3.606 \(\int (a+b x^2)^2 \sqrt {c+d x^2} \, dx\)

Optimal. Leaf size=149 \[ \frac {x \sqrt {c+d x^2} \left (8 a^2 d^2-4 a b c d+b^2 c^2\right )}{16 d^2}+\frac {c \left (8 a^2 d^2-4 a b c d+b^2 c^2\right ) \tanh ^{-1}\left (\frac {\sqrt {d} x}{\sqrt {c+d x^2}}\right )}{16 d^{5/2}}-\frac {b x \left (c+d x^2\right )^{3/2} (3 b c-8 a d)}{24 d^2}+\frac {b x \left (a+b x^2\right ) \left (c+d x^2\right )^{3/2}}{6 d} \]

[Out]

-1/24*b*(-8*a*d+3*b*c)*x*(d*x^2+c)^(3/2)/d^2+1/6*b*x*(b*x^2+a)*(d*x^2+c)^(3/2)/d+1/16*c*(8*a^2*d^2-4*a*b*c*d+b
^2*c^2)*arctanh(x*d^(1/2)/(d*x^2+c)^(1/2))/d^(5/2)+1/16*(8*a^2*d^2-4*a*b*c*d+b^2*c^2)*x*(d*x^2+c)^(1/2)/d^2

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Rubi [A]  time = 0.10, antiderivative size = 149, normalized size of antiderivative = 1.00, number of steps used = 5, number of rules used = 5, integrand size = 21, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.238, Rules used = {416, 388, 195, 217, 206} \[ \frac {x \sqrt {c+d x^2} \left (8 a^2 d^2-4 a b c d+b^2 c^2\right )}{16 d^2}+\frac {c \left (8 a^2 d^2-4 a b c d+b^2 c^2\right ) \tanh ^{-1}\left (\frac {\sqrt {d} x}{\sqrt {c+d x^2}}\right )}{16 d^{5/2}}-\frac {b x \left (c+d x^2\right )^{3/2} (3 b c-8 a d)}{24 d^2}+\frac {b x \left (a+b x^2\right ) \left (c+d x^2\right )^{3/2}}{6 d} \]

Antiderivative was successfully verified.

[In]

Int[(a + b*x^2)^2*Sqrt[c + d*x^2],x]

[Out]

((b^2*c^2 - 4*a*b*c*d + 8*a^2*d^2)*x*Sqrt[c + d*x^2])/(16*d^2) - (b*(3*b*c - 8*a*d)*x*(c + d*x^2)^(3/2))/(24*d
^2) + (b*x*(a + b*x^2)*(c + d*x^2)^(3/2))/(6*d) + (c*(b^2*c^2 - 4*a*b*c*d + 8*a^2*d^2)*ArcTanh[(Sqrt[d]*x)/Sqr
t[c + d*x^2]])/(16*d^(5/2))

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]

Rule 416

Int[((a_) + (b_.)*(x_)^(n_))^(p_)*((c_) + (d_.)*(x_)^(n_))^(q_), x_Symbol] :> Simp[(d*x*(a + b*x^n)^(p + 1)*(c
+ d*x^n)^(q - 1))/(b*(n*(p + q) + 1)), x] + Dist[1/(b*(n*(p + q) + 1)), Int[(a + b*x^n)^p*(c + d*x^n)^(q - 2)
*Simp[c*(b*c*(n*(p + q) + 1) - a*d) + d*(b*c*(n*(p + 2*q - 1) + 1) - a*d*(n*(q - 1) + 1))*x^n, x], x], x] /; F
reeQ[{a, b, c, d, n, p}, x] && NeQ[b*c - a*d, 0] && GtQ[q, 1] && NeQ[n*(p + q) + 1, 0] &&  !IGtQ[p, 1] && IntB
inomialQ[a, b, c, d, n, p, q, x]

Rubi steps

\begin {align*} \int \left (a+b x^2\right )^2 \sqrt {c+d x^2} \, dx &=\frac {b x \left (a+b x^2\right ) \left (c+d x^2\right )^{3/2}}{6 d}+\frac {\int \sqrt {c+d x^2} \left (-a (b c-6 a d)-b (3 b c-8 a d) x^2\right ) \, dx}{6 d}\\ &=-\frac {b (3 b c-8 a d) x \left (c+d x^2\right )^{3/2}}{24 d^2}+\frac {b x \left (a+b x^2\right ) \left (c+d x^2\right )^{3/2}}{6 d}+\frac {\left (b^2 c^2-4 a b c d+8 a^2 d^2\right ) \int \sqrt {c+d x^2} \, dx}{8 d^2}\\ &=\frac {\left (b^2 c^2-4 a b c d+8 a^2 d^2\right ) x \sqrt {c+d x^2}}{16 d^2}-\frac {b (3 b c-8 a d) x \left (c+d x^2\right )^{3/2}}{24 d^2}+\frac {b x \left (a+b x^2\right ) \left (c+d x^2\right )^{3/2}}{6 d}+\frac {\left (c \left (b^2 c^2-4 a b c d+8 a^2 d^2\right )\right ) \int \frac {1}{\sqrt {c+d x^2}} \, dx}{16 d^2}\\ &=\frac {\left (b^2 c^2-4 a b c d+8 a^2 d^2\right ) x \sqrt {c+d x^2}}{16 d^2}-\frac {b (3 b c-8 a d) x \left (c+d x^2\right )^{3/2}}{24 d^2}+\frac {b x \left (a+b x^2\right ) \left (c+d x^2\right )^{3/2}}{6 d}+\frac {\left (c \left (b^2 c^2-4 a b c d+8 a^2 d^2\right )\right ) \operatorname {Subst}\left (\int \frac {1}{1-d x^2} \, dx,x,\frac {x}{\sqrt {c+d x^2}}\right )}{16 d^2}\\ &=\frac {\left (b^2 c^2-4 a b c d+8 a^2 d^2\right ) x \sqrt {c+d x^2}}{16 d^2}-\frac {b (3 b c-8 a d) x \left (c+d x^2\right )^{3/2}}{24 d^2}+\frac {b x \left (a+b x^2\right ) \left (c+d x^2\right )^{3/2}}{6 d}+\frac {c \left (b^2 c^2-4 a b c d+8 a^2 d^2\right ) \tanh ^{-1}\left (\frac {\sqrt {d} x}{\sqrt {c+d x^2}}\right )}{16 d^{5/2}}\\ \end {align*}

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Mathematica [A]  time = 0.13, size = 122, normalized size = 0.82 \[ \frac {3 c \left (8 a^2 d^2-4 a b c d+b^2 c^2\right ) \log \left (\sqrt {d} \sqrt {c+d x^2}+d x\right )+\sqrt {d} x \sqrt {c+d x^2} \left (24 a^2 d^2+12 a b d \left (c+2 d x^2\right )+b^2 \left (-3 c^2+2 c d x^2+8 d^2 x^4\right )\right )}{48 d^{5/2}} \]

Antiderivative was successfully verified.

[In]

Integrate[(a + b*x^2)^2*Sqrt[c + d*x^2],x]

[Out]

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

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fricas [A]  time = 0.92, size = 262, normalized size = 1.76 \[ \left [\frac {3 \, {\left (b^{2} c^{3} - 4 \, a b c^{2} d + 8 \, a^{2} c d^{2}\right )} \sqrt {d} \log \left (-2 \, d x^{2} - 2 \, \sqrt {d x^{2} + c} \sqrt {d} x - c\right ) + 2 \, {\left (8 \, b^{2} d^{3} x^{5} + 2 \, {\left (b^{2} c d^{2} + 12 \, a b d^{3}\right )} x^{3} - 3 \, {\left (b^{2} c^{2} d - 4 \, a b c d^{2} - 8 \, a^{2} d^{3}\right )} x\right )} \sqrt {d x^{2} + c}}{96 \, d^{3}}, -\frac {3 \, {\left (b^{2} c^{3} - 4 \, a b c^{2} d + 8 \, a^{2} c d^{2}\right )} \sqrt {-d} \arctan \left (\frac {\sqrt {-d} x}{\sqrt {d x^{2} + c}}\right ) - {\left (8 \, b^{2} d^{3} x^{5} + 2 \, {\left (b^{2} c d^{2} + 12 \, a b d^{3}\right )} x^{3} - 3 \, {\left (b^{2} c^{2} d - 4 \, a b c d^{2} - 8 \, a^{2} d^{3}\right )} x\right )} \sqrt {d x^{2} + c}}{48 \, d^{3}}\right ] \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

[1/96*(3*(b^2*c^3 - 4*a*b*c^2*d + 8*a^2*c*d^2)*sqrt(d)*log(-2*d*x^2 - 2*sqrt(d*x^2 + c)*sqrt(d)*x - c) + 2*(8*
b^2*d^3*x^5 + 2*(b^2*c*d^2 + 12*a*b*d^3)*x^3 - 3*(b^2*c^2*d - 4*a*b*c*d^2 - 8*a^2*d^3)*x)*sqrt(d*x^2 + c))/d^3
, -1/48*(3*(b^2*c^3 - 4*a*b*c^2*d + 8*a^2*c*d^2)*sqrt(-d)*arctan(sqrt(-d)*x/sqrt(d*x^2 + c)) - (8*b^2*d^3*x^5
+ 2*(b^2*c*d^2 + 12*a*b*d^3)*x^3 - 3*(b^2*c^2*d - 4*a*b*c*d^2 - 8*a^2*d^3)*x)*sqrt(d*x^2 + c))/d^3]

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giac [A]  time = 0.38, size = 128, normalized size = 0.86 \[ \frac {1}{48} \, {\left (2 \, {\left (4 \, b^{2} x^{2} + \frac {b^{2} c d^{3} + 12 \, a b d^{4}}{d^{4}}\right )} x^{2} - \frac {3 \, {\left (b^{2} c^{2} d^{2} - 4 \, a b c d^{3} - 8 \, a^{2} d^{4}\right )}}{d^{4}}\right )} \sqrt {d x^{2} + c} x - \frac {{\left (b^{2} c^{3} - 4 \, a b c^{2} d + 8 \, a^{2} c d^{2}\right )} \log \left ({\left | -\sqrt {d} x + \sqrt {d x^{2} + c} \right |}\right )}{16 \, d^{\frac {5}{2}}} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

1/48*(2*(4*b^2*x^2 + (b^2*c*d^3 + 12*a*b*d^4)/d^4)*x^2 - 3*(b^2*c^2*d^2 - 4*a*b*c*d^3 - 8*a^2*d^4)/d^4)*sqrt(d
*x^2 + c)*x - 1/16*(b^2*c^3 - 4*a*b*c^2*d + 8*a^2*c*d^2)*log(abs(-sqrt(d)*x + sqrt(d*x^2 + c)))/d^(5/2)

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maple [A]  time = 0.01, size = 190, normalized size = 1.28 \[ \frac {\left (d \,x^{2}+c \right )^{\frac {3}{2}} b^{2} x^{3}}{6 d}+\frac {a^{2} c \ln \left (\sqrt {d}\, x +\sqrt {d \,x^{2}+c}\right )}{2 \sqrt {d}}-\frac {a b \,c^{2} \ln \left (\sqrt {d}\, x +\sqrt {d \,x^{2}+c}\right )}{4 d^{\frac {3}{2}}}+\frac {b^{2} c^{3} \ln \left (\sqrt {d}\, x +\sqrt {d \,x^{2}+c}\right )}{16 d^{\frac {5}{2}}}+\frac {\sqrt {d \,x^{2}+c}\, a^{2} x}{2}-\frac {\sqrt {d \,x^{2}+c}\, a b c x}{4 d}+\frac {\sqrt {d \,x^{2}+c}\, b^{2} c^{2} x}{16 d^{2}}+\frac {\left (d \,x^{2}+c \right )^{\frac {3}{2}} a b x}{2 d}-\frac {\left (d \,x^{2}+c \right )^{\frac {3}{2}} b^{2} c x}{8 d^{2}} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

1/6*b^2*x^3*(d*x^2+c)^(3/2)/d-1/8*b^2*c/d^2*x*(d*x^2+c)^(3/2)+1/16*b^2*c^2/d^2*x*(d*x^2+c)^(1/2)+1/16*b^2*c^3/
d^(5/2)*ln(d^(1/2)*x+(d*x^2+c)^(1/2))+1/2*a*b*x*(d*x^2+c)^(3/2)/d-1/4*a*b*c/d*x*(d*x^2+c)^(1/2)-1/4*a*b*c^2/d^
(3/2)*ln(d^(1/2)*x+(d*x^2+c)^(1/2))+1/2*a^2*x*(d*x^2+c)^(1/2)+1/2*a^2*c/d^(1/2)*ln(d^(1/2)*x+(d*x^2+c)^(1/2))

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maxima [A]  time = 1.00, size = 168, normalized size = 1.13 \[ \frac {{\left (d x^{2} + c\right )}^{\frac {3}{2}} b^{2} x^{3}}{6 \, d} + \frac {1}{2} \, \sqrt {d x^{2} + c} a^{2} x - \frac {{\left (d x^{2} + c\right )}^{\frac {3}{2}} b^{2} c x}{8 \, d^{2}} + \frac {\sqrt {d x^{2} + c} b^{2} c^{2} x}{16 \, d^{2}} + \frac {{\left (d x^{2} + c\right )}^{\frac {3}{2}} a b x}{2 \, d} - \frac {\sqrt {d x^{2} + c} a b c x}{4 \, d} + \frac {b^{2} c^{3} \operatorname {arsinh}\left (\frac {d x}{\sqrt {c d}}\right )}{16 \, d^{\frac {5}{2}}} - \frac {a b c^{2} \operatorname {arsinh}\left (\frac {d x}{\sqrt {c d}}\right )}{4 \, d^{\frac {3}{2}}} + \frac {a^{2} c \operatorname {arsinh}\left (\frac {d x}{\sqrt {c d}}\right )}{2 \, \sqrt {d}} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

1/6*(d*x^2 + c)^(3/2)*b^2*x^3/d + 1/2*sqrt(d*x^2 + c)*a^2*x - 1/8*(d*x^2 + c)^(3/2)*b^2*c*x/d^2 + 1/16*sqrt(d*
x^2 + c)*b^2*c^2*x/d^2 + 1/2*(d*x^2 + c)^(3/2)*a*b*x/d - 1/4*sqrt(d*x^2 + c)*a*b*c*x/d + 1/16*b^2*c^3*arcsinh(
d*x/sqrt(c*d))/d^(5/2) - 1/4*a*b*c^2*arcsinh(d*x/sqrt(c*d))/d^(3/2) + 1/2*a^2*c*arcsinh(d*x/sqrt(c*d))/sqrt(d)

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

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

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

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sympy [B]  time = 13.72, size = 291, normalized size = 1.95 \[ \frac {a^{2} \sqrt {c} x \sqrt {1 + \frac {d x^{2}}{c}}}{2} + \frac {a^{2} c \operatorname {asinh}{\left (\frac {\sqrt {d} x}{\sqrt {c}} \right )}}{2 \sqrt {d}} + \frac {a b c^{\frac {3}{2}} x}{4 d \sqrt {1 + \frac {d x^{2}}{c}}} + \frac {3 a b \sqrt {c} x^{3}}{4 \sqrt {1 + \frac {d x^{2}}{c}}} - \frac {a b c^{2} \operatorname {asinh}{\left (\frac {\sqrt {d} x}{\sqrt {c}} \right )}}{4 d^{\frac {3}{2}}} + \frac {a b d x^{5}}{2 \sqrt {c} \sqrt {1 + \frac {d x^{2}}{c}}} - \frac {b^{2} c^{\frac {5}{2}} x}{16 d^{2} \sqrt {1 + \frac {d x^{2}}{c}}} - \frac {b^{2} c^{\frac {3}{2}} x^{3}}{48 d \sqrt {1 + \frac {d x^{2}}{c}}} + \frac {5 b^{2} \sqrt {c} x^{5}}{24 \sqrt {1 + \frac {d x^{2}}{c}}} + \frac {b^{2} c^{3} \operatorname {asinh}{\left (\frac {\sqrt {d} x}{\sqrt {c}} \right )}}{16 d^{\frac {5}{2}}} + \frac {b^{2} d x^{7}}{6 \sqrt {c} \sqrt {1 + \frac {d x^{2}}{c}}} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

a**2*sqrt(c)*x*sqrt(1 + d*x**2/c)/2 + a**2*c*asinh(sqrt(d)*x/sqrt(c))/(2*sqrt(d)) + a*b*c**(3/2)*x/(4*d*sqrt(1
+ d*x**2/c)) + 3*a*b*sqrt(c)*x**3/(4*sqrt(1 + d*x**2/c)) - a*b*c**2*asinh(sqrt(d)*x/sqrt(c))/(4*d**(3/2)) + a
*b*d*x**5/(2*sqrt(c)*sqrt(1 + d*x**2/c)) - b**2*c**(5/2)*x/(16*d**2*sqrt(1 + d*x**2/c)) - b**2*c**(3/2)*x**3/(
48*d*sqrt(1 + d*x**2/c)) + 5*b**2*sqrt(c)*x**5/(24*sqrt(1 + d*x**2/c)) + b**2*c**3*asinh(sqrt(d)*x/sqrt(c))/(1
6*d**(5/2)) + b**2*d*x**7/(6*sqrt(c)*sqrt(1 + d*x**2/c))

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