\(\int \frac {(A+B x) (a+b x^2)^{5/2}}{c+d x} \, dx\) [166]

Optimal result

Integrand size = 24, antiderivative size = 346 \[ \int \frac {(A+B x) \left (a+b x^2\right )^{5/2}}{c+d x} \, dx=-\frac {\left (16 (B c-A d) \left (b c^2+a d^2\right )^2-d \left (5 a^2 B d^4+8 b^2 c^3 (B c-A d)+14 a b c d^2 (B c-A d)\right ) x\right ) \sqrt {a+b x^2}}{16 d^6}-\frac {\left (8 (B c-A d) \left (b c^2+a d^2\right )-d \left (5 a B d^2+6 b c (B c-A d)\right ) x\right ) \left (a+b x^2\right )^{3/2}}{24 d^4}-\frac {(6 (B c-A d)-5 B d x) \left (a+b x^2\right )^{5/2}}{30 d^2}+\frac {\left (5 a^3 B d^6+16 b^3 c^5 (B c-A d)+40 a b^2 c^3 d^2 (B c-A d)+30 a^2 b c d^4 (B c-A d)\right ) \text {arctanh}\left (\frac {\sqrt {b} x}{\sqrt {a+b x^2}}\right )}{16 \sqrt {b} d^7}+\frac {(B c-A d) \left (b c^2+a d^2\right )^{5/2} \text {arctanh}\left (\frac {a d-b c x}{\sqrt {b c^2+a d^2} \sqrt {a+b x^2}}\right )}{d^7} \] Output:

-1/16*(16*(-A*d+B*c)*(a*d^2+b*c^2)^2-d*(5*a^2*B*d^4+8*b^2*c^3*(-A*d+B*c)+1 
4*a*b*c*d^2*(-A*d+B*c))*x)*(b*x^2+a)^(1/2)/d^6-1/24*(8*(-A*d+B*c)*(a*d^2+b 
*c^2)-d*(5*a*B*d^2+6*b*c*(-A*d+B*c))*x)*(b*x^2+a)^(3/2)/d^4-1/30*(-5*B*d*x 
-6*A*d+6*B*c)*(b*x^2+a)^(5/2)/d^2+1/16*(5*a^3*B*d^6+16*b^3*c^5*(-A*d+B*c)+ 
40*a*b^2*c^3*d^2*(-A*d+B*c)+30*a^2*b*c*d^4*(-A*d+B*c))*arctanh(b^(1/2)*x/( 
b*x^2+a)^(1/2))/b^(1/2)/d^7+(-A*d+B*c)*(a*d^2+b*c^2)^(5/2)*arctanh((-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 = 2.12 (sec) , antiderivative size = 373, normalized size of antiderivative = 1.08 \[ \int \frac {(A+B x) \left (a+b x^2\right )^{5/2}}{c+d x} \, dx=\frac {d \sqrt {a+b x^2} \left (a^2 d^4 (-368 B c+368 A d+165 B d x)-2 a b d^2 \left (A d \left (-280 c^2+135 c d x-88 d^2 x^2\right )+B \left (280 c^3-135 c^2 d x+88 c d^2 x^2-65 d^3 x^3\right )\right )-4 b^2 \left (A d \left (-60 c^4+30 c^3 d x-20 c^2 d^2 x^2+15 c d^3 x^3-12 d^4 x^4\right )+B \left (60 c^5-30 c^4 d x+20 c^3 d^2 x^2-15 c^2 d^3 x^3+12 c d^4 x^4-10 d^5 x^5\right )\right )\right )-480 (B c-A d) \left (-b c^2-a d^2\right )^{5/2} \arctan \left (\frac {\sqrt {b} (c+d x)-d \sqrt {a+b x^2}}{\sqrt {-b c^2-a d^2}}\right )-\frac {15 \left (5 a^3 B d^6+16 b^3 c^5 (B c-A d)+40 a b^2 c^3 d^2 (B c-A d)+30 a^2 b c d^4 (B c-A d)\right ) \log \left (-\sqrt {b} x+\sqrt {a+b x^2}\right )}{\sqrt {b}}}{240 d^7} \] Input:

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

Output:

(d*Sqrt[a + b*x^2]*(a^2*d^4*(-368*B*c + 368*A*d + 165*B*d*x) - 2*a*b*d^2*( 
A*d*(-280*c^2 + 135*c*d*x - 88*d^2*x^2) + B*(280*c^3 - 135*c^2*d*x + 88*c* 
d^2*x^2 - 65*d^3*x^3)) - 4*b^2*(A*d*(-60*c^4 + 30*c^3*d*x - 20*c^2*d^2*x^2 
 + 15*c*d^3*x^3 - 12*d^4*x^4) + B*(60*c^5 - 30*c^4*d*x + 20*c^3*d^2*x^2 - 
15*c^2*d^3*x^3 + 12*c*d^4*x^4 - 10*d^5*x^5))) - 480*(B*c - A*d)*(-(b*c^2) 
- a*d^2)^(5/2)*ArcTan[(Sqrt[b]*(c + d*x) - d*Sqrt[a + b*x^2])/Sqrt[-(b*c^2 
) - a*d^2]] - (15*(5*a^3*B*d^6 + 16*b^3*c^5*(B*c - A*d) + 40*a*b^2*c^3*d^2 
*(B*c - A*d) + 30*a^2*b*c*d^4*(B*c - A*d))*Log[-(Sqrt[b]*x) + Sqrt[a + b*x 
^2]])/Sqrt[b])/(240*d^7)
 

Rubi [A] (verified)

Time = 0.81 (sec) , antiderivative size = 369, normalized size of antiderivative = 1.07, number of steps used = 13, number of rules used = 12, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.500, Rules used = {682, 25, 27, 682, 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.

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

\(\Big \downarrow \) 682

\(\displaystyle \frac {\int -\frac {b \left (a d (B c-6 A d)-\left (5 a B d^2+6 b c (B c-A d)\right ) x\right ) \left (b x^2+a\right )^{3/2}}{c+d x}dx}{6 b d^2}-\frac {\left (a+b x^2\right )^{5/2} (6 (B c-A d)-5 B d x)}{30 d^2}\)

\(\Big \downarrow \) 25

\(\displaystyle -\frac {\int \frac {b \left (a d (B c-6 A d)-\left (5 a B d^2+6 b c (B c-A d)\right ) x\right ) \left (b x^2+a\right )^{3/2}}{c+d x}dx}{6 b d^2}-\frac {\left (a+b x^2\right )^{5/2} (6 (B c-A d)-5 B d x)}{30 d^2}\)

\(\Big \downarrow \) 27

\(\displaystyle -\frac {\int \frac {\left (a d (B c-6 A d)-\left (5 a B d^2+6 b c (B c-A d)\right ) x\right ) \left (b x^2+a\right )^{3/2}}{c+d x}dx}{6 d^2}-\frac {\left (a+b x^2\right )^{5/2} (6 (B c-A d)-5 B d x)}{30 d^2}\)

\(\Big \downarrow \) 682

\(\displaystyle -\frac {\frac {\int \frac {3 b \left (a d \left (2 b (B c-A d) c^2+a d^2 (3 B c-8 A d)\right )-\left (5 a^2 B d^4+14 a b c (B c-A d) d^2+8 b^2 c^3 (B c-A d)\right ) x\right ) \sqrt {b x^2+a}}{c+d x}dx}{4 b d^2}+\frac {\left (a+b x^2\right )^{3/2} \left (8 \left (a d^2+b c^2\right ) (B c-A d)-d x \left (5 a B d^2+6 b c (B c-A d)\right )\right )}{4 d^2}}{6 d^2}-\frac {\left (a+b x^2\right )^{5/2} (6 (B c-A d)-5 B d x)}{30 d^2}\)

\(\Big \downarrow \) 27

\(\displaystyle -\frac {\frac {3 \int \frac {\left (a d \left (2 b (B c-A d) c^2+a d^2 (3 B c-8 A d)\right )-\left (5 a^2 B d^4+14 a b c (B c-A d) d^2+8 b^2 c^3 (B c-A d)\right ) x\right ) \sqrt {b x^2+a}}{c+d x}dx}{4 d^2}+\frac {\left (a+b x^2\right )^{3/2} \left (8 \left (a d^2+b c^2\right ) (B c-A d)-d x \left (5 a B d^2+6 b c (B c-A d)\right )\right )}{4 d^2}}{6 d^2}-\frac {\left (a+b x^2\right )^{5/2} (6 (B c-A d)-5 B d x)}{30 d^2}\)

\(\Big \downarrow \) 682

\(\displaystyle -\frac {\frac {3 \left (\frac {\int \frac {b \left (a d \left (8 b^2 (B c-A d) c^4+18 a b d^2 (B c-A d) c^2+a^2 d^4 (11 B c-16 A d)\right )-\left (5 a^3 B d^6+30 a^2 b c (B c-A d) d^4+40 a b^2 c^3 (B c-A d) d^2+16 b^3 c^5 (B c-A d)\right ) x\right )}{(c+d x) \sqrt {b x^2+a}}dx}{2 b d^2}+\frac {\sqrt {a+b x^2} \left (16 \left (a d^2+b c^2\right )^2 (B c-A d)-d x \left (5 a^2 B d^4+14 a b c d^2 (B c-A d)+8 b^2 c^3 (B c-A d)\right )\right )}{2 d^2}\right )}{4 d^2}+\frac {\left (a+b x^2\right )^{3/2} \left (8 \left (a d^2+b c^2\right ) (B c-A d)-d x \left (5 a B d^2+6 b c (B c-A d)\right )\right )}{4 d^2}}{6 d^2}-\frac {\left (a+b x^2\right )^{5/2} (6 (B c-A d)-5 B d x)}{30 d^2}\)

\(\Big \downarrow \) 27

\(\displaystyle -\frac {\frac {3 \left (\frac {\int \frac {a d \left (8 b^2 (B c-A d) c^4+18 a b d^2 (B c-A d) c^2+a^2 d^4 (11 B c-16 A d)\right )-\left (5 a^3 B d^6+30 a^2 b c (B c-A d) d^4+40 a b^2 c^3 (B c-A d) d^2+16 b^3 c^5 (B c-A d)\right ) x}{(c+d x) \sqrt {b x^2+a}}dx}{2 d^2}+\frac {\sqrt {a+b x^2} \left (16 \left (a d^2+b c^2\right )^2 (B c-A d)-d x \left (5 a^2 B d^4+14 a b c d^2 (B c-A d)+8 b^2 c^3 (B c-A d)\right )\right )}{2 d^2}\right )}{4 d^2}+\frac {\left (a+b x^2\right )^{3/2} \left (8 \left (a d^2+b c^2\right ) (B c-A d)-d x \left (5 a B d^2+6 b c (B c-A d)\right )\right )}{4 d^2}}{6 d^2}-\frac {\left (a+b x^2\right )^{5/2} (6 (B c-A d)-5 B d x)}{30 d^2}\)

\(\Big \downarrow \) 719

\(\displaystyle -\frac {\frac {3 \left (\frac {\frac {16 \left (a d^2+b c^2\right )^3 (B c-A d) \int \frac {1}{(c+d x) \sqrt {b x^2+a}}dx}{d}-\frac {\left (5 a^3 B d^6+30 a^2 b c d^4 (B c-A d)+40 a b^2 c^3 d^2 (B c-A d)+16 b^3 c^5 (B c-A d)\right ) \int \frac {1}{\sqrt {b x^2+a}}dx}{d}}{2 d^2}+\frac {\sqrt {a+b x^2} \left (16 \left (a d^2+b c^2\right )^2 (B c-A d)-d x \left (5 a^2 B d^4+14 a b c d^2 (B c-A d)+8 b^2 c^3 (B c-A d)\right )\right )}{2 d^2}\right )}{4 d^2}+\frac {\left (a+b x^2\right )^{3/2} \left (8 \left (a d^2+b c^2\right ) (B c-A d)-d x \left (5 a B d^2+6 b c (B c-A d)\right )\right )}{4 d^2}}{6 d^2}-\frac {\left (a+b x^2\right )^{5/2} (6 (B c-A d)-5 B d x)}{30 d^2}\)

\(\Big \downarrow \) 224

\(\displaystyle -\frac {\frac {3 \left (\frac {\frac {16 \left (a d^2+b c^2\right )^3 (B c-A d) \int \frac {1}{(c+d x) \sqrt {b x^2+a}}dx}{d}-\frac {\left (5 a^3 B d^6+30 a^2 b c d^4 (B c-A d)+40 a b^2 c^3 d^2 (B c-A d)+16 b^3 c^5 (B c-A d)\right ) \int \frac {1}{1-\frac {b x^2}{b x^2+a}}d\frac {x}{\sqrt {b x^2+a}}}{d}}{2 d^2}+\frac {\sqrt {a+b x^2} \left (16 \left (a d^2+b c^2\right )^2 (B c-A d)-d x \left (5 a^2 B d^4+14 a b c d^2 (B c-A d)+8 b^2 c^3 (B c-A d)\right )\right )}{2 d^2}\right )}{4 d^2}+\frac {\left (a+b x^2\right )^{3/2} \left (8 \left (a d^2+b c^2\right ) (B c-A d)-d x \left (5 a B d^2+6 b c (B c-A d)\right )\right )}{4 d^2}}{6 d^2}-\frac {\left (a+b x^2\right )^{5/2} (6 (B c-A d)-5 B d x)}{30 d^2}\)

\(\Big \downarrow \) 219

\(\displaystyle -\frac {\frac {3 \left (\frac {\frac {16 \left (a d^2+b c^2\right )^3 (B c-A d) \int \frac {1}{(c+d x) \sqrt {b x^2+a}}dx}{d}-\frac {\text {arctanh}\left (\frac {\sqrt {b} x}{\sqrt {a+b x^2}}\right ) \left (5 a^3 B d^6+30 a^2 b c d^4 (B c-A d)+40 a b^2 c^3 d^2 (B c-A d)+16 b^3 c^5 (B c-A d)\right )}{\sqrt {b} d}}{2 d^2}+\frac {\sqrt {a+b x^2} \left (16 \left (a d^2+b c^2\right )^2 (B c-A d)-d x \left (5 a^2 B d^4+14 a b c d^2 (B c-A d)+8 b^2 c^3 (B c-A d)\right )\right )}{2 d^2}\right )}{4 d^2}+\frac {\left (a+b x^2\right )^{3/2} \left (8 \left (a d^2+b c^2\right ) (B c-A d)-d x \left (5 a B d^2+6 b c (B c-A d)\right )\right )}{4 d^2}}{6 d^2}-\frac {\left (a+b x^2\right )^{5/2} (6 (B c-A d)-5 B d x)}{30 d^2}\)

\(\Big \downarrow \) 488

\(\displaystyle -\frac {\frac {3 \left (\frac {-\frac {16 \left (a d^2+b c^2\right )^3 (B c-A d) \int \frac {1}{b c^2+a d^2-\frac {(a d-b c x)^2}{b x^2+a}}d\frac {a d-b c x}{\sqrt {b x^2+a}}}{d}-\frac {\text {arctanh}\left (\frac {\sqrt {b} x}{\sqrt {a+b x^2}}\right ) \left (5 a^3 B d^6+30 a^2 b c d^4 (B c-A d)+40 a b^2 c^3 d^2 (B c-A d)+16 b^3 c^5 (B c-A d)\right )}{\sqrt {b} d}}{2 d^2}+\frac {\sqrt {a+b x^2} \left (16 \left (a d^2+b c^2\right )^2 (B c-A d)-d x \left (5 a^2 B d^4+14 a b c d^2 (B c-A d)+8 b^2 c^3 (B c-A d)\right )\right )}{2 d^2}\right )}{4 d^2}+\frac {\left (a+b x^2\right )^{3/2} \left (8 \left (a d^2+b c^2\right ) (B c-A d)-d x \left (5 a B d^2+6 b c (B c-A d)\right )\right )}{4 d^2}}{6 d^2}-\frac {\left (a+b x^2\right )^{5/2} (6 (B c-A d)-5 B d x)}{30 d^2}\)

\(\Big \downarrow \) 219

\(\displaystyle -\frac {\frac {3 \left (\frac {\sqrt {a+b x^2} \left (16 \left (a d^2+b c^2\right )^2 (B c-A d)-d x \left (5 a^2 B d^4+14 a b c d^2 (B c-A d)+8 b^2 c^3 (B c-A d)\right )\right )}{2 d^2}+\frac {-\frac {\text {arctanh}\left (\frac {\sqrt {b} x}{\sqrt {a+b x^2}}\right ) \left (5 a^3 B d^6+30 a^2 b c d^4 (B c-A d)+40 a b^2 c^3 d^2 (B c-A d)+16 b^3 c^5 (B c-A d)\right )}{\sqrt {b} d}-\frac {16 \left (a d^2+b c^2\right )^{5/2} (B c-A d) \text {arctanh}\left (\frac {a d-b c x}{\sqrt {a+b x^2} \sqrt {a d^2+b c^2}}\right )}{d}}{2 d^2}\right )}{4 d^2}+\frac {\left (a+b x^2\right )^{3/2} \left (8 \left (a d^2+b c^2\right ) (B c-A d)-d x \left (5 a B d^2+6 b c (B c-A d)\right )\right )}{4 d^2}}{6 d^2}-\frac {\left (a+b x^2\right )^{5/2} (6 (B c-A d)-5 B d x)}{30 d^2}\)

Input:

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

Output:

-1/30*((6*(B*c - A*d) - 5*B*d*x)*(a + b*x^2)^(5/2))/d^2 - (((8*(B*c - A*d) 
*(b*c^2 + a*d^2) - d*(5*a*B*d^2 + 6*b*c*(B*c - A*d))*x)*(a + b*x^2)^(3/2)) 
/(4*d^2) + (3*(((16*(B*c - A*d)*(b*c^2 + a*d^2)^2 - d*(5*a^2*B*d^4 + 8*b^2 
*c^3*(B*c - A*d) + 14*a*b*c*d^2*(B*c - A*d))*x)*Sqrt[a + b*x^2])/(2*d^2) + 
 (-(((5*a^3*B*d^6 + 16*b^3*c^5*(B*c - A*d) + 40*a*b^2*c^3*d^2*(B*c - A*d) 
+ 30*a^2*b*c*d^4*(B*c - A*d))*ArcTanh[(Sqrt[b]*x)/Sqrt[a + b*x^2]])/(Sqrt[ 
b]*d)) - (16*(B*c - A*d)*(b*c^2 + a*d^2)^(5/2)*ArcTanh[(a*d - b*c*x)/(Sqrt 
[b*c^2 + a*d^2]*Sqrt[a + b*x^2])])/d)/(2*d^2)))/(4*d^2))/(6*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 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 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 [A] (verified)

Time = 1.36 (sec) , antiderivative size = 585, normalized size of antiderivative = 1.69

method	result	size
risch	\(\frac {\left (40 B \,b^{2} d^{5} x^{5}+48 A \,b^{2} d^{5} x^{4}-48 B \,b^{2} c \,d^{4} x^{4}-60 A \,b^{2} c \,d^{4} x^{3}+130 B a b \,d^{5} x^{3}+60 B \,b^{2} c^{2} d^{3} x^{3}+176 A a b \,d^{5} x^{2}+80 A \,b^{2} c^{2} d^{3} x^{2}-176 B a b c \,d^{4} x^{2}-80 B \,b^{2} c^{3} d^{2} x^{2}-270 A a b c \,d^{4} x -120 A \,b^{2} c^{3} d^{2} x +165 B \,a^{2} d^{5} x +270 B a b \,c^{2} d^{3} x +120 B \,b^{2} c^{4} d x +368 a^{2} A \,d^{5}+560 A a b \,c^{2} d^{3}+240 c^{4} b^{2} A d -368 B \,a^{2} c \,d^{4}-560 B a b \,c^{3} d^{2}-240 c^{5} B \,b^{2}\right ) \sqrt {b \,x^{2}+a}}{240 d^{6}}-\frac {\frac {\left (30 A \,a^{2} b c \,d^{5}+40 A a \,b^{2} c^{3} d^{3}+16 A \,b^{3} c^{5} d -5 a^{3} B \,d^{6}-30 B \,a^{2} b \,c^{2} d^{4}-40 B a \,b^{2} c^{4} d^{2}-16 B \,b^{3} c^{6}\right ) \ln \left (\sqrt {b}\, x +\sqrt {b \,x^{2}+a}\right )}{d \sqrt {b}}+\frac {16 \left (a^{3} A \,d^{7}+3 c^{2} A \,a^{2} b \,d^{5}+3 c^{4} A a \,b^{2} d^{3}+c^{6} A \,b^{3} d -B \,a^{3} c \,d^{6}-3 c^{3} B \,a^{2} b \,d^{4}-3 c^{5} B a \,b^{2} d^{2}-c^{7} B \,b^{3}\right ) \ln \left (\frac {\frac {2 a \,d^{2}+2 b \,c^{2}}{d^{2}}-\frac {2 b c \left (x +\frac {c}{d}\right )}{d}+2 \sqrt {\frac {a \,d^{2}+b \,c^{2}}{d^{2}}}\, \sqrt {b \left (x +\frac {c}{d}\right )^{2}-\frac {2 b c \left (x +\frac {c}{d}\right )}{d}+\frac {a \,d^{2}+b \,c^{2}}{d^{2}}}}{x +\frac {c}{d}}\right )}{d^{2} \sqrt {\frac {a \,d^{2}+b \,c^{2}}{d^{2}}}}}{16 d^{6}}\)	\(585\)
default	\(\frac {B \left (\frac {x \left (b \,x^{2}+a \right )^{\frac {5}{2}}}{6}+\frac {5 a \left (\frac {x \left (b \,x^{2}+a \right )^{\frac {3}{2}}}{4}+\frac {3 a \left (\frac {x \sqrt {b \,x^{2}+a}}{2}+\frac {a \ln \left (\sqrt {b}\, x +\sqrt {b \,x^{2}+a}\right )}{2 \sqrt {b}}\right )}{4}\right )}{6}\right )}{d}+\frac {\left (A d -B c \right ) \left (\frac {\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 {5}{2}}}{5}-\frac {b c \left (\frac {\left (2 b \left (x +\frac {c}{d}\right )-\frac {2 b c}{d}\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}}}{8 b}+\frac {3 \left (\frac {4 b \left (a \,d^{2}+b \,c^{2}\right )}{d^{2}}-\frac {4 b^{2} c^{2}}{d^{2}}\right ) \left (\frac {\left (2 b \left (x +\frac {c}{d}\right )-\frac {2 b c}{d}\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}}}}{4 b}+\frac {\left (\frac {4 b \left (a \,d^{2}+b \,c^{2}\right )}{d^{2}}-\frac {4 b^{2} c^{2}}{d^{2}}\right ) \ln \left (\frac {-\frac {b c}{d}+b \left (x +\frac {c}{d}\right )}{\sqrt {b}}+\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 )}{8 b^{\frac {3}{2}}}\right )}{16 b}\right )}{d}+\frac {\left (a \,d^{2}+b \,c^{2}\right ) \left (\frac {\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}}}{3}-\frac {b c \left (\frac {\left (2 b \left (x +\frac {c}{d}\right )-\frac {2 b c}{d}\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}}}}{4 b}+\frac {\left (\frac {4 b \left (a \,d^{2}+b \,c^{2}\right )}{d^{2}}-\frac {4 b^{2} c^{2}}{d^{2}}\right ) \ln \left (\frac {-\frac {b c}{d}+b \left (x +\frac {c}{d}\right )}{\sqrt {b}}+\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 )}{8 b^{\frac {3}{2}}}\right )}{d}+\frac {\left (a \,d^{2}+b \,c^{2}\right ) \left (\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 {\sqrt {b}\, c \ln \left (\frac {-\frac {b c}{d}+b \left (x +\frac {c}{d}\right )}{\sqrt {b}}+\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 )}{d}-\frac {\left (a \,d^{2}+b \,c^{2}\right ) \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 )}{d^{2} \sqrt {\frac {a \,d^{2}+b \,c^{2}}{d^{2}}}}\right )}{d^{2}}\right )}{d^{2}}\right )}{d^{2}}\)	\(913\)

Input:

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

Output:

1/240*(40*B*b^2*d^5*x^5+48*A*b^2*d^5*x^4-48*B*b^2*c*d^4*x^4-60*A*b^2*c*d^4 
*x^3+130*B*a*b*d^5*x^3+60*B*b^2*c^2*d^3*x^3+176*A*a*b*d^5*x^2+80*A*b^2*c^2 
*d^3*x^2-176*B*a*b*c*d^4*x^2-80*B*b^2*c^3*d^2*x^2-270*A*a*b*c*d^4*x-120*A* 
b^2*c^3*d^2*x+165*B*a^2*d^5*x+270*B*a*b*c^2*d^3*x+120*B*b^2*c^4*d*x+368*A* 
a^2*d^5+560*A*a*b*c^2*d^3+240*A*b^2*c^4*d-368*B*a^2*c*d^4-560*B*a*b*c^3*d^ 
2-240*B*b^2*c^5)*(b*x^2+a)^(1/2)/d^6-1/16/d^6*((30*A*a^2*b*c*d^5+40*A*a*b^ 
2*c^3*d^3+16*A*b^3*c^5*d-5*B*a^3*d^6-30*B*a^2*b*c^2*d^4-40*B*a*b^2*c^4*d^2 
-16*B*b^3*c^6)/d*ln(b^(1/2)*x+(b*x^2+a)^(1/2))/b^(1/2)+16*(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^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 [F(-1)]

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

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

Output:

Timed out
 

Sympy [F]

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

integrate((B*x+A)*(b*x**2+a)**(5/2)/(d*x+c),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. 654 vs. \(2 (318) = 636\).

Time = 0.15 (sec) , antiderivative size = 654, normalized size of antiderivative = 1.89 \[ \int \frac {(A+B x) \left (a+b x^2\right )^{5/2}}{c+d x} \, dx =\text {Too large to display} \] Input:

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

Output:

1/2*sqrt(b*x^2 + a)*B*b^2*c^4*x/d^5 - 1/2*sqrt(b*x^2 + a)*A*b^2*c^3*x/d^4 
+ 1/4*(b*x^2 + a)^(3/2)*B*b*c^2*x/d^3 + 7/8*sqrt(b*x^2 + a)*B*a*b*c^2*x/d^ 
3 - 1/4*(b*x^2 + a)^(3/2)*A*b*c*x/d^2 - 7/8*sqrt(b*x^2 + a)*A*a*b*c*x/d^2 
+ 1/6*(b*x^2 + a)^(5/2)*B*x/d + 5/24*(b*x^2 + a)^(3/2)*B*a*x/d + 5/16*sqrt 
(b*x^2 + a)*B*a^2*x/d + B*b^(5/2)*c^6*arcsinh(b*x/sqrt(a*b))/d^7 - A*b^(5/ 
2)*c^5*arcsinh(b*x/sqrt(a*b))/d^6 + 5/2*B*a*b^(3/2)*c^4*arcsinh(b*x/sqrt(a 
*b))/d^5 - 5/2*A*a*b^(3/2)*c^3*arcsinh(b*x/sqrt(a*b))/d^4 + 15/8*B*a^2*sqr 
t(b)*c^2*arcsinh(b*x/sqrt(a*b))/d^3 - 15/8*A*a^2*sqrt(b)*c*arcsinh(b*x/sqr 
t(a*b))/d^2 + 5/16*B*a^3*arcsinh(b*x/sqrt(a*b))/(sqrt(b)*d) - B*(a + b*c^2 
/d^2)^(5/2)*c*arcsinh(b*c*x/(sqrt(a*b)*abs(d*x + c)) - a*d/(sqrt(a*b)*abs( 
d*x + c)))/d^2 + A*(a + b*c^2/d^2)^(5/2)*arcsinh(b*c*x/(sqrt(a*b)*abs(d*x 
+ c)) - a*d/(sqrt(a*b)*abs(d*x + c)))/d - sqrt(b*x^2 + a)*B*b^2*c^5/d^6 + 
sqrt(b*x^2 + a)*A*b^2*c^4/d^5 - 1/3*(b*x^2 + a)^(3/2)*B*b*c^3/d^4 - 2*sqrt 
(b*x^2 + a)*B*a*b*c^3/d^4 + 1/3*(b*x^2 + a)^(3/2)*A*b*c^2/d^3 + 2*sqrt(b*x 
^2 + a)*A*a*b*c^2/d^3 - 1/5*(b*x^2 + a)^(5/2)*B*c/d^2 - 1/3*(b*x^2 + a)^(3 
/2)*B*a*c/d^2 - sqrt(b*x^2 + a)*B*a^2*c/d^2 + 1/5*(b*x^2 + a)^(5/2)*A/d + 
1/3*(b*x^2 + a)^(3/2)*A*a/d + sqrt(b*x^2 + a)*A*a^2/d
 

Giac [F(-2)]

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

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

Output:

Exception raised: TypeError >> an error occurred running a Giac command:IN 
PUT:sage2:=int(sage0,sageVARx):;OUTPUT:index.cc index_m i_lex_is_greater E 
rror: Bad Argument Value
 

Mupad [F(-1)]

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

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

Output:

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

Reduce [F]

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

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

Output:

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