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

Optimal. Leaf size=117 \[ \frac {(d+e x)^3 (d g+e f)}{5 d e^2 \left (d^2-e^2 x^2\right )^{5/2}}+\frac {2 (d+e x) (2 e f-3 d g)}{15 d e^2 \left (d^2-e^2 x^2\right )^{3/2}}+\frac {x (2 e f-3 d g)}{15 d^3 e \sqrt {d^2-e^2 x^2}} \]

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Rubi [A]  time = 0.06, antiderivative size = 117, normalized size of antiderivative = 1.00, number of steps used = 3, number of rules used = 3, integrand size = 29, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.103, Rules used = {789, 653, 191} \begin {gather*} \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}} \end {gather*}

Antiderivative was successfully verified.

[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 191

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 653

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 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]

Rubi steps

\begin {align*} \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 {(-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*}

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Mathematica [A]  time = 0.24, size = 83, normalized size = 0.71 \begin {gather*} -\frac {(d+e x) \left (3 d^3 g-d^2 e (7 f+9 g x)+3 d e^2 x (2 f+g x)-2 e^3 f x^2\right )}{15 d^3 e^2 (d-e x)^2 \sqrt {d^2-e^2 x^2}} \end {gather*}

Antiderivative was successfully verified.

[In]

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

[Out]

-1/15*((d + e*x)*(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)))/(d^3*e^2*(d - e*x)^2*S
qrt[d^2 - e^2*x^2])

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IntegrateAlgebraic [A]  time = 0.55, size = 83, normalized size = 0.71 \begin {gather*} \frac {\sqrt {d^2-e^2 x^2} \left (-3 d^3 g+7 d^2 e f+9 d^2 e g x-6 d e^2 f x-3 d e^2 g x^2+2 e^3 f x^2\right )}{15 d^3 e^2 (d-e x)^3} \end {gather*}

Antiderivative was successfully verified.

[In]

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

[Out]

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

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fricas [A]  time = 0.39, size = 183, normalized size = 1.56 \begin {gather*} -\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 )}} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[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)

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giac [A]  time = 0.33, size = 139, normalized size = 1.19 \begin {gather*} -\frac {\sqrt {-x^{2} e^{2} + d^{2}} {\left ({\left (15 \, d f - {\left (x {\left (\frac {{\left (3 \, d^{2} g e^{7} - 2 \, d f e^{8}\right )} x^{2} e^{\left (-4\right )}}{d^{4}} - \frac {5 \, {\left (3 \, d^{4} g e^{5} - d^{3} f e^{6}\right )} e^{\left (-4\right )}}{d^{4}}\right )} - \frac {5 \, {\left (3 \, d^{5} g e^{4} + d^{4} f e^{5}\right )} e^{\left (-4\right )}}{d^{4}}\right )} x\right )} x - \frac {{\left (3 \, d^{7} g e^{2} - 7 \, d^{6} f e^{3}\right )} e^{\left (-4\right )}}{d^{4}}\right )}}{15 \, {\left (x^{2} e^{2} - d^{2}\right )}^{3}} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

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

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maple [A]  time = 0.01, size = 85, normalized size = 0.73 \begin {gather*} -\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 \left (-e^{2} x^{2}+d^{2}\right )^{\frac {7}{2}} d^{3} e^{2}} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

-1/15*(-e*x+d)*(e*x+d)^4*(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^2/(-e^2*x
^2+d^2)^(7/2)

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maxima [B]  time = 0.46, size = 373, normalized size = 3.19 \begin {gather*} \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}} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[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)

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mupad [B]  time = 2.79, size = 79, normalized size = 0.68 \begin {gather*} -\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} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[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)

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sympy [F]  time = 0.00, size = 0, normalized size = 0.00 \begin {gather*} \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 \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[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)

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