\(\int \frac {(d+e x)^3 (f+g x)}{(d^2-e^2 x^2)^{7/2}} \, dx\) [583]

Optimal result

Integrand size = 29, antiderivative size = 117 \[ \int \frac {(d+e x)^3 (f+g x)}{\left (d^2-e^2 x^2\right )^{7/2}} \, dx=\frac {(e f+d g) (d+e x)^3}{5 d e^2 \left (d^2-e^2 x^2\right )^{5/2}}+\frac {2 (2 e f-3 d g) (d+e x)}{15 d e^2 \left (d^2-e^2 x^2\right )^{3/2}}+\frac {(2 e f-3 d g) x}{15 d^3 e \sqrt {d^2-e^2 x^2}} \]

[Out]

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

Rubi [A] (verified)

Time = 0.04 (sec) , antiderivative size = 117, normalized size of antiderivative = 1.00, number of steps used = 3, number of rules used = 3, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.103, Rules used = {803, 667, 197} \[ \int \frac {(d+e x)^3 (f+g x)}{\left (d^2-e^2 x^2\right )^{7/2}} \, dx=\frac {(d+e x)^3 (d g+e f)}{5 d e^2 \left (d^2-e^2 x^2\right )^{5/2}}+\frac {2 (d+e x) (2 e f-3 d g)}{15 d e^2 \left (d^2-e^2 x^2\right )^{3/2}}+\frac {x (2 e f-3 d g)}{15 d^3 e \sqrt {d^2-e^2 x^2}} \]

[In]

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

[Out]

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

Rule 197

Int[((a_) + (b_.)*(x_)^(n_))^(p_), x_Symbol] :> Simp[x*((a + b*x^n)^(p + 1)/a), x] /; FreeQ[{a, b, n, p}, x] &
& EqQ[1/n + p + 1, 0]

Rule 667

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

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]

Rubi steps \begin{align*} \text {integral}& = \frac {(e f+d g) (d+e x)^3}{5 d e^2 \left (d^2-e^2 x^2\right )^{5/2}}-\frac {(-5 e f+3 (e f+d g)) \int \frac {(d+e x)^2}{\left (d^2-e^2 x^2\right )^{5/2}} \, dx}{5 d e} \\ & = \frac {(e f+d g) (d+e x)^3}{5 d e^2 \left (d^2-e^2 x^2\right )^{5/2}}+\frac {2 (2 e f-3 d g) (d+e x)}{15 d e^2 \left (d^2-e^2 x^2\right )^{3/2}}-\frac {(-5 e f+3 (e f+d g)) \int \frac {1}{\left (d^2-e^2 x^2\right )^{3/2}} \, dx}{15 d e} \\ & = \frac {(e f+d g) (d+e x)^3}{5 d e^2 \left (d^2-e^2 x^2\right )^{5/2}}+\frac {2 (2 e f-3 d g) (d+e x)}{15 d e^2 \left (d^2-e^2 x^2\right )^{3/2}}+\frac {(2 e f-3 d g) x}{15 d^3 e \sqrt {d^2-e^2 x^2}} \\ \end{align*}

Mathematica [A] (verified)

Time = 0.47 (sec) , antiderivative size = 77, normalized size of antiderivative = 0.66 \[ \int \frac {(d+e x)^3 (f+g x)}{\left (d^2-e^2 x^2\right )^{7/2}} \, dx=\frac {\sqrt {d^2-e^2 x^2} \left (-3 d^3 g+2 e^3 f x^2-3 d e^2 x (2 f+g x)+d^2 e (7 f+9 g x)\right )}{15 d^3 e^2 (d-e x)^3} \]

[In]

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

[Out]

(Sqrt[d^2 - e^2*x^2]*(-3*d^3*g + 2*e^3*f*x^2 - 3*d*e^2*x*(2*f + g*x) + d^2*e*(7*f + 9*g*x)))/(15*d^3*e^2*(d -
e*x)^3)

Maple [A] (verified)

Time = 0.55 (sec) , antiderivative size = 80, normalized size of antiderivative = 0.68

method	result	size
trager	\(-\frac {\left (3 d \,e^{2} g \,x^{2}-2 e^{3} f \,x^{2}-9 d^{2} e g x +6 d \,e^{2} f x +3 d^{3} g -7 d^{2} e f \right ) \sqrt {-e^{2} x^{2}+d^{2}}}{15 d^{3} \left (-e x +d \right )^{3} e^{2}}\)	\(80\)
gosper	\(-\frac {\left (-e x +d \right ) \left (e x +d \right )^{4} \left (3 d \,e^{2} g \,x^{2}-2 e^{3} f \,x^{2}-9 d^{2} e g x +6 d \,e^{2} f x +3 d^{3} g -7 d^{2} e f \right )}{15 d^{3} e^{2} \left (-e^{2} x^{2}+d^{2}\right )^{\frac {7}{2}}}\)	\(85\)
default	\(d^{3} f \left (\frac {x}{5 d^{2} \left (-e^{2} x^{2}+d^{2}\right )^{\frac {5}{2}}}+\frac {\frac {4 x}{15 d^{2} \left (-e^{2} x^{2}+d^{2}\right )^{\frac {3}{2}}}+\frac {8 x}{15 d^{4} \sqrt {-e^{2} x^{2}+d^{2}}}}{d^{2}}\right )+e^{3} g \left (\frac {x^{3}}{2 e^{2} \left (-e^{2} x^{2}+d^{2}\right )^{\frac {5}{2}}}-\frac {3 d^{2} \left (\frac {x}{4 e^{2} \left (-e^{2} x^{2}+d^{2}\right )^{\frac {5}{2}}}-\frac {d^{2} \left (\frac {x}{5 d^{2} \left (-e^{2} x^{2}+d^{2}\right )^{\frac {5}{2}}}+\frac {\frac {4 x}{15 d^{2} \left (-e^{2} x^{2}+d^{2}\right )^{\frac {3}{2}}}+\frac {8 x}{15 d^{4} \sqrt {-e^{2} x^{2}+d^{2}}}}{d^{2}}\right )}{4 e^{2}}\right )}{2 e^{2}}\right )+\left (3 d \,e^{2} g +e^{3} f \right ) \left (\frac {x^{2}}{3 e^{2} \left (-e^{2} x^{2}+d^{2}\right )^{\frac {5}{2}}}-\frac {2 d^{2}}{15 e^{4} \left (-e^{2} x^{2}+d^{2}\right )^{\frac {5}{2}}}\right )+\frac {d^{3} g +3 d^{2} e f}{5 e^{2} \left (-e^{2} x^{2}+d^{2}\right )^{\frac {5}{2}}}+\left (3 d^{2} e g +3 d \,e^{2} f \right ) \left (\frac {x}{4 e^{2} \left (-e^{2} x^{2}+d^{2}\right )^{\frac {5}{2}}}-\frac {d^{2} \left (\frac {x}{5 d^{2} \left (-e^{2} x^{2}+d^{2}\right )^{\frac {5}{2}}}+\frac {\frac {4 x}{15 d^{2} \left (-e^{2} x^{2}+d^{2}\right )^{\frac {3}{2}}}+\frac {8 x}{15 d^{4} \sqrt {-e^{2} x^{2}+d^{2}}}}{d^{2}}\right )}{4 e^{2}}\right )\)	\(409\)

[In]

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

[Out]

-1/15*(3*d*e^2*g*x^2-2*e^3*f*x^2-9*d^2*e*g*x+6*d*e^2*f*x+3*d^3*g-7*d^2*e*f)/d^3/(-e*x+d)^3/e^2*(-e^2*x^2+d^2)^
(1/2)

Fricas [A] (verification not implemented)

none

Time = 0.32 (sec) , antiderivative size = 183, normalized size of antiderivative = 1.56 \[ \int \frac {(d+e x)^3 (f+g x)}{\left (d^2-e^2 x^2\right )^{7/2}} \, dx=-\frac {7 \, d^{3} e f - 3 \, d^{4} g - {\left (7 \, e^{4} f - 3 \, d e^{3} g\right )} x^{3} + 3 \, {\left (7 \, d e^{3} f - 3 \, d^{2} e^{2} g\right )} x^{2} - 3 \, {\left (7 \, d^{2} e^{2} f - 3 \, d^{3} e g\right )} x + {\left (7 \, d^{2} e f - 3 \, d^{3} g + {\left (2 \, e^{3} f - 3 \, d e^{2} g\right )} x^{2} - 3 \, {\left (2 \, d e^{2} f - 3 \, d^{2} e g\right )} x\right )} \sqrt {-e^{2} x^{2} + d^{2}}}{15 \, {\left (d^{3} e^{5} x^{3} - 3 \, d^{4} e^{4} x^{2} + 3 \, d^{5} e^{3} x - d^{6} e^{2}\right )}} \]

[In]

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

[Out]

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

Sympy [F]

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

[In]

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

[Out]

Integral((d + e*x)**3*(f + g*x)/(-(-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. 373 vs. \(2 (105) = 210\).

Time = 0.20 (sec) , antiderivative size = 373, normalized size of antiderivative = 3.19 \[ \int \frac {(d+e x)^3 (f+g x)}{\left (d^2-e^2 x^2\right )^{7/2}} \, dx=\frac {e g x^{3}}{2 \, {\left (-e^{2} x^{2} + d^{2}\right )}^{\frac {5}{2}}} + \frac {d f x}{5 \, {\left (-e^{2} x^{2} + d^{2}\right )}^{\frac {5}{2}}} - \frac {3 \, d^{2} g x}{10 \, {\left (-e^{2} x^{2} + d^{2}\right )}^{\frac {5}{2}} e} + \frac {3 \, d^{2} f}{5 \, {\left (-e^{2} x^{2} + d^{2}\right )}^{\frac {5}{2}} e} + \frac {d^{3} g}{5 \, {\left (-e^{2} x^{2} + d^{2}\right )}^{\frac {5}{2}} e^{2}} + \frac {4 \, f x}{15 \, {\left (-e^{2} x^{2} + d^{2}\right )}^{\frac {3}{2}} d} + \frac {g x}{10 \, {\left (-e^{2} x^{2} + d^{2}\right )}^{\frac {3}{2}} e} + \frac {8 \, f x}{15 \, \sqrt {-e^{2} x^{2} + d^{2}} d^{3}} + \frac {g x}{5 \, \sqrt {-e^{2} x^{2} + d^{2}} d^{2} e} + \frac {{\left (e^{3} f + 3 \, d e^{2} g\right )} x^{2}}{3 \, {\left (-e^{2} x^{2} + d^{2}\right )}^{\frac {5}{2}} e^{2}} + \frac {3 \, {\left (d e^{2} f + d^{2} e g\right )} x}{5 \, {\left (-e^{2} x^{2} + d^{2}\right )}^{\frac {5}{2}} e^{2}} - \frac {2 \, {\left (e^{3} f + 3 \, d e^{2} g\right )} d^{2}}{15 \, {\left (-e^{2} x^{2} + d^{2}\right )}^{\frac {5}{2}} e^{4}} - \frac {{\left (d e^{2} f + d^{2} e g\right )} x}{5 \, {\left (-e^{2} x^{2} + d^{2}\right )}^{\frac {3}{2}} d^{2} e^{2}} - \frac {2 \, {\left (d e^{2} f + d^{2} e g\right )} x}{5 \, \sqrt {-e^{2} x^{2} + d^{2}} d^{4} e^{2}} \]

[In]

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

[Out]

1/2*e*g*x^3/(-e^2*x^2 + d^2)^(5/2) + 1/5*d*f*x/(-e^2*x^2 + d^2)^(5/2) - 3/10*d^2*g*x/((-e^2*x^2 + d^2)^(5/2)*e
) + 3/5*d^2*f/((-e^2*x^2 + d^2)^(5/2)*e) + 1/5*d^3*g/((-e^2*x^2 + d^2)^(5/2)*e^2) + 4/15*f*x/((-e^2*x^2 + d^2)
^(3/2)*d) + 1/10*g*x/((-e^2*x^2 + d^2)^(3/2)*e) + 8/15*f*x/(sqrt(-e^2*x^2 + d^2)*d^3) + 1/5*g*x/(sqrt(-e^2*x^2
+ d^2)*d^2*e) + 1/3*(e^3*f + 3*d*e^2*g)*x^2/((-e^2*x^2 + d^2)^(5/2)*e^2) + 3/5*(d*e^2*f + d^2*e*g)*x/((-e^2*x
^2 + d^2)^(5/2)*e^2) - 2/15*(e^3*f + 3*d*e^2*g)*d^2/((-e^2*x^2 + d^2)^(5/2)*e^4) - 1/5*(d*e^2*f + d^2*e*g)*x/(
(-e^2*x^2 + d^2)^(3/2)*d^2*e^2) - 2/5*(d*e^2*f + d^2*e*g)*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. 276 vs. \(2 (105) = 210\).

Time = 0.31 (sec) , antiderivative size = 276, normalized size of antiderivative = 2.36 \[ \int \frac {(d+e x)^3 (f+g x)}{\left (d^2-e^2 x^2\right )^{7/2}} \, dx=\frac {2 \, {\left (7 \, e f - 3 \, d g - \frac {20 \, {\left (d e + \sqrt {-e^{2} x^{2} + d^{2}} {\left | e \right |}\right )} f}{e x} + \frac {15 \, {\left (d e + \sqrt {-e^{2} x^{2} + d^{2}} {\left | e \right |}\right )} d g}{e^{2} x} + \frac {40 \, {\left (d e + \sqrt {-e^{2} x^{2} + d^{2}} {\left | e \right |}\right )}^{2} f}{e^{3} x^{2}} - \frac {15 \, {\left (d e + \sqrt {-e^{2} x^{2} + d^{2}} {\left | e \right |}\right )}^{2} d g}{e^{4} x^{2}} - \frac {30 \, {\left (d e + \sqrt {-e^{2} x^{2} + d^{2}} {\left | e \right |}\right )}^{3} f}{e^{5} x^{3}} + \frac {15 \, {\left (d e + \sqrt {-e^{2} x^{2} + d^{2}} {\left | e \right |}\right )}^{3} d g}{e^{6} x^{3}} + \frac {15 \, {\left (d e + \sqrt {-e^{2} x^{2} + d^{2}} {\left | e \right |}\right )}^{4} f}{e^{7} x^{4}}\right )}}{15 \, d^{3} e {\left (\frac {d e + \sqrt {-e^{2} x^{2} + d^{2}} {\left | e \right |}}{e^{2} x} - 1\right )}^{5} {\left | e \right |}} \]

[In]

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

[Out]

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

Mupad [B] (verification not implemented)

Time = 12.72 (sec) , antiderivative size = 79, normalized size of antiderivative = 0.68 \[ \int \frac {(d+e x)^3 (f+g x)}{\left (d^2-e^2 x^2\right )^{7/2}} \, dx=-\frac {\sqrt {d^2-e^2\,x^2}\,\left (3\,g\,d^3-9\,g\,d^2\,e\,x-7\,f\,d^2\,e+3\,g\,d\,e^2\,x^2+6\,f\,d\,e^2\,x-2\,f\,e^3\,x^2\right )}{15\,d^3\,e^2\,{\left (d-e\,x\right )}^3} \]

[In]

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

[Out]

-((d^2 - e^2*x^2)^(1/2)*(3*d^3*g - 2*e^3*f*x^2 - 7*d^2*e*f + 6*d*e^2*f*x - 9*d^2*e*g*x + 3*d*e^2*g*x^2))/(15*d
^3*e^2*(d - e*x)^3)