3.748 \(\int \frac{(a^2+2 a b x^2+b^2 x^4)^{5/2}}{(d x)^{7/2}} \, dx\)

Optimal. Leaf size=295 \[ \frac{2 b^5 (d x)^{15/2} \sqrt{a^2+2 a b x^2+b^2 x^4}}{15 d^{11} \left (a+b x^2\right )}+\frac{10 a b^4 (d x)^{11/2} \sqrt{a^2+2 a b x^2+b^2 x^4}}{11 d^9 \left (a+b x^2\right )}+\frac{20 a^2 b^3 (d x)^{7/2} \sqrt{a^2+2 a b x^2+b^2 x^4}}{7 d^7 \left (a+b x^2\right )}+\frac{20 a^3 b^2 (d x)^{3/2} \sqrt{a^2+2 a b x^2+b^2 x^4}}{3 d^5 \left (a+b x^2\right )}-\frac{10 a^4 b \sqrt{a^2+2 a b x^2+b^2 x^4}}{d^3 \sqrt{d x} \left (a+b x^2\right )}-\frac{2 a^5 \sqrt{a^2+2 a b x^2+b^2 x^4}}{5 d (d x)^{5/2} \left (a+b x^2\right )} \]

[Out]

(-2*a^5*Sqrt[a^2 + 2*a*b*x^2 + b^2*x^4])/(5*d*(d*x)^(5/2)*(a + b*x^2)) - (10*a^4*b*Sqrt[a^2 + 2*a*b*x^2 + b^2*
x^4])/(d^3*Sqrt[d*x]*(a + b*x^2)) + (20*a^3*b^2*(d*x)^(3/2)*Sqrt[a^2 + 2*a*b*x^2 + b^2*x^4])/(3*d^5*(a + b*x^2
)) + (20*a^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)) + (10*a*b^4*(d*x)^(11/2)*Sqr
t[a^2 + 2*a*b*x^2 + b^2*x^4])/(11*d^9*(a + b*x^2)) + (2*b^5*(d*x)^(15/2)*Sqrt[a^2 + 2*a*b*x^2 + b^2*x^4])/(15*
d^11*(a + b*x^2))

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Rubi [A]  time = 0.0792314, antiderivative size = 295, 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} \[ \frac{2 b^5 (d x)^{15/2} \sqrt{a^2+2 a b x^2+b^2 x^4}}{15 d^{11} \left (a+b x^2\right )}+\frac{10 a b^4 (d x)^{11/2} \sqrt{a^2+2 a b x^2+b^2 x^4}}{11 d^9 \left (a+b x^2\right )}+\frac{20 a^2 b^3 (d x)^{7/2} \sqrt{a^2+2 a b x^2+b^2 x^4}}{7 d^7 \left (a+b x^2\right )}+\frac{20 a^3 b^2 (d x)^{3/2} \sqrt{a^2+2 a b x^2+b^2 x^4}}{3 d^5 \left (a+b x^2\right )}-\frac{10 a^4 b \sqrt{a^2+2 a b x^2+b^2 x^4}}{d^3 \sqrt{d x} \left (a+b x^2\right )}-\frac{2 a^5 \sqrt{a^2+2 a b x^2+b^2 x^4}}{5 d (d x)^{5/2} \left (a+b x^2\right )} \]

Antiderivative was successfully verified.

[In]

Int[(a^2 + 2*a*b*x^2 + b^2*x^4)^(5/2)/(d*x)^(7/2),x]

[Out]

(-2*a^5*Sqrt[a^2 + 2*a*b*x^2 + b^2*x^4])/(5*d*(d*x)^(5/2)*(a + b*x^2)) - (10*a^4*b*Sqrt[a^2 + 2*a*b*x^2 + b^2*
x^4])/(d^3*Sqrt[d*x]*(a + b*x^2)) + (20*a^3*b^2*(d*x)^(3/2)*Sqrt[a^2 + 2*a*b*x^2 + b^2*x^4])/(3*d^5*(a + b*x^2
)) + (20*a^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)) + (10*a*b^4*(d*x)^(11/2)*Sqr
t[a^2 + 2*a*b*x^2 + b^2*x^4])/(11*d^9*(a + b*x^2)) + (2*b^5*(d*x)^(15/2)*Sqrt[a^2 + 2*a*b*x^2 + b^2*x^4])/(15*
d^11*(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 )^{5/2}}{(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 )^5}{(d x)^{7/2}} \, dx}{b^4 \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^5 b^5}{(d x)^{7/2}}+\frac{5 a^4 b^6}{d^2 (d x)^{3/2}}+\frac{10 a^3 b^7 \sqrt{d x}}{d^4}+\frac{10 a^2 b^8 (d x)^{5/2}}{d^6}+\frac{5 a b^9 (d x)^{9/2}}{d^8}+\frac{b^{10} (d x)^{13/2}}{d^{10}}\right ) \, dx}{b^4 \left (a b+b^2 x^2\right )}\\ &=-\frac{2 a^5 \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{10 a^4 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{20 a^3 b^2 (d x)^{3/2} \sqrt{a^2+2 a b x^2+b^2 x^4}}{3 d^5 \left (a+b x^2\right )}+\frac{20 a^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{10 a b^4 (d x)^{11/2} \sqrt{a^2+2 a b x^2+b^2 x^4}}{11 d^9 \left (a+b x^2\right )}+\frac{2 b^5 (d x)^{15/2} \sqrt{a^2+2 a b x^2+b^2 x^4}}{15 d^{11} \left (a+b x^2\right )}\\ \end{align*}

Mathematica [A]  time = 0.0355773, size = 88, normalized size = 0.3 \[ \frac{2 x \sqrt{\left (a+b x^2\right )^2} \left (1650 a^2 b^3 x^6+3850 a^3 b^2 x^4-5775 a^4 b x^2-231 a^5+525 a b^4 x^8+77 b^5 x^{10}\right )}{1155 (d x)^{7/2} \left (a+b x^2\right )} \]

Antiderivative was successfully verified.

[In]

Integrate[(a^2 + 2*a*b*x^2 + b^2*x^4)^(5/2)/(d*x)^(7/2),x]

[Out]

(2*x*Sqrt[(a + b*x^2)^2]*(-231*a^5 - 5775*a^4*b*x^2 + 3850*a^3*b^2*x^4 + 1650*a^2*b^3*x^6 + 525*a*b^4*x^8 + 77
*b^5*x^10))/(1155*(d*x)^(7/2)*(a + b*x^2))

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Maple [A]  time = 0.171, size = 83, normalized size = 0.3 \begin{align*} -{\frac{2\, \left ( -77\,{b}^{5}{x}^{10}-525\,a{b}^{4}{x}^{8}-1650\,{a}^{2}{b}^{3}{x}^{6}-3850\,{b}^{2}{a}^{3}{x}^{4}+5775\,{a}^{4}b{x}^{2}+231\,{a}^{5} \right ) x}{1155\, \left ( b{x}^{2}+a \right ) ^{5}} \left ( \left ( b{x}^{2}+a \right ) ^{2} \right ) ^{{\frac{5}{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)^(5/2)/(d*x)^(7/2),x)

[Out]

-2/1155*x*(-77*b^5*x^10-525*a*b^4*x^8-1650*a^2*b^3*x^6-3850*a^3*b^2*x^4+5775*a^4*b*x^2+231*a^5)*((b*x^2+a)^2)^
(5/2)/(b*x^2+a)^5/(d*x)^(7/2)

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Maxima [A]  time = 1.01593, size = 203, normalized size = 0.69 \begin{align*} \frac{2 \,{\left (7 \,{\left (11 \, b^{5} \sqrt{d} x^{3} + 15 \, a b^{4} \sqrt{d} x\right )} x^{\frac{9}{2}} + 60 \,{\left (7 \, a b^{4} \sqrt{d} x^{3} + 11 \, a^{2} b^{3} \sqrt{d} x\right )} x^{\frac{5}{2}} + 330 \,{\left (3 \, a^{2} b^{3} \sqrt{d} x^{3} + 7 \, a^{3} b^{2} \sqrt{d} x\right )} \sqrt{x} + \frac{1540 \,{\left (a^{3} b^{2} \sqrt{d} x^{3} - 3 \, a^{4} b \sqrt{d} x\right )}}{x^{\frac{3}{2}}} - \frac{231 \,{\left (5 \, a^{4} b \sqrt{d} x^{3} + a^{5} \sqrt{d} x\right )}}{x^{\frac{7}{2}}}\right )}}{1155 \, 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)^(5/2)/(d*x)^(7/2),x, algorithm="maxima")

[Out]

2/1155*(7*(11*b^5*sqrt(d)*x^3 + 15*a*b^4*sqrt(d)*x)*x^(9/2) + 60*(7*a*b^4*sqrt(d)*x^3 + 11*a^2*b^3*sqrt(d)*x)*
x^(5/2) + 330*(3*a^2*b^3*sqrt(d)*x^3 + 7*a^3*b^2*sqrt(d)*x)*sqrt(x) + 1540*(a^3*b^2*sqrt(d)*x^3 - 3*a^4*b*sqrt
(d)*x)/x^(3/2) - 231*(5*a^4*b*sqrt(d)*x^3 + a^5*sqrt(d)*x)/x^(7/2))/d^4

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Fricas [A]  time = 1.49153, size = 166, normalized size = 0.56 \begin{align*} \frac{2 \,{\left (77 \, b^{5} x^{10} + 525 \, a b^{4} x^{8} + 1650 \, a^{2} b^{3} x^{6} + 3850 \, a^{3} b^{2} x^{4} - 5775 \, a^{4} b x^{2} - 231 \, a^{5}\right )} \sqrt{d x}}{1155 \, 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)^(5/2)/(d*x)^(7/2),x, algorithm="fricas")

[Out]

2/1155*(77*b^5*x^10 + 525*a*b^4*x^8 + 1650*a^2*b^3*x^6 + 3850*a^3*b^2*x^4 - 5775*a^4*b*x^2 - 231*a^5)*sqrt(d*x
)/(d^4*x^3)

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Sympy [F(-1)]  time = 0., size = 0, normalized size = 0. \begin{align*} \text{Timed out} \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)**(5/2)/(d*x)**(7/2),x)

[Out]

Timed out

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Giac [A]  time = 1.26529, size = 219, normalized size = 0.74 \begin{align*} -\frac{2 \,{\left (\frac{231 \,{\left (25 \, a^{4} b d^{3} x^{2} \mathrm{sgn}\left (b x^{2} + a\right ) + a^{5} d^{3} \mathrm{sgn}\left (b x^{2} + a\right )\right )}}{\sqrt{d x} d^{2} x^{2}} - \frac{77 \, \sqrt{d x} b^{5} d^{105} x^{7} \mathrm{sgn}\left (b x^{2} + a\right ) + 525 \, \sqrt{d x} a b^{4} d^{105} x^{5} \mathrm{sgn}\left (b x^{2} + a\right ) + 1650 \, \sqrt{d x} a^{2} b^{3} d^{105} x^{3} \mathrm{sgn}\left (b x^{2} + a\right ) + 3850 \, \sqrt{d x} a^{3} b^{2} d^{105} x \mathrm{sgn}\left (b x^{2} + a\right )}{d^{105}}\right )}}{1155 \, 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)^(5/2)/(d*x)^(7/2),x, algorithm="giac")

[Out]

-2/1155*(231*(25*a^4*b*d^3*x^2*sgn(b*x^2 + a) + a^5*d^3*sgn(b*x^2 + a))/(sqrt(d*x)*d^2*x^2) - (77*sqrt(d*x)*b^
5*d^105*x^7*sgn(b*x^2 + a) + 525*sqrt(d*x)*a*b^4*d^105*x^5*sgn(b*x^2 + a) + 1650*sqrt(d*x)*a^2*b^3*d^105*x^3*s
gn(b*x^2 + a) + 3850*sqrt(d*x)*a^3*b^2*d^105*x*sgn(b*x^2 + a))/d^105)/d^4