1/2)*x/(b*x^2+a)^(1/2))/b^(5/2)
 

Mathematica [A] (verified)

Time = 1.08 (sec) , antiderivative size = 237, normalized size of antiderivative = 1.01 \[ \int \frac {(A+B x) (c+d x)^4}{\left (a+b x^2\right )^{3/2}} \, dx=\frac {-16 a^3 B d^4+6 A b^3 c^4 x+a^2 b d^2 \left (3 A d (16 c+3 d x)+4 B \left (18 c^2+9 c d x-2 d^2 x^2\right )\right )+a b^2 \left (3 A d \left (-8 c^3-12 c^2 d x+8 c d^2 x^2+d^3 x^3\right )+B \left (-6 c^4-24 c^3 d x+36 c^2 d^2 x^2+12 c d^3 x^3+2 d^4 x^4\right )\right )+3 a \sqrt {b} d \left (3 a d^2 (4 B c+A d)-4 b c^2 (2 B c+3 A d)\right ) \sqrt {a+b x^2} \log \left (-\sqrt {b} x+\sqrt {a+b x^2}\right )}{6 a b^3 \sqrt {a+b x^2}} \] Input:

Integrate[((A + B*x)*(c + d*x)^4)/(a + b*x^2)^(3/2),x]
 

Output:

(-16*a^3*B*d^4 + 6*A*b^3*c^4*x + a^2*b*d^2*(3*A*d*(16*c + 3*d*x) + 4*B*(18 
*c^2 + 9*c*d*x - 2*d^2*x^2)) + a*b^2*(3*A*d*(-8*c^3 - 12*c^2*d*x + 8*c*d^2 
*x^2 + d^3*x^3) + B*(-6*c^4 - 24*c^3*d*x + 36*c^2*d^2*x^2 + 12*c*d^3*x^3 + 
 2*d^4*x^4)) + 3*a*Sqrt[b]*d*(3*a*d^2*(4*B*c + A*d) - 4*b*c^2*(2*B*c + 3*A 
*d))*Sqrt[a + b*x^2]*Log[-(Sqrt[b]*x) + Sqrt[a + b*x^2]])/(6*a*b^3*Sqrt[a 
+ b*x^2])
 

Rubi [A] (verified)

Time = 0.44 (sec) , antiderivative size = 252, normalized size of antiderivative = 1.07, number of steps used = 8, number of rules used = 7, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.292, Rules used = {684, 27, 687, 25, 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.

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

\(\Big \downarrow \) 684

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

\(\Big \downarrow \) 27

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

\(\Big \downarrow \) 687

\(\displaystyle \frac {d \left (\frac {\int -\frac {(c+d x) \left (a \left (8 a B d^2-3 b c (4 B c+5 A d)\right )+b \left (6 A b c^2-20 a B d c-9 a A d^2\right ) x\right )}{\sqrt {b x^2+a}}dx}{3 b}-\frac {\sqrt {a+b x^2} (c+d x)^2 (3 A b c-4 a B d)}{3 b}\right )}{a b}-\frac {(c+d x)^3 (a (A d+B c)-x (A b c-a B d))}{a b \sqrt {a+b x^2}}\)

\(\Big \downarrow \) 25

\(\displaystyle \frac {d \left (-\frac {\int \frac {(c+d x) \left (a \left (8 a B d^2-3 b c (4 B c+5 A d)\right )+b \left (6 A b c^2-20 a B d c-9 a A d^2\right ) x\right )}{\sqrt {b x^2+a}}dx}{3 b}-\frac {\sqrt {a+b x^2} (c+d x)^2 (3 A b c-4 a B d)}{3 b}\right )}{a b}-\frac {(c+d x)^3 (a (A d+B c)-x (A b c-a B d))}{a b \sqrt {a+b x^2}}\)

\(\Big \downarrow \) 676

\(\displaystyle \frac {d \left (-\frac {\frac {3}{2} a \left (3 a d^2 (A d+4 B c)-4 b c^2 (3 A d+2 B c)\right ) \int \frac {1}{\sqrt {b x^2+a}}dx+\frac {2 \sqrt {a+b x^2} \left (3 A b c \left (b c^2-4 a d^2\right )-4 a B d \left (4 b c^2-a d^2\right )\right )}{b}+\frac {1}{2} d x \sqrt {a+b x^2} \left (-9 a A d^2-20 a B c d+6 A b c^2\right )}{3 b}-\frac {\sqrt {a+b x^2} (c+d x)^2 (3 A b c-4 a B d)}{3 b}\right )}{a b}-\frac {(c+d x)^3 (a (A d+B c)-x (A b c-a B d))}{a b \sqrt {a+b x^2}}\)

\(\Big \downarrow \) 224

\(\displaystyle \frac {d \left (-\frac {\frac {3}{2} a \left (3 a d^2 (A d+4 B c)-4 b c^2 (3 A d+2 B c)\right ) \int \frac {1}{1-\frac {b x^2}{b x^2+a}}d\frac {x}{\sqrt {b x^2+a}}+\frac {2 \sqrt {a+b x^2} \left (3 A b c \left (b c^2-4 a d^2\right )-4 a B d \left (4 b c^2-a d^2\right )\right )}{b}+\frac {1}{2} d x \sqrt {a+b x^2} \left (-9 a A d^2-20 a B c d+6 A b c^2\right )}{3 b}-\frac {\sqrt {a+b x^2} (c+d x)^2 (3 A b c-4 a B d)}{3 b}\right )}{a b}-\frac {(c+d x)^3 (a (A d+B c)-x (A b c-a B d))}{a b \sqrt {a+b x^2}}\)

\(\Big \downarrow \) 219

\(\displaystyle \frac {d \left (-\frac {\frac {3 a \text {arctanh}\left (\frac {\sqrt {b} x}{\sqrt {a+b x^2}}\right ) \left (3 a d^2 (A d+4 B c)-4 b c^2 (3 A d+2 B c)\right )}{2 \sqrt {b}}+\frac {2 \sqrt {a+b x^2} \left (3 A b c \left (b c^2-4 a d^2\right )-4 a B d \left (4 b c^2-a d^2\right )\right )}{b}+\frac {1}{2} d x \sqrt {a+b x^2} \left (-9 a A d^2-20 a B c d+6 A b c^2\right )}{3 b}-\frac {\sqrt {a+b x^2} (c+d x)^2 (3 A b c-4 a B d)}{3 b}\right )}{a b}-\frac {(c+d x)^3 (a (A d+B c)-x (A b c-a B d))}{a b \sqrt {a+b x^2}}\)

Input:

Int[((A + B*x)*(c + d*x)^4)/(a + b*x^2)^(3/2),x]
 

Output:

-(((c + d*x)^3*(a*(B*c + A*d) - (A*b*c - a*B*d)*x))/(a*b*Sqrt[a + b*x^2])) 
 + (d*(-1/3*((3*A*b*c - 4*a*B*d)*(c + d*x)^2*Sqrt[a + b*x^2])/b - ((2*(3*A 
*b*c*(b*c^2 - 4*a*d^2) - 4*a*B*d*(4*b*c^2 - a*d^2))*Sqrt[a + b*x^2])/b + ( 
d*(6*A*b*c^2 - 20*a*B*c*d - 9*a*A*d^2)*x*Sqrt[a + b*x^2])/2 + (3*a*(3*a*d^ 
2*(4*B*c + A*d) - 4*b*c^2*(2*B*c + 3*A*d))*ArcTanh[(Sqrt[b]*x)/Sqrt[a + b* 
x^2]])/(2*Sqrt[b]))/(3*b)))/(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])
 

rule 687

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

Maple [A] (verified)

Time = 1.31 (sec) , antiderivative size = 260, normalized size of antiderivative = 1.11

method	result	size
risch	\(\frac {d^{2} \left (2 B b \,d^{2} x^{2}+3 A b \,d^{2} x +12 B b c d x +24 A b c d -10 a B \,d^{2}+36 B b \,c^{2}\right ) \sqrt {b \,x^{2}+a}}{6 b^{3}}-\frac {b d \left (3 A \,d^{3} a -12 A b \,c^{2} d +12 a B c \,d^{2}-8 b B \,c^{3}\right ) \left (-\frac {x}{b \sqrt {b \,x^{2}+a}}+\frac {\ln \left (\sqrt {b}\, x +\sqrt {b \,x^{2}+a}\right )}{b^{\frac {3}{2}}}\right )-\frac {8 A a b c \,d^{3}-8 A \,b^{2} c^{3} d -2 a^{2} B \,d^{4}+12 B a b \,c^{2} d^{2}-2 B \,b^{2} c^{4}}{b \sqrt {b \,x^{2}+a}}+\frac {A a \,d^{4} x}{\sqrt {b \,x^{2}+a}}-\frac {2 A \,b^{2} c^{4} x}{a \sqrt {b \,x^{2}+a}}+\frac {4 B a c \,d^{3} x}{\sqrt {b \,x^{2}+a}}}{2 b^{2}}\)	\(260\)
default	\(\frac {A \,c^{4} x}{\sqrt {b \,x^{2}+a}\, a}+d^{3} \left (A d +4 B c \right ) \left (\frac {x^{3}}{2 b \sqrt {b \,x^{2}+a}}-\frac {3 a \left (-\frac {x}{b \sqrt {b \,x^{2}+a}}+\frac {\ln \left (\sqrt {b}\, x +\sqrt {b \,x^{2}+a}\right )}{b^{\frac {3}{2}}}\right )}{2 b}\right )+2 c \,d^{2} \left (2 A d +3 B c \right ) \left (\frac {x^{2}}{b \sqrt {b \,x^{2}+a}}+\frac {2 a}{b^{2} \sqrt {b \,x^{2}+a}}\right )+2 c^{2} d \left (3 A d +2 B c \right ) \left (-\frac {x}{b \sqrt {b \,x^{2}+a}}+\frac {\ln \left (\sqrt {b}\, x +\sqrt {b \,x^{2}+a}\right )}{b^{\frac {3}{2}}}\right )-\frac {c^{3} \left (4 A d +B c \right )}{b \sqrt {b \,x^{2}+a}}+B \,d^{4} \left (\frac {x^{4}}{3 b \sqrt {b \,x^{2}+a}}-\frac {4 a \left (\frac {x^{2}}{b \sqrt {b \,x^{2}+a}}+\frac {2 a}{b^{2} \sqrt {b \,x^{2}+a}}\right )}{3 b}\right )\)	\(276\)

Input:

int((B*x+A)*(d*x+c)^4/(b*x^2+a)^(3/2),x,method=_RETURNVERBOSE)
 

Output:

1/6*d^2*(2*B*b*d^2*x^2+3*A*b*d^2*x+12*B*b*c*d*x+24*A*b*c*d-10*B*a*d^2+36*B 
*b*c^2)*(b*x^2+a)^(1/2)/b^3-1/2/b^2*(b*d*(3*A*a*d^3-12*A*b*c^2*d+12*B*a*c* 
d^2-8*B*b*c^3)*(-x/b/(b*x^2+a)^(1/2)+1/b^(3/2)*ln(b^(1/2)*x+(b*x^2+a)^(1/2 
)))-(8*A*a*b*c*d^3-8*A*b^2*c^3*d-2*B*a^2*d^4+12*B*a*b*c^2*d^2-2*B*b^2*c^4) 
/b/(b*x^2+a)^(1/2)+A*a*d^4*x/(b*x^2+a)^(1/2)-2*A*b^2*c^4*x/a/(b*x^2+a)^(1/ 
2)+4*B*a*c*d^3*x/(b*x^2+a)^(1/2))
 

Fricas [A] (verification not implemented)

Time = 0.12 (sec) , antiderivative size = 692, normalized size of antiderivative = 2.94 \[ \int \frac {(A+B x) (c+d x)^4}{\left (a+b x^2\right )^{3/2}} \, dx=\left [-\frac {3 \, {\left (8 \, B a^{2} b c^{3} d + 12 \, A a^{2} b c^{2} d^{2} - 12 \, B a^{3} c d^{3} - 3 \, A a^{3} d^{4} + {\left (8 \, B a b^{2} c^{3} d + 12 \, A a b^{2} c^{2} d^{2} - 12 \, B a^{2} b c d^{3} - 3 \, A a^{2} b d^{4}\right )} x^{2}\right )} \sqrt {b} \log \left (-2 \, b x^{2} + 2 \, \sqrt {b x^{2} + a} \sqrt {b} x - a\right ) - 2 \, {\left (2 \, B a b^{2} d^{4} x^{4} - 6 \, B a b^{2} c^{4} - 24 \, A a b^{2} c^{3} d + 72 \, B a^{2} b c^{2} d^{2} + 48 \, A a^{2} b c d^{3} - 16 \, B a^{3} d^{4} + 3 \, {\left (4 \, B a b^{2} c d^{3} + A a b^{2} d^{4}\right )} x^{3} + 4 \, {\left (9 \, B a b^{2} c^{2} d^{2} + 6 \, A a b^{2} c d^{3} - 2 \, B a^{2} b d^{4}\right )} x^{2} + 3 \, {\left (2 \, A b^{3} c^{4} - 8 \, B a b^{2} c^{3} d - 12 \, A a b^{2} c^{2} d^{2} + 12 \, B a^{2} b c d^{3} + 3 \, A a^{2} b d^{4}\right )} x\right )} \sqrt {b x^{2} + a}}{12 \, {\left (a b^{4} x^{2} + a^{2} b^{3}\right )}}, -\frac {3 \, {\left (8 \, B a^{2} b c^{3} d + 12 \, A a^{2} b c^{2} d^{2} - 12 \, B a^{3} c d^{3} - 3 \, A a^{3} d^{4} + {\left (8 \, B a b^{2} c^{3} d + 12 \, A a b^{2} c^{2} d^{2} - 12 \, B a^{2} b c d^{3} - 3 \, A a^{2} b d^{4}\right )} x^{2}\right )} \sqrt {-b} \arctan \left (\frac {\sqrt {-b} x}{\sqrt {b x^{2} + a}}\right ) - {\left (2 \, B a b^{2} d^{4} x^{4} - 6 \, B a b^{2} c^{4} - 24 \, A a b^{2} c^{3} d + 72 \, B a^{2} b c^{2} d^{2} + 48 \, A a^{2} b c d^{3} - 16 \, B a^{3} d^{4} + 3 \, {\left (4 \, B a b^{2} c d^{3} + A a b^{2} d^{4}\right )} x^{3} + 4 \, {\left (9 \, B a b^{2} c^{2} d^{2} + 6 \, A a b^{2} c d^{3} - 2 \, B a^{2} b d^{4}\right )} x^{2} + 3 \, {\left (2 \, A b^{3} c^{4} - 8 \, B a b^{2} c^{3} d - 12 \, A a b^{2} c^{2} d^{2} + 12 \, B a^{2} b c d^{3} + 3 \, A a^{2} b d^{4}\right )} x\right )} \sqrt {b x^{2} + a}}{6 \, {\left (a b^{4} x^{2} + a^{2} b^{3}\right )}}\right ] \] Input:

integrate((B*x+A)*(d*x+c)^4/(b*x^2+a)^(3/2),x, algorithm="fricas")
 

Output:

[-1/12*(3*(8*B*a^2*b*c^3*d + 12*A*a^2*b*c^2*d^2 - 12*B*a^3*c*d^3 - 3*A*a^3 
*d^4 + (8*B*a*b^2*c^3*d + 12*A*a*b^2*c^2*d^2 - 12*B*a^2*b*c*d^3 - 3*A*a^2* 
b*d^4)*x^2)*sqrt(b)*log(-2*b*x^2 + 2*sqrt(b*x^2 + a)*sqrt(b)*x - a) - 2*(2 
*B*a*b^2*d^4*x^4 - 6*B*a*b^2*c^4 - 24*A*a*b^2*c^3*d + 72*B*a^2*b*c^2*d^2 + 
 48*A*a^2*b*c*d^3 - 16*B*a^3*d^4 + 3*(4*B*a*b^2*c*d^3 + A*a*b^2*d^4)*x^3 + 
 4*(9*B*a*b^2*c^2*d^2 + 6*A*a*b^2*c*d^3 - 2*B*a^2*b*d^4)*x^2 + 3*(2*A*b^3* 
c^4 - 8*B*a*b^2*c^3*d - 12*A*a*b^2*c^2*d^2 + 12*B*a^2*b*c*d^3 + 3*A*a^2*b* 
d^4)*x)*sqrt(b*x^2 + a))/(a*b^4*x^2 + a^2*b^3), -1/6*(3*(8*B*a^2*b*c^3*d + 
 12*A*a^2*b*c^2*d^2 - 12*B*a^3*c*d^3 - 3*A*a^3*d^4 + (8*B*a*b^2*c^3*d + 12 
*A*a*b^2*c^2*d^2 - 12*B*a^2*b*c*d^3 - 3*A*a^2*b*d^4)*x^2)*sqrt(-b)*arctan( 
sqrt(-b)*x/sqrt(b*x^2 + a)) - (2*B*a*b^2*d^4*x^4 - 6*B*a*b^2*c^4 - 24*A*a* 
b^2*c^3*d + 72*B*a^2*b*c^2*d^2 + 48*A*a^2*b*c*d^3 - 16*B*a^3*d^4 + 3*(4*B* 
a*b^2*c*d^3 + A*a*b^2*d^4)*x^3 + 4*(9*B*a*b^2*c^2*d^2 + 6*A*a*b^2*c*d^3 - 
2*B*a^2*b*d^4)*x^2 + 3*(2*A*b^3*c^4 - 8*B*a*b^2*c^3*d - 12*A*a*b^2*c^2*d^2 
 + 12*B*a^2*b*c*d^3 + 3*A*a^2*b*d^4)*x)*sqrt(b*x^2 + a))/(a*b^4*x^2 + a^2* 
b^3)]
 

Sympy [F]

\[ \int \frac {(A+B x) (c+d x)^4}{\left (a+b x^2\right )^{3/2}} \, dx=\int \frac {\left (A + B x\right ) \left (c + d x\right )^{4}}{\left (a + b x^{2}\right )^{\frac {3}{2}}}\, dx \] Input:

integrate((B*x+A)*(d*x+c)**4/(b*x**2+a)**(3/2),x)
 

Output:

Integral((A + B*x)*(c + d*x)**4/(a + b*x**2)**(3/2), x)
 

Maxima [A] (verification not implemented)

Time = 0.04 (sec) , antiderivative size = 336, normalized size of antiderivative = 1.43 \[ \int \frac {(A+B x) (c+d x)^4}{\left (a+b x^2\right )^{3/2}} \, dx=\frac {B d^{4} x^{4}}{3 \, \sqrt {b x^{2} + a} b} - \frac {4 \, B a d^{4} x^{2}}{3 \, \sqrt {b x^{2} + a} b^{2}} + \frac {A c^{4} x}{\sqrt {b x^{2} + a} a} - \frac {B c^{4}}{\sqrt {b x^{2} + a} b} - \frac {4 \, A c^{3} d}{\sqrt {b x^{2} + a} b} - \frac {8 \, B a^{2} d^{4}}{3 \, \sqrt {b x^{2} + a} b^{3}} + \frac {{\left (4 \, B c d^{3} + A d^{4}\right )} x^{3}}{2 \, \sqrt {b x^{2} + a} b} + \frac {2 \, {\left (3 \, B c^{2} d^{2} + 2 \, A c d^{3}\right )} x^{2}}{\sqrt {b x^{2} + a} b} + \frac {3 \, {\left (4 \, B c d^{3} + A d^{4}\right )} a x}{2 \, \sqrt {b x^{2} + a} b^{2}} - \frac {2 \, {\left (2 \, B c^{3} d + 3 \, A c^{2} d^{2}\right )} x}{\sqrt {b x^{2} + a} b} - \frac {3 \, {\left (4 \, B c d^{3} + A d^{4}\right )} a \operatorname {arsinh}\left (\frac {b x}{\sqrt {a b}}\right )}{2 \, b^{\frac {5}{2}}} + \frac {2 \, {\left (2 \, B c^{3} d + 3 \, A c^{2} d^{2}\right )} \operatorname {arsinh}\left (\frac {b x}{\sqrt {a b}}\right )}{b^{\frac {3}{2}}} + \frac {4 \, {\left (3 \, B c^{2} d^{2} + 2 \, A c d^{3}\right )} a}{\sqrt {b x^{2} + a} b^{2}} \] Input:

integrate((B*x+A)*(d*x+c)^4/(b*x^2+a)^(3/2),x, algorithm="maxima")
 

Output:

1/3*B*d^4*x^4/(sqrt(b*x^2 + a)*b) - 4/3*B*a*d^4*x^2/(sqrt(b*x^2 + a)*b^2) 
+ A*c^4*x/(sqrt(b*x^2 + a)*a) - B*c^4/(sqrt(b*x^2 + a)*b) - 4*A*c^3*d/(sqr 
t(b*x^2 + a)*b) - 8/3*B*a^2*d^4/(sqrt(b*x^2 + a)*b^3) + 1/2*(4*B*c*d^3 + A 
*d^4)*x^3/(sqrt(b*x^2 + a)*b) + 2*(3*B*c^2*d^2 + 2*A*c*d^3)*x^2/(sqrt(b*x^ 
2 + a)*b) + 3/2*(4*B*c*d^3 + A*d^4)*a*x/(sqrt(b*x^2 + a)*b^2) - 2*(2*B*c^3 
*d + 3*A*c^2*d^2)*x/(sqrt(b*x^2 + a)*b) - 3/2*(4*B*c*d^3 + A*d^4)*a*arcsin 
h(b*x/sqrt(a*b))/b^(5/2) + 2*(2*B*c^3*d + 3*A*c^2*d^2)*arcsinh(b*x/sqrt(a* 
b))/b^(3/2) + 4*(3*B*c^2*d^2 + 2*A*c*d^3)*a/(sqrt(b*x^2 + a)*b^2)
 

Giac [A] (verification not implemented)

Time = 0.13 (sec) , antiderivative size = 300, normalized size of antiderivative = 1.28 \[ \int \frac {(A+B x) (c+d x)^4}{\left (a+b x^2\right )^{3/2}} \, dx=\frac {{\left ({\left ({\left (\frac {2 \, B d^{4} x}{b} + \frac {3 \, {\left (4 \, B a b^{4} c d^{3} + A a b^{4} d^{4}\right )}}{a b^{5}}\right )} x + \frac {4 \, {\left (9 \, B a b^{4} c^{2} d^{2} + 6 \, A a b^{4} c d^{3} - 2 \, B a^{2} b^{3} d^{4}\right )}}{a b^{5}}\right )} x + \frac {3 \, {\left (2 \, A b^{5} c^{4} - 8 \, B a b^{4} c^{3} d - 12 \, A a b^{4} c^{2} d^{2} + 12 \, B a^{2} b^{3} c d^{3} + 3 \, A a^{2} b^{3} d^{4}\right )}}{a b^{5}}\right )} x - \frac {2 \, {\left (3 \, B a b^{4} c^{4} + 12 \, A a b^{4} c^{3} d - 36 \, B a^{2} b^{3} c^{2} d^{2} - 24 \, A a^{2} b^{3} c d^{3} + 8 \, B a^{3} b^{2} d^{4}\right )}}{a b^{5}}}{6 \, \sqrt {b x^{2} + a}} - \frac {{\left (8 \, B b c^{3} d + 12 \, A b c^{2} d^{2} - 12 \, B a c d^{3} - 3 \, A a d^{4}\right )} \log \left ({\left | -\sqrt {b} x + \sqrt {b x^{2} + a} \right |}\right )}{2 \, b^{\frac {5}{2}}} \] Input:

integrate((B*x+A)*(d*x+c)^4/(b*x^2+a)^(3/2),x, algorithm="giac")
 

Output:

1/6*((((2*B*d^4*x/b + 3*(4*B*a*b^4*c*d^3 + A*a*b^4*d^4)/(a*b^5))*x + 4*(9* 
B*a*b^4*c^2*d^2 + 6*A*a*b^4*c*d^3 - 2*B*a^2*b^3*d^4)/(a*b^5))*x + 3*(2*A*b 
^5*c^4 - 8*B*a*b^4*c^3*d - 12*A*a*b^4*c^2*d^2 + 12*B*a^2*b^3*c*d^3 + 3*A*a 
^2*b^3*d^4)/(a*b^5))*x - 2*(3*B*a*b^4*c^4 + 12*A*a*b^4*c^3*d - 36*B*a^2*b^ 
3*c^2*d^2 - 24*A*a^2*b^3*c*d^3 + 8*B*a^3*b^2*d^4)/(a*b^5))/sqrt(b*x^2 + a) 
 - 1/2*(8*B*b*c^3*d + 12*A*b*c^2*d^2 - 12*B*a*c*d^3 - 3*A*a*d^4)*log(abs(- 
sqrt(b)*x + sqrt(b*x^2 + a)))/b^(5/2)
 

Mupad [F(-1)]

Timed out. \[ \int \frac {(A+B x) (c+d x)^4}{\left (a+b x^2\right )^{3/2}} \, dx=\int \frac {\left (A+B\,x\right )\,{\left (c+d\,x\right )}^4}{{\left (b\,x^2+a\right )}^{3/2}} \,d x \] Input:

int(((A + B*x)*(c + d*x)^4)/(a + b*x^2)^(3/2),x)
 

Output:

int(((A + B*x)*(c + d*x)^4)/(a + b*x^2)^(3/2), x)
 

Reduce [F]

\[ \int \frac {(A+B x) (c+d x)^4}{\left (a+b x^2\right )^{3/2}} \, dx=\int \frac {\left (B x +A \right ) \left (d x +c \right )^{4}}{\left (b \,x^{2}+a \right )^{\frac {3}{2}}}d x \] Input:

int((B*x+A)*(d*x+c)^4/(b*x^2+a)^(3/2),x)
 

Output:

int((B*x+A)*(d*x+c)^4/(b*x^2+a)^(3/2),x)