\(\int \frac {(A+B x) (a+b x^2)^{3/2}}{(c+d x)^2} \, dx\) [160]

Optimal result

Integrand size = 24, antiderivative size = 231 \[ \int \frac {(A+B x) \left (a+b x^2\right )^{3/2}}{(c+d x)^2} \, dx=\frac {\left (2 \left (a B d^2+b c (4 B c-3 A d)\right )-b d (4 B c-3 A d) x\right ) \sqrt {a+b x^2}}{2 d^4}+\frac {(4 B c-3 A d+B d x) \left (a+b x^2\right )^{3/2}}{3 d^2 (c+d x)}-\frac {\sqrt {b} \left (2 a B c d^2+(4 B c-3 A d) \left (2 b c^2+a d^2\right )\right ) \text {arctanh}\left (\frac {\sqrt {b} x}{\sqrt {a+b x^2}}\right )}{2 d^5}-\frac {\sqrt {b c^2+a d^2} \left (a B d^2+b c (4 B c-3 A d)\right ) \text {arctanh}\left (\frac {a d-b c x}{\sqrt {b c^2+a d^2} \sqrt {a+b x^2}}\right )}{d^5} \] Output:

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

Mathematica [A] (verified)

Time = 1.73 (sec) , antiderivative size = 246, normalized size of antiderivative = 1.06 \[ \int \frac {(A+B x) \left (a+b x^2\right )^{3/2}}{(c+d x)^2} \, dx=\frac {\frac {d \sqrt {a+b x^2} \left (2 a d^2 (7 B c-3 A d+4 B d x)+3 A b d \left (-6 c^2-3 c d x+d^2 x^2\right )+2 b B \left (12 c^3+6 c^2 d x-2 c d^2 x^2+d^3 x^3\right )\right )}{c+d x}+12 \sqrt {-b c^2-a d^2} \left (a B d^2+b c (4 B c-3 A d)\right ) \arctan \left (\frac {\sqrt {b} (c+d x)-d \sqrt {a+b x^2}}{\sqrt {-b c^2-a d^2}}\right )+3 \sqrt {b} \left (2 b c^2 (4 B c-3 A d)-3 a d^2 (-2 B c+A d)\right ) \log \left (-\sqrt {b} x+\sqrt {a+b x^2}\right )}{6 d^5} \] Input:

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

Output:

((d*Sqrt[a + b*x^2]*(2*a*d^2*(7*B*c - 3*A*d + 4*B*d*x) + 3*A*b*d*(-6*c^2 - 
 3*c*d*x + d^2*x^2) + 2*b*B*(12*c^3 + 6*c^2*d*x - 2*c*d^2*x^2 + d^3*x^3))) 
/(c + d*x) + 12*Sqrt[-(b*c^2) - a*d^2]*(a*B*d^2 + b*c*(4*B*c - 3*A*d))*Arc 
Tan[(Sqrt[b]*(c + d*x) - d*Sqrt[a + b*x^2])/Sqrt[-(b*c^2) - a*d^2]] + 3*Sq 
rt[b]*(2*b*c^2*(4*B*c - 3*A*d) - 3*a*d^2*(-2*B*c + A*d))*Log[-(Sqrt[b]*x) 
+ Sqrt[a + b*x^2]])/(6*d^5)
 

Rubi [A] (verified)

Time = 0.50 (sec) , antiderivative size = 242, normalized size of antiderivative = 1.05, number of steps used = 10, number of rules used = 9, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.375, Rules used = {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.

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

\(\Big \downarrow \) 681

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

\(\Big \downarrow \) 27

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

\(\Big \downarrow \) 682

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

\(\Big \downarrow \) 27

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

\(\Big \downarrow \) 719

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

\(\Big \downarrow \) 224

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

\(\Big \downarrow \) 219

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

\(\Big \downarrow \) 488

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

\(\Big \downarrow \) 219

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

Input:

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

Output:

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

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. \(570\) vs. \(2(208)=416\).

Time = 1.40 (sec) , antiderivative size = 571, normalized size of antiderivative = 2.47

method	result	size
risch	\(-\frac {\left (-2 B b \,d^{2} x^{2}-3 A b \,d^{2} x +6 B b c d x +12 A b c d -8 a B \,d^{2}-18 B b \,c^{2}\right ) \sqrt {b \,x^{2}+a}}{6 d^{4}}+\frac {\frac {\sqrt {b}\, \left (3 A \,d^{3} a +6 A b \,c^{2} d -6 a B c \,d^{2}-8 b B \,c^{3}\right ) \ln \left (\sqrt {b}\, x +\sqrt {b \,x^{2}+a}\right )}{d}+\frac {2 \left (4 A a b c \,d^{3}+4 A \,b^{2} c^{3} d -a^{2} B \,d^{4}-6 B a b \,c^{2} d^{2}-5 B \,b^{2} c^{4}\right ) \ln \left (\frac {\frac {2 a \,d^{2}+2 b \,c^{2}}{d^{2}}-\frac {2 b c \left (x +\frac {c}{d}\right )}{d}+2 \sqrt {\frac {a \,d^{2}+b \,c^{2}}{d^{2}}}\, \sqrt {b \left (x +\frac {c}{d}\right )^{2}-\frac {2 b c \left (x +\frac {c}{d}\right )}{d}+\frac {a \,d^{2}+b \,c^{2}}{d^{2}}}}{x +\frac {c}{d}}\right )}{d^{2} \sqrt {\frac {a \,d^{2}+b \,c^{2}}{d^{2}}}}+\frac {2 \left (a^{2} A \,d^{5}+2 A a b \,c^{2} d^{3}+c^{4} b^{2} A d -B \,a^{2} c \,d^{4}-2 B a b \,c^{3} d^{2}-c^{5} B \,b^{2}\right ) \left (-\frac {d^{2} \sqrt {b \left (x +\frac {c}{d}\right )^{2}-\frac {2 b c \left (x +\frac {c}{d}\right )}{d}+\frac {a \,d^{2}+b \,c^{2}}{d^{2}}}}{\left (a \,d^{2}+b \,c^{2}\right ) \left (x +\frac {c}{d}\right )}-\frac {b c d \ln \left (\frac {\frac {2 a \,d^{2}+2 b \,c^{2}}{d^{2}}-\frac {2 b c \left (x +\frac {c}{d}\right )}{d}+2 \sqrt {\frac {a \,d^{2}+b \,c^{2}}{d^{2}}}\, \sqrt {b \left (x +\frac {c}{d}\right )^{2}-\frac {2 b c \left (x +\frac {c}{d}\right )}{d}+\frac {a \,d^{2}+b \,c^{2}}{d^{2}}}}{x +\frac {c}{d}}\right )}{\left (a \,d^{2}+b \,c^{2}\right ) \sqrt {\frac {a \,d^{2}+b \,c^{2}}{d^{2}}}}\right )}{d^{3}}}{2 d^{4}}\)	\(571\)
default	\(\text {Expression too large to display}\)	\(1377\)

Input:

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

Output:

-1/6*(-2*B*b*d^2*x^2-3*A*b*d^2*x+6*B*b*c*d*x+12*A*b*c*d-8*B*a*d^2-18*B*b*c 
^2)*(b*x^2+a)^(1/2)/d^4+1/2/d^4*(b^(1/2)*(3*A*a*d^3+6*A*b*c^2*d-6*B*a*c*d^ 
2-8*B*b*c^3)/d*ln(b^(1/2)*x+(b*x^2+a)^(1/2))+2/d^2*(4*A*a*b*c*d^3+4*A*b^2* 
c^3*d-B*a^2*d^4-6*B*a*b*c^2*d^2-5*B*b^2*c^4)/((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))+2*(A*a^2*d^5+2*A*a* 
b*c^2*d^3+A*b^2*c^4*d-B*a^2*c*d^4-2*B*a*b*c^3*d^2-B*b^2*c^5)/d^3*(-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))))
 

Fricas [F(-1)]

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

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

Output:

Timed out
 

Sympy [F]

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

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

Output:

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

Maxima [A] (verification not implemented)

Time = 0.09 (sec) , antiderivative size = 398, normalized size of antiderivative = 1.72 \[ \int \frac {(A+B x) \left (a+b x^2\right )^{3/2}}{(c+d x)^2} \, dx=\frac {{\left (b x^{2} + a\right )}^{\frac {3}{2}} B c}{d^{3} x + c d^{2}} - \frac {{\left (b x^{2} + a\right )}^{\frac {3}{2}} A}{d^{2} x + c d} - \frac {2 \, \sqrt {b x^{2} + a} B b c x}{d^{3}} + \frac {3 \, \sqrt {b x^{2} + a} A b x}{2 \, d^{2}} - \frac {4 \, B b^{\frac {3}{2}} c^{3} \operatorname {arsinh}\left (\frac {b x}{\sqrt {a b}}\right )}{d^{5}} + \frac {3 \, A b^{\frac {3}{2}} c^{2} \operatorname {arsinh}\left (\frac {b x}{\sqrt {a b}}\right )}{d^{4}} - \frac {3 \, B a \sqrt {b} c \operatorname {arsinh}\left (\frac {b x}{\sqrt {a b}}\right )}{d^{3}} + \frac {3 \, A a \sqrt {b} \operatorname {arsinh}\left (\frac {b x}{\sqrt {a b}}\right )}{2 \, d^{2}} + \frac {3 \, B \sqrt {a + \frac {b c^{2}}{d^{2}}} b c^{2} \operatorname {arsinh}\left (\frac {b c x}{\sqrt {a b} {\left | d x + c \right |}} - \frac {a d}{\sqrt {a b} {\left | d x + c \right |}}\right )}{d^{4}} - \frac {3 \, A \sqrt {a + \frac {b c^{2}}{d^{2}}} b c \operatorname {arsinh}\left (\frac {b c x}{\sqrt {a b} {\left | d x + c \right |}} - \frac {a d}{\sqrt {a b} {\left | d x + c \right |}}\right )}{d^{3}} + \frac {B {\left (a + \frac {b c^{2}}{d^{2}}\right )}^{\frac {3}{2}} \operatorname {arsinh}\left (\frac {b c x}{\sqrt {a b} {\left | d x + c \right |}} - \frac {a d}{\sqrt {a b} {\left | d x + c \right |}}\right )}{d^{2}} + \frac {4 \, \sqrt {b x^{2} + a} B b c^{2}}{d^{4}} - \frac {3 \, \sqrt {b x^{2} + a} A b c}{d^{3}} + \frac {{\left (b x^{2} + a\right )}^{\frac {3}{2}} B}{3 \, d^{2}} + \frac {\sqrt {b x^{2} + a} B a}{d^{2}} \] Input:

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

Output:

(b*x^2 + a)^(3/2)*B*c/(d^3*x + c*d^2) - (b*x^2 + a)^(3/2)*A/(d^2*x + c*d) 
- 2*sqrt(b*x^2 + a)*B*b*c*x/d^3 + 3/2*sqrt(b*x^2 + a)*A*b*x/d^2 - 4*B*b^(3 
/2)*c^3*arcsinh(b*x/sqrt(a*b))/d^5 + 3*A*b^(3/2)*c^2*arcsinh(b*x/sqrt(a*b) 
)/d^4 - 3*B*a*sqrt(b)*c*arcsinh(b*x/sqrt(a*b))/d^3 + 3/2*A*a*sqrt(b)*arcsi 
nh(b*x/sqrt(a*b))/d^2 + 3*B*sqrt(a + b*c^2/d^2)*b*c^2*arcsinh(b*c*x/(sqrt( 
a*b)*abs(d*x + c)) - a*d/(sqrt(a*b)*abs(d*x + c)))/d^4 - 3*A*sqrt(a + b*c^ 
2/d^2)*b*c*arcsinh(b*c*x/(sqrt(a*b)*abs(d*x + c)) - a*d/(sqrt(a*b)*abs(d*x 
 + c)))/d^3 + B*(a + b*c^2/d^2)^(3/2)*arcsinh(b*c*x/(sqrt(a*b)*abs(d*x + c 
)) - a*d/(sqrt(a*b)*abs(d*x + c)))/d^2 + 4*sqrt(b*x^2 + a)*B*b*c^2/d^4 - 3 
*sqrt(b*x^2 + a)*A*b*c/d^3 + 1/3*(b*x^2 + a)^(3/2)*B/d^2 + sqrt(b*x^2 + a) 
*B*a/d^2
 

Giac [F(-1)]

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

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

Output:

Timed out
 

Mupad [F(-1)]

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

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

Output:

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

Reduce [B] (verification not implemented)

Time = 0.29 (sec) , antiderivative size = 1076, normalized size of antiderivative = 4.66 \[ \int \frac {(A+B x) \left (a+b x^2\right )^{3/2}}{(c+d x)^2} \, dx =\text {Too large to display} \] Input:

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

Output:

(36*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*c**2*d + 36*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*c*d**2*x - 12*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*c*d** 
2 - 12*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*d**3*x - 48*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**2*c**3 - 48*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**2*c**2 
*d*x - 36*sqrt(a*d**2 + b*c**2)*log(c + d*x)*a*b*c**2*d - 36*sqrt(a*d**2 + 
 b*c**2)*log(c + d*x)*a*b*c*d**2*x + 12*sqrt(a*d**2 + b*c**2)*log(c + d*x) 
*a*b*c*d**2 + 12*sqrt(a*d**2 + b*c**2)*log(c + d*x)*a*b*d**3*x + 48*sqrt(a 
*d**2 + b*c**2)*log(c + d*x)*b**2*c**3 + 48*sqrt(a*d**2 + b*c**2)*log(c + 
d*x)*b**2*c**2*d*x - 12*sqrt(a + b*x**2)*a**2*d**4 - 36*sqrt(a + b*x**2)*a 
*b*c**2*d**2 - 18*sqrt(a + b*x**2)*a*b*c*d**3*x + 28*sqrt(a + b*x**2)*a*b* 
c*d**3 + 6*sqrt(a + b*x**2)*a*b*d**4*x**2 + 16*sqrt(a + b*x**2)*a*b*d**4*x 
 + 48*sqrt(a + b*x**2)*b**2*c**3*d + 24*sqrt(a + b*x**2)*b**2*c**2*d**2*x 
- 8*sqrt(a + b*x**2)*b**2*c*d**3*x**2 + 4*sqrt(a + b*x**2)*b**2*d**4*x**3 
- 9*sqrt(b)*log(sqrt(a + b*x**2) - sqrt(b)*x)*a**2*c*d**3 - 9*sqrt(b)*log( 
sqrt(a + b*x**2) - sqrt(b)*x)*a**2*d**4*x - 18*sqrt(b)*log(sqrt(a + b*x**2 
) - sqrt(b)*x)*a*b*c**3*d - 18*sqrt(b)*log(sqrt(a + b*x**2) - sqrt(b)*x...