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

Optimal. Leaf size=191 \[ \frac {6 a b^2 (d x)^{7/2} \sqrt {a^2+2 a b x^2+b^2 x^4}}{7 d^5 \left (a+b x^2\right )}+\frac {2 a^2 b (d x)^{3/2} \sqrt {a^2+2 a b x^2+b^2 x^4}}{d^3 \left (a+b x^2\right )}+\frac {2 b^3 (d x)^{11/2} \sqrt {a^2+2 a b x^2+b^2 x^4}}{11 d^7 \left (a+b x^2\right )}-\frac {2 a^3 \sqrt {a^2+2 a b x^2+b^2 x^4}}{d \sqrt {d x} \left (a+b x^2\right )} \]

[Out]

2*a^2*b*(d*x)^(3/2)*((b*x^2+a)^2)^(1/2)/d^3/(b*x^2+a)+6/7*a*b^2*(d*x)^(7/2)*((b*x^2+a)^2)^(1/2)/d^5/(b*x^2+a)+
2/11*b^3*(d*x)^(11/2)*((b*x^2+a)^2)^(1/2)/d^7/(b*x^2+a)-2*a^3*((b*x^2+a)^2)^(1/2)/d/(b*x^2+a)/(d*x)^(1/2)

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Rubi [A]  time = 0.06, antiderivative size = 191, 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} \[ \frac {2 b^3 (d x)^{11/2} \sqrt {a^2+2 a b x^2+b^2 x^4}}{11 d^7 \left (a+b x^2\right )}+\frac {6 a b^2 (d x)^{7/2} \sqrt {a^2+2 a b x^2+b^2 x^4}}{7 d^5 \left (a+b x^2\right )}+\frac {2 a^2 b (d x)^{3/2} \sqrt {a^2+2 a b x^2+b^2 x^4}}{d^3 \left (a+b x^2\right )}-\frac {2 a^3 \sqrt {a^2+2 a b x^2+b^2 x^4}}{d \sqrt {d x} \left (a+b x^2\right )} \]

Antiderivative was successfully verified.

[In]

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

[Out]

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

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Mathematica [A]  time = 0.02, size = 66, normalized size = 0.35 \[ \frac {2 x \sqrt {\left (a+b x^2\right )^2} \left (-77 a^3+77 a^2 b x^2+33 a b^2 x^4+7 b^3 x^6\right )}{77 (d x)^{3/2} \left (a+b x^2\right )} \]

Antiderivative was successfully verified.

[In]

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

[Out]

(2*x*Sqrt[(a + b*x^2)^2]*(-77*a^3 + 77*a^2*b*x^2 + 33*a*b^2*x^4 + 7*b^3*x^6))/(77*(d*x)^(3/2)*(a + b*x^2))

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fricas [A]  time = 1.06, size = 45, normalized size = 0.24 \[ \frac {2 \, {\left (7 \, b^{3} x^{6} + 33 \, a b^{2} x^{4} + 77 \, a^{2} b x^{2} - 77 \, a^{3}\right )} \sqrt {d x}}{77 \, d^{2} x} \]

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)^(3/2),x, algorithm="fricas")

[Out]

2/77*(7*b^3*x^6 + 33*a*b^2*x^4 + 77*a^2*b*x^2 - 77*a^3)*sqrt(d*x)/(d^2*x)

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giac [A]  time = 0.17, size = 102, normalized size = 0.53 \[ -\frac {2 \, {\left (\frac {77 \, a^{3} \mathrm {sgn}\left (b x^{2} + a\right )}{\sqrt {d x}} - \frac {7 \, \sqrt {d x} b^{3} d^{65} x^{5} \mathrm {sgn}\left (b x^{2} + a\right ) + 33 \, \sqrt {d x} a b^{2} d^{65} x^{3} \mathrm {sgn}\left (b x^{2} + a\right ) + 77 \, \sqrt {d x} a^{2} b d^{65} x \mathrm {sgn}\left (b x^{2} + a\right )}{d^{66}}\right )}}{77 \, d} \]

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)^(3/2),x, algorithm="giac")

[Out]

-2/77*(77*a^3*sgn(b*x^2 + a)/sqrt(d*x) - (7*sqrt(d*x)*b^3*d^65*x^5*sgn(b*x^2 + a) + 33*sqrt(d*x)*a*b^2*d^65*x^
3*sgn(b*x^2 + a) + 77*sqrt(d*x)*a^2*b*d^65*x*sgn(b*x^2 + a))/d^66)/d

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maple [A]  time = 0.01, size = 61, normalized size = 0.32 \[ -\frac {2 \left (-7 b^{3} x^{6}-33 a \,b^{2} x^{4}-77 a^{2} b \,x^{2}+77 a^{3}\right ) \left (\left (b \,x^{2}+a \right )^{2}\right )^{\frac {3}{2}} x}{77 \left (b \,x^{2}+a \right )^{3} \left (d x \right )^{\frac {3}{2}}} \]

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)^(3/2),x)

[Out]

-2/77*x*(-7*b^3*x^6-33*a*b^2*x^4-77*a^2*b*x^2+77*a^3)*((b*x^2+a)^2)^(3/2)/(b*x^2+a)^3/(d*x)^(3/2)

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maxima [A]  time = 1.42, size = 87, normalized size = 0.46 \[ \frac {2 \, {\left (3 \, {\left (7 \, b^{3} \sqrt {d} x^{3} + 11 \, a b^{2} \sqrt {d} x\right )} x^{\frac {5}{2}} + 22 \, {\left (3 \, a b^{2} \sqrt {d} x^{3} + 7 \, a^{2} b \sqrt {d} x\right )} \sqrt {x} + \frac {77 \, {\left (a^{2} b \sqrt {d} x^{3} - 3 \, a^{3} \sqrt {d} x\right )}}{x^{\frac {3}{2}}}\right )}}{231 \, d^{2}} \]

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)^(3/2),x, algorithm="maxima")

[Out]

2/231*(3*(7*b^3*sqrt(d)*x^3 + 11*a*b^2*sqrt(d)*x)*x^(5/2) + 22*(3*a*b^2*sqrt(d)*x^3 + 7*a^2*b*sqrt(d)*x)*sqrt(
x) + 77*(a^2*b*sqrt(d)*x^3 - 3*a^3*sqrt(d)*x)/x^(3/2))/d^2

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mupad [B]  time = 4.53, size = 87, normalized size = 0.46 \[ \frac {\sqrt {a^2+2\,a\,b\,x^2+b^2\,x^4}\,\left (\frac {2\,a^2\,x^2}{d}-\frac {2\,a^3}{b\,d}+\frac {2\,b^2\,x^6}{11\,d}+\frac {6\,a\,b\,x^4}{7\,d}\right )}{x^2\,\sqrt {d\,x}+\frac {a\,\sqrt {d\,x}}{b}} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

((a^2 + b^2*x^4 + 2*a*b*x^2)^(1/2)*((2*a^2*x^2)/d - (2*a^3)/(b*d) + (2*b^2*x^6)/(11*d) + (6*a*b*x^4)/(7*d)))/(
x^2*(d*x)^(1/2) + (a*(d*x)^(1/2))/b)

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

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)**(3/2),x)

[Out]

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

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