3.1.55 \(\int (a+b x^2)^{3/2} (c+d x^2) \, dx\) [55]

Optimal. Leaf size=118 \[ \frac {a (6 b c-a d) x \sqrt {a+b x^2}}{16 b}+\frac {(6 b c-a d) x \left (a+b x^2\right )^{3/2}}{24 b}+\frac {d x \left (a+b x^2\right )^{5/2}}{6 b}+\frac {a^2 (6 b c-a d) \tanh ^{-1}\left (\frac {\sqrt {b} x}{\sqrt {a+b x^2}}\right )}{16 b^{3/2}} \]

[Out]

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

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Rubi [A]
time = 0.03, antiderivative size = 118, normalized size of antiderivative = 1.00, number of steps used = 5, number of rules used = 4, integrand size = 19, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.210, Rules used = {396, 201, 223, 212} \begin {gather*} \frac {a^2 (6 b c-a d) \tanh ^{-1}\left (\frac {\sqrt {b} x}{\sqrt {a+b x^2}}\right )}{16 b^{3/2}}+\frac {x \left (a+b x^2\right )^{3/2} (6 b c-a d)}{24 b}+\frac {a x \sqrt {a+b x^2} (6 b c-a d)}{16 b}+\frac {d x \left (a+b x^2\right )^{5/2}}{6 b} \end {gather*}

Antiderivative was successfully verified.

[In]

Int[(a + b*x^2)^(3/2)*(c + d*x^2),x]

[Out]

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

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]

Rubi steps

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

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Mathematica [A]
time = 0.15, size = 99, normalized size = 0.84 \begin {gather*} \frac {x \sqrt {a+b x^2} \left (30 a b c+3 a^2 d+12 b^2 c x^2+14 a b d x^2+8 b^2 d x^4\right )}{48 b}+\frac {a^2 (-6 b c+a d) \log \left (-\sqrt {b} x+\sqrt {a+b x^2}\right )}{16 b^{3/2}} \end {gather*}

Antiderivative was successfully verified.

[In]

Integrate[(a + b*x^2)^(3/2)*(c + d*x^2),x]

[Out]

(x*Sqrt[a + b*x^2]*(30*a*b*c + 3*a^2*d + 12*b^2*c*x^2 + 14*a*b*d*x^2 + 8*b^2*d*x^4))/(48*b) + (a^2*(-6*b*c + a
*d)*Log[-(Sqrt[b]*x) + Sqrt[a + b*x^2]])/(16*b^(3/2))

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Maple [A]
time = 0.05, size = 130, normalized size = 1.10

method	result	size
risch	\(\frac {x \left (8 b^{2} d \,x^{4}+14 x^{2} a b d +12 b^{2} c \,x^{2}+3 a^{2} d +30 a b c \right ) \sqrt {b \,x^{2}+a}}{48 b}-\frac {a^{3} \ln \left (x \sqrt {b}+\sqrt {b \,x^{2}+a}\right ) d}{16 b^{\frac {3}{2}}}+\frac {3 a^{2} \ln \left (x \sqrt {b}+\sqrt {b \,x^{2}+a}\right ) c}{8 \sqrt {b}}\)	\(105\)
default	\(d \left (\frac {x \left (b \,x^{2}+a \right )^{\frac {5}{2}}}{6 b}-\frac {a \left (\frac {x \left (b \,x^{2}+a \right )^{\frac {3}{2}}}{4}+\frac {3 a \left (\frac {x \sqrt {b \,x^{2}+a}}{2}+\frac {a \ln \left (x \sqrt {b}+\sqrt {b \,x^{2}+a}\right )}{2 \sqrt {b}}\right )}{4}\right )}{6 b}\right )+c \left (\frac {x \left (b \,x^{2}+a \right )^{\frac {3}{2}}}{4}+\frac {3 a \left (\frac {x \sqrt {b \,x^{2}+a}}{2}+\frac {a \ln \left (x \sqrt {b}+\sqrt {b \,x^{2}+a}\right )}{2 \sqrt {b}}\right )}{4}\right )\)	\(130\)

Verification of antiderivative is not currently implemented for this CAS.

[In]

int((b*x^2+a)^(3/2)*(d*x^2+c),x,method=_RETURNVERBOSE)

[Out]

d*(1/6*x*(b*x^2+a)^(5/2)/b-1/6*a/b*(1/4*x*(b*x^2+a)^(3/2)+3/4*a*(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*(1/4*x*(b*x^2+a)^(3/2)+3/4*a*(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 = 0.27, size = 116, normalized size = 0.98 \begin {gather*} \frac {1}{4} \, {\left (b x^{2} + a\right )}^{\frac {3}{2}} c x + \frac {3}{8} \, \sqrt {b x^{2} + a} a c x + \frac {{\left (b x^{2} + a\right )}^{\frac {5}{2}} d x}{6 \, b} - \frac {{\left (b x^{2} + a\right )}^{\frac {3}{2}} a d x}{24 \, b} - \frac {\sqrt {b x^{2} + a} a^{2} d x}{16 \, b} + \frac {3 \, a^{2} c \operatorname {arsinh}\left (\frac {b x}{\sqrt {a b}}\right )}{8 \, \sqrt {b}} - \frac {a^{3} d \operatorname {arsinh}\left (\frac {b x}{\sqrt {a b}}\right )}{16 \, b^{\frac {3}{2}}} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate((b*x^2+a)^(3/2)*(d*x^2+c),x, algorithm="maxima")

[Out]

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

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Fricas [A]
time = 0.55, size = 210, normalized size = 1.78 \begin {gather*} \left [-\frac {3 \, {\left (6 \, a^{2} b c - a^{3} d\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 x^{5} + 2 \, {\left (6 \, b^{3} c + 7 \, a b^{2} d\right )} x^{3} + 3 \, {\left (10 \, a b^{2} c + a^{2} b d\right )} x\right )} \sqrt {b x^{2} + a}}{96 \, b^{2}}, -\frac {3 \, {\left (6 \, a^{2} b c - a^{3} d\right )} \sqrt {-b} \arctan \left (\frac {\sqrt {-b} x}{\sqrt {b x^{2} + a}}\right ) - {\left (8 \, b^{3} d x^{5} + 2 \, {\left (6 \, b^{3} c + 7 \, a b^{2} d\right )} x^{3} + 3 \, {\left (10 \, a b^{2} c + a^{2} b d\right )} x\right )} \sqrt {b x^{2} + a}}{48 \, b^{2}}\right ] \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate((b*x^2+a)^(3/2)*(d*x^2+c),x, algorithm="fricas")

[Out]

[-1/96*(3*(6*a^2*b*c - a^3*d)*sqrt(b)*log(-2*b*x^2 + 2*sqrt(b*x^2 + a)*sqrt(b)*x - a) - 2*(8*b^3*d*x^5 + 2*(6*
b^3*c + 7*a*b^2*d)*x^3 + 3*(10*a*b^2*c + a^2*b*d)*x)*sqrt(b*x^2 + a))/b^2, -1/48*(3*(6*a^2*b*c - a^3*d)*sqrt(-
b)*arctan(sqrt(-b)*x/sqrt(b*x^2 + a)) - (8*b^3*d*x^5 + 2*(6*b^3*c + 7*a*b^2*d)*x^3 + 3*(10*a*b^2*c + a^2*b*d)*
x)*sqrt(b*x^2 + a))/b^2]

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Sympy [B] Leaf count of result is larger than twice the leaf count of optimal. 253 vs. \(2 (102) = 204\).
time = 10.12, size = 253, normalized size = 2.14 \begin {gather*} \frac {a^{\frac {5}{2}} d x}{16 b \sqrt {1 + \frac {b x^{2}}{a}}} + \frac {a^{\frac {3}{2}} c x \sqrt {1 + \frac {b x^{2}}{a}}}{2} + \frac {a^{\frac {3}{2}} c x}{8 \sqrt {1 + \frac {b x^{2}}{a}}} + \frac {17 a^{\frac {3}{2}} d x^{3}}{48 \sqrt {1 + \frac {b x^{2}}{a}}} + \frac {3 \sqrt {a} b c x^{3}}{8 \sqrt {1 + \frac {b x^{2}}{a}}} + \frac {11 \sqrt {a} b d x^{5}}{24 \sqrt {1 + \frac {b x^{2}}{a}}} - \frac {a^{3} d \operatorname {asinh}{\left (\frac {\sqrt {b} x}{\sqrt {a}} \right )}}{16 b^{\frac {3}{2}}} + \frac {3 a^{2} c \operatorname {asinh}{\left (\frac {\sqrt {b} x}{\sqrt {a}} \right )}}{8 \sqrt {b}} + \frac {b^{2} c x^{5}}{4 \sqrt {a} \sqrt {1 + \frac {b x^{2}}{a}}} + \frac {b^{2} d 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)**(3/2)*(d*x**2+c),x)

[Out]

a**(5/2)*d*x/(16*b*sqrt(1 + b*x**2/a)) + a**(3/2)*c*x*sqrt(1 + b*x**2/a)/2 + a**(3/2)*c*x/(8*sqrt(1 + b*x**2/a
)) + 17*a**(3/2)*d*x**3/(48*sqrt(1 + b*x**2/a)) + 3*sqrt(a)*b*c*x**3/(8*sqrt(1 + b*x**2/a)) + 11*sqrt(a)*b*d*x
**5/(24*sqrt(1 + b*x**2/a)) - a**3*d*asinh(sqrt(b)*x/sqrt(a))/(16*b**(3/2)) + 3*a**2*c*asinh(sqrt(b)*x/sqrt(a)
)/(8*sqrt(b)) + b**2*c*x**5/(4*sqrt(a)*sqrt(1 + b*x**2/a)) + b**2*d*x**7/(6*sqrt(a)*sqrt(1 + b*x**2/a))

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Giac [A]
time = 0.60, size = 103, normalized size = 0.87 \begin {gather*} \frac {1}{48} \, {\left (2 \, {\left (4 \, b d x^{2} + \frac {6 \, b^{5} c + 7 \, a b^{4} d}{b^{4}}\right )} x^{2} + \frac {3 \, {\left (10 \, a b^{4} c + a^{2} b^{3} d\right )}}{b^{4}}\right )} \sqrt {b x^{2} + a} x - \frac {{\left (6 \, a^{2} b c - a^{3} d\right )} \log \left ({\left | -\sqrt {b} x + \sqrt {b x^{2} + a} \right |}\right )}{16 \, b^{\frac {3}{2}}} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate((b*x^2+a)^(3/2)*(d*x^2+c),x, algorithm="giac")

[Out]

1/48*(2*(4*b*d*x^2 + (6*b^5*c + 7*a*b^4*d)/b^4)*x^2 + 3*(10*a*b^4*c + a^2*b^3*d)/b^4)*sqrt(b*x^2 + a)*x - 1/16
*(6*a^2*b*c - a^3*d)*log(abs(-sqrt(b)*x + sqrt(b*x^2 + a)))/b^(3/2)

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Mupad [F]
time = 0.00, size = -1, normalized size = -0.01 \begin {gather*} \int {\left (b\,x^2+a\right )}^{3/2}\,\left (d\,x^2+c\right ) \,d x \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

int((a + b*x^2)^(3/2)*(c + d*x^2),x)

[Out]

int((a + b*x^2)^(3/2)*(c + d*x^2), x)

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