3.8.29 \(\int (d x)^{3/2} \sqrt {a^2+2 a b x^2+b^2 x^4} \, dx\) [729]

Optimal. Leaf size=93 \[ \frac {2 a (d x)^{5/2} \sqrt {a^2+2 a b x^2+b^2 x^4}}{5 d \left (a+b x^2\right )}+\frac {2 b (d x)^{9/2} \sqrt {a^2+2 a b x^2+b^2 x^4}}{9 d^3 \left (a+b x^2\right )} \]

[Out]

2/5*a*(d*x)^(5/2)*((b*x^2+a)^2)^(1/2)/d/(b*x^2+a)+2/9*b*(d*x)^(9/2)*((b*x^2+a)^2)^(1/2)/d^3/(b*x^2+a)

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Rubi [A]
time = 0.02, antiderivative size = 93, normalized size of antiderivative = 1.00, number of steps used = 3, number of rules used = 2, integrand size = 30, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.067, Rules used = {1126, 14} \begin {gather*} \frac {2 b (d x)^{9/2} \sqrt {a^2+2 a b x^2+b^2 x^4}}{9 d^3 \left (a+b x^2\right )}+\frac {2 a (d x)^{5/2} \sqrt {a^2+2 a b x^2+b^2 x^4}}{5 d \left (a+b x^2\right )} \end {gather*}

Antiderivative was successfully verified.

[In]

Int[(d*x)^(3/2)*Sqrt[a^2 + 2*a*b*x^2 + b^2*x^4],x]

[Out]

(2*a*(d*x)^(5/2)*Sqrt[a^2 + 2*a*b*x^2 + b^2*x^4])/(5*d*(a + b*x^2)) + (2*b*(d*x)^(9/2)*Sqrt[a^2 + 2*a*b*x^2 +
b^2*x^4])/(9*d^3*(a + b*x^2))

Rule 14

Int[(u_)*((c_.)*(x_))^(m_.), x_Symbol] :> Int[ExpandIntegrand[(c*x)^m*u, x], x] /; FreeQ[{c, m}, x] && SumQ[u]
&&  !LinearQ[u, x] &&  !MatchQ[u, (a_) + (b_.)*(v_) /; FreeQ[{a, b}, x] && InverseFunctionQ[v]]

Rule 1126

Int[((d_.)*(x_))^(m_.)*((a_) + (b_.)*(x_)^2 + (c_.)*(x_)^4)^(p_), x_Symbol] :> Dist[(a + b*x^2 + c*x^4)^FracPa
rt[p]/(c^IntPart[p]*(b/2 + c*x^2)^(2*FracPart[p])), Int[(d*x)^m*(b/2 + c*x^2)^(2*p), x], x] /; FreeQ[{a, b, c,
d, m, p}, x] && EqQ[b^2 - 4*a*c, 0] && IntegerQ[p - 1/2]

Rubi steps

\begin {align*} \int (d x)^{3/2} \sqrt {a^2+2 a b x^2+b^2 x^4} \, dx &=\frac {\sqrt {a^2+2 a b x^2+b^2 x^4} \int (d x)^{3/2} \left (a b+b^2 x^2\right ) \, dx}{a b+b^2 x^2}\\ &=\frac {\sqrt {a^2+2 a b x^2+b^2 x^4} \int \left (a b (d x)^{3/2}+\frac {b^2 (d x)^{7/2}}{d^2}\right ) \, dx}{a b+b^2 x^2}\\ &=\frac {2 a (d x)^{5/2} \sqrt {a^2+2 a b x^2+b^2 x^4}}{5 d \left (a+b x^2\right )}+\frac {2 b (d x)^{9/2} \sqrt {a^2+2 a b x^2+b^2 x^4}}{9 d^3 \left (a+b x^2\right )}\\ \end {align*}

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Mathematica [A]
time = 0.02, size = 44, normalized size = 0.47 \begin {gather*} \frac {2 x (d x)^{3/2} \sqrt {\left (a+b x^2\right )^2} \left (9 a+5 b x^2\right )}{45 \left (a+b x^2\right )} \end {gather*}

Antiderivative was successfully verified.

[In]

Integrate[(d*x)^(3/2)*Sqrt[a^2 + 2*a*b*x^2 + b^2*x^4],x]

[Out]

(2*x*(d*x)^(3/2)*Sqrt[(a + b*x^2)^2]*(9*a + 5*b*x^2))/(45*(a + b*x^2))

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Maple [A]
time = 0.04, size = 41, normalized size = 0.44

method	result	size
gosper	\(\frac {2 x \left (5 b \,x^{2}+9 a \right ) \left (d x \right )^{\frac {3}{2}} \sqrt {\left (b \,x^{2}+a \right )^{2}}}{45 \left (b \,x^{2}+a \right )}\)	\(39\)
default	\(\frac {2 \sqrt {\left (b \,x^{2}+a \right )^{2}}\, \left (d x \right )^{\frac {5}{2}} \left (5 b \,x^{2}+9 a \right )}{45 d \left (b \,x^{2}+a \right )}\)	\(41\)
risch	\(\frac {2 d^{2} \sqrt {\left (b \,x^{2}+a \right )^{2}}\, x^{3} \left (5 b \,x^{2}+9 a \right )}{45 \left (b \,x^{2}+a \right ) \sqrt {d x}}\)	\(44\)

Verification of antiderivative is not currently implemented for this CAS.

[In]

int((d*x)^(3/2)*((b*x^2+a)^2)^(1/2),x,method=_RETURNVERBOSE)

[Out]

2/45*((b*x^2+a)^2)^(1/2)*(d*x)^(5/2)/d*(5*b*x^2+9*a)/(b*x^2+a)

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Maxima [A]
time = 0.29, size = 25, normalized size = 0.27 \begin {gather*} \frac {2 \, {\left (5 \, \left (d x\right )^{\frac {9}{2}} b + 9 \, \left (d x\right )^{\frac {5}{2}} a d^{2}\right )}}{45 \, d^{3}} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate((d*x)^(3/2)*((b*x^2+a)^2)^(1/2),x, algorithm="maxima")

[Out]

2/45*(5*(d*x)^(9/2)*b + 9*(d*x)^(5/2)*a*d^2)/d^3

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Fricas [A]
time = 0.35, size = 22, normalized size = 0.24 \begin {gather*} \frac {2}{45} \, {\left (5 \, b d x^{4} + 9 \, a d x^{2}\right )} \sqrt {d x} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate((d*x)^(3/2)*((b*x^2+a)^2)^(1/2),x, algorithm="fricas")

[Out]

2/45*(5*b*d*x^4 + 9*a*d*x^2)*sqrt(d*x)

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Sympy [F(-1)] Timed out
time = 0.00, size = 0, normalized size = 0.00 \begin {gather*} \text {Timed out} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate((d*x)**(3/2)*((b*x**2+a)**2)**(1/2),x)

[Out]

Timed out

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Giac [A]
time = 4.56, size = 42, normalized size = 0.45 \begin {gather*} \frac {2}{45} \, {\left (5 \, \sqrt {d x} b x^{4} \mathrm {sgn}\left (b x^{2} + a\right ) + 9 \, \sqrt {d x} a x^{2} \mathrm {sgn}\left (b x^{2} + a\right )\right )} d \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate((d*x)^(3/2)*((b*x^2+a)^2)^(1/2),x, algorithm="giac")

[Out]

2/45*(5*sqrt(d*x)*b*x^4*sgn(b*x^2 + a) + 9*sqrt(d*x)*a*x^2*sgn(b*x^2 + a))*d

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

Verification of antiderivative is not currently implemented for this CAS.

[In]

int((d*x)^(3/2)*((a + b*x^2)^2)^(1/2),x)

[Out]

int((d*x)^(3/2)*((a + b*x^2)^2)^(1/2), x)

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