3.5.33 \(\int \frac {(d+e x)^4}{(a+c x^2)^3} \, dx\)

Optimal. Leaf size=120 \[ \frac {3 \left (a e^2+c d^2\right )^2 \tan ^{-1}\left (\frac {\sqrt {c} x}{\sqrt {a}}\right )}{8 a^{5/2} c^{5/2}}-\frac {3 (d+e x) \left (a e^2+c d^2\right ) (a e-c d x)}{8 a^2 c^2 \left (a+c x^2\right )}-\frac {(d+e x)^3 (a e-c d x)}{4 a c \left (a+c x^2\right )^2} \]

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Rubi [A]  time = 0.05, antiderivative size = 120, normalized size of antiderivative = 1.00, number of steps used = 3, number of rules used = 2, integrand size = 17, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.118, Rules used = {723, 205} \begin {gather*} -\frac {3 (d+e x) \left (a e^2+c d^2\right ) (a e-c d x)}{8 a^2 c^2 \left (a+c x^2\right )}+\frac {3 \left (a e^2+c d^2\right )^2 \tan ^{-1}\left (\frac {\sqrt {c} x}{\sqrt {a}}\right )}{8 a^{5/2} c^{5/2}}-\frac {(d+e x)^3 (a e-c d x)}{4 a c \left (a+c x^2\right )^2} \end {gather*}

Antiderivative was successfully verified.

[In]

Int[(d + e*x)^4/(a + c*x^2)^3,x]

[Out]

-((a*e - c*d*x)*(d + e*x)^3)/(4*a*c*(a + c*x^2)^2) - (3*(c*d^2 + a*e^2)*(a*e - c*d*x)*(d + e*x))/(8*a^2*c^2*(a
+ c*x^2)) + (3*(c*d^2 + a*e^2)^2*ArcTan[(Sqrt[c]*x)/Sqrt[a]])/(8*a^(5/2)*c^(5/2))

Rule 205

Int[((a_) + (b_.)*(x_)^2)^(-1), x_Symbol] :> Simp[(Rt[a/b, 2]*ArcTan[x/Rt[a/b, 2]])/a, x] /; FreeQ[{a, b}, x]
&& PosQ[a/b]

Rule 723

Int[((d_) + (e_.)*(x_))^(m_)*((a_) + (c_.)*(x_)^2)^(p_), x_Symbol] :> Simp[((d + e*x)^(m - 1)*(a*e - c*d*x)*(a
+ c*x^2)^(p + 1))/(2*a*c*(p + 1)), x] + Dist[((2*p + 3)*(c*d^2 + a*e^2))/(2*a*c*(p + 1)), Int[(d + e*x)^(m -
2)*(a + c*x^2)^(p + 1), x], x] /; FreeQ[{a, c, d, e}, x] && NeQ[c*d^2 + a*e^2, 0] && EqQ[m + 2*p + 2, 0] && Lt
Q[p, -1]

Rubi steps

\begin {align*} \int \frac {(d+e x)^4}{\left (a+c x^2\right )^3} \, dx &=-\frac {(a e-c d x) (d+e x)^3}{4 a c \left (a+c x^2\right )^2}+\frac {\left (3 \left (c d^2+a e^2\right )\right ) \int \frac {(d+e x)^2}{\left (a+c x^2\right )^2} \, dx}{4 a c}\\ &=-\frac {(a e-c d x) (d+e x)^3}{4 a c \left (a+c x^2\right )^2}-\frac {3 \left (c d^2+a e^2\right ) (a e-c d x) (d+e x)}{8 a^2 c^2 \left (a+c x^2\right )}+\frac {\left (3 \left (c d^2+a e^2\right )^2\right ) \int \frac {1}{a+c x^2} \, dx}{8 a^2 c^2}\\ &=-\frac {(a e-c d x) (d+e x)^3}{4 a c \left (a+c x^2\right )^2}-\frac {3 \left (c d^2+a e^2\right ) (a e-c d x) (d+e x)}{8 a^2 c^2 \left (a+c x^2\right )}+\frac {3 \left (c d^2+a e^2\right )^2 \tan ^{-1}\left (\frac {\sqrt {c} x}{\sqrt {a}}\right )}{8 a^{5/2} c^{5/2}}\\ \end {align*}

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Mathematica [A]  time = 0.10, size = 148, normalized size = 1.23 \begin {gather*} \frac {3 \left (a e^2+c d^2\right )^2 \tan ^{-1}\left (\frac {\sqrt {c} x}{\sqrt {a}}\right )}{8 a^{5/2} c^{5/2}}+\frac {-a^3 e^3 (8 d+3 e x)-a^2 c e \left (8 d^3+6 d^2 e x+16 d e^2 x^2+5 e^3 x^3\right )+a c^2 d^2 x \left (5 d^2+6 e^2 x^2\right )+3 c^3 d^4 x^3}{8 a^2 c^2 \left (a+c x^2\right )^2} \end {gather*}

Antiderivative was successfully verified.

[In]

Integrate[(d + e*x)^4/(a + c*x^2)^3,x]

[Out]

(3*c^3*d^4*x^3 - a^3*e^3*(8*d + 3*e*x) + a*c^2*d^2*x*(5*d^2 + 6*e^2*x^2) - a^2*c*e*(8*d^3 + 6*d^2*e*x + 16*d*e
^2*x^2 + 5*e^3*x^3))/(8*a^2*c^2*(a + c*x^2)^2) + (3*(c*d^2 + a*e^2)^2*ArcTan[(Sqrt[c]*x)/Sqrt[a]])/(8*a^(5/2)*
c^(5/2))

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IntegrateAlgebraic [F]  time = 0.00, size = 0, normalized size = 0.00 \begin {gather*} \int \frac {(d+e x)^4}{\left (a+c x^2\right )^3} \, dx \end {gather*}

Verification is not applicable to the result.

[In]

IntegrateAlgebraic[(d + e*x)^4/(a + c*x^2)^3,x]

[Out]

IntegrateAlgebraic[(d + e*x)^4/(a + c*x^2)^3, x]

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fricas [B]  time = 0.41, size = 554, normalized size = 4.62 \begin {gather*} \left [-\frac {32 \, a^{3} c^{2} d e^{3} x^{2} + 16 \, a^{3} c^{2} d^{3} e + 16 \, a^{4} c d e^{3} - 2 \, {\left (3 \, a c^{4} d^{4} + 6 \, a^{2} c^{3} d^{2} e^{2} - 5 \, a^{3} c^{2} e^{4}\right )} x^{3} + 3 \, {\left (a^{2} c^{2} d^{4} + 2 \, a^{3} c d^{2} e^{2} + a^{4} e^{4} + {\left (c^{4} d^{4} + 2 \, a c^{3} d^{2} e^{2} + a^{2} c^{2} e^{4}\right )} x^{4} + 2 \, {\left (a c^{3} d^{4} + 2 \, a^{2} c^{2} d^{2} e^{2} + a^{3} c e^{4}\right )} x^{2}\right )} \sqrt {-a c} \log \left (\frac {c x^{2} - 2 \, \sqrt {-a c} x - a}{c x^{2} + a}\right ) - 2 \, {\left (5 \, a^{2} c^{3} d^{4} - 6 \, a^{3} c^{2} d^{2} e^{2} - 3 \, a^{4} c e^{4}\right )} x}{16 \, {\left (a^{3} c^{5} x^{4} + 2 \, a^{4} c^{4} x^{2} + a^{5} c^{3}\right )}}, -\frac {16 \, a^{3} c^{2} d e^{3} x^{2} + 8 \, a^{3} c^{2} d^{3} e + 8 \, a^{4} c d e^{3} - {\left (3 \, a c^{4} d^{4} + 6 \, a^{2} c^{3} d^{2} e^{2} - 5 \, a^{3} c^{2} e^{4}\right )} x^{3} - 3 \, {\left (a^{2} c^{2} d^{4} + 2 \, a^{3} c d^{2} e^{2} + a^{4} e^{4} + {\left (c^{4} d^{4} + 2 \, a c^{3} d^{2} e^{2} + a^{2} c^{2} e^{4}\right )} x^{4} + 2 \, {\left (a c^{3} d^{4} + 2 \, a^{2} c^{2} d^{2} e^{2} + a^{3} c e^{4}\right )} x^{2}\right )} \sqrt {a c} \arctan \left (\frac {\sqrt {a c} x}{a}\right ) - {\left (5 \, a^{2} c^{3} d^{4} - 6 \, a^{3} c^{2} d^{2} e^{2} - 3 \, a^{4} c e^{4}\right )} x}{8 \, {\left (a^{3} c^{5} x^{4} + 2 \, a^{4} c^{4} x^{2} + a^{5} c^{3}\right )}}\right ] \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate((e*x+d)^4/(c*x^2+a)^3,x, algorithm="fricas")

[Out]

[-1/16*(32*a^3*c^2*d*e^3*x^2 + 16*a^3*c^2*d^3*e + 16*a^4*c*d*e^3 - 2*(3*a*c^4*d^4 + 6*a^2*c^3*d^2*e^2 - 5*a^3*
c^2*e^4)*x^3 + 3*(a^2*c^2*d^4 + 2*a^3*c*d^2*e^2 + a^4*e^4 + (c^4*d^4 + 2*a*c^3*d^2*e^2 + a^2*c^2*e^4)*x^4 + 2*
(a*c^3*d^4 + 2*a^2*c^2*d^2*e^2 + a^3*c*e^4)*x^2)*sqrt(-a*c)*log((c*x^2 - 2*sqrt(-a*c)*x - a)/(c*x^2 + a)) - 2*
(5*a^2*c^3*d^4 - 6*a^3*c^2*d^2*e^2 - 3*a^4*c*e^4)*x)/(a^3*c^5*x^4 + 2*a^4*c^4*x^2 + a^5*c^3), -1/8*(16*a^3*c^2
*d*e^3*x^2 + 8*a^3*c^2*d^3*e + 8*a^4*c*d*e^3 - (3*a*c^4*d^4 + 6*a^2*c^3*d^2*e^2 - 5*a^3*c^2*e^4)*x^3 - 3*(a^2*
c^2*d^4 + 2*a^3*c*d^2*e^2 + a^4*e^4 + (c^4*d^4 + 2*a*c^3*d^2*e^2 + a^2*c^2*e^4)*x^4 + 2*(a*c^3*d^4 + 2*a^2*c^2
*d^2*e^2 + a^3*c*e^4)*x^2)*sqrt(a*c)*arctan(sqrt(a*c)*x/a) - (5*a^2*c^3*d^4 - 6*a^3*c^2*d^2*e^2 - 3*a^4*c*e^4)
*x)/(a^3*c^5*x^4 + 2*a^4*c^4*x^2 + a^5*c^3)]

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giac [A]  time = 0.19, size = 161, normalized size = 1.34 \begin {gather*} \frac {3 \, {\left (c^{2} d^{4} + 2 \, a c d^{2} e^{2} + a^{2} e^{4}\right )} \arctan \left (\frac {c x}{\sqrt {a c}}\right )}{8 \, \sqrt {a c} a^{2} c^{2}} + \frac {3 \, c^{3} d^{4} x^{3} + 6 \, a c^{2} d^{2} x^{3} e^{2} + 5 \, a c^{2} d^{4} x - 5 \, a^{2} c x^{3} e^{4} - 16 \, a^{2} c d x^{2} e^{3} - 6 \, a^{2} c d^{2} x e^{2} - 8 \, a^{2} c d^{3} e - 3 \, a^{3} x e^{4} - 8 \, a^{3} d e^{3}}{8 \, {\left (c x^{2} + a\right )}^{2} a^{2} c^{2}} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate((e*x+d)^4/(c*x^2+a)^3,x, algorithm="giac")

[Out]

3/8*(c^2*d^4 + 2*a*c*d^2*e^2 + a^2*e^4)*arctan(c*x/sqrt(a*c))/(sqrt(a*c)*a^2*c^2) + 1/8*(3*c^3*d^4*x^3 + 6*a*c
^2*d^2*x^3*e^2 + 5*a*c^2*d^4*x - 5*a^2*c*x^3*e^4 - 16*a^2*c*d*x^2*e^3 - 6*a^2*c*d^2*x*e^2 - 8*a^2*c*d^3*e - 3*
a^3*x*e^4 - 8*a^3*d*e^3)/((c*x^2 + a)^2*a^2*c^2)

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maple [A]  time = 0.05, size = 189, normalized size = 1.58 \begin {gather*} \frac {3 d^{2} e^{2} \arctan \left (\frac {c x}{\sqrt {a c}}\right )}{4 \sqrt {a c}\, a c}+\frac {3 d^{4} \arctan \left (\frac {c x}{\sqrt {a c}}\right )}{8 \sqrt {a c}\, a^{2}}+\frac {3 e^{4} \arctan \left (\frac {c x}{\sqrt {a c}}\right )}{8 \sqrt {a c}\, c^{2}}+\frac {-\frac {2 d \,e^{3} x^{2}}{c}-\frac {\left (5 a^{2} e^{4}-6 a c \,d^{2} e^{2}-3 c^{2} d^{4}\right ) x^{3}}{8 a^{2} c}-\frac {\left (a \,e^{2}+c \,d^{2}\right ) d e}{c^{2}}-\frac {\left (3 a^{2} e^{4}+6 a c \,d^{2} e^{2}-5 c^{2} d^{4}\right ) x}{8 a \,c^{2}}}{\left (c \,x^{2}+a \right )^{2}} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

int((e*x+d)^4/(c*x^2+a)^3,x)

[Out]

(-1/8*(5*a^2*e^4-6*a*c*d^2*e^2-3*c^2*d^4)/a^2/c*x^3-2/c*d*e^3*x^2-1/8*(3*a^2*e^4+6*a*c*d^2*e^2-5*c^2*d^4)/a/c^
2*x-d*e*(a*e^2+c*d^2)/c^2)/(c*x^2+a)^2+3/8/c^2/(a*c)^(1/2)*arctan(1/(a*c)^(1/2)*c*x)*e^4+3/4/a/c/(a*c)^(1/2)*a
rctan(1/(a*c)^(1/2)*c*x)*d^2*e^2+3/8/a^2/(a*c)^(1/2)*arctan(1/(a*c)^(1/2)*c*x)*d^4

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maxima [A]  time = 3.01, size = 182, normalized size = 1.52 \begin {gather*} -\frac {16 \, a^{2} c d e^{3} x^{2} + 8 \, a^{2} c d^{3} e + 8 \, a^{3} d e^{3} - {\left (3 \, c^{3} d^{4} + 6 \, a c^{2} d^{2} e^{2} - 5 \, a^{2} c e^{4}\right )} x^{3} - {\left (5 \, a c^{2} d^{4} - 6 \, a^{2} c d^{2} e^{2} - 3 \, a^{3} e^{4}\right )} x}{8 \, {\left (a^{2} c^{4} x^{4} + 2 \, a^{3} c^{3} x^{2} + a^{4} c^{2}\right )}} + \frac {3 \, {\left (c^{2} d^{4} + 2 \, a c d^{2} e^{2} + a^{2} e^{4}\right )} \arctan \left (\frac {c x}{\sqrt {a c}}\right )}{8 \, \sqrt {a c} a^{2} c^{2}} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate((e*x+d)^4/(c*x^2+a)^3,x, algorithm="maxima")

[Out]

-1/8*(16*a^2*c*d*e^3*x^2 + 8*a^2*c*d^3*e + 8*a^3*d*e^3 - (3*c^3*d^4 + 6*a*c^2*d^2*e^2 - 5*a^2*c*e^4)*x^3 - (5*
a*c^2*d^4 - 6*a^2*c*d^2*e^2 - 3*a^3*e^4)*x)/(a^2*c^4*x^4 + 2*a^3*c^3*x^2 + a^4*c^2) + 3/8*(c^2*d^4 + 2*a*c*d^2
*e^2 + a^2*e^4)*arctan(c*x/sqrt(a*c))/(sqrt(a*c)*a^2*c^2)

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mupad [B]  time = 0.16, size = 197, normalized size = 1.64 \begin {gather*} \frac {3\,\mathrm {atan}\left (\frac {\sqrt {c}\,x\,{\left (c\,d^2+a\,e^2\right )}^2}{\sqrt {a}\,\left (a^2\,e^4+2\,a\,c\,d^2\,e^2+c^2\,d^4\right )}\right )\,{\left (c\,d^2+a\,e^2\right )}^2}{8\,a^{5/2}\,c^{5/2}}-\frac {\frac {2\,d\,e^3\,x^2}{c}+\frac {x\,\left (3\,a^2\,e^4+6\,a\,c\,d^2\,e^2-5\,c^2\,d^4\right )}{8\,a\,c^2}+\frac {d\,e\,\left (c\,d^2+a\,e^2\right )}{c^2}-\frac {x^3\,\left (-5\,a^2\,e^4+6\,a\,c\,d^2\,e^2+3\,c^2\,d^4\right )}{8\,a^2\,c}}{a^2+2\,a\,c\,x^2+c^2\,x^4} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

int((d + e*x)^4/(a + c*x^2)^3,x)

[Out]

(3*atan((c^(1/2)*x*(a*e^2 + c*d^2)^2)/(a^(1/2)*(a^2*e^4 + c^2*d^4 + 2*a*c*d^2*e^2)))*(a*e^2 + c*d^2)^2)/(8*a^(
5/2)*c^(5/2)) - ((2*d*e^3*x^2)/c + (x*(3*a^2*e^4 - 5*c^2*d^4 + 6*a*c*d^2*e^2))/(8*a*c^2) + (d*e*(a*e^2 + c*d^2
))/c^2 - (x^3*(3*c^2*d^4 - 5*a^2*e^4 + 6*a*c*d^2*e^2))/(8*a^2*c))/(a^2 + c^2*x^4 + 2*a*c*x^2)

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sympy [B]  time = 2.15, size = 328, normalized size = 2.73 \begin {gather*} - \frac {3 \sqrt {- \frac {1}{a^{5} c^{5}}} \left (a e^{2} + c d^{2}\right )^{2} \log {\left (- \frac {3 a^{3} c^{2} \sqrt {- \frac {1}{a^{5} c^{5}}} \left (a e^{2} + c d^{2}\right )^{2}}{3 a^{2} e^{4} + 6 a c d^{2} e^{2} + 3 c^{2} d^{4}} + x \right )}}{16} + \frac {3 \sqrt {- \frac {1}{a^{5} c^{5}}} \left (a e^{2} + c d^{2}\right )^{2} \log {\left (\frac {3 a^{3} c^{2} \sqrt {- \frac {1}{a^{5} c^{5}}} \left (a e^{2} + c d^{2}\right )^{2}}{3 a^{2} e^{4} + 6 a c d^{2} e^{2} + 3 c^{2} d^{4}} + x \right )}}{16} + \frac {- 8 a^{3} d e^{3} - 8 a^{2} c d^{3} e - 16 a^{2} c d e^{3} x^{2} + x^{3} \left (- 5 a^{2} c e^{4} + 6 a c^{2} d^{2} e^{2} + 3 c^{3} d^{4}\right ) + x \left (- 3 a^{3} e^{4} - 6 a^{2} c d^{2} e^{2} + 5 a c^{2} d^{4}\right )}{8 a^{4} c^{2} + 16 a^{3} c^{3} x^{2} + 8 a^{2} c^{4} x^{4}} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate((e*x+d)**4/(c*x**2+a)**3,x)

[Out]

-3*sqrt(-1/(a**5*c**5))*(a*e**2 + c*d**2)**2*log(-3*a**3*c**2*sqrt(-1/(a**5*c**5))*(a*e**2 + c*d**2)**2/(3*a**
2*e**4 + 6*a*c*d**2*e**2 + 3*c**2*d**4) + x)/16 + 3*sqrt(-1/(a**5*c**5))*(a*e**2 + c*d**2)**2*log(3*a**3*c**2*
sqrt(-1/(a**5*c**5))*(a*e**2 + c*d**2)**2/(3*a**2*e**4 + 6*a*c*d**2*e**2 + 3*c**2*d**4) + x)/16 + (-8*a**3*d*e
**3 - 8*a**2*c*d**3*e - 16*a**2*c*d*e**3*x**2 + x**3*(-5*a**2*c*e**4 + 6*a*c**2*d**2*e**2 + 3*c**3*d**4) + x*(
-3*a**3*e**4 - 6*a**2*c*d**2*e**2 + 5*a*c**2*d**4))/(8*a**4*c**2 + 16*a**3*c**3*x**2 + 8*a**2*c**4*x**4)

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