\(\int \frac {(A+B x) (a+b x^2)^{3/2}}{c+d x} \, dx\) [159]

Optimal result

Integrand size = 24, antiderivative size = 236 \[ \int \frac {(A+B x) \left (a+b x^2\right )^{3/2}}{c+d x} \, dx=-\frac {\left (8 (B c-A d) \left (b c^2+a d^2\right )-d \left (3 a B d^2+4 b c (B c-A d)\right ) x\right ) \sqrt {a+b x^2}}{8 d^4}-\frac {(4 (B c-A d)-3 B d x) \left (a+b x^2\right )^{3/2}}{12 d^2}+\frac {\left (3 a^2 B d^4+8 b^2 c^3 (B c-A d)+12 a b c d^2 (B c-A d)\right ) \text {arctanh}\left (\frac {\sqrt {b} x}{\sqrt {a+b x^2}}\right )}{8 \sqrt {b} d^5}+\frac {(B c-A d) \left (b c^2+a d^2\right )^{3/2} \text {arctanh}\left (\frac {a d-b c x}{\sqrt {b c^2+a d^2} \sqrt {a+b x^2}}\right )}{d^5} \] Output:

-1/8*(8*(-A*d+B*c)*(a*d^2+b*c^2)-d*(3*a*B*d^2+4*b*c*(-A*d+B*c))*x)*(b*x^2+ 
a)^(1/2)/d^4-1/12*(-3*B*d*x-4*A*d+4*B*c)*(b*x^2+a)^(3/2)/d^2+1/8*(3*a^2*B* 
d^4+8*b^2*c^3*(-A*d+B*c)+12*a*b*c*d^2*(-A*d+B*c))*arctanh(b^(1/2)*x/(b*x^2 
+a)^(1/2))/b^(1/2)/d^5+(-A*d+B*c)*(a*d^2+b*c^2)^(3/2)*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 = 1.61 (sec) , antiderivative size = 240, normalized size of antiderivative = 1.02 \[ \int \frac {(A+B x) \left (a+b x^2\right )^{3/2}}{c+d x} \, dx=\frac {d \sqrt {a+b x^2} \left (a d^2 (-32 B c+32 A d+15 B d x)+4 A b d \left (6 c^2-3 c d x+2 d^2 x^2\right )+b B \left (-24 c^3+12 c^2 d x-8 c d^2 x^2+6 d^3 x^3\right )\right )+48 (B c-A d) \left (-b c^2-a d^2\right )^{3/2} \arctan \left (\frac {\sqrt {b} (c+d x)-d \sqrt {a+b x^2}}{\sqrt {-b c^2-a d^2}}\right )-\frac {3 \left (3 a^2 B d^4+8 b^2 c^3 (B c-A d)+12 a b c d^2 (B c-A d)\right ) \log \left (-\sqrt {b} x+\sqrt {a+b x^2}\right )}{\sqrt {b}}}{24 d^5} \] Input:

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

Output:

(d*Sqrt[a + b*x^2]*(a*d^2*(-32*B*c + 32*A*d + 15*B*d*x) + 4*A*b*d*(6*c^2 - 
 3*c*d*x + 2*d^2*x^2) + b*B*(-24*c^3 + 12*c^2*d*x - 8*c*d^2*x^2 + 6*d^3*x^ 
3)) + 48*(B*c - A*d)*(-(b*c^2) - a*d^2)^(3/2)*ArcTan[(Sqrt[b]*(c + d*x) - 
d*Sqrt[a + b*x^2])/Sqrt[-(b*c^2) - a*d^2]] - (3*(3*a^2*B*d^4 + 8*b^2*c^3*( 
B*c - A*d) + 12*a*b*c*d^2*(B*c - A*d))*Log[-(Sqrt[b]*x) + Sqrt[a + b*x^2]] 
)/Sqrt[b])/(24*d^5)
 

Rubi [A] (verified)

Time = 0.55 (sec) , antiderivative size = 251, normalized size of antiderivative = 1.06, number of steps used = 11, number of rules used = 10, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.417, Rules used = {682, 25, 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.

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

\(\Big \downarrow \) 682

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

\(\Big \downarrow \) 25

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

\(\Big \downarrow \) 27

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

\(\Big \downarrow \) 682

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

\(\Big \downarrow \) 27

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

\(\Big \downarrow \) 719

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

\(\Big \downarrow \) 224

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

\(\Big \downarrow \) 219

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

\(\Big \downarrow \) 488

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

\(\Big \downarrow \) 219

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

Input:

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

Output:

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

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 = 367, normalized size of antiderivative = 1.56

method	result	size
risch	\(\frac {\left (6 B b \,d^{3} x^{3}+8 A b \,d^{3} x^{2}-8 B b c \,d^{2} x^{2}-12 A b c \,d^{2} x +15 B a \,d^{3} x +12 B b \,c^{2} d x +32 A \,d^{3} a +24 A b \,c^{2} d -32 a B c \,d^{2}-24 b B \,c^{3}\right ) \sqrt {b \,x^{2}+a}}{24 d^{4}}-\frac {\frac {\left (12 A a b c \,d^{3}+8 A \,b^{2} c^{3} d -3 a^{2} B \,d^{4}-12 B a b \,c^{2} d^{2}-8 B \,b^{2} c^{4}\right ) \ln \left (\sqrt {b}\, x +\sqrt {b \,x^{2}+a}\right )}{d \sqrt {b}}+\frac {8 \left (a^{2} A \,d^{5}+2 A a b \,c^{2} d^{3}+c^{4} b^{2} A d -B \,a^{2} c \,d^{4}-2 B a b \,c^{3} d^{2}-c^{5} B \,b^{2}\right ) \ln \left (\frac {\frac {2 a \,d^{2}+2 b \,c^{2}}{d^{2}}-\frac {2 b c \left (x +\frac {c}{d}\right )}{d}+2 \sqrt {\frac {a \,d^{2}+b \,c^{2}}{d^{2}}}\, \sqrt {b \left (x +\frac {c}{d}\right )^{2}-\frac {2 b c \left (x +\frac {c}{d}\right )}{d}+\frac {a \,d^{2}+b \,c^{2}}{d^{2}}}}{x +\frac {c}{d}}\right )}{d^{2} \sqrt {\frac {a \,d^{2}+b \,c^{2}}{d^{2}}}}}{8 d^{4}}\)	\(367\)
default	\(\frac {B \left (\frac {x \left (b \,x^{2}+a \right )^{\frac {3}{2}}}{4}+\frac {3 a \left (\frac {x \sqrt {b \,x^{2}+a}}{2}+\frac {a \ln \left (\sqrt {b}\, x +\sqrt {b \,x^{2}+a}\right )}{2 \sqrt {b}}\right )}{4}\right )}{d}+\frac {\left (A d -B c \right ) \left (\frac {\left (b \left (x +\frac {c}{d}\right )^{2}-\frac {2 b c \left (x +\frac {c}{d}\right )}{d}+\frac {a \,d^{2}+b \,c^{2}}{d^{2}}\right )^{\frac {3}{2}}}{3}-\frac {b c \left (\frac {\left (2 b \left (x +\frac {c}{d}\right )-\frac {2 b c}{d}\right ) \sqrt {b \left (x +\frac {c}{d}\right )^{2}-\frac {2 b c \left (x +\frac {c}{d}\right )}{d}+\frac {a \,d^{2}+b \,c^{2}}{d^{2}}}}{4 b}+\frac {\left (\frac {4 b \left (a \,d^{2}+b \,c^{2}\right )}{d^{2}}-\frac {4 b^{2} c^{2}}{d^{2}}\right ) \ln \left (\frac {-\frac {b c}{d}+b \left (x +\frac {c}{d}\right )}{\sqrt {b}}+\sqrt {b \left (x +\frac {c}{d}\right )^{2}-\frac {2 b c \left (x +\frac {c}{d}\right )}{d}+\frac {a \,d^{2}+b \,c^{2}}{d^{2}}}\right )}{8 b^{\frac {3}{2}}}\right )}{d}+\frac {\left (a \,d^{2}+b \,c^{2}\right ) \left (\sqrt {b \left (x +\frac {c}{d}\right )^{2}-\frac {2 b c \left (x +\frac {c}{d}\right )}{d}+\frac {a \,d^{2}+b \,c^{2}}{d^{2}}}-\frac {\sqrt {b}\, c \ln \left (\frac {-\frac {b c}{d}+b \left (x +\frac {c}{d}\right )}{\sqrt {b}}+\sqrt {b \left (x +\frac {c}{d}\right )^{2}-\frac {2 b c \left (x +\frac {c}{d}\right )}{d}+\frac {a \,d^{2}+b \,c^{2}}{d^{2}}}\right )}{d}-\frac {\left (a \,d^{2}+b \,c^{2}\right ) \ln \left (\frac {\frac {2 a \,d^{2}+2 b \,c^{2}}{d^{2}}-\frac {2 b c \left (x +\frac {c}{d}\right )}{d}+2 \sqrt {\frac {a \,d^{2}+b \,c^{2}}{d^{2}}}\, \sqrt {b \left (x +\frac {c}{d}\right )^{2}-\frac {2 b c \left (x +\frac {c}{d}\right )}{d}+\frac {a \,d^{2}+b \,c^{2}}{d^{2}}}}{x +\frac {c}{d}}\right )}{d^{2} \sqrt {\frac {a \,d^{2}+b \,c^{2}}{d^{2}}}}\right )}{d^{2}}\right )}{d^{2}}\)	\(561\)

Input:

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

Output:

1/24*(6*B*b*d^3*x^3+8*A*b*d^3*x^2-8*B*b*c*d^2*x^2-12*A*b*c*d^2*x+15*B*a*d^ 
3*x+12*B*b*c^2*d*x+32*A*a*d^3+24*A*b*c^2*d-32*B*a*c*d^2-24*B*b*c^3)*(b*x^2 
+a)^(1/2)/d^4-1/8/d^4*((12*A*a*b*c*d^3+8*A*b^2*c^3*d-3*B*a^2*d^4-12*B*a*b* 
c^2*d^2-8*B*b^2*c^4)/d*ln(b^(1/2)*x+(b*x^2+a)^(1/2))/b^(1/2)+8*(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^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. \[ \int \frac {(A+B x) \left (a+b x^2\right )^{3/2}}{c+d x} \, dx=\text {Timed out} \] Input:

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

Output:

Timed out
 

Sympy [F]

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

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

Output:

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

Maxima [A] (verification not implemented)

Time = 0.09 (sec) , antiderivative size = 384, normalized size of antiderivative = 1.63 \[ \int \frac {(A+B x) \left (a+b x^2\right )^{3/2}}{c+d x} \, dx=\frac {\sqrt {b x^{2} + a} B b c^{2} x}{2 \, d^{3}} - \frac {\sqrt {b x^{2} + a} A b c x}{2 \, d^{2}} + \frac {{\left (b x^{2} + a\right )}^{\frac {3}{2}} B x}{4 \, d} + \frac {3 \, \sqrt {b x^{2} + a} B a x}{8 \, d} + \frac {B b^{\frac {3}{2}} c^{4} \operatorname {arsinh}\left (\frac {b x}{\sqrt {a b}}\right )}{d^{5}} - \frac {A b^{\frac {3}{2}} c^{3} \operatorname {arsinh}\left (\frac {b x}{\sqrt {a b}}\right )}{d^{4}} + \frac {3 \, B a \sqrt {b} c^{2} \operatorname {arsinh}\left (\frac {b x}{\sqrt {a b}}\right )}{2 \, d^{3}} - \frac {3 \, A a \sqrt {b} c \operatorname {arsinh}\left (\frac {b x}{\sqrt {a b}}\right )}{2 \, d^{2}} + \frac {3 \, B a^{2} \operatorname {arsinh}\left (\frac {b x}{\sqrt {a b}}\right )}{8 \, \sqrt {b} d} - \frac {B {\left (a + \frac {b c^{2}}{d^{2}}\right )}^{\frac {3}{2}} c \operatorname {arsinh}\left (\frac {b c x}{\sqrt {a b} {\left | d x + c \right |}} - \frac {a d}{\sqrt {a b} {\left | d x + c \right |}}\right )}{d^{2}} + \frac {A {\left (a + \frac {b c^{2}}{d^{2}}\right )}^{\frac {3}{2}} \operatorname {arsinh}\left (\frac {b c x}{\sqrt {a b} {\left | d x + c \right |}} - \frac {a d}{\sqrt {a b} {\left | d x + c \right |}}\right )}{d} - \frac {\sqrt {b x^{2} + a} B b c^{3}}{d^{4}} + \frac {\sqrt {b x^{2} + a} A b c^{2}}{d^{3}} - \frac {{\left (b x^{2} + a\right )}^{\frac {3}{2}} B c}{3 \, d^{2}} - \frac {\sqrt {b x^{2} + a} B a c}{d^{2}} + \frac {{\left (b x^{2} + a\right )}^{\frac {3}{2}} A}{3 \, d} + \frac {\sqrt {b x^{2} + a} A a}{d} \] Input:

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

Output:

1/2*sqrt(b*x^2 + a)*B*b*c^2*x/d^3 - 1/2*sqrt(b*x^2 + a)*A*b*c*x/d^2 + 1/4* 
(b*x^2 + a)^(3/2)*B*x/d + 3/8*sqrt(b*x^2 + a)*B*a*x/d + B*b^(3/2)*c^4*arcs 
inh(b*x/sqrt(a*b))/d^5 - A*b^(3/2)*c^3*arcsinh(b*x/sqrt(a*b))/d^4 + 3/2*B* 
a*sqrt(b)*c^2*arcsinh(b*x/sqrt(a*b))/d^3 - 3/2*A*a*sqrt(b)*c*arcsinh(b*x/s 
qrt(a*b))/d^2 + 3/8*B*a^2*arcsinh(b*x/sqrt(a*b))/(sqrt(b)*d) - B*(a + b*c^ 
2/d^2)^(3/2)*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)^(3/2)*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*c^3/d^4 + s 
qrt(b*x^2 + a)*A*b*c^2/d^3 - 1/3*(b*x^2 + a)^(3/2)*B*c/d^2 - sqrt(b*x^2 + 
a)*B*a*c/d^2 + 1/3*(b*x^2 + a)^(3/2)*A/d + sqrt(b*x^2 + a)*A*a/d
 

Giac [F(-2)]

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

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

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

Output:

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

Reduce [F]

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

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

Output:

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