3.741 \(\int \frac{(a^2+2 a b x^2+b^2 x^4)^{3/2}}{(d x)^{7/2}} \, dx\)

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

[Out]

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

________________________________________________________________________________________

Rubi [A]  time = 0.0552309, antiderivative size = 191, 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} \[ \frac{2 b^3 (d x)^{7/2} \sqrt{a^2+2 a b x^2+b^2 x^4}}{7 d^7 \left (a+b x^2\right )}+\frac{2 a b^2 (d x)^{3/2} \sqrt{a^2+2 a b x^2+b^2 x^4}}{d^5 \left (a+b x^2\right )}-\frac{6 a^2 b \sqrt{a^2+2 a b x^2+b^2 x^4}}{d^3 \sqrt{d x} \left (a+b x^2\right )}-\frac{2 a^3 \sqrt{a^2+2 a b x^2+b^2 x^4}}{5 d (d x)^{5/2} \left (a+b x^2\right )} \]

Antiderivative was successfully verified.

[In]

Int[(a^2 + 2*a*b*x^2 + b^2*x^4)^(3/2)/(d*x)^(7/2),x]

[Out]

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

Mathematica [A]  time = 0.0260261, size = 66, normalized size = 0.35 \[ \frac{2 x \sqrt{\left (a+b x^2\right )^2} \left (-105 a^2 b x^2-7 a^3+35 a b^2 x^4+5 b^3 x^6\right )}{35 (d x)^{7/2} \left (a+b x^2\right )} \]

Antiderivative was successfully verified.

[In]

Integrate[(a^2 + 2*a*b*x^2 + b^2*x^4)^(3/2)/(d*x)^(7/2),x]

[Out]

(2*x*Sqrt[(a + b*x^2)^2]*(-7*a^3 - 105*a^2*b*x^2 + 35*a*b^2*x^4 + 5*b^3*x^6))/(35*(d*x)^(7/2)*(a + b*x^2))

________________________________________________________________________________________

Maple [A]  time = 0.17, size = 61, normalized size = 0.3 \begin{align*} -{\frac{2\, \left ( -5\,{b}^{3}{x}^{6}-35\,a{x}^{4}{b}^{2}+105\,{a}^{2}b{x}^{2}+7\,{a}^{3} \right ) x}{35\, \left ( b{x}^{2}+a \right ) ^{3}} \left ( \left ( b{x}^{2}+a \right ) ^{2} \right ) ^{{\frac{3}{2}}} \left ( dx \right ) ^{-{\frac{7}{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)^(7/2),x)

[Out]

-2/35*x*(-5*b^3*x^6-35*a*b^2*x^4+105*a^2*b*x^2+7*a^3)*((b*x^2+a)^2)^(3/2)/(b*x^2+a)^3/(d*x)^(7/2)

________________________________________________________________________________________

Maxima [A]  time = 1.00227, size = 116, normalized size = 0.61 \begin{align*} \frac{2 \,{\left (5 \,{\left (3 \, b^{3} \sqrt{d} x^{3} + 7 \, a b^{2} \sqrt{d} x\right )} \sqrt{x} + \frac{70 \,{\left (a b^{2} \sqrt{d} x^{3} - 3 \, a^{2} b \sqrt{d} x\right )}}{x^{\frac{3}{2}}} - \frac{21 \,{\left (5 \, a^{2} b \sqrt{d} x^{3} + a^{3} \sqrt{d} x\right )}}{x^{\frac{7}{2}}}\right )}}{105 \, d^{4}} \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)^(7/2),x, algorithm="maxima")

[Out]

2/105*(5*(3*b^3*sqrt(d)*x^3 + 7*a*b^2*sqrt(d)*x)*sqrt(x) + 70*(a*b^2*sqrt(d)*x^3 - 3*a^2*b*sqrt(d)*x)/x^(3/2)
- 21*(5*a^2*b*sqrt(d)*x^3 + a^3*sqrt(d)*x)/x^(7/2))/d^4

________________________________________________________________________________________

Fricas [A]  time = 1.4689, size = 104, normalized size = 0.54 \begin{align*} \frac{2 \,{\left (5 \, b^{3} x^{6} + 35 \, a b^{2} x^{4} - 105 \, a^{2} b x^{2} - 7 \, a^{3}\right )} \sqrt{d x}}{35 \, d^{4} x^{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)^(7/2),x, algorithm="fricas")

[Out]

2/35*(5*b^3*x^6 + 35*a*b^2*x^4 - 105*a^2*b*x^2 - 7*a^3)*sqrt(d*x)/(d^4*x^3)

________________________________________________________________________________________

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{7}{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)**(7/2),x)

[Out]

Integral(((a + b*x**2)**2)**(3/2)/(d*x)**(7/2), x)

________________________________________________________________________________________

Giac [A]  time = 1.2064, size = 144, normalized size = 0.75 \begin{align*} -\frac{2 \,{\left (\frac{7 \,{\left (15 \, a^{2} b d^{3} x^{2} \mathrm{sgn}\left (b x^{2} + a\right ) + a^{3} d^{3} \mathrm{sgn}\left (b x^{2} + a\right )\right )}}{\sqrt{d x} d^{2} x^{2}} - \frac{5 \,{\left (\sqrt{d x} b^{3} d^{21} x^{3} \mathrm{sgn}\left (b x^{2} + a\right ) + 7 \, \sqrt{d x} a b^{2} d^{21} x \mathrm{sgn}\left (b x^{2} + a\right )\right )}}{d^{21}}\right )}}{35 \, d^{4}} \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)^(7/2),x, algorithm="giac")

[Out]

-2/35*(7*(15*a^2*b*d^3*x^2*sgn(b*x^2 + a) + a^3*d^3*sgn(b*x^2 + a))/(sqrt(d*x)*d^2*x^2) - 5*(sqrt(d*x)*b^3*d^2
1*x^3*sgn(b*x^2 + a) + 7*sqrt(d*x)*a*b^2*d^21*x*sgn(b*x^2 + a))/d^21)/d^4