\(\int (a+b x^2)^{5/2} (A+B x^2) \, dx\) [88]

Optimal result

Integrand size = 19, antiderivative size = 149 \[ \int \left (a+b x^2\right )^{5/2} \left (A+B x^2\right ) \, dx=\frac {5 a^2 (8 A b-a B) x \sqrt {a+b x^2}}{128 b}+\frac {5 a (8 A b-a B) x \left (a+b x^2\right )^{3/2}}{192 b}+\frac {(8 A b-a B) x \left (a+b x^2\right )^{5/2}}{48 b}+\frac {B x \left (a+b x^2\right )^{7/2}}{8 b}+\frac {5 a^3 (8 A b-a B) \text {arctanh}\left (\frac {\sqrt {b} x}{\sqrt {a+b x^2}}\right )}{128 b^{3/2}} \] Output:

5/128*a^2*(8*A*b-B*a)*x*(b*x^2+a)^(1/2)/b+5/192*a*(8*A*b-B*a)*x*(b*x^2+a)^ 
(3/2)/b+1/48*(8*A*b-B*a)*x*(b*x^2+a)^(5/2)/b+1/8*B*x*(b*x^2+a)^(7/2)/b+5/1 
28*a^3*(8*A*b-B*a)*arctanh(b^(1/2)*x/(b*x^2+a)^(1/2))/b^(3/2)
 

Mathematica [A] (verified)

Time = 0.21 (sec) , antiderivative size = 123, normalized size of antiderivative = 0.83 \[ \int \left (a+b x^2\right )^{5/2} \left (A+B x^2\right ) \, dx=\frac {x \sqrt {a+b x^2} \left (264 a^2 A b+15 a^3 B+208 a A b^2 x^2+118 a^2 b B x^2+64 A b^3 x^4+136 a b^2 B x^4+48 b^3 B x^6\right )}{384 b}+\frac {5 a^3 (-8 A b+a B) \log \left (-\sqrt {b} x+\sqrt {a+b x^2}\right )}{128 b^{3/2}} \] Input:

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

Output:

(x*Sqrt[a + b*x^2]*(264*a^2*A*b + 15*a^3*B + 208*a*A*b^2*x^2 + 118*a^2*b*B 
*x^2 + 64*A*b^3*x^4 + 136*a*b^2*B*x^4 + 48*b^3*B*x^6))/(384*b) + (5*a^3*(- 
8*A*b + a*B)*Log[-(Sqrt[b]*x) + Sqrt[a + b*x^2]])/(128*b^(3/2))
 

Rubi [A] (verified)

Time = 0.21 (sec) , antiderivative size = 127, normalized size of antiderivative = 0.85, number of steps used = 7, number of rules used = 6, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.316, Rules used = {299, 211, 211, 211, 224, 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.

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

\(\Big \downarrow \) 299

\(\displaystyle \frac {(8 A b-a B) \int \left (b x^2+a\right )^{5/2}dx}{8 b}+\frac {B x \left (a+b x^2\right )^{7/2}}{8 b}\)

\(\Big \downarrow \) 211

\(\displaystyle \frac {(8 A b-a B) \left (\frac {5}{6} a \int \left (b x^2+a\right )^{3/2}dx+\frac {1}{6} x \left (a+b x^2\right )^{5/2}\right )}{8 b}+\frac {B x \left (a+b x^2\right )^{7/2}}{8 b}\)

\(\Big \downarrow \) 211

\(\displaystyle \frac {(8 A b-a B) \left (\frac {5}{6} a \left (\frac {3}{4} a \int \sqrt {b x^2+a}dx+\frac {1}{4} x \left (a+b x^2\right )^{3/2}\right )+\frac {1}{6} x \left (a+b x^2\right )^{5/2}\right )}{8 b}+\frac {B x \left (a+b x^2\right )^{7/2}}{8 b}\)

\(\Big \downarrow \) 211

\(\displaystyle \frac {(8 A b-a B) \left (\frac {5}{6} a \left (\frac {3}{4} a \left (\frac {1}{2} a \int \frac {1}{\sqrt {b x^2+a}}dx+\frac {1}{2} x \sqrt {a+b x^2}\right )+\frac {1}{4} x \left (a+b x^2\right )^{3/2}\right )+\frac {1}{6} x \left (a+b x^2\right )^{5/2}\right )}{8 b}+\frac {B x \left (a+b x^2\right )^{7/2}}{8 b}\)

\(\Big \downarrow \) 224

\(\displaystyle \frac {(8 A b-a B) \left (\frac {5}{6} a \left (\frac {3}{4} a \left (\frac {1}{2} a \int \frac {1}{1-\frac {b x^2}{b x^2+a}}d\frac {x}{\sqrt {b x^2+a}}+\frac {1}{2} x \sqrt {a+b x^2}\right )+\frac {1}{4} x \left (a+b x^2\right )^{3/2}\right )+\frac {1}{6} x \left (a+b x^2\right )^{5/2}\right )}{8 b}+\frac {B x \left (a+b x^2\right )^{7/2}}{8 b}\)

\(\Big \downarrow \) 219

\(\displaystyle \frac {(8 A b-a B) \left (\frac {5}{6} a \left (\frac {3}{4} a \left (\frac {a \text {arctanh}\left (\frac {\sqrt {b} x}{\sqrt {a+b x^2}}\right )}{2 \sqrt {b}}+\frac {1}{2} x \sqrt {a+b x^2}\right )+\frac {1}{4} x \left (a+b x^2\right )^{3/2}\right )+\frac {1}{6} x \left (a+b x^2\right )^{5/2}\right )}{8 b}+\frac {B x \left (a+b x^2\right )^{7/2}}{8 b}\)

Input:

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

Output:

(B*x*(a + b*x^2)^(7/2))/(8*b) + ((8*A*b - a*B)*((x*(a + b*x^2)^(5/2))/6 + 
(5*a*((x*(a + b*x^2)^(3/2))/4 + (3*a*((x*Sqrt[a + b*x^2])/2 + (a*ArcTanh[( 
Sqrt[b]*x)/Sqrt[a + b*x^2]])/(2*Sqrt[b])))/4))/6))/(8*b)
 

Defintions of rubi rules used

rule 211

Int[((a_) + (b_.)*(x_)^2)^(p_), x_Symbol] :> Simp[x*((a + b*x^2)^p/(2*p + 1 
)), x] + Simp[2*a*(p/(2*p + 1))   Int[(a + b*x^2)^(p - 1), x], x] /; FreeQ[ 
{a, b}, x] && GtQ[p, 0] && (IntegerQ[4*p] || IntegerQ[6*p])
 

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 299

Int[((a_) + (b_.)*(x_)^2)^(p_)*((c_) + (d_.)*(x_)^2), x_Symbol] :> Simp[d*x 
*((a + b*x^2)^(p + 1)/(b*(2*p + 3))), x] - Simp[(a*d - b*c*(2*p + 3))/(b*(2 
*p + 3))   Int[(a + b*x^2)^p, x], x] /; FreeQ[{a, b, c, d}, x] && NeQ[b*c - 
 a*d, 0] && NeQ[2*p + 3, 0]
 

Maple [A] (verified)

Time = 0.56 (sec) , antiderivative size = 106, normalized size of antiderivative = 0.71

method	result	size
pseudoelliptic	\(\frac {\frac {5 \left (A b -\frac {B a}{8}\right ) a^{3} \operatorname {arctanh}\left (\frac {\sqrt {b \,x^{2}+a}}{x \sqrt {b}}\right )}{16}+\frac {13 \left (\frac {33 a^{2} \left (\frac {59 x^{2} B}{132}+A \right ) b^{\frac {3}{2}}}{26}+x^{2} a \left (\frac {17 x^{2} B}{26}+A \right ) b^{\frac {5}{2}}+\frac {4 x^{4} \left (\frac {3 x^{2} B}{4}+A \right ) b^{\frac {7}{2}}}{13}+\frac {15 B \,a^{3} \sqrt {b}}{208}\right ) \sqrt {b \,x^{2}+a}\, x}{24}}{b^{\frac {3}{2}}}\)	\(106\)
risch	\(\frac {x \left (48 b^{3} B \,x^{6}+64 A \,x^{4} b^{3}+136 B \,x^{4} a \,b^{2}+208 a A \,b^{2} x^{2}+118 B \,a^{2} b \,x^{2}+264 a^{2} b A +15 a^{3} B \right ) \sqrt {b \,x^{2}+a}}{384 b}+\frac {5 a^{3} \left (8 A b -B a \right ) \ln \left (\sqrt {b}\, x +\sqrt {b \,x^{2}+a}\right )}{128 b^{\frac {3}{2}}}\)	\(112\)
default	\(A \left (\frac {x \left (b \,x^{2}+a \right )^{\frac {5}{2}}}{6}+\frac {5 a \left (\frac {x \left (b \,x^{2}+a \right )^{\frac {3}{2}}}{4}+\frac {3 a \left (\frac {x \sqrt {b \,x^{2}+a}}{2}+\frac {a \ln \left (\sqrt {b}\, x +\sqrt {b \,x^{2}+a}\right )}{2 \sqrt {b}}\right )}{4}\right )}{6}\right )+B \left (\frac {x \left (b \,x^{2}+a \right )^{\frac {7}{2}}}{8 b}-\frac {a \left (\frac {x \left (b \,x^{2}+a \right )^{\frac {5}{2}}}{6}+\frac {5 a \left (\frac {x \left (b \,x^{2}+a \right )^{\frac {3}{2}}}{4}+\frac {3 a \left (\frac {x \sqrt {b \,x^{2}+a}}{2}+\frac {a \ln \left (\sqrt {b}\, x +\sqrt {b \,x^{2}+a}\right )}{2 \sqrt {b}}\right )}{4}\right )}{6}\right )}{8 b}\right )\)	\(162\)

Input:

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

Output:

13/24*(15/26*(A*b-1/8*B*a)*a^3*arctanh((b*x^2+a)^(1/2)/x/b^(1/2))+(33/26*a 
^2*(59/132*x^2*B+A)*b^(3/2)+x^2*a*(17/26*x^2*B+A)*b^(5/2)+4/13*x^4*(3/4*x^ 
2*B+A)*b^(7/2)+15/208*B*a^3*b^(1/2))*(b*x^2+a)^(1/2)*x)/b^(3/2)
 

Fricas [A] (verification not implemented)

Time = 0.10 (sec) , antiderivative size = 257, normalized size of antiderivative = 1.72 \[ \int \left (a+b x^2\right )^{5/2} \left (A+B x^2\right ) \, dx=\left [-\frac {15 \, {\left (B a^{4} - 8 \, A a^{3} b\right )} \sqrt {b} \log \left (-2 \, b x^{2} - 2 \, \sqrt {b x^{2} + a} \sqrt {b} x - a\right ) - 2 \, {\left (48 \, B b^{4} x^{7} + 8 \, {\left (17 \, B a b^{3} + 8 \, A b^{4}\right )} x^{5} + 2 \, {\left (59 \, B a^{2} b^{2} + 104 \, A a b^{3}\right )} x^{3} + 3 \, {\left (5 \, B a^{3} b + 88 \, A a^{2} b^{2}\right )} x\right )} \sqrt {b x^{2} + a}}{768 \, b^{2}}, \frac {15 \, {\left (B a^{4} - 8 \, A a^{3} b\right )} \sqrt {-b} \arctan \left (\frac {\sqrt {-b} x}{\sqrt {b x^{2} + a}}\right ) + {\left (48 \, B b^{4} x^{7} + 8 \, {\left (17 \, B a b^{3} + 8 \, A b^{4}\right )} x^{5} + 2 \, {\left (59 \, B a^{2} b^{2} + 104 \, A a b^{3}\right )} x^{3} + 3 \, {\left (5 \, B a^{3} b + 88 \, A a^{2} b^{2}\right )} x\right )} \sqrt {b x^{2} + a}}{384 \, b^{2}}\right ] \] Input:

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

Output:

[-1/768*(15*(B*a^4 - 8*A*a^3*b)*sqrt(b)*log(-2*b*x^2 - 2*sqrt(b*x^2 + a)*s 
qrt(b)*x - a) - 2*(48*B*b^4*x^7 + 8*(17*B*a*b^3 + 8*A*b^4)*x^5 + 2*(59*B*a 
^2*b^2 + 104*A*a*b^3)*x^3 + 3*(5*B*a^3*b + 88*A*a^2*b^2)*x)*sqrt(b*x^2 + a 
))/b^2, 1/384*(15*(B*a^4 - 8*A*a^3*b)*sqrt(-b)*arctan(sqrt(-b)*x/sqrt(b*x^ 
2 + a)) + (48*B*b^4*x^7 + 8*(17*B*a*b^3 + 8*A*b^4)*x^5 + 2*(59*B*a^2*b^2 + 
 104*A*a*b^3)*x^3 + 3*(5*B*a^3*b + 88*A*a^2*b^2)*x)*sqrt(b*x^2 + a))/b^2]
 

Sympy [B] (verification not implemented)

Leaf count of result is larger than twice the leaf count of optimal. 279 vs. \(2 (134) = 268\).

Time = 0.39 (sec) , antiderivative size = 279, normalized size of antiderivative = 1.87 \[ \int \left (a+b x^2\right )^{5/2} \left (A+B x^2\right ) \, dx=\begin {cases} \sqrt {a + b x^{2}} \left (\frac {B b^{2} x^{7}}{8} + \frac {x^{5} \left (A b^{3} + \frac {17 B a b^{2}}{8}\right )}{6 b} + \frac {x^{3} \cdot \left (3 A a b^{2} + 3 B a^{2} b - \frac {5 a \left (A b^{3} + \frac {17 B a b^{2}}{8}\right )}{6 b}\right )}{4 b} + \frac {x \left (3 A a^{2} b + B a^{3} - \frac {3 a \left (3 A a b^{2} + 3 B a^{2} b - \frac {5 a \left (A b^{3} + \frac {17 B a b^{2}}{8}\right )}{6 b}\right )}{4 b}\right )}{2 b}\right ) + \left (A a^{3} - \frac {a \left (3 A a^{2} b + B a^{3} - \frac {3 a \left (3 A a b^{2} + 3 B a^{2} b - \frac {5 a \left (A b^{3} + \frac {17 B a b^{2}}{8}\right )}{6 b}\right )}{4 b}\right )}{2 b}\right ) \left (\begin {cases} \frac {\log {\left (2 \sqrt {b} \sqrt {a + b x^{2}} + 2 b x \right )}}{\sqrt {b}} & \text {for}\: a \neq 0 \\\frac {x \log {\left (x \right )}}{\sqrt {b x^{2}}} & \text {otherwise} \end {cases}\right ) & \text {for}\: b \neq 0 \\a^{\frac {5}{2}} \left (A x + \frac {B x^{3}}{3}\right ) & \text {otherwise} \end {cases} \] Input:

integrate((b*x**2+a)**(5/2)*(B*x**2+A),x)
 

Output:

Piecewise((sqrt(a + b*x**2)*(B*b**2*x**7/8 + x**5*(A*b**3 + 17*B*a*b**2/8) 
/(6*b) + x**3*(3*A*a*b**2 + 3*B*a**2*b - 5*a*(A*b**3 + 17*B*a*b**2/8)/(6*b 
))/(4*b) + x*(3*A*a**2*b + B*a**3 - 3*a*(3*A*a*b**2 + 3*B*a**2*b - 5*a*(A* 
b**3 + 17*B*a*b**2/8)/(6*b))/(4*b))/(2*b)) + (A*a**3 - a*(3*A*a**2*b + B*a 
**3 - 3*a*(3*A*a*b**2 + 3*B*a**2*b - 5*a*(A*b**3 + 17*B*a*b**2/8)/(6*b))/( 
4*b))/(2*b))*Piecewise((log(2*sqrt(b)*sqrt(a + b*x**2) + 2*b*x)/sqrt(b), N 
e(a, 0)), (x*log(x)/sqrt(b*x**2), True)), Ne(b, 0)), (a**(5/2)*(A*x + B*x* 
*3/3), True))
 

Maxima [A] (verification not implemented)

Time = 0.05 (sec) , antiderivative size = 151, normalized size of antiderivative = 1.01 \[ \int \left (a+b x^2\right )^{5/2} \left (A+B x^2\right ) \, dx=\frac {1}{6} \, {\left (b x^{2} + a\right )}^{\frac {5}{2}} A x + \frac {5}{24} \, {\left (b x^{2} + a\right )}^{\frac {3}{2}} A a x + \frac {5}{16} \, \sqrt {b x^{2} + a} A a^{2} x + \frac {{\left (b x^{2} + a\right )}^{\frac {7}{2}} B x}{8 \, b} - \frac {{\left (b x^{2} + a\right )}^{\frac {5}{2}} B a x}{48 \, b} - \frac {5 \, {\left (b x^{2} + a\right )}^{\frac {3}{2}} B a^{2} x}{192 \, b} - \frac {5 \, \sqrt {b x^{2} + a} B a^{3} x}{128 \, b} - \frac {5 \, B a^{4} \operatorname {arsinh}\left (\frac {b x}{\sqrt {a b}}\right )}{128 \, b^{\frac {3}{2}}} + \frac {5 \, A a^{3} \operatorname {arsinh}\left (\frac {b x}{\sqrt {a b}}\right )}{16 \, \sqrt {b}} \] Input:

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

Output:

1/6*(b*x^2 + a)^(5/2)*A*x + 5/24*(b*x^2 + a)^(3/2)*A*a*x + 5/16*sqrt(b*x^2 
 + a)*A*a^2*x + 1/8*(b*x^2 + a)^(7/2)*B*x/b - 1/48*(b*x^2 + a)^(5/2)*B*a*x 
/b - 5/192*(b*x^2 + a)^(3/2)*B*a^2*x/b - 5/128*sqrt(b*x^2 + a)*B*a^3*x/b - 
 5/128*B*a^4*arcsinh(b*x/sqrt(a*b))/b^(3/2) + 5/16*A*a^3*arcsinh(b*x/sqrt( 
a*b))/sqrt(b)
 

Giac [A] (verification not implemented)

Time = 0.14 (sec) , antiderivative size = 134, normalized size of antiderivative = 0.90 \[ \int \left (a+b x^2\right )^{5/2} \left (A+B x^2\right ) \, dx=\frac {1}{384} \, {\left (2 \, {\left (4 \, {\left (6 \, B b^{2} x^{2} + \frac {17 \, B a b^{7} + 8 \, A b^{8}}{b^{6}}\right )} x^{2} + \frac {59 \, B a^{2} b^{6} + 104 \, A a b^{7}}{b^{6}}\right )} x^{2} + \frac {3 \, {\left (5 \, B a^{3} b^{5} + 88 \, A a^{2} b^{6}\right )}}{b^{6}}\right )} \sqrt {b x^{2} + a} x + \frac {5 \, {\left (B a^{4} - 8 \, A a^{3} b\right )} \log \left ({\left | -\sqrt {b} x + \sqrt {b x^{2} + a} \right |}\right )}{128 \, b^{\frac {3}{2}}} \] Input:

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

Output:

1/384*(2*(4*(6*B*b^2*x^2 + (17*B*a*b^7 + 8*A*b^8)/b^6)*x^2 + (59*B*a^2*b^6 
 + 104*A*a*b^7)/b^6)*x^2 + 3*(5*B*a^3*b^5 + 88*A*a^2*b^6)/b^6)*sqrt(b*x^2 
+ a)*x + 5/128*(B*a^4 - 8*A*a^3*b)*log(abs(-sqrt(b)*x + sqrt(b*x^2 + a)))/ 
b^(3/2)
 

Mupad [F(-1)]

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

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

Output:

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

Reduce [B] (verification not implemented)

Time = 0.17 (sec) , antiderivative size = 99, normalized size of antiderivative = 0.66 \[ \int \left (a+b x^2\right )^{5/2} \left (A+B x^2\right ) \, dx=\frac {279 \sqrt {b \,x^{2}+a}\, a^{3} b x +326 \sqrt {b \,x^{2}+a}\, a^{2} b^{2} x^{3}+200 \sqrt {b \,x^{2}+a}\, a \,b^{3} x^{5}+48 \sqrt {b \,x^{2}+a}\, b^{4} x^{7}+105 \sqrt {b}\, \mathrm {log}\left (\frac {\sqrt {b \,x^{2}+a}+\sqrt {b}\, x}{\sqrt {a}}\right ) a^{4}}{384 b} \] Input:

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

Output:

(279*sqrt(a + b*x**2)*a**3*b*x + 326*sqrt(a + b*x**2)*a**2*b**2*x**3 + 200 
*sqrt(a + b*x**2)*a*b**3*x**5 + 48*sqrt(a + b*x**2)*b**4*x**7 + 105*sqrt(b 
)*log((sqrt(a + b*x**2) + sqrt(b)*x)/sqrt(a))*a**4)/(384*b)