3.582 \(\int \frac {(d+e x)^3 (f+g x)^2}{(d^2-e^2 x^2)^{7/2}} \, dx\)

Optimal. Leaf size=145 \[ \frac {(d+e x)^3 (d g+e f)^2}{5 d e^3 \left (d^2-e^2 x^2\right )^{5/2}}+\frac {2 (d+e x)^2 (e f-4 d g) (d g+e f)}{15 d^2 e^3 \left (d^2-e^2 x^2\right )^{3/2}}+\frac {(d+e x) \left (7 d^2 g^2-6 d e f g+2 e^2 f^2\right )}{15 d^3 e^3 \sqrt {d^2-e^2 x^2}} \]

[Out]

1/5*(d*g+e*f)^2*(e*x+d)^3/d/e^3/(-e^2*x^2+d^2)^(5/2)+2/15*(-4*d*g+e*f)*(d*g+e*f)*(e*x+d)^2/d^2/e^3/(-e^2*x^2+d
^2)^(3/2)+1/15*(7*d^2*g^2-6*d*e*f*g+2*e^2*f^2)*(e*x+d)/d^3/e^3/(-e^2*x^2+d^2)^(1/2)

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Rubi [A]  time = 0.22, antiderivative size = 145, normalized size of antiderivative = 1.00, number of steps used = 3, number of rules used = 3, integrand size = 31, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.097, Rules used = {1635, 789, 637} \[ \frac {(d+e x) \left (7 d^2 g^2-6 d e f g+2 e^2 f^2\right )}{15 d^3 e^3 \sqrt {d^2-e^2 x^2}}+\frac {(d+e x)^3 (d g+e f)^2}{5 d e^3 \left (d^2-e^2 x^2\right )^{5/2}}+\frac {2 (d+e x)^2 (e f-4 d g) (d g+e f)}{15 d^2 e^3 \left (d^2-e^2 x^2\right )^{3/2}} \]

Antiderivative was successfully verified.

[In]

Int[((d + e*x)^3*(f + g*x)^2)/(d^2 - e^2*x^2)^(7/2),x]

[Out]

((e*f + d*g)^2*(d + e*x)^3)/(5*d*e^3*(d^2 - e^2*x^2)^(5/2)) + (2*(e*f - 4*d*g)*(e*f + d*g)*(d + e*x)^2)/(15*d^
2*e^3*(d^2 - e^2*x^2)^(3/2)) + ((2*e^2*f^2 - 6*d*e*f*g + 7*d^2*g^2)*(d + e*x))/(15*d^3*e^3*Sqrt[d^2 - e^2*x^2]
)

Rule 637

Int[((d_) + (e_.)*(x_))/((a_) + (c_.)*(x_)^2)^(3/2), x_Symbol] :> Simp[(-(a*e) + c*d*x)/(a*c*Sqrt[a + c*x^2]),
x] /; FreeQ[{a, c, d, e}, x]

Rule 789

Int[((d_.) + (e_.)*(x_))^(m_)*((f_.) + (g_.)*(x_))*((a_) + (c_.)*(x_)^2)^(p_), x_Symbol] :> Simp[((d*g + e*f)*
(d + e*x)^m*(a + c*x^2)^(p + 1))/(2*c*d*(p + 1)), x] - Dist[(e*(m*(d*g + e*f) + 2*e*f*(p + 1)))/(2*c*d*(p + 1)
), Int[(d + e*x)^(m - 1)*(a + c*x^2)^(p + 1), x], x] /; FreeQ[{a, c, d, e, f, g}, x] && EqQ[c*d^2 + a*e^2, 0]
&& LtQ[p, -1] && GtQ[m, 0]

Rule 1635

Int[(Pq_)*((d_) + (e_.)*(x_))^(m_.)*((a_) + (c_.)*(x_)^2)^(p_), x_Symbol] :> With[{Q = PolynomialQuotient[Pq,
a*e + c*d*x, x], f = PolynomialRemainder[Pq, a*e + c*d*x, x]}, -Simp[(d*f*(d + e*x)^m*(a + c*x^2)^(p + 1))/(2*
a*e*(p + 1)), x] + Dist[d/(2*a*(p + 1)), Int[(d + e*x)^(m - 1)*(a + c*x^2)^(p + 1)*ExpandToSum[2*a*e*(p + 1)*Q
+ f*(m + 2*p + 2), x], x], x]] /; FreeQ[{a, c, d, e}, x] && PolyQ[Pq, x] && EqQ[c*d^2 + a*e^2, 0] && ILtQ[p +
1/2, 0] && GtQ[m, 0]

Rubi steps

\begin {align*} \int \frac {(d+e x)^3 (f+g x)^2}{\left (d^2-e^2 x^2\right )^{7/2}} \, dx &=\frac {(e f+d g)^2 (d+e x)^3}{5 d e^3 \left (d^2-e^2 x^2\right )^{5/2}}-\frac {\int \frac {(d+e x)^2 \left (-2 f^2+\frac {6 d f g}{e}+\frac {3 d^2 g^2}{e^2}+\frac {5 d g^2 x}{e}\right )}{\left (d^2-e^2 x^2\right )^{5/2}} \, dx}{5 d}\\ &=\frac {(e f+d g)^2 (d+e x)^3}{5 d e^3 \left (d^2-e^2 x^2\right )^{5/2}}+\frac {2 (e f-4 d g) (e f+d g) (d+e x)^2}{15 d^2 e^3 \left (d^2-e^2 x^2\right )^{3/2}}+\frac {\left (2 e^2 f^2-6 d e f g+7 d^2 g^2\right ) \int \frac {d+e x}{\left (d^2-e^2 x^2\right )^{3/2}} \, dx}{15 d^2 e^2}\\ &=\frac {(e f+d g)^2 (d+e x)^3}{5 d e^3 \left (d^2-e^2 x^2\right )^{5/2}}+\frac {2 (e f-4 d g) (e f+d g) (d+e x)^2}{15 d^2 e^3 \left (d^2-e^2 x^2\right )^{3/2}}+\frac {\left (2 e^2 f^2-6 d e f g+7 d^2 g^2\right ) (d+e x)}{15 d^3 e^3 \sqrt {d^2-e^2 x^2}}\\ \end {align*}

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Mathematica [A]  time = 0.39, size = 110, normalized size = 0.76 \[ \frac {(d+e x) \left (2 d^4 g^2-6 d^3 e g (f+g x)+d^2 e^2 \left (7 f^2+18 f g x+7 g^2 x^2\right )-6 d e^3 f x (f+g x)+2 e^4 f^2 x^2\right )}{15 d^3 e^3 (d-e x)^2 \sqrt {d^2-e^2 x^2}} \]

Antiderivative was successfully verified.

[In]

Integrate[((d + e*x)^3*(f + g*x)^2)/(d^2 - e^2*x^2)^(7/2),x]

[Out]

((d + e*x)*(2*d^4*g^2 + 2*e^4*f^2*x^2 - 6*d^3*e*g*(f + g*x) - 6*d*e^3*f*x*(f + g*x) + d^2*e^2*(7*f^2 + 18*f*g*
x + 7*g^2*x^2)))/(15*d^3*e^3*(d - e*x)^2*Sqrt[d^2 - e^2*x^2])

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fricas [B]  time = 0.96, size = 279, normalized size = 1.92 \[ -\frac {7 \, d^{3} e^{2} f^{2} - 6 \, d^{4} e f g + 2 \, d^{5} g^{2} - {\left (7 \, e^{5} f^{2} - 6 \, d e^{4} f g + 2 \, d^{2} e^{3} g^{2}\right )} x^{3} + 3 \, {\left (7 \, d e^{4} f^{2} - 6 \, d^{2} e^{3} f g + 2 \, d^{3} e^{2} g^{2}\right )} x^{2} - 3 \, {\left (7 \, d^{2} e^{3} f^{2} - 6 \, d^{3} e^{2} f g + 2 \, d^{4} e g^{2}\right )} x + {\left (7 \, d^{2} e^{2} f^{2} - 6 \, d^{3} e f g + 2 \, d^{4} g^{2} + {\left (2 \, e^{4} f^{2} - 6 \, d e^{3} f g + 7 \, d^{2} e^{2} g^{2}\right )} x^{2} - 6 \, {\left (d e^{3} f^{2} - 3 \, d^{2} e^{2} f g + d^{3} e g^{2}\right )} x\right )} \sqrt {-e^{2} x^{2} + d^{2}}}{15 \, {\left (d^{3} e^{6} x^{3} - 3 \, d^{4} e^{5} x^{2} + 3 \, d^{5} e^{4} x - d^{6} e^{3}\right )}} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate((e*x+d)^3*(g*x+f)^2/(-e^2*x^2+d^2)^(7/2),x, algorithm="fricas")

[Out]

-1/15*(7*d^3*e^2*f^2 - 6*d^4*e*f*g + 2*d^5*g^2 - (7*e^5*f^2 - 6*d*e^4*f*g + 2*d^2*e^3*g^2)*x^3 + 3*(7*d*e^4*f^
2 - 6*d^2*e^3*f*g + 2*d^3*e^2*g^2)*x^2 - 3*(7*d^2*e^3*f^2 - 6*d^3*e^2*f*g + 2*d^4*e*g^2)*x + (7*d^2*e^2*f^2 -
6*d^3*e*f*g + 2*d^4*g^2 + (2*e^4*f^2 - 6*d*e^3*f*g + 7*d^2*e^2*g^2)*x^2 - 6*(d*e^3*f^2 - 3*d^2*e^2*f*g + d^3*e
*g^2)*x)*sqrt(-e^2*x^2 + d^2))/(d^3*e^6*x^3 - 3*d^4*e^5*x^2 + 3*d^5*e^4*x - d^6*e^3)

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giac [A]  time = 0.34, size = 198, normalized size = 1.37 \[ -\frac {\sqrt {-x^{2} e^{2} + d^{2}} {\left ({\left (15 \, d f^{2} + {\left ({\left ({\left (15 \, g^{2} e + \frac {{\left (7 \, d^{3} g^{2} e^{6} - 6 \, d^{2} f g e^{7} + 2 \, d f^{2} e^{8}\right )} x e^{\left (-4\right )}}{d^{4}}\right )} x + \frac {5 \, {\left (d^{5} g^{2} e^{4} + 6 \, d^{4} f g e^{5} - d^{3} f^{2} e^{6}\right )} e^{\left (-4\right )}}{d^{4}}\right )} x - \frac {5 \, {\left (d^{6} g^{2} e^{3} - 6 \, d^{5} f g e^{4} - d^{4} f^{2} e^{5}\right )} e^{\left (-4\right )}}{d^{4}}\right )} x\right )} x + \frac {{\left (2 \, d^{8} g^{2} e - 6 \, d^{7} f g e^{2} + 7 \, d^{6} f^{2} e^{3}\right )} e^{\left (-4\right )}}{d^{4}}\right )}}{15 \, {\left (x^{2} e^{2} - d^{2}\right )}^{3}} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate((e*x+d)^3*(g*x+f)^2/(-e^2*x^2+d^2)^(7/2),x, algorithm="giac")

[Out]

-1/15*sqrt(-x^2*e^2 + d^2)*((15*d*f^2 + (((15*g^2*e + (7*d^3*g^2*e^6 - 6*d^2*f*g*e^7 + 2*d*f^2*e^8)*x*e^(-4)/d
^4)*x + 5*(d^5*g^2*e^4 + 6*d^4*f*g*e^5 - d^3*f^2*e^6)*e^(-4)/d^4)*x - 5*(d^6*g^2*e^3 - 6*d^5*f*g*e^4 - d^4*f^2
*e^5)*e^(-4)/d^4)*x)*x + (2*d^8*g^2*e - 6*d^7*f*g*e^2 + 7*d^6*f^2*e^3)*e^(-4)/d^4)/(x^2*e^2 - d^2)^3

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maple [A]  time = 0.01, size = 131, normalized size = 0.90 \[ \frac {\left (-e x +d \right ) \left (e x +d \right )^{4} \left (7 d^{2} e^{2} g^{2} x^{2}-6 d \,e^{3} f g \,x^{2}+2 e^{4} f^{2} x^{2}-6 d^{3} e \,g^{2} x +18 d^{2} e^{2} f g x -6 d \,e^{3} f^{2} x +2 d^{4} g^{2}-6 d^{3} e f g +7 d^{2} e^{2} f^{2}\right )}{15 \left (-e^{2} x^{2}+d^{2}\right )^{\frac {7}{2}} d^{3} e^{3}} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

int((e*x+d)^3*(g*x+f)^2/(-e^2*x^2+d^2)^(7/2),x)

[Out]

1/15*(-e*x+d)*(e*x+d)^4*(7*d^2*e^2*g^2*x^2-6*d*e^3*f*g*x^2+2*e^4*f^2*x^2-6*d^3*e*g^2*x+18*d^2*e^2*f*g*x-6*d*e^
3*f^2*x+2*d^4*g^2-6*d^3*e*f*g+7*d^2*e^2*f^2)/d^3/e^3/(-e^2*x^2+d^2)^(7/2)

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maxima [B]  time = 0.47, size = 583, normalized size = 4.02 \[ \frac {e g^{2} x^{4}}{{\left (-e^{2} x^{2} + d^{2}\right )}^{\frac {5}{2}}} - \frac {4 \, d^{2} g^{2} x^{2}}{3 \, {\left (-e^{2} x^{2} + d^{2}\right )}^{\frac {5}{2}} e} + \frac {d f^{2} x}{5 \, {\left (-e^{2} x^{2} + d^{2}\right )}^{\frac {5}{2}}} + \frac {3 \, d^{2} f^{2}}{5 \, {\left (-e^{2} x^{2} + d^{2}\right )}^{\frac {5}{2}} e} + \frac {2 \, d^{3} f g}{5 \, {\left (-e^{2} x^{2} + d^{2}\right )}^{\frac {5}{2}} e^{2}} + \frac {8 \, d^{4} g^{2}}{15 \, {\left (-e^{2} x^{2} + d^{2}\right )}^{\frac {5}{2}} e^{3}} + \frac {4 \, f^{2} x}{15 \, {\left (-e^{2} x^{2} + d^{2}\right )}^{\frac {3}{2}} d} + \frac {8 \, f^{2} x}{15 \, \sqrt {-e^{2} x^{2} + d^{2}} d^{3}} + \frac {{\left (2 \, e^{3} f g + 3 \, d e^{2} g^{2}\right )} x^{3}}{2 \, {\left (-e^{2} x^{2} + d^{2}\right )}^{\frac {5}{2}} e^{2}} + \frac {{\left (e^{3} f^{2} + 6 \, d e^{2} f g + 3 \, d^{2} e g^{2}\right )} x^{2}}{3 \, {\left (-e^{2} x^{2} + d^{2}\right )}^{\frac {5}{2}} e^{2}} - \frac {3 \, {\left (2 \, e^{3} f g + 3 \, d e^{2} g^{2}\right )} d^{2} x}{10 \, {\left (-e^{2} x^{2} + d^{2}\right )}^{\frac {5}{2}} e^{4}} + \frac {{\left (3 \, d e^{2} f^{2} + 6 \, d^{2} e f g + d^{3} g^{2}\right )} x}{5 \, {\left (-e^{2} x^{2} + d^{2}\right )}^{\frac {5}{2}} e^{2}} - \frac {2 \, {\left (e^{3} f^{2} + 6 \, d e^{2} f g + 3 \, d^{2} e g^{2}\right )} d^{2}}{15 \, {\left (-e^{2} x^{2} + d^{2}\right )}^{\frac {5}{2}} e^{4}} + \frac {{\left (2 \, e^{3} f g + 3 \, d e^{2} g^{2}\right )} x}{10 \, {\left (-e^{2} x^{2} + d^{2}\right )}^{\frac {3}{2}} e^{4}} - \frac {{\left (3 \, d e^{2} f^{2} + 6 \, d^{2} e f g + d^{3} g^{2}\right )} x}{15 \, {\left (-e^{2} x^{2} + d^{2}\right )}^{\frac {3}{2}} d^{2} e^{2}} + \frac {{\left (2 \, e^{3} f g + 3 \, d e^{2} g^{2}\right )} x}{5 \, \sqrt {-e^{2} x^{2} + d^{2}} d^{2} e^{4}} - \frac {2 \, {\left (3 \, d e^{2} f^{2} + 6 \, d^{2} e f g + d^{3} g^{2}\right )} x}{15 \, \sqrt {-e^{2} x^{2} + d^{2}} d^{4} e^{2}} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate((e*x+d)^3*(g*x+f)^2/(-e^2*x^2+d^2)^(7/2),x, algorithm="maxima")

[Out]

e*g^2*x^4/(-e^2*x^2 + d^2)^(5/2) - 4/3*d^2*g^2*x^2/((-e^2*x^2 + d^2)^(5/2)*e) + 1/5*d*f^2*x/(-e^2*x^2 + d^2)^(
5/2) + 3/5*d^2*f^2/((-e^2*x^2 + d^2)^(5/2)*e) + 2/5*d^3*f*g/((-e^2*x^2 + d^2)^(5/2)*e^2) + 8/15*d^4*g^2/((-e^2
*x^2 + d^2)^(5/2)*e^3) + 4/15*f^2*x/((-e^2*x^2 + d^2)^(3/2)*d) + 8/15*f^2*x/(sqrt(-e^2*x^2 + d^2)*d^3) + 1/2*(
2*e^3*f*g + 3*d*e^2*g^2)*x^3/((-e^2*x^2 + d^2)^(5/2)*e^2) + 1/3*(e^3*f^2 + 6*d*e^2*f*g + 3*d^2*e*g^2)*x^2/((-e
^2*x^2 + d^2)^(5/2)*e^2) - 3/10*(2*e^3*f*g + 3*d*e^2*g^2)*d^2*x/((-e^2*x^2 + d^2)^(5/2)*e^4) + 1/5*(3*d*e^2*f^
2 + 6*d^2*e*f*g + d^3*g^2)*x/((-e^2*x^2 + d^2)^(5/2)*e^2) - 2/15*(e^3*f^2 + 6*d*e^2*f*g + 3*d^2*e*g^2)*d^2/((-
e^2*x^2 + d^2)^(5/2)*e^4) + 1/10*(2*e^3*f*g + 3*d*e^2*g^2)*x/((-e^2*x^2 + d^2)^(3/2)*e^4) - 1/15*(3*d*e^2*f^2
+ 6*d^2*e*f*g + d^3*g^2)*x/((-e^2*x^2 + d^2)^(3/2)*d^2*e^2) + 1/5*(2*e^3*f*g + 3*d*e^2*g^2)*x/(sqrt(-e^2*x^2 +
d^2)*d^2*e^4) - 2/15*(3*d*e^2*f^2 + 6*d^2*e*f*g + d^3*g^2)*x/(sqrt(-e^2*x^2 + d^2)*d^4*e^2)

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mupad [B]  time = 2.87, size = 125, normalized size = 0.86 \[ \frac {\sqrt {d^2-e^2\,x^2}\,\left (2\,d^4\,g^2-6\,d^3\,e\,f\,g-6\,d^3\,e\,g^2\,x+7\,d^2\,e^2\,f^2+18\,d^2\,e^2\,f\,g\,x+7\,d^2\,e^2\,g^2\,x^2-6\,d\,e^3\,f^2\,x-6\,d\,e^3\,f\,g\,x^2+2\,e^4\,f^2\,x^2\right )}{15\,d^3\,e^3\,{\left (d-e\,x\right )}^3} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

int(((f + g*x)^2*(d + e*x)^3)/(d^2 - e^2*x^2)^(7/2),x)

[Out]

((d^2 - e^2*x^2)^(1/2)*(2*d^4*g^2 + 7*d^2*e^2*f^2 + 2*e^4*f^2*x^2 - 6*d^3*e*f*g + 7*d^2*e^2*g^2*x^2 - 6*d*e^3*
f^2*x - 6*d^3*e*g^2*x + 18*d^2*e^2*f*g*x - 6*d*e^3*f*g*x^2))/(15*d^3*e^3*(d - e*x)^3)

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sympy [F]  time = 0.00, size = 0, normalized size = 0.00 \[ \int \frac {\left (d + e x\right )^{3} \left (f + g x\right )^{2}}{\left (- \left (- d + e x\right ) \left (d + e x\right )\right )^{\frac {7}{2}}}\, dx \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate((e*x+d)**3*(g*x+f)**2/(-e**2*x**2+d**2)**(7/2),x)

[Out]

Integral((d + e*x)**3*(f + g*x)**2/(-(-d + e*x)*(d + e*x))**(7/2), x)

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