3.6.59 \(\int \sqrt {d x} (a^2+2 a b x^2+b^2 x^4)^{3/2} \, dx\)

Optimal. Leaf size=195 \[ \frac {6 a b^2 (d x)^{11/2} \sqrt {a^2+2 a b x^2+b^2 x^4}}{11 d^5 \left (a+b x^2\right )}+\frac {6 a^2 b (d x)^{7/2} \sqrt {a^2+2 a b x^2+b^2 x^4}}{7 d^3 \left (a+b x^2\right )}+\frac {2 b^3 (d x)^{15/2} \sqrt {a^2+2 a b x^2+b^2 x^4}}{15 d^7 \left (a+b x^2\right )}+\frac {2 a^3 (d x)^{3/2} \sqrt {a^2+2 a b x^2+b^2 x^4}}{3 d \left (a+b x^2\right )} \]

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Rubi [A]  time = 0.05, antiderivative size = 195, 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 = {1112, 270} \begin {gather*} \frac {2 b^3 (d x)^{15/2} \sqrt {a^2+2 a b x^2+b^2 x^4}}{15 d^7 \left (a+b x^2\right )}+\frac {6 a b^2 (d x)^{11/2} \sqrt {a^2+2 a b x^2+b^2 x^4}}{11 d^5 \left (a+b x^2\right )}+\frac {6 a^2 b (d x)^{7/2} \sqrt {a^2+2 a b x^2+b^2 x^4}}{7 d^3 \left (a+b x^2\right )}+\frac {2 a^3 (d x)^{3/2} \sqrt {a^2+2 a b x^2+b^2 x^4}}{3 d \left (a+b x^2\right )} \end {gather*}

Antiderivative was successfully verified.

[In]

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

[Out]

(2*a^3*(d*x)^(3/2)*Sqrt[a^2 + 2*a*b*x^2 + b^2*x^4])/(3*d*(a + b*x^2)) + (6*a^2*b*(d*x)^(7/2)*Sqrt[a^2 + 2*a*b*
x^2 + b^2*x^4])/(7*d^3*(a + b*x^2)) + (6*a*b^2*(d*x)^(11/2)*Sqrt[a^2 + 2*a*b*x^2 + b^2*x^4])/(11*d^5*(a + b*x^
2)) + (2*b^3*(d*x)^(15/2)*Sqrt[a^2 + 2*a*b*x^2 + b^2*x^4])/(15*d^7*(a + b*x^2))

Rule 270

Int[((c_.)*(x_))^(m_.)*((a_) + (b_.)*(x_)^(n_))^(p_.), x_Symbol] :> Int[ExpandIntegrand[(c*x)^m*(a + b*x^n)^p,
x], x] /; FreeQ[{a, b, c, m, n}, x] && IGtQ[p, 0]

Rule 1112

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 \sqrt {d x} \left (a^2+2 a b x^2+b^2 x^4\right )^{3/2} \, dx &=\frac {\sqrt {a^2+2 a b x^2+b^2 x^4} \int \sqrt {d x} \left (a b+b^2 x^2\right )^3 \, dx}{b^2 \left (a b+b^2 x^2\right )}\\ &=\frac {\sqrt {a^2+2 a b x^2+b^2 x^4} \int \left (a^3 b^3 \sqrt {d x}+\frac {3 a^2 b^4 (d x)^{5/2}}{d^2}+\frac {3 a b^5 (d x)^{9/2}}{d^4}+\frac {b^6 (d x)^{13/2}}{d^6}\right ) \, dx}{b^2 \left (a b+b^2 x^2\right )}\\ &=\frac {2 a^3 (d x)^{3/2} \sqrt {a^2+2 a b x^2+b^2 x^4}}{3 d \left (a+b x^2\right )}+\frac {6 a^2 b (d x)^{7/2} \sqrt {a^2+2 a b x^2+b^2 x^4}}{7 d^3 \left (a+b x^2\right )}+\frac {6 a b^2 (d x)^{11/2} \sqrt {a^2+2 a b x^2+b^2 x^4}}{11 d^5 \left (a+b x^2\right )}+\frac {2 b^3 (d x)^{15/2} \sqrt {a^2+2 a b x^2+b^2 x^4}}{15 d^7 \left (a+b x^2\right )}\\ \end {align*}

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Mathematica [A]  time = 0.02, size = 66, normalized size = 0.34 \begin {gather*} \frac {2 \sqrt {d x} \sqrt {\left (a+b x^2\right )^2} \left (385 a^3 x+495 a^2 b x^3+315 a b^2 x^5+77 b^3 x^7\right )}{1155 \left (a+b x^2\right )} \end {gather*}

Antiderivative was successfully verified.

[In]

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

[Out]

(2*Sqrt[d*x]*Sqrt[(a + b*x^2)^2]*(385*a^3*x + 495*a^2*b*x^3 + 315*a*b^2*x^5 + 77*b^3*x^7))/(1155*(a + b*x^2))

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IntegrateAlgebraic [A]  time = 83.78, size = 96, normalized size = 0.49 \begin {gather*} \frac {2 (d x)^{3/2} \left (a d^2+b d^2 x^2\right ) \left (385 a^3 d^6+495 a^2 b d^6 x^2+315 a b^2 d^6 x^4+77 b^3 d^6 x^6\right )}{1155 d^9 \sqrt {\frac {\left (a d^2+b d^2 x^2\right )^2}{d^4}}} \end {gather*}

Antiderivative was successfully verified.

[In]

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

[Out]

(2*(d*x)^(3/2)*(a*d^2 + b*d^2*x^2)*(385*a^3*d^6 + 495*a^2*b*d^6*x^2 + 315*a*b^2*d^6*x^4 + 77*b^3*d^6*x^6))/(11
55*d^9*Sqrt[(a*d^2 + b*d^2*x^2)^2/d^4])

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fricas [A]  time = 1.77, size = 40, normalized size = 0.21 \begin {gather*} \frac {2}{1155} \, {\left (77 \, b^{3} x^{7} + 315 \, a b^{2} x^{5} + 495 \, a^{2} b x^{3} + 385 \, a^{3} x\right )} \sqrt {d x} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate((b^2*x^4+2*a*b*x^2+a^2)^(3/2)*(d*x)^(1/2),x, algorithm="fricas")

[Out]

2/1155*(77*b^3*x^7 + 315*a*b^2*x^5 + 495*a^2*b*x^3 + 385*a^3*x)*sqrt(d*x)

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giac [A]  time = 0.16, size = 85, normalized size = 0.44 \begin {gather*} \frac {2}{15} \, \sqrt {d x} b^{3} x^{7} \mathrm {sgn}\left (b x^{2} + a\right ) + \frac {6}{11} \, \sqrt {d x} a b^{2} x^{5} \mathrm {sgn}\left (b x^{2} + a\right ) + \frac {6}{7} \, \sqrt {d x} a^{2} b x^{3} \mathrm {sgn}\left (b x^{2} + a\right ) + \frac {2}{3} \, \sqrt {d x} a^{3} x \mathrm {sgn}\left (b x^{2} + a\right ) \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate((b^2*x^4+2*a*b*x^2+a^2)^(3/2)*(d*x)^(1/2),x, algorithm="giac")

[Out]

2/15*sqrt(d*x)*b^3*x^7*sgn(b*x^2 + a) + 6/11*sqrt(d*x)*a*b^2*x^5*sgn(b*x^2 + a) + 6/7*sqrt(d*x)*a^2*b*x^3*sgn(
b*x^2 + a) + 2/3*sqrt(d*x)*a^3*x*sgn(b*x^2 + a)

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maple [A]  time = 0.01, size = 61, normalized size = 0.31 \begin {gather*} \frac {2 \left (77 b^{3} x^{6}+315 a \,b^{2} x^{4}+495 a^{2} b \,x^{2}+385 a^{3}\right ) \left (\left (b \,x^{2}+a \right )^{2}\right )^{\frac {3}{2}} \sqrt {d x}\, x}{1155 \left (b \,x^{2}+a \right )^{3}} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

int((b^2*x^4+2*a*b*x^2+a^2)^(3/2)*(d*x)^(1/2),x)

[Out]

2/1155*x*(77*b^3*x^6+315*a*b^2*x^4+495*a^2*b*x^2+385*a^3)*((b*x^2+a)^2)^(3/2)*(d*x)^(1/2)/(b*x^2+a)^3

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maxima [A]  time = 1.44, size = 83, normalized size = 0.43 \begin {gather*} \frac {2}{165} \, {\left (11 \, b^{3} \sqrt {d} x^{3} + 15 \, a b^{2} \sqrt {d} x\right )} x^{\frac {9}{2}} + \frac {4}{77} \, {\left (7 \, a b^{2} \sqrt {d} x^{3} + 11 \, a^{2} b \sqrt {d} x\right )} x^{\frac {5}{2}} + \frac {2}{21} \, {\left (3 \, a^{2} b \sqrt {d} x^{3} + 7 \, a^{3} \sqrt {d} x\right )} \sqrt {x} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate((b^2*x^4+2*a*b*x^2+a^2)^(3/2)*(d*x)^(1/2),x, algorithm="maxima")

[Out]

2/165*(11*b^3*sqrt(d)*x^3 + 15*a*b^2*sqrt(d)*x)*x^(9/2) + 4/77*(7*a*b^2*sqrt(d)*x^3 + 11*a^2*b*sqrt(d)*x)*x^(5
/2) + 2/21*(3*a^2*b*sqrt(d)*x^3 + 7*a^3*sqrt(d)*x)*sqrt(x)

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mupad [F]  time = 0.00, size = -1, normalized size = -0.01 \begin {gather*} \int \sqrt {d\,x}\,{\left (a^2+2\,a\,b\,x^2+b^2\,x^4\right )}^{3/2} \,d x \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

int((d*x)^(1/2)*(a^2 + b^2*x^4 + 2*a*b*x^2)^(3/2),x)

[Out]

int((d*x)^(1/2)*(a^2 + b^2*x^4 + 2*a*b*x^2)^(3/2), x)

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sympy [F]  time = 0.00, size = 0, normalized size = 0.00 \begin {gather*} \int \sqrt {d x} \left (\left (a + b x^{2}\right )^{2}\right )^{\frac {3}{2}}\, dx \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate((b**2*x**4+2*a*b*x**2+a**2)**(3/2)*(d*x)**(1/2),x)

[Out]

Integral(sqrt(d*x)*((a + b*x**2)**2)**(3/2), x)

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