3.740 \(\int \frac{(a^2+2 a b x^2+b^2 x^4)^{3/2}}{(d x)^{5/2}} \, dx\)

Optimal. Leaf size=193 \[ \frac{2 b^3 (d x)^{9/2} \sqrt{a^2+2 a b x^2+b^2 x^4}}{9 d^7 \left (a+b x^2\right )}+\frac{6 a b^2 (d x)^{5/2} \sqrt{a^2+2 a b x^2+b^2 x^4}}{5 d^5 \left (a+b x^2\right )}+\frac{6 a^2 b \sqrt{d x} \sqrt{a^2+2 a b x^2+b^2 x^4}}{d^3 \left (a+b x^2\right )}-\frac{2 a^3 \sqrt{a^2+2 a b x^2+b^2 x^4}}{3 d (d x)^{3/2} \left (a+b x^2\right )} \]

[Out]

(-2*a^3*Sqrt[a^2 + 2*a*b*x^2 + b^2*x^4])/(3*d*(d*x)^(3/2)*(a + b*x^2)) + (6*a^2*b*Sqrt[d*x]*Sqrt[a^2 + 2*a*b*x
^2 + b^2*x^4])/(d^3*(a + b*x^2)) + (6*a*b^2*(d*x)^(5/2)*Sqrt[a^2 + 2*a*b*x^2 + b^2*x^4])/(5*d^5*(a + b*x^2)) +
(2*b^3*(d*x)^(9/2)*Sqrt[a^2 + 2*a*b*x^2 + b^2*x^4])/(9*d^7*(a + b*x^2))

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Rubi [A]  time = 0.0547985, antiderivative size = 193, normalized size of antiderivative = 1., 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} \[ \frac{2 b^3 (d x)^{9/2} \sqrt{a^2+2 a b x^2+b^2 x^4}}{9 d^7 \left (a+b x^2\right )}+\frac{6 a b^2 (d x)^{5/2} \sqrt{a^2+2 a b x^2+b^2 x^4}}{5 d^5 \left (a+b x^2\right )}+\frac{6 a^2 b \sqrt{d x} \sqrt{a^2+2 a b x^2+b^2 x^4}}{d^3 \left (a+b x^2\right )}-\frac{2 a^3 \sqrt{a^2+2 a b x^2+b^2 x^4}}{3 d (d x)^{3/2} \left (a+b x^2\right )} \]

Antiderivative was successfully verified.

[In]

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

[Out]

(-2*a^3*Sqrt[a^2 + 2*a*b*x^2 + b^2*x^4])/(3*d*(d*x)^(3/2)*(a + b*x^2)) + (6*a^2*b*Sqrt[d*x]*Sqrt[a^2 + 2*a*b*x
^2 + b^2*x^4])/(d^3*(a + b*x^2)) + (6*a*b^2*(d*x)^(5/2)*Sqrt[a^2 + 2*a*b*x^2 + b^2*x^4])/(5*d^5*(a + b*x^2)) +
(2*b^3*(d*x)^(9/2)*Sqrt[a^2 + 2*a*b*x^2 + b^2*x^4])/(9*d^7*(a + b*x^2))

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]

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]

Rubi steps

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

Mathematica [A]  time = 0.0269476, size = 66, normalized size = 0.34 \[ \frac{2 x \sqrt{\left (a+b x^2\right )^2} \left (135 a^2 b x^2-15 a^3+27 a b^2 x^4+5 b^3 x^6\right )}{45 (d x)^{5/2} \left (a+b x^2\right )} \]

Antiderivative was successfully verified.

[In]

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

[Out]

(2*x*Sqrt[(a + b*x^2)^2]*(-15*a^3 + 135*a^2*b*x^2 + 27*a*b^2*x^4 + 5*b^3*x^6))/(45*(d*x)^(5/2)*(a + b*x^2))

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Maple [A]  time = 0.17, size = 61, normalized size = 0.3 \begin{align*} -{\frac{2\, \left ( -5\,{b}^{3}{x}^{6}-27\,a{x}^{4}{b}^{2}-135\,{a}^{2}b{x}^{2}+15\,{a}^{3} \right ) x}{45\, \left ( b{x}^{2}+a \right ) ^{3}} \left ( \left ( b{x}^{2}+a \right ) ^{2} \right ) ^{{\frac{3}{2}}} \left ( dx \right ) ^{-{\frac{5}{2}}}} \end{align*}

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)^(5/2),x)

[Out]

-2/45*x*(-5*b^3*x^6-27*a*b^2*x^4-135*a^2*b*x^2+15*a^3)*((b*x^2+a)^2)^(3/2)/(b*x^2+a)^3/(d*x)^(5/2)

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Maxima [A]  time = 1.02564, size = 116, normalized size = 0.6 \begin{align*} \frac{2 \,{\left ({\left (5 \, b^{3} \sqrt{d} x^{3} + 9 \, a b^{2} \sqrt{d} x\right )} x^{\frac{3}{2}} + \frac{18 \,{\left (a b^{2} \sqrt{d} x^{3} + 5 \, a^{2} b \sqrt{d} x\right )}}{\sqrt{x}} + \frac{15 \,{\left (3 \, a^{2} b \sqrt{d} x^{3} - a^{3} \sqrt{d} x\right )}}{x^{\frac{5}{2}}}\right )}}{45 \, d^{3}} \end{align*}

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)^(5/2),x, algorithm="maxima")

[Out]

2/45*((5*b^3*sqrt(d)*x^3 + 9*a*b^2*sqrt(d)*x)*x^(3/2) + 18*(a*b^2*sqrt(d)*x^3 + 5*a^2*b*sqrt(d)*x)/sqrt(x) + 1
5*(3*a^2*b*sqrt(d)*x^3 - a^3*sqrt(d)*x)/x^(5/2))/d^3

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Fricas [A]  time = 1.49424, size = 105, normalized size = 0.54 \begin{align*} \frac{2 \,{\left (5 \, b^{3} x^{6} + 27 \, a b^{2} x^{4} + 135 \, a^{2} b x^{2} - 15 \, a^{3}\right )} \sqrt{d x}}{45 \, d^{3} x^{2}} \end{align*}

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)^(5/2),x, algorithm="fricas")

[Out]

2/45*(5*b^3*x^6 + 27*a*b^2*x^4 + 135*a^2*b*x^2 - 15*a^3)*sqrt(d*x)/(d^3*x^2)

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Sympy [F]  time = 0., size = 0, normalized size = 0. \begin{align*} \int \frac{\left (\left (a + b x^{2}\right )^{2}\right )^{\frac{3}{2}}}{\left (d x\right )^{\frac{5}{2}}}\, dx \end{align*}

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)**(5/2),x)

[Out]

Integral(((a + b*x**2)**2)**(3/2)/(d*x)**(5/2), x)

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Giac [A]  time = 1.28162, size = 142, normalized size = 0.74 \begin{align*} -\frac{2 \,{\left (\frac{15 \, a^{3} d \mathrm{sgn}\left (b x^{2} + a\right )}{\sqrt{d x} x} - \frac{5 \, \sqrt{d x} b^{3} d^{36} x^{4} \mathrm{sgn}\left (b x^{2} + a\right ) + 27 \, \sqrt{d x} a b^{2} d^{36} x^{2} \mathrm{sgn}\left (b x^{2} + a\right ) + 135 \, \sqrt{d x} a^{2} b d^{36} \mathrm{sgn}\left (b x^{2} + a\right )}{d^{36}}\right )}}{45 \, d^{3}} \end{align*}

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)^(5/2),x, algorithm="giac")

[Out]

-2/45*(15*a^3*d*sgn(b*x^2 + a)/(sqrt(d*x)*x) - (5*sqrt(d*x)*b^3*d^36*x^4*sgn(b*x^2 + a) + 27*sqrt(d*x)*a*b^2*d
^36*x^2*sgn(b*x^2 + a) + 135*sqrt(d*x)*a^2*b*d^36*sgn(b*x^2 + a))/d^36)/d^3