\(\int \frac {(A+B x) (a+b x^2)^{5/2}}{(c+d x)^3} \, dx\) [168]

Optimal result

Integrand size = 24, antiderivative size = 341 \[ \int \frac {(A+B x) \left (a+b x^2\right )^{5/2}}{(c+d x)^3} \, dx=-\frac {5 b \left (4 \left (a d^2 (3 B c-A d)+b \left (6 B c^3-4 A c^2 d\right )\right )-d \left (3 a B d^2+4 b c (3 B c-2 A d)\right ) x\right ) \sqrt {a+b x^2}}{8 d^6}-\frac {5 \left (3 a B d^2+4 b c (3 B c-2 A d)+b d (3 B c-2 A d) x\right ) \left (a+b x^2\right )^{3/2}}{12 d^4 (c+d x)}+\frac {(3 B c-2 A d+B d x) \left (a+b x^2\right )^{5/2}}{4 d^2 (c+d x)^2}+\frac {5 \sqrt {b} \left (3 a^2 B d^4+8 b^2 c^3 (3 B c-2 A d)+12 a b c d^2 (2 B c-A d)\right ) \text {arctanh}\left (\frac {\sqrt {b} x}{\sqrt {a+b x^2}}\right )}{8 d^7}+\frac {5 b \sqrt {b c^2+a d^2} \left (2 b c^2 (3 B c-2 A d)+a d^2 (3 B c-A d)\right ) \text {arctanh}\left (\frac {a d-b c x}{\sqrt {b c^2+a d^2} \sqrt {a+b x^2}}\right )}{2 d^7} \] Output:

-5/8*b*(4*a*d^2*(-A*d+3*B*c)+4*b*(-4*A*c^2*d+6*B*c^3)-d*(3*a*B*d^2+4*b*c*( 
-2*A*d+3*B*c))*x)*(b*x^2+a)^(1/2)/d^6-5/12*(3*a*B*d^2+4*b*c*(-2*A*d+3*B*c) 
+b*d*(-2*A*d+3*B*c)*x)*(b*x^2+a)^(3/2)/d^4/(d*x+c)+1/4*(B*d*x-2*A*d+3*B*c) 
*(b*x^2+a)^(5/2)/d^2/(d*x+c)^2+5/8*b^(1/2)*(3*a^2*B*d^4+8*b^2*c^3*(-2*A*d+ 
3*B*c)+12*a*b*c*d^2*(-A*d+2*B*c))*arctanh(b^(1/2)*x/(b*x^2+a)^(1/2))/d^7+5 
/2*b*(a*d^2+b*c^2)^(1/2)*(2*b*c^2*(-2*A*d+3*B*c)+a*d^2*(-A*d+3*B*c))*arcta 
nh((-b*c*x+a*d)/(a*d^2+b*c^2)^(1/2)/(b*x^2+a)^(1/2))/d^7
 

Mathematica [A] (verified)

Time = 5.17 (sec) , antiderivative size = 385, normalized size of antiderivative = 1.13 \[ \int \frac {(A+B x) \left (a+b x^2\right )^{5/2}}{(c+d x)^3} \, dx=-\frac {\frac {d \sqrt {a+b x^2} \left (12 a^2 d^4 (A d+B (c+2 d x))+a b d^2 \left (-4 A d \left (35 c^2+55 c d x+14 d^2 x^2\right )+3 B \left (100 c^3+155 c^2 d x+38 c d^2 x^2-9 d^3 x^3\right )\right )+2 b^2 \left (-2 A d \left (60 c^4+90 c^3 d x+20 c^2 d^2 x^2-5 c d^3 x^3+2 d^4 x^4\right )+3 B \left (60 c^5+90 c^4 d x+20 c^3 d^2 x^2-5 c^2 d^3 x^3+2 c d^4 x^4-d^5 x^5\right )\right )\right )}{(c+d x)^2}+120 b \sqrt {-b c^2-a d^2} \left (2 b c^2 (3 B c-2 A d)+a d^2 (3 B c-A d)\right ) \arctan \left (\frac {\sqrt {b} (c+d x)-d \sqrt {a+b x^2}}{\sqrt {-b c^2-a d^2}}\right )+15 \sqrt {b} \left (3 a^2 B d^4+8 b^2 c^3 (3 B c-2 A d)+12 a b c d^2 (2 B c-A d)\right ) \log \left (-\sqrt {b} x+\sqrt {a+b x^2}\right )}{24 d^7} \] Input:

Integrate[((A + B*x)*(a + b*x^2)^(5/2))/(c + d*x)^3,x]
 

Output:

-1/24*((d*Sqrt[a + b*x^2]*(12*a^2*d^4*(A*d + B*(c + 2*d*x)) + a*b*d^2*(-4* 
A*d*(35*c^2 + 55*c*d*x + 14*d^2*x^2) + 3*B*(100*c^3 + 155*c^2*d*x + 38*c*d 
^2*x^2 - 9*d^3*x^3)) + 2*b^2*(-2*A*d*(60*c^4 + 90*c^3*d*x + 20*c^2*d^2*x^2 
 - 5*c*d^3*x^3 + 2*d^4*x^4) + 3*B*(60*c^5 + 90*c^4*d*x + 20*c^3*d^2*x^2 - 
5*c^2*d^3*x^3 + 2*c*d^4*x^4 - d^5*x^5))))/(c + d*x)^2 + 120*b*Sqrt[-(b*c^2 
) - a*d^2]*(2*b*c^2*(3*B*c - 2*A*d) + a*d^2*(3*B*c - A*d))*ArcTan[(Sqrt[b] 
*(c + d*x) - d*Sqrt[a + b*x^2])/Sqrt[-(b*c^2) - a*d^2]] + 15*Sqrt[b]*(3*a^ 
2*B*d^4 + 8*b^2*c^3*(3*B*c - 2*A*d) + 12*a*b*c*d^2*(2*B*c - A*d))*Log[-(Sq 
rt[b]*x) + Sqrt[a + b*x^2]])/d^7
 

Rubi [A] (verified)

Time = 0.68 (sec) , antiderivative size = 358, normalized size of antiderivative = 1.05, number of steps used = 12, number of rules used = 11, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.458, Rules used = {681, 27, 681, 27, 682, 27, 719, 224, 219, 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.

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

\(\Big \downarrow \) 681

\(\displaystyle \frac {\left (a+b x^2\right )^{5/2} (-2 A d+3 B c+B d x)}{4 d^2 (c+d x)^2}-\frac {5 \int -\frac {4 (a B d-b (3 B c-2 A d) x) \left (b x^2+a\right )^{3/2}}{(c+d x)^2}dx}{16 d^2}\)

\(\Big \downarrow \) 27

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

\(\Big \downarrow \) 681

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

\(\Big \downarrow \) 27

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

\(\Big \downarrow \) 682

\(\displaystyle \frac {5 \left (-\frac {b \left (\frac {\int \frac {b \left (a d \left (4 b (3 B c-2 A d) c^2+a d^2 (9 B c-4 A d)\right )-\left (3 a^2 B d^4+12 a b c (2 B c-A d) d^2+8 b^2 c^3 (3 B c-2 A d)\right ) x\right )}{(c+d x) \sqrt {b x^2+a}}dx}{2 b d^2}+\frac {\sqrt {a+b x^2} \left (4 \left (a d^2 (3 B c-A d)+b \left (6 B c^3-4 A c^2 d\right )\right )-d x \left (3 a B d^2+4 b c (3 B c-2 A d)\right )\right )}{2 d^2}\right )}{d^2}-\frac {\left (a+b x^2\right )^{3/2} \left (3 a B d^2+b d x (3 B c-2 A d)+4 b c (3 B c-2 A d)\right )}{3 d^2 (c+d x)}\right )}{4 d^2}+\frac {\left (a+b x^2\right )^{5/2} (-2 A d+3 B c+B d x)}{4 d^2 (c+d x)^2}\)

\(\Big \downarrow \) 27

\(\displaystyle \frac {5 \left (-\frac {b \left (\frac {\int \frac {a d \left (4 b (3 B c-2 A d) c^2+a d^2 (9 B c-4 A d)\right )-\left (3 a^2 B d^4+12 a b c (2 B c-A d) d^2+8 b^2 c^3 (3 B c-2 A d)\right ) x}{(c+d x) \sqrt {b x^2+a}}dx}{2 d^2}+\frac {\sqrt {a+b x^2} \left (4 \left (a d^2 (3 B c-A d)+b \left (6 B c^3-4 A c^2 d\right )\right )-d x \left (3 a B d^2+4 b c (3 B c-2 A d)\right )\right )}{2 d^2}\right )}{d^2}-\frac {\left (a+b x^2\right )^{3/2} \left (3 a B d^2+b d x (3 B c-2 A d)+4 b c (3 B c-2 A d)\right )}{3 d^2 (c+d x)}\right )}{4 d^2}+\frac {\left (a+b x^2\right )^{5/2} (-2 A d+3 B c+B d x)}{4 d^2 (c+d x)^2}\)

\(\Big \downarrow \) 719

\(\displaystyle \frac {5 \left (-\frac {b \left (\frac {\frac {4 \left (a d^2+b c^2\right ) \left (a d^2 (3 B c-A d)+2 b c^2 (3 B c-2 A d)\right ) \int \frac {1}{(c+d x) \sqrt {b x^2+a}}dx}{d}-\frac {\left (3 a^2 B d^4+12 a b c d^2 (2 B c-A d)+8 b^2 c^3 (3 B c-2 A d)\right ) \int \frac {1}{\sqrt {b x^2+a}}dx}{d}}{2 d^2}+\frac {\sqrt {a+b x^2} \left (4 \left (a d^2 (3 B c-A d)+b \left (6 B c^3-4 A c^2 d\right )\right )-d x \left (3 a B d^2+4 b c (3 B c-2 A d)\right )\right )}{2 d^2}\right )}{d^2}-\frac {\left (a+b x^2\right )^{3/2} \left (3 a B d^2+b d x (3 B c-2 A d)+4 b c (3 B c-2 A d)\right )}{3 d^2 (c+d x)}\right )}{4 d^2}+\frac {\left (a+b x^2\right )^{5/2} (-2 A d+3 B c+B d x)}{4 d^2 (c+d x)^2}\)

\(\Big \downarrow \) 224

\(\displaystyle \frac {5 \left (-\frac {b \left (\frac {\frac {4 \left (a d^2+b c^2\right ) \left (a d^2 (3 B c-A d)+2 b c^2 (3 B c-2 A d)\right ) \int \frac {1}{(c+d x) \sqrt {b x^2+a}}dx}{d}-\frac {\left (3 a^2 B d^4+12 a b c d^2 (2 B c-A d)+8 b^2 c^3 (3 B c-2 A d)\right ) \int \frac {1}{1-\frac {b x^2}{b x^2+a}}d\frac {x}{\sqrt {b x^2+a}}}{d}}{2 d^2}+\frac {\sqrt {a+b x^2} \left (4 \left (a d^2 (3 B c-A d)+b \left (6 B c^3-4 A c^2 d\right )\right )-d x \left (3 a B d^2+4 b c (3 B c-2 A d)\right )\right )}{2 d^2}\right )}{d^2}-\frac {\left (a+b x^2\right )^{3/2} \left (3 a B d^2+b d x (3 B c-2 A d)+4 b c (3 B c-2 A d)\right )}{3 d^2 (c+d x)}\right )}{4 d^2}+\frac {\left (a+b x^2\right )^{5/2} (-2 A d+3 B c+B d x)}{4 d^2 (c+d x)^2}\)

\(\Big \downarrow \) 219

\(\displaystyle \frac {5 \left (-\frac {b \left (\frac {\frac {4 \left (a d^2+b c^2\right ) \left (a d^2 (3 B c-A d)+2 b c^2 (3 B c-2 A d)\right ) \int \frac {1}{(c+d x) \sqrt {b x^2+a}}dx}{d}-\frac {\text {arctanh}\left (\frac {\sqrt {b} x}{\sqrt {a+b x^2}}\right ) \left (3 a^2 B d^4+12 a b c d^2 (2 B c-A d)+8 b^2 c^3 (3 B c-2 A d)\right )}{\sqrt {b} d}}{2 d^2}+\frac {\sqrt {a+b x^2} \left (4 \left (a d^2 (3 B c-A d)+b \left (6 B c^3-4 A c^2 d\right )\right )-d x \left (3 a B d^2+4 b c (3 B c-2 A d)\right )\right )}{2 d^2}\right )}{d^2}-\frac {\left (a+b x^2\right )^{3/2} \left (3 a B d^2+b d x (3 B c-2 A d)+4 b c (3 B c-2 A d)\right )}{3 d^2 (c+d x)}\right )}{4 d^2}+\frac {\left (a+b x^2\right )^{5/2} (-2 A d+3 B c+B d x)}{4 d^2 (c+d x)^2}\)

\(\Big \downarrow \) 488

\(\displaystyle \frac {5 \left (-\frac {b \left (\frac {-\frac {4 \left (a d^2+b c^2\right ) \left (a d^2 (3 B c-A d)+2 b c^2 (3 B c-2 A d)\right ) \int \frac {1}{b c^2+a d^2-\frac {(a d-b c x)^2}{b x^2+a}}d\frac {a d-b c x}{\sqrt {b x^2+a}}}{d}-\frac {\text {arctanh}\left (\frac {\sqrt {b} x}{\sqrt {a+b x^2}}\right ) \left (3 a^2 B d^4+12 a b c d^2 (2 B c-A d)+8 b^2 c^3 (3 B c-2 A d)\right )}{\sqrt {b} d}}{2 d^2}+\frac {\sqrt {a+b x^2} \left (4 \left (a d^2 (3 B c-A d)+b \left (6 B c^3-4 A c^2 d\right )\right )-d x \left (3 a B d^2+4 b c (3 B c-2 A d)\right )\right )}{2 d^2}\right )}{d^2}-\frac {\left (a+b x^2\right )^{3/2} \left (3 a B d^2+b d x (3 B c-2 A d)+4 b c (3 B c-2 A d)\right )}{3 d^2 (c+d x)}\right )}{4 d^2}+\frac {\left (a+b x^2\right )^{5/2} (-2 A d+3 B c+B d x)}{4 d^2 (c+d x)^2}\)

\(\Big \downarrow \) 219

\(\displaystyle \frac {5 \left (-\frac {b \left (\frac {-\frac {\text {arctanh}\left (\frac {\sqrt {b} x}{\sqrt {a+b x^2}}\right ) \left (3 a^2 B d^4+12 a b c d^2 (2 B c-A d)+8 b^2 c^3 (3 B c-2 A d)\right )}{\sqrt {b} d}-\frac {4 \sqrt {a d^2+b c^2} \left (a d^2 (3 B c-A d)+2 b c^2 (3 B c-2 A d)\right ) \text {arctanh}\left (\frac {a d-b c x}{\sqrt {a+b x^2} \sqrt {a d^2+b c^2}}\right )}{d}}{2 d^2}+\frac {\sqrt {a+b x^2} \left (4 \left (a d^2 (3 B c-A d)+b \left (6 B c^3-4 A c^2 d\right )\right )-d x \left (3 a B d^2+4 b c (3 B c-2 A d)\right )\right )}{2 d^2}\right )}{d^2}-\frac {\left (a+b x^2\right )^{3/2} \left (3 a B d^2+b d x (3 B c-2 A d)+4 b c (3 B c-2 A d)\right )}{3 d^2 (c+d x)}\right )}{4 d^2}+\frac {\left (a+b x^2\right )^{5/2} (-2 A d+3 B c+B d x)}{4 d^2 (c+d x)^2}\)

Input:

Int[((A + B*x)*(a + b*x^2)^(5/2))/(c + d*x)^3,x]
 

Output:

((3*B*c - 2*A*d + B*d*x)*(a + b*x^2)^(5/2))/(4*d^2*(c + d*x)^2) + (5*(-1/3 
*((3*a*B*d^2 + 4*b*c*(3*B*c - 2*A*d) + b*d*(3*B*c - 2*A*d)*x)*(a + b*x^2)^ 
(3/2))/(d^2*(c + d*x)) - (b*(((4*(a*d^2*(3*B*c - A*d) + b*(6*B*c^3 - 4*A*c 
^2*d)) - d*(3*a*B*d^2 + 4*b*c*(3*B*c - 2*A*d))*x)*Sqrt[a + b*x^2])/(2*d^2) 
 + (-(((3*a^2*B*d^4 + 8*b^2*c^3*(3*B*c - 2*A*d) + 12*a*b*c*d^2*(2*B*c - A* 
d))*ArcTanh[(Sqrt[b]*x)/Sqrt[a + b*x^2]])/(Sqrt[b]*d)) - (4*Sqrt[b*c^2 + a 
*d^2]*(2*b*c^2*(3*B*c - 2*A*d) + a*d^2*(3*B*c - A*d))*ArcTanh[(a*d - b*c*x 
)/(Sqrt[b*c^2 + a*d^2]*Sqrt[a + b*x^2])])/d)/(2*d^2)))/d^2))/(4*d^2)
 

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 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 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 681

Int[((d_.) + (e_.)*(x_))^(m_)*((f_.) + (g_.)*(x_))*((a_) + (c_.)*(x_)^2)^(p 
_.), x_Symbol] :> Simp[(d + e*x)^(m + 1)*(e*f*(m + 2*p + 2) - d*g*(2*p + 1) 
 + e*g*(m + 1)*x)*((a + c*x^2)^p/(e^2*(m + 1)*(m + 2*p + 2))), x] + Simp[p/ 
(e^2*(m + 1)*(m + 2*p + 2))   Int[(d + e*x)^(m + 1)*(a + c*x^2)^(p - 1)*Sim 
p[g*(2*a*e + 2*a*e*m) + (g*(2*c*d + 4*c*d*p) - 2*c*e*f*(m + 2*p + 2))*x, x] 
, x], x] /; FreeQ[{a, c, d, e, f, g, m}, x] && GtQ[p, 0] && (LtQ[m, -1] || 
EqQ[p, 1] || (IntegerQ[p] &&  !RationalQ[m])) && NeQ[m, -1] &&  !ILtQ[m + 2 
*p + 1, 0] && (IntegerQ[m] || IntegerQ[p] || IntegersQ[2*m, 2*p])
 

rule 682

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

rule 719

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

Maple [B] (verified)

Leaf count of result is larger than twice the leaf count of optimal. \(1197\) vs. \(2(312)=624\).

Time = 1.54 (sec) , antiderivative size = 1198, normalized size of antiderivative = 3.51

method	result	size
risch	\(\text {Expression too large to display}\)	\(1198\)
default	\(\text {Expression too large to display}\)	\(3562\)

Input:

int((B*x+A)*(b*x^2+a)^(5/2)/(d*x+c)^3,x,method=_RETURNVERBOSE)
 

Output:

1/24*b*(6*B*b*d^3*x^3+8*A*b*d^3*x^2-24*B*b*c*d^2*x^2-36*A*b*c*d^2*x+27*B*a 
*d^3*x+72*B*b*c^2*d*x+56*A*a*d^3+144*A*b*c^2*d-168*B*a*c*d^2-240*B*b*c^3)* 
(b*x^2+a)^(1/2)/d^6-1/8/d^6*(8/d^3*(6*A*a^2*b*c*d^5+12*A*a*b^2*c^3*d^3+6*A 
*b^3*c^5*d-B*a^3*d^6-9*B*a^2*b*c^2*d^4-15*B*a*b^2*c^4*d^2-7*B*b^3*c^6)*(-1 
/(a*d^2+b*c^2)*d^2/(x+c/d)*(b*(x+c/d)^2-2*b*c/d*(x+c/d)+(a*d^2+b*c^2)/d^2) 
^(1/2)-b*c*d/(a*d^2+b*c^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)))-8*(A*a^3*d^7+3*A*a^2*b*c^2*d^5+3*A*a 
*b^2*c^4*d^3+A*b^3*c^6*d-B*a^3*c*d^6-3*B*a^2*b*c^3*d^4-3*B*a*b^2*c^5*d^2-B 
*b^3*c^7)/d^4*(-1/2/(a*d^2+b*c^2)*d^2/(x+c/d)^2*(b*(x+c/d)^2-2*b*c/d*(x+c/ 
d)+(a*d^2+b*c^2)/d^2)^(1/2)+3/2*b*c*d/(a*d^2+b*c^2)*(-1/(a*d^2+b*c^2)*d^2/ 
(x+c/d)*(b*(x+c/d)^2-2*b*c/d*(x+c/d)+(a*d^2+b*c^2)/d^2)^(1/2)-b*c*d/(a*d^2 
+b*c^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)))+1/2*b/(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)))+5*b^(1/2)*(12*A*a*b* 
c*d^3+16*A*b^2*c^3*d-3*B*a^2*d^4-24*B*a*b*c^2*d^2-24*B*b^2*c^4)/d*ln(b^(1/ 
2)*x+(b*x^2+a)^(1/2))+24*b/d^2*(A*a^2*d^5+6*A*a*b*c^2*d^3+5*A*b^2*c^4*d-3* 
B*a^2*c*d^4-10*B*a*b*c^3*d^2-7*B*b^2*c^5)/((a*d^2+b*c^2)/d^2)^(1/2)*ln(...
 

Fricas [F(-1)]

Timed out. \[ \int \frac {(A+B x) \left (a+b x^2\right )^{5/2}}{(c+d x)^3} \, dx=\text {Timed out} \] Input:

integrate((B*x+A)*(b*x^2+a)^(5/2)/(d*x+c)^3,x, algorithm="fricas")
 

Output:

Timed out
 

Sympy [F]

\[ \int \frac {(A+B x) \left (a+b x^2\right )^{5/2}}{(c+d x)^3} \, dx=\int \frac {\left (A + B x\right ) \left (a + b x^{2}\right )^{\frac {5}{2}}}{\left (c + d x\right )^{3}}\, dx \] Input:

integrate((B*x+A)*(b*x**2+a)**(5/2)/(d*x+c)**3,x)
 

Output:

Integral((A + B*x)*(a + b*x**2)**(5/2)/(c + d*x)**3, x)
 

Maxima [B] (verification not implemented)

Leaf count of result is larger than twice the leaf count of optimal. 1257 vs. \(2 (313) = 626\).

Time = 0.21 (sec) , antiderivative size = 1257, normalized size of antiderivative = 3.69 \[ \int \frac {(A+B x) \left (a+b x^2\right )^{5/2}}{(c+d x)^3} \, dx=\text {Too large to display} \] Input:

integrate((B*x+A)*(b*x^2+a)^(5/2)/(d*x+c)^3,x, algorithm="maxima")
 

Output:

-15/4*B*b^4*c^6*arcsinh(b*x/sqrt(a*b))/(b^(3/2)*c^2*d^7 + a*sqrt(b)*d^9) + 
 15/4*A*b^4*c^5*arcsinh(b*x/sqrt(a*b))/(b^(3/2)*c^2*d^6 + a*sqrt(b)*d^8) - 
 15/4*B*a*b^3*c^4*arcsinh(b*x/sqrt(a*b))/(b^(3/2)*c^2*d^5 + a*sqrt(b)*d^7) 
 + 15/4*sqrt(b*x^2 + a)*B*b^3*c^4*x/(b*c^2*d^5 + a*d^7) + 15/4*A*a*b^3*c^3 
*arcsinh(b*x/sqrt(a*b))/(b^(3/2)*c^2*d^4 + a*sqrt(b)*d^6) - 15/4*sqrt(b*x^ 
2 + a)*A*b^3*c^3*x/(b*c^2*d^4 + a*d^6) - 5/2*(b*x^2 + a)^(3/2)*B*b^2*c^3/( 
b*c^2*d^4 + a*d^6) + 5/2*(b*x^2 + a)^(3/2)*B*b^2*c^2*x/(b*c^2*d^3 + a*d^5) 
 + 15/4*sqrt(b*x^2 + a)*B*a*b^2*c^2*x/(b*c^2*d^3 + a*d^5) - 3/2*(b*x^2 + a 
)^(5/2)*B*b*c^2/(b*c^2*d^3*x + a*d^5*x + b*c^3*d^2 + a*c*d^4) + 5/2*(b*x^2 
 + a)^(3/2)*A*b^2*c^2/(b*c^2*d^3 + a*d^5) - 5/2*(b*x^2 + a)^(3/2)*A*b^2*c* 
x/(b*c^2*d^2 + a*d^4) - 15/4*sqrt(b*x^2 + a)*A*a*b^2*c*x/(b*c^2*d^2 + a*d^ 
4) + 1/2*(b*x^2 + a)^(7/2)*B*c/(b*c^2*d^2*x^2 + a*d^4*x^2 + 2*b*c^3*d*x + 
2*a*c*d^3*x + b*c^4 + a*c^2*d^2) + 3/2*(b*x^2 + a)^(5/2)*A*b*c/(b*c^2*d^2* 
x + a*d^4*x + b*c^3*d + a*c*d^3) - 1/2*(b*x^2 + a)^(5/2)*B*b*c/(b*c^2*d^2 
+ a*d^4) - 1/2*(b*x^2 + a)^(7/2)*A/(b*c^2*d*x^2 + a*d^3*x^2 + 2*b*c^3*x + 
2*a*c*d^2*x + b*c^4/d + a*c^2*d) + 1/2*(b*x^2 + a)^(5/2)*A*b/(b*c^2*d + a* 
d^3) - (b*x^2 + a)^(5/2)*B/(d^3*x + c*d^2) + 15/4*sqrt(b*x^2 + a)*B*b^2*c^ 
2*x/d^5 - 5/4*sqrt(b*x^2 + a)*A*b^2*c*x/d^4 + 5/4*(b*x^2 + a)^(3/2)*B*b*x/ 
d^3 + 15/8*sqrt(b*x^2 + a)*B*a*b*x/d^3 + 75/4*B*b^(5/2)*c^4*arcsinh(b*x/sq 
rt(a*b))/d^7 - 55/4*A*b^(5/2)*c^3*arcsinh(b*x/sqrt(a*b))/d^6 + 15*B*a*b...
 

Giac [B] (verification not implemented)

Leaf count of result is larger than twice the leaf count of optimal. 1067 vs. \(2 (313) = 626\).

Time = 0.49 (sec) , antiderivative size = 1067, normalized size of antiderivative = 3.13 \[ \int \frac {(A+B x) \left (a+b x^2\right )^{5/2}}{(c+d x)^3} \, dx =\text {Too large to display} \] Input:

integrate((B*x+A)*(b*x^2+a)^(5/2)/(d*x+c)^3,x, algorithm="giac")
 

Output:

1/24*sqrt(b*x^2 + a)*((2*(3*B*b^2*x/d^3 - 4*(3*B*b^4*c*d^21 - A*b^4*d^22)/ 
(b^2*d^25))*x + 9*(8*B*b^4*c^2*d^20 - 4*A*b^4*c*d^21 + 3*B*a*b^3*d^22)/(b^ 
2*d^25))*x - 8*(30*B*b^4*c^3*d^19 - 18*A*b^4*c^2*d^20 + 21*B*a*b^3*c*d^21 
- 7*A*a*b^3*d^22)/(b^2*d^25)) - 5/8*(24*B*b^(5/2)*c^4 - 16*A*b^(5/2)*c^3*d 
 + 24*B*a*b^(3/2)*c^2*d^2 - 12*A*a*b^(3/2)*c*d^3 + 3*B*a^2*sqrt(b)*d^4)*lo 
g(abs(-sqrt(b)*x + sqrt(b*x^2 + a)))/d^7 - 5*(6*B*b^3*c^5 - 4*A*b^3*c^4*d 
+ 9*B*a*b^2*c^3*d^2 - 5*A*a*b^2*c^2*d^3 + 3*B*a^2*b*c*d^4 - A*a^2*b*d^5)*a 
rctan(-((sqrt(b)*x - sqrt(b*x^2 + a))*d + sqrt(b)*c)/sqrt(-b*c^2 - a*d^2)) 
/(sqrt(-b*c^2 - a*d^2)*d^7) - (12*(sqrt(b)*x - sqrt(b*x^2 + a))^3*B*b^3*c^ 
5*d - 10*(sqrt(b)*x - sqrt(b*x^2 + a))^3*A*b^3*c^4*d^2 + 15*(sqrt(b)*x - s 
qrt(b*x^2 + a))^3*B*a*b^2*c^3*d^3 - 11*(sqrt(b)*x - sqrt(b*x^2 + a))^3*A*a 
*b^2*c^2*d^4 + 3*(sqrt(b)*x - sqrt(b*x^2 + a))^3*B*a^2*b*c*d^5 - (sqrt(b)* 
x - sqrt(b*x^2 + a))^3*A*a^2*b*d^6 + 22*(sqrt(b)*x - sqrt(b*x^2 + a))^2*B* 
b^(7/2)*c^6 - 18*(sqrt(b)*x - sqrt(b*x^2 + a))^2*A*b^(7/2)*c^5*d + 15*(sqr 
t(b)*x - sqrt(b*x^2 + a))^2*B*a*b^(5/2)*c^4*d^2 - 9*(sqrt(b)*x - sqrt(b*x^ 
2 + a))^2*A*a*b^(5/2)*c^3*d^3 - 9*(sqrt(b)*x - sqrt(b*x^2 + a))^2*B*a^2*b^ 
(3/2)*c^2*d^4 + 9*(sqrt(b)*x - sqrt(b*x^2 + a))^2*A*a^2*b^(3/2)*c*d^5 - 2* 
(sqrt(b)*x - sqrt(b*x^2 + a))^2*B*a^3*sqrt(b)*d^6 - 32*(sqrt(b)*x - sqrt(b 
*x^2 + a))*B*a*b^3*c^5*d + 26*(sqrt(b)*x - sqrt(b*x^2 + a))*A*a*b^3*c^4*d^ 
2 - 37*(sqrt(b)*x - sqrt(b*x^2 + a))*B*a^2*b^2*c^3*d^3 + 25*(sqrt(b)*x ...
 

Mupad [F(-1)]

Timed out. \[ \int \frac {(A+B x) \left (a+b x^2\right )^{5/2}}{(c+d x)^3} \, dx=\int \frac {{\left (b\,x^2+a\right )}^{5/2}\,\left (A+B\,x\right )}{{\left (c+d\,x\right )}^3} \,d x \] Input:

int(((a + b*x^2)^(5/2)*(A + B*x))/(c + d*x)^3,x)
 

Output:

int(((a + b*x^2)^(5/2)*(A + B*x))/(c + d*x)^3, x)
 

Reduce [B] (verification not implemented)

Time = 0.24 (sec) , antiderivative size = 2321, normalized size of antiderivative = 6.81 \[ \int \frac {(A+B x) \left (a+b x^2\right )^{5/2}}{(c+d x)^3} \, dx =\text {Too large to display} \] Input:

int((B*x+A)*(b*x^2+a)^(5/2)/(d*x+c)^3,x)
 

Output:

(120*sqrt(a*d**2 + b*c**2)*log(sqrt(a + b*x**2)*sqrt(a*d**2 + b*c**2) - a* 
d + b*c*x)*a**2*b*c**2*d**3 + 240*sqrt(a*d**2 + b*c**2)*log(sqrt(a + b*x** 
2)*sqrt(a*d**2 + b*c**2) - a*d + b*c*x)*a**2*b*c*d**4*x + 120*sqrt(a*d**2 
+ b*c**2)*log(sqrt(a + b*x**2)*sqrt(a*d**2 + b*c**2) - a*d + b*c*x)*a**2*b 
*d**5*x**2 + 480*sqrt(a*d**2 + b*c**2)*log(sqrt(a + b*x**2)*sqrt(a*d**2 + 
b*c**2) - a*d + b*c*x)*a*b**2*c**4*d + 960*sqrt(a*d**2 + b*c**2)*log(sqrt( 
a + b*x**2)*sqrt(a*d**2 + b*c**2) - a*d + b*c*x)*a*b**2*c**3*d**2*x - 360* 
sqrt(a*d**2 + b*c**2)*log(sqrt(a + b*x**2)*sqrt(a*d**2 + b*c**2) - a*d + b 
*c*x)*a*b**2*c**3*d**2 + 480*sqrt(a*d**2 + b*c**2)*log(sqrt(a + b*x**2)*sq 
rt(a*d**2 + b*c**2) - a*d + b*c*x)*a*b**2*c**2*d**3*x**2 - 720*sqrt(a*d**2 
 + b*c**2)*log(sqrt(a + b*x**2)*sqrt(a*d**2 + b*c**2) - a*d + b*c*x)*a*b** 
2*c**2*d**3*x - 360*sqrt(a*d**2 + b*c**2)*log(sqrt(a + b*x**2)*sqrt(a*d**2 
 + b*c**2) - a*d + b*c*x)*a*b**2*c*d**4*x**2 - 720*sqrt(a*d**2 + b*c**2)*l 
og(sqrt(a + b*x**2)*sqrt(a*d**2 + b*c**2) - a*d + b*c*x)*b**3*c**5 - 1440* 
sqrt(a*d**2 + b*c**2)*log(sqrt(a + b*x**2)*sqrt(a*d**2 + b*c**2) - a*d + b 
*c*x)*b**3*c**4*d*x - 720*sqrt(a*d**2 + b*c**2)*log(sqrt(a + b*x**2)*sqrt( 
a*d**2 + b*c**2) - a*d + b*c*x)*b**3*c**3*d**2*x**2 - 120*sqrt(a*d**2 + b* 
c**2)*log(c + d*x)*a**2*b*c**2*d**3 - 240*sqrt(a*d**2 + b*c**2)*log(c + d* 
x)*a**2*b*c*d**4*x - 120*sqrt(a*d**2 + b*c**2)*log(c + d*x)*a**2*b*d**5*x* 
*2 - 480*sqrt(a*d**2 + b*c**2)*log(c + d*x)*a*b**2*c**4*d - 960*sqrt(a*...