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

Optimal. Leaf size=149 \[ \frac {5 a^3 (8 A b-a B) \tanh ^{-1}\left (\frac {\sqrt {b} x}{\sqrt {a+b x^2}}\right )}{128 b^{3/2}}+\frac {5 a^2 x \sqrt {a+b x^2} (8 A b-a B)}{128 b}+\frac {x \left (a+b x^2\right )^{5/2} (8 A b-a B)}{48 b}+\frac {5 a x \left (a+b x^2\right )^{3/2} (8 A b-a B)}{192 b}+\frac {B x \left (a+b x^2\right )^{7/2}}{8 b} \]

[Out]

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

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Rubi [A]  time = 0.05, antiderivative size = 149, normalized size of antiderivative = 1.00, number of steps used = 6, number of rules used = 4, integrand size = 19, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.210, Rules used = {388, 195, 217, 206} \[ \frac {5 a^3 (8 A b-a B) \tanh ^{-1}\left (\frac {\sqrt {b} x}{\sqrt {a+b x^2}}\right )}{128 b^{3/2}}+\frac {5 a^2 x \sqrt {a+b x^2} (8 A b-a B)}{128 b}+\frac {x \left (a+b x^2\right )^{5/2} (8 A b-a B)}{48 b}+\frac {5 a x \left (a+b x^2\right )^{3/2} (8 A b-a B)}{192 b}+\frac {B x \left (a+b x^2\right )^{7/2}}{8 b} \]

Antiderivative was successfully verified.

[In]

Int[(a + b*x^2)^(5/2)*(A + B*x^2),x]

[Out]

(5*a^2*(8*A*b - a*B)*x*Sqrt[a + b*x^2])/(128*b) + (5*a*(8*A*b - a*B)*x*(a + b*x^2)^(3/2))/(192*b) + ((8*A*b -
a*B)*x*(a + b*x^2)^(5/2))/(48*b) + (B*x*(a + b*x^2)^(7/2))/(8*b) + (5*a^3*(8*A*b - a*B)*ArcTanh[(Sqrt[b]*x)/Sq
rt[a + b*x^2]])/(128*b^(3/2))

Rule 195

Int[((a_) + (b_.)*(x_)^(n_))^(p_), x_Symbol] :> Simp[(x*(a + b*x^n)^p)/(n*p + 1), x] + Dist[(a*n*p)/(n*p + 1),
Int[(a + b*x^n)^(p - 1), x], x] /; FreeQ[{a, b}, x] && IGtQ[n, 0] && GtQ[p, 0] && (IntegerQ[2*p] || (EqQ[n, 2
] && IntegerQ[4*p]) || (EqQ[n, 2] && IntegerQ[3*p]) || LtQ[Denominator[p + 1/n], Denominator[p]])

Rule 206

Int[((a_) + (b_.)*(x_)^2)^(-1), x_Symbol] :> Simp[(1*ArcTanh[(Rt[-b, 2]*x)/Rt[a, 2]])/(Rt[a, 2]*Rt[-b, 2]), x]
/; FreeQ[{a, b}, x] && NegQ[a/b] && (GtQ[a, 0] || LtQ[b, 0])

Rule 217

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 388

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

Rubi steps

\begin {align*} \int \left (a+b x^2\right )^{5/2} \left (A+B x^2\right ) \, dx &=\frac {B x \left (a+b x^2\right )^{7/2}}{8 b}-\frac {(-8 A b+a B) \int \left (a+b x^2\right )^{5/2} \, dx}{8 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 (8 A b-a B)) \int \left (a+b x^2\right )^{3/2} \, dx}{48 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 {\left (5 a^2 (8 A b-a B)\right ) \int \sqrt {a+b x^2} \, dx}{64 b}\\ &=\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 {\left (5 a^3 (8 A b-a B)\right ) \int \frac {1}{\sqrt {a+b x^2}} \, dx}{128 b}\\ &=\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 {\left (5 a^3 (8 A b-a B)\right ) \operatorname {Subst}\left (\int \frac {1}{1-b x^2} \, dx,x,\frac {x}{\sqrt {a+b x^2}}\right )}{128 b}\\ &=\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) \tanh ^{-1}\left (\frac {\sqrt {b} x}{\sqrt {a+b x^2}}\right )}{128 b^{3/2}}\\ \end {align*}

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Mathematica [A]  time = 0.23, size = 130, normalized size = 0.87 \[ \frac {\sqrt {a+b x^2} \left (\sqrt {b} x \left (15 a^3 B+2 a^2 b \left (132 A+59 B x^2\right )+8 a b^2 x^2 \left (26 A+17 B x^2\right )+16 b^3 x^4 \left (4 A+3 B x^2\right )\right )-\frac {15 a^{5/2} (a B-8 A b) \sinh ^{-1}\left (\frac {\sqrt {b} x}{\sqrt {a}}\right )}{\sqrt {\frac {b x^2}{a}+1}}\right )}{384 b^{3/2}} \]

Antiderivative was successfully verified.

[In]

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

[Out]

(Sqrt[a + b*x^2]*(Sqrt[b]*x*(15*a^3*B + 16*b^3*x^4*(4*A + 3*B*x^2) + 8*a*b^2*x^2*(26*A + 17*B*x^2) + 2*a^2*b*(
132*A + 59*B*x^2)) - (15*a^(5/2)*(-8*A*b + a*B)*ArcSinh[(Sqrt[b]*x)/Sqrt[a]])/Sqrt[1 + (b*x^2)/a]))/(384*b^(3/
2))

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fricas [A]  time = 0.81, size = 257, normalized size = 1.72 \[ \left [-\frac {15 \, {\left (B a^{4} - 8 \, A a^{3} b\right )} \sqrt {b} \log \left (-2 \, b x^{2} - 2 \, \sqrt {b x^{2} + a} \sqrt {b} x - a\right ) - 2 \, {\left (48 \, B b^{4} x^{7} + 8 \, {\left (17 \, B a b^{3} + 8 \, A b^{4}\right )} x^{5} + 2 \, {\left (59 \, B a^{2} b^{2} + 104 \, A a b^{3}\right )} x^{3} + 3 \, {\left (5 \, B a^{3} b + 88 \, A a^{2} b^{2}\right )} x\right )} \sqrt {b x^{2} + a}}{768 \, b^{2}}, \frac {15 \, {\left (B a^{4} - 8 \, A a^{3} b\right )} \sqrt {-b} \arctan \left (\frac {\sqrt {-b} x}{\sqrt {b x^{2} + a}}\right ) + {\left (48 \, B b^{4} x^{7} + 8 \, {\left (17 \, B a b^{3} + 8 \, A b^{4}\right )} x^{5} + 2 \, {\left (59 \, B a^{2} b^{2} + 104 \, A a b^{3}\right )} x^{3} + 3 \, {\left (5 \, B a^{3} b + 88 \, A a^{2} b^{2}\right )} x\right )} \sqrt {b x^{2} + a}}{384 \, b^{2}}\right ] \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

[-1/768*(15*(B*a^4 - 8*A*a^3*b)*sqrt(b)*log(-2*b*x^2 - 2*sqrt(b*x^2 + a)*sqrt(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]

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giac [A]  time = 0.51, size = 134, normalized size = 0.90 \[ \frac {1}{384} \, {\left (2 \, {\left (4 \, {\left (6 \, B b^{2} x^{2} + \frac {17 \, B a b^{7} + 8 \, A b^{8}}{b^{6}}\right )} x^{2} + \frac {59 \, B a^{2} b^{6} + 104 \, A a b^{7}}{b^{6}}\right )} x^{2} + \frac {3 \, {\left (5 \, B a^{3} b^{5} + 88 \, A a^{2} b^{6}\right )}}{b^{6}}\right )} \sqrt {b x^{2} + a} x + \frac {5 \, {\left (B a^{4} - 8 \, A a^{3} b\right )} \log \left ({\left | -\sqrt {b} x + \sqrt {b x^{2} + a} \right |}\right )}{128 \, b^{\frac {3}{2}}} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

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)

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maple [A]  time = 0.01, size = 166, normalized size = 1.11 \[ \frac {5 A \,a^{3} \ln \left (\sqrt {b}\, x +\sqrt {b \,x^{2}+a}\right )}{16 \sqrt {b}}-\frac {5 B \,a^{4} \ln \left (\sqrt {b}\, x +\sqrt {b \,x^{2}+a}\right )}{128 b^{\frac {3}{2}}}+\frac {5 \sqrt {b \,x^{2}+a}\, A \,a^{2} x}{16}-\frac {5 \sqrt {b \,x^{2}+a}\, B \,a^{3} x}{128 b}+\frac {5 \left (b \,x^{2}+a \right )^{\frac {3}{2}} A a x}{24}-\frac {5 \left (b \,x^{2}+a \right )^{\frac {3}{2}} B \,a^{2} x}{192 b}+\frac {\left (b \,x^{2}+a \right )^{\frac {5}{2}} A x}{6}-\frac {\left (b \,x^{2}+a \right )^{\frac {5}{2}} B a x}{48 b}+\frac {\left (b \,x^{2}+a \right )^{\frac {7}{2}} B x}{8 b} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

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

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maxima [A]  time = 1.13, size = 151, normalized size = 1.01 \[ \frac {1}{6} \, {\left (b x^{2} + a\right )}^{\frac {5}{2}} A x + \frac {5}{24} \, {\left (b x^{2} + a\right )}^{\frac {3}{2}} A a x + \frac {5}{16} \, \sqrt {b x^{2} + a} A a^{2} x + \frac {{\left (b x^{2} + a\right )}^{\frac {7}{2}} B x}{8 \, b} - \frac {{\left (b x^{2} + a\right )}^{\frac {5}{2}} B a x}{48 \, b} - \frac {5 \, {\left (b x^{2} + a\right )}^{\frac {3}{2}} B a^{2} x}{192 \, b} - \frac {5 \, \sqrt {b x^{2} + a} B a^{3} x}{128 \, b} - \frac {5 \, B a^{4} \operatorname {arsinh}\left (\frac {b x}{\sqrt {a b}}\right )}{128 \, b^{\frac {3}{2}}} + \frac {5 \, A a^{3} \operatorname {arsinh}\left (\frac {b x}{\sqrt {a b}}\right )}{16 \, \sqrt {b}} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

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)

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mupad [F]  time = 0.00, size = -1, normalized size = -0.01 \[ \int \left (B\,x^2+A\right )\,{\left (b\,x^2+a\right )}^{5/2} \,d x \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

int((A + B*x^2)*(a + b*x^2)^(5/2),x)

[Out]

int((A + B*x^2)*(a + b*x^2)^(5/2), x)

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sympy [B]  time = 32.84, size = 316, normalized size = 2.12 \[ \frac {A a^{\frac {5}{2}} x \sqrt {1 + \frac {b x^{2}}{a}}}{2} + \frac {3 A a^{\frac {5}{2}} x}{16 \sqrt {1 + \frac {b x^{2}}{a}}} + \frac {35 A a^{\frac {3}{2}} b x^{3}}{48 \sqrt {1 + \frac {b x^{2}}{a}}} + \frac {17 A \sqrt {a} b^{2} x^{5}}{24 \sqrt {1 + \frac {b x^{2}}{a}}} + \frac {5 A a^{3} \operatorname {asinh}{\left (\frac {\sqrt {b} x}{\sqrt {a}} \right )}}{16 \sqrt {b}} + \frac {A b^{3} x^{7}}{6 \sqrt {a} \sqrt {1 + \frac {b x^{2}}{a}}} + \frac {5 B a^{\frac {7}{2}} x}{128 b \sqrt {1 + \frac {b x^{2}}{a}}} + \frac {133 B a^{\frac {5}{2}} x^{3}}{384 \sqrt {1 + \frac {b x^{2}}{a}}} + \frac {127 B a^{\frac {3}{2}} b x^{5}}{192 \sqrt {1 + \frac {b x^{2}}{a}}} + \frac {23 B \sqrt {a} b^{2} x^{7}}{48 \sqrt {1 + \frac {b x^{2}}{a}}} - \frac {5 B a^{4} \operatorname {asinh}{\left (\frac {\sqrt {b} x}{\sqrt {a}} \right )}}{128 b^{\frac {3}{2}}} + \frac {B b^{3} x^{9}}{8 \sqrt {a} \sqrt {1 + \frac {b x^{2}}{a}}} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

A*a**(5/2)*x*sqrt(1 + b*x**2/a)/2 + 3*A*a**(5/2)*x/(16*sqrt(1 + b*x**2/a)) + 35*A*a**(3/2)*b*x**3/(48*sqrt(1 +
b*x**2/a)) + 17*A*sqrt(a)*b**2*x**5/(24*sqrt(1 + b*x**2/a)) + 5*A*a**3*asinh(sqrt(b)*x/sqrt(a))/(16*sqrt(b))
+ A*b**3*x**7/(6*sqrt(a)*sqrt(1 + b*x**2/a)) + 5*B*a**(7/2)*x/(128*b*sqrt(1 + b*x**2/a)) + 133*B*a**(5/2)*x**3
/(384*sqrt(1 + b*x**2/a)) + 127*B*a**(3/2)*b*x**5/(192*sqrt(1 + b*x**2/a)) + 23*B*sqrt(a)*b**2*x**7/(48*sqrt(1
+ b*x**2/a)) - 5*B*a**4*asinh(sqrt(b)*x/sqrt(a))/(128*b**(3/2)) + B*b**3*x**9/(8*sqrt(a)*sqrt(1 + b*x**2/a))

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