3.1.46 \(\int \sqrt {a+b x^2} (c+d x^2)^2 \, dx\) [46]

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

[Out]

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

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Rubi [A]
time = 0.06, 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 = {427, 396, 201, 223, 212} \begin {gather*} \frac {x \sqrt {a+b x^2} \left (a^2 d^2-4 a b c d+8 b^2 c^2\right )}{16 b^2}+\frac {a \left (a^2 d^2-4 a b c d+8 b^2 c^2\right ) \tanh ^{-1}\left (\frac {\sqrt {b} x}{\sqrt {a+b x^2}}\right )}{16 b^{5/2}}+\frac {d x \left (a+b x^2\right )^{3/2} (8 b c-3 a d)}{24 b^2}+\frac {d x \left (a+b x^2\right )^{3/2} \left (c+d x^2\right )}{6 b} \end {gather*}

Antiderivative was successfully verified.

[In]

Int[Sqrt[a + b*x^2]*(c + d*x^2)^2,x]

[Out]

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

Rule 201

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 212

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] && (GtQ[a, 0] || LtQ[b, 0])

Rule 223

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 396

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 427

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

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Mathematica [A]
time = 0.16, size = 122, normalized size = 0.82 \begin {gather*} \frac {\sqrt {b} x \sqrt {a+b x^2} \left (-3 a^2 d^2+2 a b d \left (6 c+d x^2\right )+8 b^2 \left (3 c^2+3 c d x^2+d^2 x^4\right )\right )-3 a \left (8 b^2 c^2-4 a b c d+a^2 d^2\right ) \log \left (-\sqrt {b} x+\sqrt {a+b x^2}\right )}{48 b^{5/2}} \end {gather*}

Antiderivative was successfully verified.

[In]

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

[Out]

(Sqrt[b]*x*Sqrt[a + b*x^2]*(-3*a^2*d^2 + 2*a*b*d*(6*c + d*x^2) + 8*b^2*(3*c^2 + 3*c*d*x^2 + d^2*x^4)) - 3*a*(8
*b^2*c^2 - 4*a*b*c*d + a^2*d^2)*Log[-(Sqrt[b]*x) + Sqrt[a + b*x^2]])/(48*b^(5/2))

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Maple [A]
time = 0.06, size = 187, normalized size = 1.26

method	result	size
risch	\(-\frac {x \left (-8 b^{2} d^{2} x^{4}-2 a b \,d^{2} x^{2}-24 b^{2} c d \,x^{2}+3 a^{2} d^{2}-12 a b c d -24 b^{2} c^{2}\right ) \sqrt {b \,x^{2}+a}}{48 b^{2}}+\frac {a^{3} \ln \left (x \sqrt {b}+\sqrt {b \,x^{2}+a}\right ) d^{2}}{16 b^{\frac {5}{2}}}-\frac {a^{2} \ln \left (x \sqrt {b}+\sqrt {b \,x^{2}+a}\right ) c d}{4 b^{\frac {3}{2}}}+\frac {a \ln \left (x \sqrt {b}+\sqrt {b \,x^{2}+a}\right ) c^{2}}{2 \sqrt {b}}\)	\(149\)
default	\(d^{2} \left (\frac {x^{3} \left (b \,x^{2}+a \right )^{\frac {3}{2}}}{6 b}-\frac {a \left (\frac {x \left (b \,x^{2}+a \right )^{\frac {3}{2}}}{4 b}-\frac {a \left (\frac {x \sqrt {b \,x^{2}+a}}{2}+\frac {a \ln \left (x \sqrt {b}+\sqrt {b \,x^{2}+a}\right )}{2 \sqrt {b}}\right )}{4 b}\right )}{2 b}\right )+2 c d \left (\frac {x \left (b \,x^{2}+a \right )^{\frac {3}{2}}}{4 b}-\frac {a \left (\frac {x \sqrt {b \,x^{2}+a}}{2}+\frac {a \ln \left (x \sqrt {b}+\sqrt {b \,x^{2}+a}\right )}{2 \sqrt {b}}\right )}{4 b}\right )+c^{2} \left (\frac {x \sqrt {b \,x^{2}+a}}{2}+\frac {a \ln \left (x \sqrt {b}+\sqrt {b \,x^{2}+a}\right )}{2 \sqrt {b}}\right )\)	\(187\)

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

d^2*(1/6*x^3*(b*x^2+a)^(3/2)/b-1/2*a/b*(1/4*x*(b*x^2+a)^(3/2)/b-1/4*a/b*(1/2*x*(b*x^2+a)^(1/2)+1/2*a/b^(1/2)*l
n(x*b^(1/2)+(b*x^2+a)^(1/2)))))+2*c*d*(1/4*x*(b*x^2+a)^(3/2)/b-1/4*a/b*(1/2*x*(b*x^2+a)^(1/2)+1/2*a/b^(1/2)*ln
(x*b^(1/2)+(b*x^2+a)^(1/2))))+c^2*(1/2*x*(b*x^2+a)^(1/2)+1/2*a/b^(1/2)*ln(x*b^(1/2)+(b*x^2+a)^(1/2)))

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Maxima [A]
time = 0.28, size = 168, normalized size = 1.13 \begin {gather*} \frac {{\left (b x^{2} + a\right )}^{\frac {3}{2}} d^{2} x^{3}}{6 \, b} + \frac {1}{2} \, \sqrt {b x^{2} + a} c^{2} x + \frac {{\left (b x^{2} + a\right )}^{\frac {3}{2}} c d x}{2 \, b} - \frac {\sqrt {b x^{2} + a} a c d x}{4 \, b} - \frac {{\left (b x^{2} + a\right )}^{\frac {3}{2}} a d^{2} x}{8 \, b^{2}} + \frac {\sqrt {b x^{2} + a} a^{2} d^{2} x}{16 \, b^{2}} + \frac {a c^{2} \operatorname {arsinh}\left (\frac {b x}{\sqrt {a b}}\right )}{2 \, \sqrt {b}} - \frac {a^{2} c d \operatorname {arsinh}\left (\frac {b x}{\sqrt {a b}}\right )}{4 \, b^{\frac {3}{2}}} + \frac {a^{3} d^{2} \operatorname {arsinh}\left (\frac {b x}{\sqrt {a b}}\right )}{16 \, b^{\frac {5}{2}}} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

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

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Fricas [A]
time = 3.46, size = 264, normalized size = 1.77 \begin {gather*} \left [\frac {3 \, {\left (8 \, a b^{2} c^{2} - 4 \, a^{2} b c d + a^{3} d^{2}\right )} \sqrt {b} \log \left (-2 \, b x^{2} - 2 \, \sqrt {b x^{2} + a} \sqrt {b} x - a\right ) + 2 \, {\left (8 \, b^{3} d^{2} x^{5} + 2 \, {\left (12 \, b^{3} c d + a b^{2} d^{2}\right )} x^{3} + 3 \, {\left (8 \, b^{3} c^{2} + 4 \, a b^{2} c d - a^{2} b d^{2}\right )} x\right )} \sqrt {b x^{2} + a}}{96 \, b^{3}}, -\frac {3 \, {\left (8 \, a b^{2} c^{2} - 4 \, a^{2} b c d + a^{3} d^{2}\right )} \sqrt {-b} \arctan \left (\frac {\sqrt {-b} x}{\sqrt {b x^{2} + a}}\right ) - {\left (8 \, b^{3} d^{2} x^{5} + 2 \, {\left (12 \, b^{3} c d + a b^{2} d^{2}\right )} x^{3} + 3 \, {\left (8 \, b^{3} c^{2} + 4 \, a b^{2} c d - a^{2} b d^{2}\right )} x\right )} \sqrt {b x^{2} + a}}{48 \, b^{3}}\right ] \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

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

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Sympy [B] Leaf count of result is larger than twice the leaf count of optimal. 291 vs. \(2 (143) = 286\).
time = 8.46, size = 291, normalized size = 1.95 \begin {gather*} - \frac {a^{\frac {5}{2}} d^{2} x}{16 b^{2} \sqrt {1 + \frac {b x^{2}}{a}}} + \frac {a^{\frac {3}{2}} c d x}{4 b \sqrt {1 + \frac {b x^{2}}{a}}} - \frac {a^{\frac {3}{2}} d^{2} x^{3}}{48 b \sqrt {1 + \frac {b x^{2}}{a}}} + \frac {\sqrt {a} c^{2} x \sqrt {1 + \frac {b x^{2}}{a}}}{2} + \frac {3 \sqrt {a} c d x^{3}}{4 \sqrt {1 + \frac {b x^{2}}{a}}} + \frac {5 \sqrt {a} d^{2} x^{5}}{24 \sqrt {1 + \frac {b x^{2}}{a}}} + \frac {a^{3} d^{2} \operatorname {asinh}{\left (\frac {\sqrt {b} x}{\sqrt {a}} \right )}}{16 b^{\frac {5}{2}}} - \frac {a^{2} c d \operatorname {asinh}{\left (\frac {\sqrt {b} x}{\sqrt {a}} \right )}}{4 b^{\frac {3}{2}}} + \frac {a c^{2} \operatorname {asinh}{\left (\frac {\sqrt {b} x}{\sqrt {a}} \right )}}{2 \sqrt {b}} + \frac {b c d x^{5}}{2 \sqrt {a} \sqrt {1 + \frac {b x^{2}}{a}}} + \frac {b d^{2} x^{7}}{6 \sqrt {a} \sqrt {1 + \frac {b x^{2}}{a}}} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

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

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Giac [A]
time = 0.73, size = 129, normalized size = 0.87 \begin {gather*} \frac {1}{48} \, {\left (2 \, {\left (4 \, d^{2} x^{2} + \frac {12 \, b^{4} c d + a b^{3} d^{2}}{b^{4}}\right )} x^{2} + \frac {3 \, {\left (8 \, b^{4} c^{2} + 4 \, a b^{3} c d - a^{2} b^{2} d^{2}\right )}}{b^{4}}\right )} \sqrt {b x^{2} + a} x - \frac {{\left (8 \, a b^{2} c^{2} - 4 \, a^{2} b c d + a^{3} d^{2}\right )} \log \left ({\left | -\sqrt {b} x + \sqrt {b x^{2} + a} \right |}\right )}{16 \, b^{\frac {5}{2}}} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

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

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

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

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

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