\(\int \frac {1}{(c+d x)^{5/2} (a-b x^2)} \, dx\) [601]

Optimal result

Integrand size = 20, antiderivative size = 190 \[ \int \frac {1}{(c+d x)^{5/2} \left (a-b x^2\right )} \, dx=\frac {2 d}{3 \left (b c^2-a d^2\right ) (c+d x)^{3/2}}+\frac {4 b c d}{\left (b c^2-a d^2\right )^2 \sqrt {c+d x}}-\frac {b^{3/4} \text {arctanh}\left (\frac {\sqrt [4]{b} \sqrt {c+d x}}{\sqrt {\sqrt {b} c-\sqrt {a} d}}\right )}{\sqrt {a} \left (\sqrt {b} c-\sqrt {a} d\right )^{5/2}}+\frac {b^{3/4} \text {arctanh}\left (\frac {\sqrt [4]{b} \sqrt {c+d x}}{\sqrt {\sqrt {b} c+\sqrt {a} d}}\right )}{\sqrt {a} \left (\sqrt {b} c+\sqrt {a} d\right )^{5/2}} \] Output:

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

Mathematica [A] (verified)

Time = 0.72 (sec) , antiderivative size = 241, normalized size of antiderivative = 1.27 \[ \int \frac {1}{(c+d x)^{5/2} \left (a-b x^2\right )} \, dx=\frac {-2 a d^3+2 b c d (7 c+6 d x)}{3 \left (b c^2-a d^2\right )^2 (c+d x)^{3/2}}+\frac {b \arctan \left (\frac {\sqrt {-b c-\sqrt {a} \sqrt {b} d} \sqrt {c+d x}}{\sqrt {b} c+\sqrt {a} d}\right )}{\sqrt {a} \left (\sqrt {b} c+\sqrt {a} d\right )^2 \sqrt {-b c-\sqrt {a} \sqrt {b} d}}-\frac {b \arctan \left (\frac {\sqrt {-b c+\sqrt {a} \sqrt {b} d} \sqrt {c+d x}}{\sqrt {b} c-\sqrt {a} d}\right )}{\sqrt {a} \left (\sqrt {b} c-\sqrt {a} d\right )^2 \sqrt {-b c+\sqrt {a} \sqrt {b} d}} \] Input:

Integrate[1/((c + d*x)^(5/2)*(a - b*x^2)),x]
 

Output:

(-2*a*d^3 + 2*b*c*d*(7*c + 6*d*x))/(3*(b*c^2 - a*d^2)^2*(c + d*x)^(3/2)) + 
 (b*ArcTan[(Sqrt[-(b*c) - Sqrt[a]*Sqrt[b]*d]*Sqrt[c + d*x])/(Sqrt[b]*c + S 
qrt[a]*d)])/(Sqrt[a]*(Sqrt[b]*c + Sqrt[a]*d)^2*Sqrt[-(b*c) - Sqrt[a]*Sqrt[ 
b]*d]) - (b*ArcTan[(Sqrt[-(b*c) + Sqrt[a]*Sqrt[b]*d]*Sqrt[c + d*x])/(Sqrt[ 
b]*c - Sqrt[a]*d)])/(Sqrt[a]*(Sqrt[b]*c - Sqrt[a]*d)^2*Sqrt[-(b*c) + Sqrt[ 
a]*Sqrt[b]*d])
 

Rubi [A] (verified)

Time = 0.80 (sec) , antiderivative size = 270, normalized size of antiderivative = 1.42, number of steps used = 9, number of rules used = 8, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.400, Rules used = {482, 655, 25, 654, 25, 27, 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.

\(\displaystyle \int \frac {1}{\left (a-b x^2\right ) (c+d x)^{5/2}} \, dx\)

\(\Big \downarrow \) 482

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

\(\Big \downarrow \) 655

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

\(\Big \downarrow \) 25

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

\(\Big \downarrow \) 654

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

\(\Big \downarrow \) 25

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

\(\Big \downarrow \) 27

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

\(\Big \downarrow \) 1480

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

\(\Big \downarrow \) 221

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

Input:

Int[1/((c + d*x)^(5/2)*(a - b*x^2)),x]
 

Output:

(2*d)/(3*(b*c^2 - a*d^2)*(c + d*x)^(3/2)) + (b*((4*c*d)/((b*c^2 - a*d^2)*S 
qrt[c + d*x]) - (2*d*(((Sqrt[b]*c + Sqrt[a]*d)^2*ArcTanh[(b^(1/4)*Sqrt[c + 
 d*x])/Sqrt[Sqrt[b]*c - Sqrt[a]*d]])/(2*Sqrt[a]*b^(1/4)*d*Sqrt[Sqrt[b]*c - 
 Sqrt[a]*d]) - ((Sqrt[b]*c - Sqrt[a]*d)^2*ArcTanh[(b^(1/4)*Sqrt[c + d*x])/ 
Sqrt[Sqrt[b]*c + Sqrt[a]*d]])/(2*Sqrt[a]*b^(1/4)*d*Sqrt[Sqrt[b]*c + Sqrt[a 
]*d])))/(b*c^2 - a*d^2)))/(b*c^2 - a*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 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 482

Int[((c_) + (d_.)*(x_))^(n_)/((a_) + (b_.)*(x_)^2), x_Symbol] :> Simp[d*((c 
 + d*x)^(n + 1)/((n + 1)*(b*c^2 + a*d^2))), x] + Simp[b/(b*c^2 + a*d^2)   I 
nt[(c + d*x)^(n + 1)*((c - d*x)/(a + b*x^2)), x], x] /; FreeQ[{a, b, c, d, 
n}, x] && LtQ[n, -1]
 

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 = 0.56 (sec) , antiderivative size = 224, normalized size of antiderivative = 1.18

method	result	size
derivativedivides	\(-2 d \left (\frac {b^{2} \left (-\frac {\left (a \,d^{2}+b \,c^{2}-2 \sqrt {a b \,d^{2}}\, c \right ) \operatorname {arctanh}\left (\frac {b \sqrt {d x +c}}{\sqrt {\left (b c +\sqrt {a b \,d^{2}}\right ) b}}\right )}{2 \sqrt {a b \,d^{2}}\, \sqrt {\left (b c +\sqrt {a b \,d^{2}}\right ) b}}+\frac {\left (-a \,d^{2}-b \,c^{2}-2 \sqrt {a b \,d^{2}}\, c \right ) \arctan \left (\frac {b \sqrt {d x +c}}{\sqrt {\left (-b c +\sqrt {a b \,d^{2}}\right ) b}}\right )}{2 \sqrt {a b \,d^{2}}\, \sqrt {\left (-b c +\sqrt {a b \,d^{2}}\right ) b}}\right )}{\left (a \,d^{2}-b \,c^{2}\right )^{2}}+\frac {1}{3 \left (a \,d^{2}-b \,c^{2}\right ) \left (d x +c \right )^{\frac {3}{2}}}-\frac {2 b c}{\left (a \,d^{2}-b \,c^{2}\right )^{2} \sqrt {d x +c}}\right )\)	\(224\)
default	\(2 d \left (-\frac {b^{2} \left (-\frac {\left (a \,d^{2}+b \,c^{2}-2 \sqrt {a b \,d^{2}}\, c \right ) \operatorname {arctanh}\left (\frac {b \sqrt {d x +c}}{\sqrt {\left (b c +\sqrt {a b \,d^{2}}\right ) b}}\right )}{2 \sqrt {a b \,d^{2}}\, \sqrt {\left (b c +\sqrt {a b \,d^{2}}\right ) b}}+\frac {\left (-a \,d^{2}-b \,c^{2}-2 \sqrt {a b \,d^{2}}\, c \right ) \arctan \left (\frac {b \sqrt {d x +c}}{\sqrt {\left (-b c +\sqrt {a b \,d^{2}}\right ) b}}\right )}{2 \sqrt {a b \,d^{2}}\, \sqrt {\left (-b c +\sqrt {a b \,d^{2}}\right ) b}}\right )}{\left (a \,d^{2}-b \,c^{2}\right )^{2}}-\frac {1}{3 \left (a \,d^{2}-b \,c^{2}\right ) \left (d x +c \right )^{\frac {3}{2}}}+\frac {2 b c}{\left (a \,d^{2}-b \,c^{2}\right )^{2} \sqrt {d x +c}}\right )\)	\(225\)
pseudoelliptic	\(\frac {d \left (\left (a \,d^{2}+b \,c^{2}+2 \sqrt {a b \,d^{2}}\, c \right ) b^{2} \left (d x +c \right )^{\frac {3}{2}} \sqrt {\left (b c +\sqrt {a b \,d^{2}}\right ) b}\, \arctan \left (\frac {b \sqrt {d x +c}}{\sqrt {\left (-b c +\sqrt {a b \,d^{2}}\right ) b}}\right )+\sqrt {\left (-b c +\sqrt {a b \,d^{2}}\right ) b}\, \left (\left (a \,d^{2}+b \,c^{2}-2 \sqrt {a b \,d^{2}}\, c \right ) b^{2} \left (d x +c \right )^{\frac {3}{2}} \operatorname {arctanh}\left (\frac {b \sqrt {d x +c}}{\sqrt {\left (b c +\sqrt {a b \,d^{2}}\right ) b}}\right )-\frac {2 \sqrt {a b \,d^{2}}\, \left (\left (-6 c d x -7 c^{2}\right ) b +a \,d^{2}\right ) \sqrt {\left (b c +\sqrt {a b \,d^{2}}\right ) b}}{3}\right )\right )}{\sqrt {a b \,d^{2}}\, \left (d x +c \right )^{\frac {3}{2}} \sqrt {\left (-b c +\sqrt {a b \,d^{2}}\right ) b}\, \sqrt {\left (b c +\sqrt {a b \,d^{2}}\right ) b}\, \left (a \,d^{2}-b \,c^{2}\right )^{2}}\)	\(265\)

Input:

int(1/(d*x+c)^(5/2)/(-b*x^2+a),x,method=_RETURNVERBOSE)
 

Output:

-2*d*(b^2/(a*d^2-b*c^2)^2*(-1/2*(a*d^2+b*c^2-2*(a*b*d^2)^(1/2)*c)/(a*b*d^2 
)^(1/2)/((b*c+(a*b*d^2)^(1/2))*b)^(1/2)*arctanh(b*(d*x+c)^(1/2)/((b*c+(a*b 
*d^2)^(1/2))*b)^(1/2))+1/2*(-a*d^2-b*c^2-2*(a*b*d^2)^(1/2)*c)/(a*b*d^2)^(1 
/2)/((-b*c+(a*b*d^2)^(1/2))*b)^(1/2)*arctan(b*(d*x+c)^(1/2)/((-b*c+(a*b*d^ 
2)^(1/2))*b)^(1/2)))+1/3/(a*d^2-b*c^2)/(d*x+c)^(3/2)-2*b*c/(a*d^2-b*c^2)^2 
/(d*x+c)^(1/2))
 

Fricas [B] (verification not implemented)

Leaf count of result is larger than twice the leaf count of optimal. 5142 vs. \(2 (144) = 288\).

Time = 0.21 (sec) , antiderivative size = 5142, normalized size of antiderivative = 27.06 \[ \int \frac {1}{(c+d x)^{5/2} \left (a-b x^2\right )} \, dx=\text {Too large to display} \] Input:

integrate(1/(d*x+c)^(5/2)/(-b*x^2+a),x, algorithm="fricas")
 

Output:

Too large to include
 

Sympy [F]

\[ \int \frac {1}{(c+d x)^{5/2} \left (a-b x^2\right )} \, dx=- \int \frac {1}{- a c^{2} \sqrt {c + d x} - 2 a c d x \sqrt {c + d x} - a d^{2} x^{2} \sqrt {c + d x} + b c^{2} x^{2} \sqrt {c + d x} + 2 b c d x^{3} \sqrt {c + d x} + b d^{2} x^{4} \sqrt {c + d x}}\, dx \] Input:

integrate(1/(d*x+c)**(5/2)/(-b*x**2+a),x)
 

Output:

-Integral(1/(-a*c**2*sqrt(c + d*x) - 2*a*c*d*x*sqrt(c + d*x) - a*d**2*x**2 
*sqrt(c + d*x) + b*c**2*x**2*sqrt(c + d*x) + 2*b*c*d*x**3*sqrt(c + d*x) + 
b*d**2*x**4*sqrt(c + d*x)), x)
 

Maxima [F]

\[ \int \frac {1}{(c+d x)^{5/2} \left (a-b x^2\right )} \, dx=\int { -\frac {1}{{\left (b x^{2} - a\right )} {\left (d x + c\right )}^{\frac {5}{2}}} \,d x } \] Input:

integrate(1/(d*x+c)^(5/2)/(-b*x^2+a),x, algorithm="maxima")
 

Output:

-integrate(1/((b*x^2 - a)*(d*x + c)^(5/2)), x)
 

Giac [B] (verification not implemented)

Leaf count of result is larger than twice the leaf count of optimal. 1165 vs. \(2 (144) = 288\).

Time = 0.20 (sec) , antiderivative size = 1165, normalized size of antiderivative = 6.13 \[ \int \frac {1}{(c+d x)^{5/2} \left (a-b x^2\right )} \, dx =\text {Too large to display} \] Input:

integrate(1/(d*x+c)^(5/2)/(-b*x^2+a),x, algorithm="giac")
 

Output:

(2*(b^2*c^4*d - 2*a*b*c^2*d^3 + a^2*d^5)^2*sqrt(-b^2*c - sqrt(a*b)*b*d)*sq 
rt(a*b)*a*c*d*abs(b) - (3*a*b^3*c^6*d - 5*a^2*b^2*c^4*d^3 + a^3*b*c^2*d^5 
+ a^4*d^7)*sqrt(-b^2*c - sqrt(a*b)*b*d)*abs(b^2*c^4*d - 2*a*b*c^2*d^3 + a^ 
2*d^5)*abs(b) + (sqrt(a*b)*b^5*c^11*d - 3*sqrt(a*b)*a*b^4*c^9*d^3 + 2*sqrt 
(a*b)*a^2*b^3*c^7*d^5 + 2*sqrt(a*b)*a^3*b^2*c^5*d^7 - 3*sqrt(a*b)*a^4*b*c^ 
3*d^9 + sqrt(a*b)*a^5*c*d^11)*sqrt(-b^2*c - sqrt(a*b)*b*d)*abs(b))*arctan( 
sqrt(d*x + c)/sqrt(-(b^3*c^5 - 2*a*b^2*c^3*d^2 + a^2*b*c*d^4 + sqrt((b^3*c 
^5 - 2*a*b^2*c^3*d^2 + a^2*b*c*d^4)^2 - (b^3*c^6 - 3*a*b^2*c^4*d^2 + 3*a^2 
*b*c^2*d^4 - a^3*d^6)*(b^3*c^4 - 2*a*b^2*c^2*d^2 + a^2*b*d^4)))/(b^3*c^4 - 
 2*a*b^2*c^2*d^2 + a^2*b*d^4)))/((a*b^6*c^10 - 5*a^2*b^5*c^8*d^2 + 10*a^3* 
b^4*c^6*d^4 - 10*a^4*b^3*c^4*d^6 + 5*a^5*b^2*c^2*d^8 - a^6*b*d^10)*abs(b^2 
*c^4*d - 2*a*b*c^2*d^3 + a^2*d^5)) - (2*(b^2*c^4*d - 2*a*b*c^2*d^3 + a^2*d 
^5)^2*sqrt(-b^2*c + sqrt(a*b)*b*d)*sqrt(a*b)*a*c*d*abs(b) + (3*a*b^3*c^6*d 
 - 5*a^2*b^2*c^4*d^3 + a^3*b*c^2*d^5 + a^4*d^7)*sqrt(-b^2*c + sqrt(a*b)*b* 
d)*abs(b^2*c^4*d - 2*a*b*c^2*d^3 + a^2*d^5)*abs(b) + (sqrt(a*b)*b^5*c^11*d 
 - 3*sqrt(a*b)*a*b^4*c^9*d^3 + 2*sqrt(a*b)*a^2*b^3*c^7*d^5 + 2*sqrt(a*b)*a 
^3*b^2*c^5*d^7 - 3*sqrt(a*b)*a^4*b*c^3*d^9 + sqrt(a*b)*a^5*c*d^11)*sqrt(-b 
^2*c + sqrt(a*b)*b*d)*abs(b))*arctan(sqrt(d*x + c)/sqrt(-(b^3*c^5 - 2*a*b^ 
2*c^3*d^2 + a^2*b*c*d^4 - sqrt((b^3*c^5 - 2*a*b^2*c^3*d^2 + a^2*b*c*d^4)^2 
 - (b^3*c^6 - 3*a*b^2*c^4*d^2 + 3*a^2*b*c^2*d^4 - a^3*d^6)*(b^3*c^4 - 2...
 

Mupad [B] (verification not implemented)

Time = 10.49 (sec) , antiderivative size = 7829, normalized size of antiderivative = 41.21 \[ \int \frac {1}{(c+d x)^{5/2} \left (a-b x^2\right )} \, dx=\text {Too large to display} \] Input:

int(1/((a - b*x^2)*(c + d*x)^(5/2)),x)
 

Output:

atan((((-(a^2*d^5*(a^3*b^3)^(1/2) + a*b^4*c^5 + 5*a^3*b^2*c*d^4 + 10*a^2*b 
^3*c^3*d^2 + 5*b^2*c^4*d*(a^3*b^3)^(1/2) + 10*a*b*c^2*d^3*(a^3*b^3)^(1/2)) 
/(4*(a^7*d^10 - a^2*b^5*c^10 - 5*a^6*b*c^2*d^8 + 5*a^3*b^4*c^8*d^2 - 10*a^ 
4*b^3*c^6*d^4 + 10*a^5*b^2*c^4*d^6)))^(1/2)*(32*a^10*b^4*d^21 + (c + d*x)^ 
(1/2)*(-(a^2*d^5*(a^3*b^3)^(1/2) + a*b^4*c^5 + 5*a^3*b^2*c*d^4 + 10*a^2*b^ 
3*c^3*d^2 + 5*b^2*c^4*d*(a^3*b^3)^(1/2) + 10*a*b*c^2*d^3*(a^3*b^3)^(1/2))/ 
(4*(a^7*d^10 - a^2*b^5*c^10 - 5*a^6*b*c^2*d^8 + 5*a^3*b^4*c^8*d^2 - 10*a^4 
*b^3*c^6*d^4 + 10*a^5*b^2*c^4*d^6)))^(1/2)*(64*a*b^14*c^21*d^2 + 64*a^11*b 
^4*c*d^22 - 640*a^2*b^13*c^19*d^4 + 2880*a^3*b^12*c^17*d^6 - 7680*a^4*b^11 
*c^15*d^8 + 13440*a^5*b^10*c^13*d^10 - 16128*a^6*b^9*c^11*d^12 + 13440*a^7 
*b^8*c^9*d^14 - 7680*a^8*b^7*c^7*d^16 + 2880*a^9*b^6*c^5*d^18 - 640*a^10*b 
^5*c^3*d^20) + 96*a*b^13*c^18*d^3 - 736*a^2*b^12*c^16*d^5 + 2432*a^3*b^11* 
c^14*d^7 - 4480*a^4*b^10*c^12*d^9 + 4928*a^5*b^9*c^10*d^11 - 3136*a^6*b^8* 
c^8*d^13 + 896*a^7*b^7*c^6*d^15 + 128*a^8*b^6*c^4*d^17 - 160*a^9*b^5*c^2*d 
^19) - (c + d*x)^(1/2)*(16*a^8*b^5*d^18 + 16*b^13*c^16*d^2 - 320*a^2*b^11* 
c^12*d^6 + 1024*a^3*b^10*c^10*d^8 - 1440*a^4*b^9*c^8*d^10 + 1024*a^5*b^8*c 
^6*d^12 - 320*a^6*b^7*c^4*d^14))*(-(a^2*d^5*(a^3*b^3)^(1/2) + a*b^4*c^5 + 
5*a^3*b^2*c*d^4 + 10*a^2*b^3*c^3*d^2 + 5*b^2*c^4*d*(a^3*b^3)^(1/2) + 10*a* 
b*c^2*d^3*(a^3*b^3)^(1/2))/(4*(a^7*d^10 - a^2*b^5*c^10 - 5*a^6*b*c^2*d^8 + 
 5*a^3*b^4*c^8*d^2 - 10*a^4*b^3*c^6*d^4 + 10*a^5*b^2*c^4*d^6)))^(1/2)*1...
 

Reduce [B] (verification not implemented)

Time = 0.19 (sec) , antiderivative size = 1425, normalized size of antiderivative = 7.50 \[ \int \frac {1}{(c+d x)^{5/2} \left (a-b x^2\right )} \, dx =\text {Too large to display} \] Input:

int(1/(d*x+c)^(5/2)/(-b*x^2+a),x)
 

Output:

(18*sqrt(a)*sqrt(c + d*x)*sqrt(sqrt(b)*sqrt(a)*d - b*c)*atan((sqrt(c + d*x 
)*b)/(sqrt(b)*sqrt(sqrt(b)*sqrt(a)*d - b*c)))*a*b*c**2*d**2 + 18*sqrt(a)*s 
qrt(c + d*x)*sqrt(sqrt(b)*sqrt(a)*d - b*c)*atan((sqrt(c + d*x)*b)/(sqrt(b) 
*sqrt(sqrt(b)*sqrt(a)*d - b*c)))*a*b*c*d**3*x + 6*sqrt(a)*sqrt(c + d*x)*sq 
rt(sqrt(b)*sqrt(a)*d - b*c)*atan((sqrt(c + d*x)*b)/(sqrt(b)*sqrt(sqrt(b)*s 
qrt(a)*d - b*c)))*b**2*c**4 + 6*sqrt(a)*sqrt(c + d*x)*sqrt(sqrt(b)*sqrt(a) 
*d - b*c)*atan((sqrt(c + d*x)*b)/(sqrt(b)*sqrt(sqrt(b)*sqrt(a)*d - b*c)))* 
b**2*c**3*d*x + 6*sqrt(b)*sqrt(c + d*x)*sqrt(sqrt(b)*sqrt(a)*d - b*c)*atan 
((sqrt(c + d*x)*b)/(sqrt(b)*sqrt(sqrt(b)*sqrt(a)*d - b*c)))*a**2*c*d**3 + 
6*sqrt(b)*sqrt(c + d*x)*sqrt(sqrt(b)*sqrt(a)*d - b*c)*atan((sqrt(c + d*x)* 
b)/(sqrt(b)*sqrt(sqrt(b)*sqrt(a)*d - b*c)))*a**2*d**4*x + 18*sqrt(b)*sqrt( 
c + d*x)*sqrt(sqrt(b)*sqrt(a)*d - b*c)*atan((sqrt(c + d*x)*b)/(sqrt(b)*sqr 
t(sqrt(b)*sqrt(a)*d - b*c)))*a*b*c**3*d + 18*sqrt(b)*sqrt(c + d*x)*sqrt(sq 
rt(b)*sqrt(a)*d - b*c)*atan((sqrt(c + d*x)*b)/(sqrt(b)*sqrt(sqrt(b)*sqrt(a 
)*d - b*c)))*a*b*c**2*d**2*x + 9*sqrt(a)*sqrt(c + d*x)*sqrt(sqrt(b)*sqrt(a 
)*d + b*c)*log( - sqrt(sqrt(b)*sqrt(a)*d + b*c) + sqrt(b)*sqrt(c + d*x))*a 
*b*c**2*d**2 + 9*sqrt(a)*sqrt(c + d*x)*sqrt(sqrt(b)*sqrt(a)*d + b*c)*log( 
- sqrt(sqrt(b)*sqrt(a)*d + b*c) + sqrt(b)*sqrt(c + d*x))*a*b*c*d**3*x + 3* 
sqrt(a)*sqrt(c + d*x)*sqrt(sqrt(b)*sqrt(a)*d + b*c)*log( - sqrt(sqrt(b)*sq 
rt(a)*d + b*c) + sqrt(b)*sqrt(c + d*x))*b**2*c**4 + 3*sqrt(a)*sqrt(c + ...