3.625 \(\int x^3 (a+b x^2)^2 (c+d x^2)^{5/2} \, dx\)

Optimal. Leaf size=114 \[ -\frac{b \left (c+d x^2\right )^{11/2} (3 b c-2 a d)}{11 d^4}+\frac{\left (c+d x^2\right )^{9/2} (b c-a d) (3 b c-a d)}{9 d^4}-\frac{c \left (c+d x^2\right )^{7/2} (b c-a d)^2}{7 d^4}+\frac{b^2 \left (c+d x^2\right )^{13/2}}{13 d^4} \]

[Out]

-(c*(b*c - a*d)^2*(c + d*x^2)^(7/2))/(7*d^4) + ((b*c - a*d)*(3*b*c - a*d)*(c + d*x^2)^(9/2))/(9*d^4) - (b*(3*b
*c - 2*a*d)*(c + d*x^2)^(11/2))/(11*d^4) + (b^2*(c + d*x^2)^(13/2))/(13*d^4)

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Rubi [A]  time = 0.0897296, antiderivative size = 114, normalized size of antiderivative = 1., number of steps used = 3, number of rules used = 2, integrand size = 24, \(\frac{\text{number of rules}}{\text{integrand size}}\) = 0.083, Rules used = {446, 77} \[ -\frac{b \left (c+d x^2\right )^{11/2} (3 b c-2 a d)}{11 d^4}+\frac{\left (c+d x^2\right )^{9/2} (b c-a d) (3 b c-a d)}{9 d^4}-\frac{c \left (c+d x^2\right )^{7/2} (b c-a d)^2}{7 d^4}+\frac{b^2 \left (c+d x^2\right )^{13/2}}{13 d^4} \]

Antiderivative was successfully verified.

[In]

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

[Out]

-(c*(b*c - a*d)^2*(c + d*x^2)^(7/2))/(7*d^4) + ((b*c - a*d)*(3*b*c - a*d)*(c + d*x^2)^(9/2))/(9*d^4) - (b*(3*b
*c - 2*a*d)*(c + d*x^2)^(11/2))/(11*d^4) + (b^2*(c + d*x^2)^(13/2))/(13*d^4)

Rule 446

Int[(x_)^(m_.)*((a_) + (b_.)*(x_)^(n_))^(p_.)*((c_) + (d_.)*(x_)^(n_))^(q_.), x_Symbol] :> Dist[1/n, Subst[Int
[x^(Simplify[(m + 1)/n] - 1)*(a + b*x)^p*(c + d*x)^q, x], x, x^n], x] /; FreeQ[{a, b, c, d, m, n, p, q}, x] &&
NeQ[b*c - a*d, 0] && IntegerQ[Simplify[(m + 1)/n]]

Rule 77

Int[((a_.) + (b_.)*(x_))*((c_) + (d_.)*(x_))^(n_.)*((e_.) + (f_.)*(x_))^(p_.), x_Symbol] :> Int[ExpandIntegran
d[(a + b*x)*(c + d*x)^n*(e + f*x)^p, x], x] /; FreeQ[{a, b, c, d, e, f, n}, x] && NeQ[b*c - a*d, 0] && ((ILtQ[
n, 0] && ILtQ[p, 0]) || EqQ[p, 1] || (IGtQ[p, 0] && ( !IntegerQ[n] || LeQ[9*p + 5*(n + 2), 0] || GeQ[n + p + 1
, 0] || (GeQ[n + p + 2, 0] && RationalQ[a, b, c, d, e, f]))))

Rubi steps

\begin{align*} \int x^3 \left (a+b x^2\right )^2 \left (c+d x^2\right )^{5/2} \, dx &=\frac{1}{2} \operatorname{Subst}\left (\int x (a+b x)^2 (c+d x)^{5/2} \, dx,x,x^2\right )\\ &=\frac{1}{2} \operatorname{Subst}\left (\int \left (-\frac{c (b c-a d)^2 (c+d x)^{5/2}}{d^3}+\frac{(b c-a d) (3 b c-a d) (c+d x)^{7/2}}{d^3}-\frac{b (3 b c-2 a d) (c+d x)^{9/2}}{d^3}+\frac{b^2 (c+d x)^{11/2}}{d^3}\right ) \, dx,x,x^2\right )\\ &=-\frac{c (b c-a d)^2 \left (c+d x^2\right )^{7/2}}{7 d^4}+\frac{(b c-a d) (3 b c-a d) \left (c+d x^2\right )^{9/2}}{9 d^4}-\frac{b (3 b c-2 a d) \left (c+d x^2\right )^{11/2}}{11 d^4}+\frac{b^2 \left (c+d x^2\right )^{13/2}}{13 d^4}\\ \end{align*}

Mathematica [A]  time = 0.0982566, size = 99, normalized size = 0.87 \[ \frac{\left (c+d x^2\right )^{7/2} \left (143 a^2 d^2 \left (7 d x^2-2 c\right )+26 a b d \left (8 c^2-28 c d x^2+63 d^2 x^4\right )+b^2 \left (168 c^2 d x^2-48 c^3-378 c d^2 x^4+693 d^3 x^6\right )\right )}{9009 d^4} \]

Antiderivative was successfully verified.

[In]

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

[Out]

((c + d*x^2)^(7/2)*(143*a^2*d^2*(-2*c + 7*d*x^2) + 26*a*b*d*(8*c^2 - 28*c*d*x^2 + 63*d^2*x^4) + b^2*(-48*c^3 +
168*c^2*d*x^2 - 378*c*d^2*x^4 + 693*d^3*x^6)))/(9009*d^4)

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Maple [A]  time = 0.007, size = 108, normalized size = 1. \begin{align*} -{\frac{-693\,{b}^{2}{x}^{6}{d}^{3}-1638\,ab{d}^{3}{x}^{4}+378\,{b}^{2}c{d}^{2}{x}^{4}-1001\,{a}^{2}{d}^{3}{x}^{2}+728\,abc{d}^{2}{x}^{2}-168\,{b}^{2}{c}^{2}d{x}^{2}+286\,{a}^{2}c{d}^{2}-208\,ab{c}^{2}d+48\,{b}^{2}{c}^{3}}{9009\,{d}^{4}} \left ( d{x}^{2}+c \right ) ^{{\frac{7}{2}}}} \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

int(x^3*(b*x^2+a)^2*(d*x^2+c)^(5/2),x)

[Out]

-1/9009*(d*x^2+c)^(7/2)*(-693*b^2*d^3*x^6-1638*a*b*d^3*x^4+378*b^2*c*d^2*x^4-1001*a^2*d^3*x^2+728*a*b*c*d^2*x^
2-168*b^2*c^2*d*x^2+286*a^2*c*d^2-208*a*b*c^2*d+48*b^2*c^3)/d^4

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Maxima [F(-2)]  time = 0., size = 0, normalized size = 0. \begin{align*} \text{Exception raised: ValueError} \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate(x^3*(b*x^2+a)^2*(d*x^2+c)^(5/2),x, algorithm="maxima")

[Out]

Exception raised: ValueError

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Fricas [B]  time = 1.43238, size = 490, normalized size = 4.3 \begin{align*} \frac{{\left (693 \, b^{2} d^{6} x^{12} + 63 \,{\left (27 \, b^{2} c d^{5} + 26 \, a b d^{6}\right )} x^{10} + 7 \,{\left (159 \, b^{2} c^{2} d^{4} + 598 \, a b c d^{5} + 143 \, a^{2} d^{6}\right )} x^{8} - 48 \, b^{2} c^{6} + 208 \, a b c^{5} d - 286 \, a^{2} c^{4} d^{2} +{\left (15 \, b^{2} c^{3} d^{3} + 2938 \, a b c^{2} d^{4} + 2717 \, a^{2} c d^{5}\right )} x^{6} - 3 \,{\left (6 \, b^{2} c^{4} d^{2} - 26 \, a b c^{3} d^{3} - 715 \, a^{2} c^{2} d^{4}\right )} x^{4} +{\left (24 \, b^{2} c^{5} d - 104 \, a b c^{4} d^{2} + 143 \, a^{2} c^{3} d^{3}\right )} x^{2}\right )} \sqrt{d x^{2} + c}}{9009 \, d^{4}} \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate(x^3*(b*x^2+a)^2*(d*x^2+c)^(5/2),x, algorithm="fricas")

[Out]

1/9009*(693*b^2*d^6*x^12 + 63*(27*b^2*c*d^5 + 26*a*b*d^6)*x^10 + 7*(159*b^2*c^2*d^4 + 598*a*b*c*d^5 + 143*a^2*
d^6)*x^8 - 48*b^2*c^6 + 208*a*b*c^5*d - 286*a^2*c^4*d^2 + (15*b^2*c^3*d^3 + 2938*a*b*c^2*d^4 + 2717*a^2*c*d^5)
*x^6 - 3*(6*b^2*c^4*d^2 - 26*a*b*c^3*d^3 - 715*a^2*c^2*d^4)*x^4 + (24*b^2*c^5*d - 104*a*b*c^4*d^2 + 143*a^2*c^
3*d^3)*x^2)*sqrt(d*x^2 + c)/d^4

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Sympy [A]  time = 9.42471, size = 468, normalized size = 4.11 \begin{align*} \begin{cases} - \frac{2 a^{2} c^{4} \sqrt{c + d x^{2}}}{63 d^{2}} + \frac{a^{2} c^{3} x^{2} \sqrt{c + d x^{2}}}{63 d} + \frac{5 a^{2} c^{2} x^{4} \sqrt{c + d x^{2}}}{21} + \frac{19 a^{2} c d x^{6} \sqrt{c + d x^{2}}}{63} + \frac{a^{2} d^{2} x^{8} \sqrt{c + d x^{2}}}{9} + \frac{16 a b c^{5} \sqrt{c + d x^{2}}}{693 d^{3}} - \frac{8 a b c^{4} x^{2} \sqrt{c + d x^{2}}}{693 d^{2}} + \frac{2 a b c^{3} x^{4} \sqrt{c + d x^{2}}}{231 d} + \frac{226 a b c^{2} x^{6} \sqrt{c + d x^{2}}}{693} + \frac{46 a b c d x^{8} \sqrt{c + d x^{2}}}{99} + \frac{2 a b d^{2} x^{10} \sqrt{c + d x^{2}}}{11} - \frac{16 b^{2} c^{6} \sqrt{c + d x^{2}}}{3003 d^{4}} + \frac{8 b^{2} c^{5} x^{2} \sqrt{c + d x^{2}}}{3003 d^{3}} - \frac{2 b^{2} c^{4} x^{4} \sqrt{c + d x^{2}}}{1001 d^{2}} + \frac{5 b^{2} c^{3} x^{6} \sqrt{c + d x^{2}}}{3003 d} + \frac{53 b^{2} c^{2} x^{8} \sqrt{c + d x^{2}}}{429} + \frac{27 b^{2} c d x^{10} \sqrt{c + d x^{2}}}{143} + \frac{b^{2} d^{2} x^{12} \sqrt{c + d x^{2}}}{13} & \text{for}\: d \neq 0 \\c^{\frac{5}{2}} \left (\frac{a^{2} x^{4}}{4} + \frac{a b x^{6}}{3} + \frac{b^{2} x^{8}}{8}\right ) & \text{otherwise} \end{cases} \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

Piecewise((-2*a**2*c**4*sqrt(c + d*x**2)/(63*d**2) + a**2*c**3*x**2*sqrt(c + d*x**2)/(63*d) + 5*a**2*c**2*x**4
*sqrt(c + d*x**2)/21 + 19*a**2*c*d*x**6*sqrt(c + d*x**2)/63 + a**2*d**2*x**8*sqrt(c + d*x**2)/9 + 16*a*b*c**5*
sqrt(c + d*x**2)/(693*d**3) - 8*a*b*c**4*x**2*sqrt(c + d*x**2)/(693*d**2) + 2*a*b*c**3*x**4*sqrt(c + d*x**2)/(
231*d) + 226*a*b*c**2*x**6*sqrt(c + d*x**2)/693 + 46*a*b*c*d*x**8*sqrt(c + d*x**2)/99 + 2*a*b*d**2*x**10*sqrt(
c + d*x**2)/11 - 16*b**2*c**6*sqrt(c + d*x**2)/(3003*d**4) + 8*b**2*c**5*x**2*sqrt(c + d*x**2)/(3003*d**3) - 2
*b**2*c**4*x**4*sqrt(c + d*x**2)/(1001*d**2) + 5*b**2*c**3*x**6*sqrt(c + d*x**2)/(3003*d) + 53*b**2*c**2*x**8*
sqrt(c + d*x**2)/429 + 27*b**2*c*d*x**10*sqrt(c + d*x**2)/143 + b**2*d**2*x**12*sqrt(c + d*x**2)/13, Ne(d, 0))
, (c**(5/2)*(a**2*x**4/4 + a*b*x**6/3 + b**2*x**8/8), True))

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Giac [B]  time = 1.14423, size = 749, normalized size = 6.57 \begin{align*} \text{result too large to display} \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate(x^3*(b*x^2+a)^2*(d*x^2+c)^(5/2),x, algorithm="giac")

[Out]

1/45045*(3003*(3*(d*x^2 + c)^(5/2) - 5*(d*x^2 + c)^(3/2)*c)*a^2*c^2/d + 858*(15*(d*x^2 + c)^(7/2) - 42*(d*x^2
+ c)^(5/2)*c + 35*(d*x^2 + c)^(3/2)*c^2)*a*b*c^2/d^2 + 858*(15*(d*x^2 + c)^(7/2) - 42*(d*x^2 + c)^(5/2)*c + 35
*(d*x^2 + c)^(3/2)*c^2)*a^2*c/d + 143*(35*(d*x^2 + c)^(9/2) - 135*(d*x^2 + c)^(7/2)*c + 189*(d*x^2 + c)^(5/2)*
c^2 - 105*(d*x^2 + c)^(3/2)*c^3)*b^2*c^2/d^3 + 572*(35*(d*x^2 + c)^(9/2) - 135*(d*x^2 + c)^(7/2)*c + 189*(d*x^
2 + c)^(5/2)*c^2 - 105*(d*x^2 + c)^(3/2)*c^3)*a*b*c/d^2 + 143*(35*(d*x^2 + c)^(9/2) - 135*(d*x^2 + c)^(7/2)*c
+ 189*(d*x^2 + c)^(5/2)*c^2 - 105*(d*x^2 + c)^(3/2)*c^3)*a^2/d + 26*(315*(d*x^2 + c)^(11/2) - 1540*(d*x^2 + c)
^(9/2)*c + 2970*(d*x^2 + c)^(7/2)*c^2 - 2772*(d*x^2 + c)^(5/2)*c^3 + 1155*(d*x^2 + c)^(3/2)*c^4)*b^2*c/d^3 + 2
6*(315*(d*x^2 + c)^(11/2) - 1540*(d*x^2 + c)^(9/2)*c + 2970*(d*x^2 + c)^(7/2)*c^2 - 2772*(d*x^2 + c)^(5/2)*c^3
+ 1155*(d*x^2 + c)^(3/2)*c^4)*a*b/d^2 + 5*(693*(d*x^2 + c)^(13/2) - 4095*(d*x^2 + c)^(11/2)*c + 10010*(d*x^2
+ c)^(9/2)*c^2 - 12870*(d*x^2 + c)^(7/2)*c^3 + 9009*(d*x^2 + c)^(5/2)*c^4 - 3003*(d*x^2 + c)^(3/2)*c^5)*b^2/d^
3)/d