\(\int \frac {(d+e x)^3 (f+g x)^2}{(d^2-e^2 x^2)^{7/2}} \, dx\) [582]

Optimal result

Integrand size = 31, antiderivative size = 145 \[ \int \frac {(d+e x)^3 (f+g x)^2}{\left (d^2-e^2 x^2\right )^{7/2}} \, dx=\frac {(e f+d g)^2 (d+e x)^3}{5 d e^3 \left (d^2-e^2 x^2\right )^{5/2}}+\frac {2 (e f-4 d g) (e f+d g) (d+e x)^2}{15 d^2 e^3 \left (d^2-e^2 x^2\right )^{3/2}}+\frac {\left (2 e^2 f^2-6 d e f g+7 d^2 g^2\right ) (d+e x)}{15 d^3 e^3 \sqrt {d^2-e^2 x^2}} \]

[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)

Rubi [A] (verified)

Time = 0.15 (sec) , antiderivative size = 145, normalized size of antiderivative = 1.00, number of steps used = 3, number of rules used = 3, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.097, Rules used = {1649, 803, 651} \[ \int \frac {(d+e x)^3 (f+g x)^2}{\left (d^2-e^2 x^2\right )^{7/2}} \, dx=\frac {(d+e x)^3 (d g+e f)^2}{5 d e^3 \left (d^2-e^2 x^2\right )^{5/2}}+\frac {2 (d+e x)^2 (e f-4 d g) (d g+e f)}{15 d^2 e^3 \left (d^2-e^2 x^2\right )^{3/2}}+\frac {(d+e x) \left (7 d^2 g^2-6 d e f g+2 e^2 f^2\right )}{15 d^3 e^3 \sqrt {d^2-e^2 x^2}} \]

[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 651

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 803

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 1649

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*} \text {integral}& = \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*}

Mathematica [A] (verified)

Time = 0.56 (sec) , antiderivative size = 105, normalized size of antiderivative = 0.72 \[ \int \frac {(d+e x)^3 (f+g x)^2}{\left (d^2-e^2 x^2\right )^{7/2}} \, dx=\frac {\sqrt {d^2-e^2 x^2} \left (2 d^4 g^2+2 e^4 f^2 x^2-6 d^3 e g (f+g x)-6 d e^3 f x (f+g x)+d^2 e^2 \left (7 f^2+18 f g x+7 g^2 x^2\right )\right )}{15 d^3 e^3 (d-e x)^3} \]

[In]

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

[Out]

(Sqrt[d^2 - e^2*x^2]*(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)^3)

Maple [A] (verified)

Time = 0.64 (sec) , antiderivative size = 126, normalized size of antiderivative = 0.87

method	result	size
trager	\(\frac {\left (7 x^{2} d^{2} e^{2} g^{2}-6 x^{2} d \,e^{3} f g +2 x^{2} e^{4} f^{2}-6 x \,d^{3} e \,g^{2}+18 x \,d^{2} e^{2} f g -6 x d \,e^{3} f^{2}+2 d^{4} g^{2}-6 f g e \,d^{3}+7 d^{2} e^{2} f^{2}\right ) \sqrt {-e^{2} x^{2}+d^{2}}}{15 d^{3} e^{3} \left (-e x +d \right )^{3}}\)	\(126\)
gosper	\(\frac {\left (-e x +d \right ) \left (e x +d \right )^{4} \left (7 x^{2} d^{2} e^{2} g^{2}-6 x^{2} d \,e^{3} f g +2 x^{2} e^{4} f^{2}-6 x \,d^{3} e \,g^{2}+18 x \,d^{2} e^{2} f g -6 x d \,e^{3} f^{2}+2 d^{4} g^{2}-6 f g e \,d^{3}+7 d^{2} e^{2} f^{2}\right )}{15 d^{3} e^{3} \left (-e^{2} x^{2}+d^{2}\right )^{\frac {7}{2}}}\)	\(131\)
default	\(d^{3} f^{2} \left (\frac {x}{5 d^{2} \left (-e^{2} x^{2}+d^{2}\right )^{\frac {5}{2}}}+\frac {\frac {4 x}{15 d^{2} \left (-e^{2} x^{2}+d^{2}\right )^{\frac {3}{2}}}+\frac {8 x}{15 d^{4} \sqrt {-e^{2} x^{2}+d^{2}}}}{d^{2}}\right )+g^{2} e^{3} \left (\frac {x^{4}}{e^{2} \left (-e^{2} x^{2}+d^{2}\right )^{\frac {5}{2}}}-\frac {4 d^{2} \left (\frac {x^{2}}{3 e^{2} \left (-e^{2} x^{2}+d^{2}\right )^{\frac {5}{2}}}-\frac {2 d^{2}}{15 e^{4} \left (-e^{2} x^{2}+d^{2}\right )^{\frac {5}{2}}}\right )}{e^{2}}\right )+\left (3 d \,g^{2} e^{2}+2 f g \,e^{3}\right ) \left (\frac {x^{3}}{2 e^{2} \left (-e^{2} x^{2}+d^{2}\right )^{\frac {5}{2}}}-\frac {3 d^{2} \left (\frac {x}{4 e^{2} \left (-e^{2} x^{2}+d^{2}\right )^{\frac {5}{2}}}-\frac {d^{2} \left (\frac {x}{5 d^{2} \left (-e^{2} x^{2}+d^{2}\right )^{\frac {5}{2}}}+\frac {\frac {4 x}{15 d^{2} \left (-e^{2} x^{2}+d^{2}\right )^{\frac {3}{2}}}+\frac {8 x}{15 d^{4} \sqrt {-e^{2} x^{2}+d^{2}}}}{d^{2}}\right )}{4 e^{2}}\right )}{2 e^{2}}\right )+\frac {2 d^{3} f g +3 d^{2} e \,f^{2}}{5 e^{2} \left (-e^{2} x^{2}+d^{2}\right )^{\frac {5}{2}}}+\left (3 d^{2} g^{2} e +6 d f g \,e^{2}+f^{2} e^{3}\right ) \left (\frac {x^{2}}{3 e^{2} \left (-e^{2} x^{2}+d^{2}\right )^{\frac {5}{2}}}-\frac {2 d^{2}}{15 e^{4} \left (-e^{2} x^{2}+d^{2}\right )^{\frac {5}{2}}}\right )+\left (d^{3} g^{2}+6 d^{2} e f g +3 d \,e^{2} f^{2}\right ) \left (\frac {x}{4 e^{2} \left (-e^{2} x^{2}+d^{2}\right )^{\frac {5}{2}}}-\frac {d^{2} \left (\frac {x}{5 d^{2} \left (-e^{2} x^{2}+d^{2}\right )^{\frac {5}{2}}}+\frac {\frac {4 x}{15 d^{2} \left (-e^{2} x^{2}+d^{2}\right )^{\frac {3}{2}}}+\frac {8 x}{15 d^{4} \sqrt {-e^{2} x^{2}+d^{2}}}}{d^{2}}\right )}{4 e^{2}}\right )\)	\(532\)

[In]

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

[Out]

1/15*(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*x+d)^3*(-e^2*x^2+d^2)^(1/2)

Fricas [B] (verification not implemented)

Leaf count of result is larger than twice the leaf count of optimal. 279 vs. \(2 (133) = 266\).

Time = 0.31 (sec) , antiderivative size = 279, normalized size of antiderivative = 1.92 \[ \int \frac {(d+e x)^3 (f+g x)^2}{\left (d^2-e^2 x^2\right )^{7/2}} \, dx=-\frac {7 \, d^{3} e^{2} f^{2} - 6 \, d^{4} e f g + 2 \, d^{5} g^{2} - {\left (7 \, e^{5} f^{2} - 6 \, d e^{4} f g + 2 \, d^{2} e^{3} g^{2}\right )} x^{3} + 3 \, {\left (7 \, d e^{4} f^{2} - 6 \, d^{2} e^{3} f g + 2 \, d^{3} e^{2} g^{2}\right )} x^{2} - 3 \, {\left (7 \, d^{2} e^{3} f^{2} - 6 \, d^{3} e^{2} f g + 2 \, d^{4} e g^{2}\right )} x + {\left (7 \, d^{2} e^{2} f^{2} - 6 \, d^{3} e f g + 2 \, d^{4} g^{2} + {\left (2 \, e^{4} f^{2} - 6 \, d e^{3} f g + 7 \, d^{2} e^{2} g^{2}\right )} x^{2} - 6 \, {\left (d e^{3} f^{2} - 3 \, d^{2} e^{2} f g + d^{3} e g^{2}\right )} x\right )} \sqrt {-e^{2} x^{2} + d^{2}}}{15 \, {\left (d^{3} e^{6} x^{3} - 3 \, d^{4} e^{5} x^{2} + 3 \, d^{5} e^{4} x - d^{6} e^{3}\right )}} \]

[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)

Sympy [F]

\[ \int \frac {(d+e x)^3 (f+g x)^2}{\left (d^2-e^2 x^2\right )^{7/2}} \, dx=\int \frac {\left (d + e x\right )^{3} \left (f + g x\right )^{2}}{\left (- \left (- d + e x\right ) \left (d + e x\right )\right )^{\frac {7}{2}}}\, dx \]

[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)

Maxima [B] (verification not implemented)

Leaf count of result is larger than twice the leaf count of optimal. 583 vs. \(2 (133) = 266\).

Time = 0.19 (sec) , antiderivative size = 583, normalized size of antiderivative = 4.02 \[ \int \frac {(d+e x)^3 (f+g x)^2}{\left (d^2-e^2 x^2\right )^{7/2}} \, dx=\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}} \]

[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)

Giac [B] (verification not implemented)

Leaf count of result is larger than twice the leaf count of optimal. 370 vs. \(2 (133) = 266\).

Time = 0.30 (sec) , antiderivative size = 370, normalized size of antiderivative = 2.55 \[ \int \frac {(d+e x)^3 (f+g x)^2}{\left (d^2-e^2 x^2\right )^{7/2}} \, dx=\frac {2 \, {\left (7 \, e^{2} f^{2} - 6 \, d e f g + 2 \, d^{2} g^{2} - \frac {20 \, {\left (d e + \sqrt {-e^{2} x^{2} + d^{2}} {\left | e \right |}\right )} f^{2}}{x} + \frac {30 \, {\left (d e + \sqrt {-e^{2} x^{2} + d^{2}} {\left | e \right |}\right )} d f g}{e x} - \frac {10 \, {\left (d e + \sqrt {-e^{2} x^{2} + d^{2}} {\left | e \right |}\right )} d^{2} g^{2}}{e^{2} x} + \frac {40 \, {\left (d e + \sqrt {-e^{2} x^{2} + d^{2}} {\left | e \right |}\right )}^{2} f^{2}}{e^{2} x^{2}} - \frac {30 \, {\left (d e + \sqrt {-e^{2} x^{2} + d^{2}} {\left | e \right |}\right )}^{2} d f g}{e^{3} x^{2}} + \frac {20 \, {\left (d e + \sqrt {-e^{2} x^{2} + d^{2}} {\left | e \right |}\right )}^{2} d^{2} g^{2}}{e^{4} x^{2}} - \frac {30 \, {\left (d e + \sqrt {-e^{2} x^{2} + d^{2}} {\left | e \right |}\right )}^{3} f^{2}}{e^{4} x^{3}} + \frac {30 \, {\left (d e + \sqrt {-e^{2} x^{2} + d^{2}} {\left | e \right |}\right )}^{3} d f g}{e^{5} x^{3}} + \frac {15 \, {\left (d e + \sqrt {-e^{2} x^{2} + d^{2}} {\left | e \right |}\right )}^{4} f^{2}}{e^{6} x^{4}}\right )}}{15 \, d^{3} e^{2} {\left (\frac {d e + \sqrt {-e^{2} x^{2} + d^{2}} {\left | e \right |}}{e^{2} x} - 1\right )}^{5} {\left | e \right |}} \]

[In]

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

[Out]

2/15*(7*e^2*f^2 - 6*d*e*f*g + 2*d^2*g^2 - 20*(d*e + sqrt(-e^2*x^2 + d^2)*abs(e))*f^2/x + 30*(d*e + sqrt(-e^2*x
^2 + d^2)*abs(e))*d*f*g/(e*x) - 10*(d*e + sqrt(-e^2*x^2 + d^2)*abs(e))*d^2*g^2/(e^2*x) + 40*(d*e + sqrt(-e^2*x
^2 + d^2)*abs(e))^2*f^2/(e^2*x^2) - 30*(d*e + sqrt(-e^2*x^2 + d^2)*abs(e))^2*d*f*g/(e^3*x^2) + 20*(d*e + sqrt(
-e^2*x^2 + d^2)*abs(e))^2*d^2*g^2/(e^4*x^2) - 30*(d*e + sqrt(-e^2*x^2 + d^2)*abs(e))^3*f^2/(e^4*x^3) + 30*(d*e
+ sqrt(-e^2*x^2 + d^2)*abs(e))^3*d*f*g/(e^5*x^3) + 15*(d*e + sqrt(-e^2*x^2 + d^2)*abs(e))^4*f^2/(e^6*x^4))/(d
^3*e^2*((d*e + sqrt(-e^2*x^2 + d^2)*abs(e))/(e^2*x) - 1)^5*abs(e))

Mupad [B] (verification not implemented)

Time = 12.12 (sec) , antiderivative size = 125, normalized size of antiderivative = 0.86 \[ \int \frac {(d+e x)^3 (f+g x)^2}{\left (d^2-e^2 x^2\right )^{7/2}} \, dx=\frac {\sqrt {d^2-e^2\,x^2}\,\left (2\,d^4\,g^2-6\,d^3\,e\,f\,g-6\,d^3\,e\,g^2\,x+7\,d^2\,e^2\,f^2+18\,d^2\,e^2\,f\,g\,x+7\,d^2\,e^2\,g^2\,x^2-6\,d\,e^3\,f^2\,x-6\,d\,e^3\,f\,g\,x^2+2\,e^4\,f^2\,x^2\right )}{15\,d^3\,e^3\,{\left (d-e\,x\right )}^3} \]

[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)