\(\int \frac {A+B x}{(d+e x)^{3/2} (a-c x^2)} \, dx\) [127]

Optimal result

Integrand size = 25, antiderivative size = 197 \[ \int \frac {A+B x}{(d+e x)^{3/2} \left (a-c x^2\right )} \, dx=-\frac {2 (B d-A e)}{\left (c d^2-a e^2\right ) \sqrt {d+e x}}+\frac {\left (\sqrt {a} B-A \sqrt {c}\right ) \text {arctanh}\left (\frac {\sqrt [4]{c} \sqrt {d+e x}}{\sqrt {\sqrt {c} d-\sqrt {a} e}}\right )}{\sqrt {a} \sqrt [4]{c} \left (\sqrt {c} d-\sqrt {a} e\right )^{3/2}}+\frac {\left (\sqrt {a} B+A \sqrt {c}\right ) \text {arctanh}\left (\frac {\sqrt [4]{c} \sqrt {d+e x}}{\sqrt {\sqrt {c} d+\sqrt {a} e}}\right )}{\sqrt {a} \sqrt [4]{c} \left (\sqrt {c} d+\sqrt {a} e\right )^{3/2}} \] Output:

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

Mathematica [A] (verified)

Time = 0.94 (sec) , antiderivative size = 255, normalized size of antiderivative = 1.29 \[ \int \frac {A+B x}{(d+e x)^{3/2} \left (a-c x^2\right )} \, dx=\frac {-2 B d+2 A e}{\left (c d^2-a e^2\right ) \sqrt {d+e x}}+\frac {\left (\sqrt {a} B+A \sqrt {c}\right ) \arctan \left (\frac {\sqrt {-c d-\sqrt {a} \sqrt {c} e} \sqrt {d+e x}}{\sqrt {c} d+\sqrt {a} e}\right )}{\sqrt {a} \left (\sqrt {c} d+\sqrt {a} e\right ) \sqrt {-c d-\sqrt {a} \sqrt {c} e}}+\frac {\left (\sqrt {a} B-A \sqrt {c}\right ) \arctan \left (\frac {\sqrt {-c d+\sqrt {a} \sqrt {c} e} \sqrt {d+e x}}{\sqrt {c} d-\sqrt {a} e}\right )}{\sqrt {a} \left (\sqrt {c} d-\sqrt {a} e\right ) \sqrt {-c d+\sqrt {a} \sqrt {c} e}} \] Input:

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

Output:

(-2*B*d + 2*A*e)/((c*d^2 - a*e^2)*Sqrt[d + e*x]) + ((Sqrt[a]*B + A*Sqrt[c] 
)*ArcTan[(Sqrt[-(c*d) - Sqrt[a]*Sqrt[c]*e]*Sqrt[d + e*x])/(Sqrt[c]*d + Sqr 
t[a]*e)])/(Sqrt[a]*(Sqrt[c]*d + Sqrt[a]*e)*Sqrt[-(c*d) - Sqrt[a]*Sqrt[c]*e 
]) + ((Sqrt[a]*B - A*Sqrt[c])*ArcTan[(Sqrt[-(c*d) + Sqrt[a]*Sqrt[c]*e]*Sqr 
t[d + e*x])/(Sqrt[c]*d - Sqrt[a]*e)])/(Sqrt[a]*(Sqrt[c]*d - Sqrt[a]*e)*Sqr 
t[-(c*d) + Sqrt[a]*Sqrt[c]*e])
 

Rubi [A] (verified)

Time = 0.45 (sec) , antiderivative size = 251, normalized size of antiderivative = 1.27, number of steps used = 7, number of rules used = 6, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.240, Rules used = {655, 25, 654, 25, 1480, 221}

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.

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

\(\Big \downarrow \) 655

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

\(\Big \downarrow \) 25

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

\(\Big \downarrow \) 654

\(\displaystyle \frac {2 \int -\frac {2 A c d e-B \left (c d^2+a e^2\right )+c (B d-A e) (d+e x)}{c d^2-2 c (d+e x) d-a e^2+c (d+e x)^2}d\sqrt {d+e x}}{c d^2-a e^2}-\frac {2 (B d-A e)}{\sqrt {d+e x} \left (c d^2-a e^2\right )}\)

\(\Big \downarrow \) 25

\(\displaystyle -\frac {2 \int \frac {2 A c d e-B \left (c d^2+a e^2\right )+c (B d-A e) (d+e x)}{c d^2-2 c (d+e x) d-a e^2+c (d+e x)^2}d\sqrt {d+e x}}{c d^2-a e^2}-\frac {2 (B d-A e)}{\sqrt {d+e x} \left (c d^2-a e^2\right )}\)

\(\Big \downarrow \) 1480

\(\displaystyle \frac {2 \left (-\frac {\sqrt {c} \left (\sqrt {a} B-A \sqrt {c}\right ) \left (\sqrt {a} e+\sqrt {c} d\right ) \int \frac {1}{c (d+e x)-\sqrt {c} \left (\sqrt {c} d-\sqrt {a} e\right )}d\sqrt {d+e x}}{2 \sqrt {a}}-\frac {\sqrt {c} \left (\sqrt {a} B+A \sqrt {c}\right ) \left (\sqrt {c} d-\sqrt {a} e\right ) \int \frac {1}{c (d+e x)-\sqrt {c} \left (\sqrt {c} d+\sqrt {a} e\right )}d\sqrt {d+e x}}{2 \sqrt {a}}\right )}{c d^2-a e^2}-\frac {2 (B d-A e)}{\sqrt {d+e x} \left (c d^2-a e^2\right )}\)

\(\Big \downarrow \) 221

\(\displaystyle \frac {2 \left (\frac {\left (\sqrt {a} B-A \sqrt {c}\right ) \left (\sqrt {a} e+\sqrt {c} d\right ) \text {arctanh}\left (\frac {\sqrt [4]{c} \sqrt {d+e x}}{\sqrt {\sqrt {c} d-\sqrt {a} e}}\right )}{2 \sqrt {a} \sqrt [4]{c} \sqrt {\sqrt {c} d-\sqrt {a} e}}+\frac {\left (\sqrt {a} B+A \sqrt {c}\right ) \left (\sqrt {c} d-\sqrt {a} e\right ) \text {arctanh}\left (\frac {\sqrt [4]{c} \sqrt {d+e x}}{\sqrt {\sqrt {a} e+\sqrt {c} d}}\right )}{2 \sqrt {a} \sqrt [4]{c} \sqrt {\sqrt {a} e+\sqrt {c} d}}\right )}{c d^2-a e^2}-\frac {2 (B d-A e)}{\sqrt {d+e x} \left (c d^2-a e^2\right )}\)

Input:

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

Output:

(-2*(B*d - A*e))/((c*d^2 - a*e^2)*Sqrt[d + e*x]) + (2*(((Sqrt[a]*B - A*Sqr 
t[c])*(Sqrt[c]*d + Sqrt[a]*e)*ArcTanh[(c^(1/4)*Sqrt[d + e*x])/Sqrt[Sqrt[c] 
*d - Sqrt[a]*e]])/(2*Sqrt[a]*c^(1/4)*Sqrt[Sqrt[c]*d - Sqrt[a]*e]) + ((Sqrt 
[a]*B + A*Sqrt[c])*(Sqrt[c]*d - Sqrt[a]*e)*ArcTanh[(c^(1/4)*Sqrt[d + e*x]) 
/Sqrt[Sqrt[c]*d + Sqrt[a]*e]])/(2*Sqrt[a]*c^(1/4)*Sqrt[Sqrt[c]*d + Sqrt[a] 
*e])))/(c*d^2 - a*e^2)
 

Defintions of rubi rules used

rule 25

Int[-(Fx_), x_Symbol] :> Simp[Identity[-1]   Int[Fx, x], x]
 

rule 221

Int[((a_) + (b_.)*(x_)^2)^(-1), x_Symbol] :> Simp[(Rt[-a/b, 2]/a)*ArcTanh[x 
/Rt[-a/b, 2]], x] /; FreeQ[{a, b}, x] && NegQ[a/b]
 

rule 654

Int[((f_.) + (g_.)*(x_))/(Sqrt[(d_.) + (e_.)*(x_)]*((a_) + (c_.)*(x_)^2)), 
x_Symbol] :> Simp[2   Subst[Int[(e*f - d*g + g*x^2)/(c*d^2 + a*e^2 - 2*c*d* 
x^2 + c*x^4), x], x, Sqrt[d + e*x]], x] /; FreeQ[{a, c, d, e, f, g}, x]
 

rule 655

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

rule 1480

Int[((d_) + (e_.)*(x_)^2)/((a_) + (b_.)*(x_)^2 + (c_.)*(x_)^4), x_Symbol] : 
> With[{q = Rt[b^2 - 4*a*c, 2]}, Simp[(e/2 + (2*c*d - b*e)/(2*q))   Int[1/( 
b/2 - q/2 + c*x^2), x], x] + Simp[(e/2 - (2*c*d - b*e)/(2*q))   Int[1/(b/2 
+ q/2 + c*x^2), x], x]] /; FreeQ[{a, b, c, d, e}, x] && NeQ[b^2 - 4*a*c, 0] 
 && NeQ[c*d^2 - a*e^2, 0] && PosQ[b^2 - 4*a*c]
 

Maple [A] (verified)

Time = 1.44 (sec) , antiderivative size = 229, normalized size of antiderivative = 1.16

method	result	size
derivativedivides	\(-\frac {2 c \left (-\frac {\left (-A c d e +B a \,e^{2}+A \sqrt {a c \,e^{2}}\, e -B \sqrt {a c \,e^{2}}\, d \right ) \operatorname {arctanh}\left (\frac {c \sqrt {e x +d}}{\sqrt {\left (c d +\sqrt {a c \,e^{2}}\right ) c}}\right )}{2 \sqrt {a c \,e^{2}}\, \sqrt {\left (c d +\sqrt {a c \,e^{2}}\right ) c}}+\frac {\left (A c d e -B a \,e^{2}+A \sqrt {a c \,e^{2}}\, e -B \sqrt {a c \,e^{2}}\, d \right ) \arctan \left (\frac {c \sqrt {e x +d}}{\sqrt {\left (-c d +\sqrt {a c \,e^{2}}\right ) c}}\right )}{2 \sqrt {a c \,e^{2}}\, \sqrt {\left (-c d +\sqrt {a c \,e^{2}}\right ) c}}\right )}{a \,e^{2}-c \,d^{2}}+\frac {-2 A e +2 B d}{\left (a \,e^{2}-c \,d^{2}\right ) \sqrt {e x +d}}\)	\(229\)
default	\(\frac {2 c \left (-\frac {\left (A c d e -B a \,e^{2}-A \sqrt {a c \,e^{2}}\, e +B \sqrt {a c \,e^{2}}\, d \right ) \operatorname {arctanh}\left (\frac {c \sqrt {e x +d}}{\sqrt {\left (c d +\sqrt {a c \,e^{2}}\right ) c}}\right )}{2 \sqrt {a c \,e^{2}}\, \sqrt {\left (c d +\sqrt {a c \,e^{2}}\right ) c}}+\frac {\left (-A c d e +B a \,e^{2}-A \sqrt {a c \,e^{2}}\, e +B \sqrt {a c \,e^{2}}\, d \right ) \arctan \left (\frac {c \sqrt {e x +d}}{\sqrt {\left (-c d +\sqrt {a c \,e^{2}}\right ) c}}\right )}{2 \sqrt {a c \,e^{2}}\, \sqrt {\left (-c d +\sqrt {a c \,e^{2}}\right ) c}}\right )}{a \,e^{2}-c \,d^{2}}-\frac {2 \left (A e -B d \right )}{\left (a \,e^{2}-c \,d^{2}\right ) \sqrt {e x +d}}\)	\(229\)
pseudoelliptic	\(-\frac {2 \left (\frac {c \left (\left (A e -B d \right ) \sqrt {a c \,e^{2}}+e \left (A c d -B a e \right )\right ) \sqrt {\left (c d +\sqrt {a c \,e^{2}}\right ) c}\, \sqrt {e x +d}\, \arctan \left (\frac {c \sqrt {e x +d}}{\sqrt {\left (-c d +\sqrt {a c \,e^{2}}\right ) c}}\right )}{2}+\left (\frac {c \left (\left (-A e +B d \right ) \sqrt {a c \,e^{2}}+e \left (A c d -B a e \right )\right ) \sqrt {e x +d}\, \operatorname {arctanh}\left (\frac {c \sqrt {e x +d}}{\sqrt {\left (c d +\sqrt {a c \,e^{2}}\right ) c}}\right )}{2}+\left (A e -B d \right ) \sqrt {\left (c d +\sqrt {a c \,e^{2}}\right ) c}\, \sqrt {a c \,e^{2}}\right ) \sqrt {\left (-c d +\sqrt {a c \,e^{2}}\right ) c}\right )}{\sqrt {\left (-c d +\sqrt {a c \,e^{2}}\right ) c}\, \sqrt {a c \,e^{2}}\, \sqrt {\left (c d +\sqrt {a c \,e^{2}}\right ) c}\, \sqrt {e x +d}\, \left (a \,e^{2}-c \,d^{2}\right )}\)	\(267\)

Input:

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

Output:

-2/(a*e^2-c*d^2)*c*(-1/2*(-A*c*d*e+B*a*e^2+A*(a*c*e^2)^(1/2)*e-B*(a*c*e^2) 
^(1/2)*d)/(a*c*e^2)^(1/2)/((c*d+(a*c*e^2)^(1/2))*c)^(1/2)*arctanh(c*(e*x+d 
)^(1/2)/((c*d+(a*c*e^2)^(1/2))*c)^(1/2))+1/2*(A*c*d*e-B*a*e^2+A*(a*c*e^2)^ 
(1/2)*e-B*(a*c*e^2)^(1/2)*d)/(a*c*e^2)^(1/2)/((-c*d+(a*c*e^2)^(1/2))*c)^(1 
/2)*arctan(c*(e*x+d)^(1/2)/((-c*d+(a*c*e^2)^(1/2))*c)^(1/2)))+2*(-A*e+B*d) 
/(a*e^2-c*d^2)/(e*x+d)^(1/2)
 

Fricas [B] (verification not implemented)

Leaf count of result is larger than twice the leaf count of optimal. 6448 vs. \(2 (147) = 294\).

Time = 3.45 (sec) , antiderivative size = 6448, normalized size of antiderivative = 32.73 \[ \int \frac {A+B x}{(d+e x)^{3/2} \left (a-c x^2\right )} \, dx=\text {Too large to display} \] Input:

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

Output:

Too large to include
 

Sympy [F]

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

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

Output:

-Integral(A/(-a*d*sqrt(d + e*x) - a*e*x*sqrt(d + e*x) + c*d*x**2*sqrt(d + 
e*x) + c*e*x**3*sqrt(d + e*x)), x) - Integral(B*x/(-a*d*sqrt(d + e*x) - a* 
e*x*sqrt(d + e*x) + c*d*x**2*sqrt(d + e*x) + c*e*x**3*sqrt(d + e*x)), x)
 

Maxima [F]

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

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

Output:

-integrate((B*x + A)/((c*x^2 - a)*(e*x + d)^(3/2)), x)
 

Giac [B] (verification not implemented)

Leaf count of result is larger than twice the leaf count of optimal. 941 vs. \(2 (147) = 294\).

Time = 0.24 (sec) , antiderivative size = 941, normalized size of antiderivative = 4.78 \[ \int \frac {A+B x}{(d+e x)^{3/2} \left (a-c x^2\right )} \, dx =\text {Too large to display} \] Input:

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

Output:

-2*(B*d - A*e)/((c*d^2 - a*e^2)*sqrt(e*x + d)) + ((c*d^2*e - a*e^3)^2*sqrt 
(a*c)*B*a*d*abs(c) - (c*d^2*e - a*e^3)^2*sqrt(a*c)*A*a*e*abs(c) + 2*(a*c^2 
*d^3*e - a^2*c*d*e^3)*A*abs(c*d^2*e - a*e^3)*abs(c) - (a*c^2*d^4 - a^3*e^4 
)*B*abs(c*d^2*e - a*e^3)*abs(c) - (sqrt(a*c)*c^3*d^6*e - 2*sqrt(a*c)*a*c^2 
*d^4*e^3 + sqrt(a*c)*a^2*c*d^2*e^5)*A*abs(c) + (sqrt(a*c)*a*c^2*d^5*e^2 - 
2*sqrt(a*c)*a^2*c*d^3*e^4 + sqrt(a*c)*a^3*d*e^6)*B*abs(c))*arctan(sqrt(e*x 
 + d)/sqrt(-(c^2*d^3 - a*c*d*e^2 + sqrt((c^2*d^3 - a*c*d*e^2)^2 - (c^2*d^4 
 - 2*a*c*d^2*e^2 + a^2*e^4)*(c^2*d^2 - a*c*e^2)))/(c^2*d^2 - a*c*e^2)))/(( 
a*c^3*d^5 - 2*a^2*c^2*d^3*e^2 + a^3*c*d*e^4 - sqrt(a*c)*a*c^2*d^4*e + 2*sq 
rt(a*c)*a^2*c*d^2*e^3 - sqrt(a*c)*a^3*e^5)*sqrt(-c^2*d - sqrt(a*c)*c*e)*ab 
s(c*d^2*e - a*e^3)) - ((c*d^2*e - a*e^3)^2*sqrt(a*c)*B*a*d*abs(c) - (c*d^2 
*e - a*e^3)^2*sqrt(a*c)*A*a*e*abs(c) - 2*(a*c^2*d^3*e - a^2*c*d*e^3)*A*abs 
(c*d^2*e - a*e^3)*abs(c) + (a*c^2*d^4 - a^3*e^4)*B*abs(c*d^2*e - a*e^3)*ab 
s(c) - (sqrt(a*c)*c^3*d^6*e - 2*sqrt(a*c)*a*c^2*d^4*e^3 + sqrt(a*c)*a^2*c* 
d^2*e^5)*A*abs(c) + (sqrt(a*c)*a*c^2*d^5*e^2 - 2*sqrt(a*c)*a^2*c*d^3*e^4 + 
 sqrt(a*c)*a^3*d*e^6)*B*abs(c))*arctan(sqrt(e*x + d)/sqrt(-(c^2*d^3 - a*c* 
d*e^2 - sqrt((c^2*d^3 - a*c*d*e^2)^2 - (c^2*d^4 - 2*a*c*d^2*e^2 + a^2*e^4) 
*(c^2*d^2 - a*c*e^2)))/(c^2*d^2 - a*c*e^2)))/((a*c^3*d^5 - 2*a^2*c^2*d^3*e 
^2 + a^3*c*d*e^4 + sqrt(a*c)*a*c^2*d^4*e - 2*sqrt(a*c)*a^2*c*d^2*e^3 + sqr 
t(a*c)*a^3*e^5)*sqrt(-c^2*d + sqrt(a*c)*c*e)*abs(c*d^2*e - a*e^3))
 

Mupad [B] (verification not implemented)

Time = 8.59 (sec) , antiderivative size = 10288, normalized size of antiderivative = 52.22 \[ \int \frac {A+B x}{(d+e x)^{3/2} \left (a-c x^2\right )} \, dx=\text {Too large to display} \] Input:

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

Output:

atan((((-(B^2*a^2*c^2*d^3 + B^2*a^2*e^3*(a^3*c)^(1/2) + A^2*a*c^3*d^3 - 2* 
A*B*c^2*d^3*(a^3*c)^(1/2) + 3*B^2*a^3*c*d*e^2 + A^2*a*c*e^3*(a^3*c)^(1/2) 
+ 3*A^2*a^2*c^2*d*e^2 - 2*A*B*a^3*c*e^3 + 3*A^2*c^2*d^2*e*(a^3*c)^(1/2) - 
6*A*B*a^2*c^2*d^2*e + 3*B^2*a*c*d^2*e*(a^3*c)^(1/2) - 6*A*B*a*c*d*e^2*(a^3 
*c)^(1/2))/(4*(a^5*c*e^6 - a^2*c^4*d^6 + 3*a^3*c^3*d^4*e^2 - 3*a^4*c^2*d^2 
*e^4)))^(1/2)*((d + e*x)^(1/2)*(-(B^2*a^2*c^2*d^3 + B^2*a^2*e^3*(a^3*c)^(1 
/2) + A^2*a*c^3*d^3 - 2*A*B*c^2*d^3*(a^3*c)^(1/2) + 3*B^2*a^3*c*d*e^2 + A^ 
2*a*c*e^3*(a^3*c)^(1/2) + 3*A^2*a^2*c^2*d*e^2 - 2*A*B*a^3*c*e^3 + 3*A^2*c^ 
2*d^2*e*(a^3*c)^(1/2) - 6*A*B*a^2*c^2*d^2*e + 3*B^2*a*c*d^2*e*(a^3*c)^(1/2 
) - 6*A*B*a*c*d*e^2*(a^3*c)^(1/2))/(4*(a^5*c*e^6 - a^2*c^4*d^6 + 3*a^3*c^3 
*d^4*e^2 - 3*a^4*c^2*d^2*e^4)))^(1/2)*(64*a*c^9*d^11*e^2 - 64*a^6*c^4*d*e^ 
12 - 320*a^2*c^8*d^9*e^4 + 640*a^3*c^7*d^7*e^6 - 640*a^4*c^6*d^5*e^8 + 320 
*a^5*c^5*d^3*e^10) - 32*B*a^6*c^3*e^12 + 64*A*a*c^8*d^9*e^3 + 64*A*a^5*c^4 
*d*e^11 - 32*B*a*c^8*d^10*e^2 - 256*A*a^2*c^7*d^7*e^5 + 384*A*a^3*c^6*d^5* 
e^7 - 256*A*a^4*c^5*d^3*e^9 + 96*B*a^2*c^7*d^8*e^4 - 64*B*a^3*c^6*d^6*e^6 
- 64*B*a^4*c^5*d^4*e^8 + 96*B*a^5*c^4*d^2*e^10) + (d + e*x)^(1/2)*(16*A^2* 
a^4*c^4*e^10 + 16*B^2*a^5*c^3*e^10 - 16*A^2*c^8*d^8*e^2 - 32*A^2*a^3*c^5*d 
^2*e^8 + 32*B^2*a^2*c^6*d^6*e^4 - 32*B^2*a^4*c^4*d^2*e^8 + 32*A^2*a*c^7*d^ 
6*e^4 - 16*B^2*a*c^7*d^8*e^2 + 64*A*B*a*c^7*d^7*e^3 - 64*A*B*a^4*c^4*d*e^9 
 - 192*A*B*a^2*c^6*d^5*e^5 + 192*A*B*a^3*c^5*d^3*e^7))*(-(B^2*a^2*c^2*d...
 

Reduce [B] (verification not implemented)

Time = 0.31 (sec) , antiderivative size = 983, normalized size of antiderivative = 4.99 \[ \int \frac {A+B x}{(d+e x)^{3/2} \left (a-c x^2\right )} \, dx =\text {Too large to display} \] Input:

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

Output:

( - 2*sqrt(a)*sqrt(d + e*x)*sqrt(sqrt(c)*sqrt(a)*e - c*d)*atan((sqrt(d + e 
*x)*c)/(sqrt(c)*sqrt(sqrt(c)*sqrt(a)*e - c*d)))*a*c*e**2 + 4*sqrt(a)*sqrt( 
d + e*x)*sqrt(sqrt(c)*sqrt(a)*e - c*d)*atan((sqrt(d + e*x)*c)/(sqrt(c)*sqr 
t(sqrt(c)*sqrt(a)*e - c*d)))*b*c*d*e - 2*sqrt(a)*sqrt(d + e*x)*sqrt(sqrt(c 
)*sqrt(a)*e - c*d)*atan((sqrt(d + e*x)*c)/(sqrt(c)*sqrt(sqrt(c)*sqrt(a)*e 
- c*d)))*c**2*d**2 + 2*sqrt(c)*sqrt(d + e*x)*sqrt(sqrt(c)*sqrt(a)*e - c*d) 
*atan((sqrt(d + e*x)*c)/(sqrt(c)*sqrt(sqrt(c)*sqrt(a)*e - c*d)))*a*b*e**2 
- 4*sqrt(c)*sqrt(d + e*x)*sqrt(sqrt(c)*sqrt(a)*e - c*d)*atan((sqrt(d + e*x 
)*c)/(sqrt(c)*sqrt(sqrt(c)*sqrt(a)*e - c*d)))*a*c*d*e + 2*sqrt(c)*sqrt(d + 
 e*x)*sqrt(sqrt(c)*sqrt(a)*e - c*d)*atan((sqrt(d + e*x)*c)/(sqrt(c)*sqrt(s 
qrt(c)*sqrt(a)*e - c*d)))*b*c*d**2 - sqrt(a)*sqrt(d + e*x)*sqrt(sqrt(c)*sq 
rt(a)*e + c*d)*log( - sqrt(sqrt(c)*sqrt(a)*e + c*d) + sqrt(c)*sqrt(d + e*x 
))*a*c*e**2 + 2*sqrt(a)*sqrt(d + e*x)*sqrt(sqrt(c)*sqrt(a)*e + c*d)*log( - 
 sqrt(sqrt(c)*sqrt(a)*e + c*d) + sqrt(c)*sqrt(d + e*x))*b*c*d*e - sqrt(a)* 
sqrt(d + e*x)*sqrt(sqrt(c)*sqrt(a)*e + c*d)*log( - sqrt(sqrt(c)*sqrt(a)*e 
+ c*d) + sqrt(c)*sqrt(d + e*x))*c**2*d**2 + sqrt(a)*sqrt(d + e*x)*sqrt(sqr 
t(c)*sqrt(a)*e + c*d)*log(sqrt(sqrt(c)*sqrt(a)*e + c*d) + sqrt(c)*sqrt(d + 
 e*x))*a*c*e**2 - 2*sqrt(a)*sqrt(d + e*x)*sqrt(sqrt(c)*sqrt(a)*e + c*d)*lo 
g(sqrt(sqrt(c)*sqrt(a)*e + c*d) + sqrt(c)*sqrt(d + e*x))*b*c*d*e + sqrt(a) 
*sqrt(d + e*x)*sqrt(sqrt(c)*sqrt(a)*e + c*d)*log(sqrt(sqrt(c)*sqrt(a)*e...