3.6.20 \(\int \frac {(d+e x)^3}{(a+c x^2)^4} \, dx\) [520]

Optimal. Leaf size=156 \[ \frac {x (d+e x)^3}{6 a \left (a+c x^2\right )^3}-\frac {(2 a e-5 c d x) (d+e x)^2}{24 a^2 c \left (a+c x^2\right )^2}-\frac {4 a e \left (5 c d^2+a e^2\right )-c d \left (15 c d^2-a e^2\right ) x}{48 a^3 c^2 \left (a+c x^2\right )}+\frac {d \left (5 c d^2+3 a e^2\right ) \tan ^{-1}\left (\frac {\sqrt {c} x}{\sqrt {a}}\right )}{16 a^{7/2} c^{3/2}} \]

[Out]

1/6*x*(e*x+d)^3/a/(c*x^2+a)^3-1/24*(-5*c*d*x+2*a*e)*(e*x+d)^2/a^2/c/(c*x^2+a)^2+1/48*(-4*a*e*(a*e^2+5*c*d^2)+c
*d*(-a*e^2+15*c*d^2)*x)/a^3/c^2/(c*x^2+a)+1/16*d*(3*a*e^2+5*c*d^2)*arctan(x*c^(1/2)/a^(1/2))/a^(7/2)/c^(3/2)

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Rubi [A]
time = 0.08, antiderivative size = 156, normalized size of antiderivative = 1.00, number of steps used = 4, number of rules used = 4, integrand size = 17, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.235, Rules used = {751, 835, 792, 211} \begin {gather*} \frac {d \text {ArcTan}\left (\frac {\sqrt {c} x}{\sqrt {a}}\right ) \left (3 a e^2+5 c d^2\right )}{16 a^{7/2} c^{3/2}}-\frac {4 a e \left (a e^2+5 c d^2\right )-c d x \left (15 c d^2-a e^2\right )}{48 a^3 c^2 \left (a+c x^2\right )}-\frac {(d+e x)^2 (2 a e-5 c d x)}{24 a^2 c \left (a+c x^2\right )^2}+\frac {x (d+e x)^3}{6 a \left (a+c x^2\right )^3} \end {gather*}

Antiderivative was successfully verified.

[In]

Int[(d + e*x)^3/(a + c*x^2)^4,x]

[Out]

(x*(d + e*x)^3)/(6*a*(a + c*x^2)^3) - ((2*a*e - 5*c*d*x)*(d + e*x)^2)/(24*a^2*c*(a + c*x^2)^2) - (4*a*e*(5*c*d
^2 + a*e^2) - c*d*(15*c*d^2 - a*e^2)*x)/(48*a^3*c^2*(a + c*x^2)) + (d*(5*c*d^2 + 3*a*e^2)*ArcTan[(Sqrt[c]*x)/S
qrt[a]])/(16*a^(7/2)*c^(3/2))

Rule 211

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

Rule 751

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

Rule 792

Int[((d_.) + (e_.)*(x_))*((f_.) + (g_.)*(x_))*((a_) + (c_.)*(x_)^2)^(p_), x_Symbol] :> Simp[(a*(e*f + d*g) - (
c*d*f - a*e*g)*x)*((a + c*x^2)^(p + 1)/(2*a*c*(p + 1))), x] - Dist[(a*e*g - c*d*f*(2*p + 3))/(2*a*c*(p + 1)),
Int[(a + c*x^2)^(p + 1), x], x] /; FreeQ[{a, c, d, e, f, g}, x] && LtQ[p, -1]

Rule 835

Int[((d_.) + (e_.)*(x_))^(m_)*((f_.) + (g_.)*(x_))*((a_) + (c_.)*(x_)^2)^(p_), x_Symbol] :> Simp[(d + e*x)^m*(
a + c*x^2)^(p + 1)*((a*g - c*f*x)/(2*a*c*(p + 1))), x] - Dist[1/(2*a*c*(p + 1)), Int[(d + e*x)^(m - 1)*(a + c*
x^2)^(p + 1)*Simp[a*e*g*m - c*d*f*(2*p + 3) - c*e*f*(m + 2*p + 3)*x, x], x], x] /; FreeQ[{a, c, d, e, f, g}, x
] && NeQ[c*d^2 + a*e^2, 0] && LtQ[p, -1] && GtQ[m, 0] && (IntegerQ[m] || IntegerQ[p] || IntegersQ[2*m, 2*p])

Rubi steps

\begin {align*} \int \frac {(d+e x)^3}{\left (a+c x^2\right )^4} \, dx &=\frac {x (d+e x)^3}{6 a \left (a+c x^2\right )^3}-\frac {\int \frac {(-5 d-2 e x) (d+e x)^2}{\left (a+c x^2\right )^3} \, dx}{6 a}\\ &=\frac {x (d+e x)^3}{6 a \left (a+c x^2\right )^3}-\frac {(2 a e-5 c d x) (d+e x)^2}{24 a^2 c \left (a+c x^2\right )^2}-\frac {\int \frac {(d+e x) \left (-15 c d^2-4 a e^2-5 c d e x\right )}{\left (a+c x^2\right )^2} \, dx}{24 a^2 c}\\ &=\frac {x (d+e x)^3}{6 a \left (a+c x^2\right )^3}-\frac {(2 a e-5 c d x) (d+e x)^2}{24 a^2 c \left (a+c x^2\right )^2}-\frac {4 a e \left (5 c d^2+a e^2\right )-c d \left (15 c d^2-a e^2\right ) x}{48 a^3 c^2 \left (a+c x^2\right )}+\frac {\left (d \left (5 c d^2+3 a e^2\right )\right ) \int \frac {1}{a+c x^2} \, dx}{16 a^3 c}\\ &=\frac {x (d+e x)^3}{6 a \left (a+c x^2\right )^3}-\frac {(2 a e-5 c d x) (d+e x)^2}{24 a^2 c \left (a+c x^2\right )^2}-\frac {4 a e \left (5 c d^2+a e^2\right )-c d \left (15 c d^2-a e^2\right ) x}{48 a^3 c^2 \left (a+c x^2\right )}+\frac {d \left (5 c d^2+3 a e^2\right ) \tan ^{-1}\left (\frac {\sqrt {c} x}{\sqrt {a}}\right )}{16 a^{7/2} c^{3/2}}\\ \end {align*}

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Mathematica [A]
time = 0.10, size = 155, normalized size = 0.99 \begin {gather*} \frac {\frac {\sqrt {a} \left (-4 a^4 e^3+15 c^4 d^3 x^5-3 a^3 c e \left (8 d^2+3 d e x+4 e^2 x^2\right )+3 a^2 c^2 d x \left (11 d^2+8 e^2 x^2\right )+a c^3 d x^3 \left (40 d^2+9 e^2 x^2\right )\right )}{\left (a+c x^2\right )^3}+3 \sqrt {c} d \left (5 c d^2+3 a e^2\right ) \tan ^{-1}\left (\frac {\sqrt {c} x}{\sqrt {a}}\right )}{48 a^{7/2} c^2} \end {gather*}

Antiderivative was successfully verified.

[In]

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

[Out]

((Sqrt[a]*(-4*a^4*e^3 + 15*c^4*d^3*x^5 - 3*a^3*c*e*(8*d^2 + 3*d*e*x + 4*e^2*x^2) + 3*a^2*c^2*d*x*(11*d^2 + 8*e
^2*x^2) + a*c^3*d*x^3*(40*d^2 + 9*e^2*x^2)))/(a + c*x^2)^3 + 3*Sqrt[c]*d*(5*c*d^2 + 3*a*e^2)*ArcTan[(Sqrt[c]*x
)/Sqrt[a]])/(48*a^(7/2)*c^2)

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Maple [A]
time = 0.46, size = 146, normalized size = 0.94

method	result	size
default	\(\frac {\frac {d \left (3 e^{2} a +5 c \,d^{2}\right ) c \,x^{5}}{16 a^{3}}+\frac {d \left (3 e^{2} a +5 c \,d^{2}\right ) x^{3}}{6 a^{2}}-\frac {e^{3} x^{2}}{4 c}-\frac {d \left (3 e^{2} a -11 c \,d^{2}\right ) x}{16 a c}-\frac {e \left (e^{2} a +6 c \,d^{2}\right )}{12 c^{2}}}{\left (c \,x^{2}+a \right )^{3}}+\frac {d \left (3 e^{2} a +5 c \,d^{2}\right ) \arctan \left (\frac {c x}{\sqrt {a c}}\right )}{16 c \,a^{3} \sqrt {a c}}\)	\(146\)
risch	\(\frac {\frac {d \left (3 e^{2} a +5 c \,d^{2}\right ) c \,x^{5}}{16 a^{3}}+\frac {d \left (3 e^{2} a +5 c \,d^{2}\right ) x^{3}}{6 a^{2}}-\frac {e^{3} x^{2}}{4 c}-\frac {d \left (3 e^{2} a -11 c \,d^{2}\right ) x}{16 a c}-\frac {e \left (e^{2} a +6 c \,d^{2}\right )}{12 c^{2}}}{\left (c \,x^{2}+a \right )^{3}}-\frac {3 d \ln \left (c x +\sqrt {-a c}\right ) e^{2}}{32 \sqrt {-a c}\, c \,a^{2}}-\frac {5 d^{3} \ln \left (c x +\sqrt {-a c}\right )}{32 \sqrt {-a c}\, a^{3}}+\frac {3 d \ln \left (-c x +\sqrt {-a c}\right ) e^{2}}{32 \sqrt {-a c}\, c \,a^{2}}+\frac {5 d^{3} \ln \left (-c x +\sqrt {-a c}\right )}{32 \sqrt {-a c}\, a^{3}}\)	\(220\)

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

(1/16*d*(3*a*e^2+5*c*d^2)*c/a^3*x^5+1/6/a^2*d*(3*a*e^2+5*c*d^2)*x^3-1/4*e^3*x^2/c-1/16*d*(3*a*e^2-11*c*d^2)/a/
c*x-1/12*e*(a*e^2+6*c*d^2)/c^2)/(c*x^2+a)^3+1/16*d*(3*a*e^2+5*c*d^2)/c/a^3/(a*c)^(1/2)*arctan(c*x/(a*c)^(1/2))

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Maxima [A]
time = 0.49, size = 183, normalized size = 1.17 \begin {gather*} -\frac {12 \, a^{3} c x^{2} e^{3} + 24 \, a^{3} c d^{2} e - 3 \, {\left (5 \, c^{4} d^{3} + 3 \, a c^{3} d e^{2}\right )} x^{5} + 4 \, a^{4} e^{3} - 8 \, {\left (5 \, a c^{3} d^{3} + 3 \, a^{2} c^{2} d e^{2}\right )} x^{3} - 3 \, {\left (11 \, a^{2} c^{2} d^{3} - 3 \, a^{3} c d e^{2}\right )} x}{48 \, {\left (a^{3} c^{5} x^{6} + 3 \, a^{4} c^{4} x^{4} + 3 \, a^{5} c^{3} x^{2} + a^{6} c^{2}\right )}} + \frac {{\left (5 \, c d^{3} + 3 \, a d e^{2}\right )} \arctan \left (\frac {c x}{\sqrt {a c}}\right )}{16 \, \sqrt {a c} a^{3} c} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

-1/48*(12*a^3*c*x^2*e^3 + 24*a^3*c*d^2*e - 3*(5*c^4*d^3 + 3*a*c^3*d*e^2)*x^5 + 4*a^4*e^3 - 8*(5*a*c^3*d^3 + 3*
a^2*c^2*d*e^2)*x^3 - 3*(11*a^2*c^2*d^3 - 3*a^3*c*d*e^2)*x)/(a^3*c^5*x^6 + 3*a^4*c^4*x^4 + 3*a^5*c^3*x^2 + a^6*
c^2) + 1/16*(5*c*d^3 + 3*a*d*e^2)*arctan(c*x/sqrt(a*c))/(sqrt(a*c)*a^3*c)

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Fricas [A]
time = 2.00, size = 528, normalized size = 3.38 \begin {gather*} \left [\frac {30 \, a c^{4} d^{3} x^{5} + 80 \, a^{2} c^{3} d^{3} x^{3} + 66 \, a^{3} c^{2} d^{3} x - 48 \, a^{4} c d^{2} e - 3 \, {\left (5 \, c^{4} d^{3} x^{6} + 15 \, a c^{3} d^{3} x^{4} + 15 \, a^{2} c^{2} d^{3} x^{2} + 5 \, a^{3} c d^{3} + 3 \, {\left (a c^{3} d x^{6} + 3 \, a^{2} c^{2} d x^{4} + 3 \, a^{3} c d x^{2} + a^{4} d\right )} e^{2}\right )} \sqrt {-a c} \log \left (\frac {c x^{2} - 2 \, \sqrt {-a c} x - a}{c x^{2} + a}\right ) - 8 \, {\left (3 \, a^{4} c x^{2} + a^{5}\right )} e^{3} + 6 \, {\left (3 \, a^{2} c^{3} d x^{5} + 8 \, a^{3} c^{2} d x^{3} - 3 \, a^{4} c d x\right )} e^{2}}{96 \, {\left (a^{4} c^{5} x^{6} + 3 \, a^{5} c^{4} x^{4} + 3 \, a^{6} c^{3} x^{2} + a^{7} c^{2}\right )}}, \frac {15 \, a c^{4} d^{3} x^{5} + 40 \, a^{2} c^{3} d^{3} x^{3} + 33 \, a^{3} c^{2} d^{3} x - 24 \, a^{4} c d^{2} e + 3 \, {\left (5 \, c^{4} d^{3} x^{6} + 15 \, a c^{3} d^{3} x^{4} + 15 \, a^{2} c^{2} d^{3} x^{2} + 5 \, a^{3} c d^{3} + 3 \, {\left (a c^{3} d x^{6} + 3 \, a^{2} c^{2} d x^{4} + 3 \, a^{3} c d x^{2} + a^{4} d\right )} e^{2}\right )} \sqrt {a c} \arctan \left (\frac {\sqrt {a c} x}{a}\right ) - 4 \, {\left (3 \, a^{4} c x^{2} + a^{5}\right )} e^{3} + 3 \, {\left (3 \, a^{2} c^{3} d x^{5} + 8 \, a^{3} c^{2} d x^{3} - 3 \, a^{4} c d x\right )} e^{2}}{48 \, {\left (a^{4} c^{5} x^{6} + 3 \, a^{5} c^{4} x^{4} + 3 \, a^{6} c^{3} x^{2} + a^{7} c^{2}\right )}}\right ] \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

[1/96*(30*a*c^4*d^3*x^5 + 80*a^2*c^3*d^3*x^3 + 66*a^3*c^2*d^3*x - 48*a^4*c*d^2*e - 3*(5*c^4*d^3*x^6 + 15*a*c^3
*d^3*x^4 + 15*a^2*c^2*d^3*x^2 + 5*a^3*c*d^3 + 3*(a*c^3*d*x^6 + 3*a^2*c^2*d*x^4 + 3*a^3*c*d*x^2 + a^4*d)*e^2)*s
qrt(-a*c)*log((c*x^2 - 2*sqrt(-a*c)*x - a)/(c*x^2 + a)) - 8*(3*a^4*c*x^2 + a^5)*e^3 + 6*(3*a^2*c^3*d*x^5 + 8*a
^3*c^2*d*x^3 - 3*a^4*c*d*x)*e^2)/(a^4*c^5*x^6 + 3*a^5*c^4*x^4 + 3*a^6*c^3*x^2 + a^7*c^2), 1/48*(15*a*c^4*d^3*x
^5 + 40*a^2*c^3*d^3*x^3 + 33*a^3*c^2*d^3*x - 24*a^4*c*d^2*e + 3*(5*c^4*d^3*x^6 + 15*a*c^3*d^3*x^4 + 15*a^2*c^2
*d^3*x^2 + 5*a^3*c*d^3 + 3*(a*c^3*d*x^6 + 3*a^2*c^2*d*x^4 + 3*a^3*c*d*x^2 + a^4*d)*e^2)*sqrt(a*c)*arctan(sqrt(
a*c)*x/a) - 4*(3*a^4*c*x^2 + a^5)*e^3 + 3*(3*a^2*c^3*d*x^5 + 8*a^3*c^2*d*x^3 - 3*a^4*c*d*x)*e^2)/(a^4*c^5*x^6
+ 3*a^5*c^4*x^4 + 3*a^6*c^3*x^2 + a^7*c^2)]

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Sympy [B] Leaf count of result is larger than twice the leaf count of optimal. 320 vs. \(2 (143) = 286\).
time = 0.96, size = 320, normalized size = 2.05 \begin {gather*} - \frac {d \sqrt {- \frac {1}{a^{7} c^{3}}} \cdot \left (3 a e^{2} + 5 c d^{2}\right ) \log {\left (- \frac {a^{4} c d \sqrt {- \frac {1}{a^{7} c^{3}}} \cdot \left (3 a e^{2} + 5 c d^{2}\right )}{3 a d e^{2} + 5 c d^{3}} + x \right )}}{32} + \frac {d \sqrt {- \frac {1}{a^{7} c^{3}}} \cdot \left (3 a e^{2} + 5 c d^{2}\right ) \log {\left (\frac {a^{4} c d \sqrt {- \frac {1}{a^{7} c^{3}}} \cdot \left (3 a e^{2} + 5 c d^{2}\right )}{3 a d e^{2} + 5 c d^{3}} + x \right )}}{32} + \frac {- 4 a^{4} e^{3} - 24 a^{3} c d^{2} e - 12 a^{3} c e^{3} x^{2} + x^{5} \cdot \left (9 a c^{3} d e^{2} + 15 c^{4} d^{3}\right ) + x^{3} \cdot \left (24 a^{2} c^{2} d e^{2} + 40 a c^{3} d^{3}\right ) + x \left (- 9 a^{3} c d e^{2} + 33 a^{2} c^{2} d^{3}\right )}{48 a^{6} c^{2} + 144 a^{5} c^{3} x^{2} + 144 a^{4} c^{4} x^{4} + 48 a^{3} c^{5} x^{6}} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

-d*sqrt(-1/(a**7*c**3))*(3*a*e**2 + 5*c*d**2)*log(-a**4*c*d*sqrt(-1/(a**7*c**3))*(3*a*e**2 + 5*c*d**2)/(3*a*d*
e**2 + 5*c*d**3) + x)/32 + d*sqrt(-1/(a**7*c**3))*(3*a*e**2 + 5*c*d**2)*log(a**4*c*d*sqrt(-1/(a**7*c**3))*(3*a
*e**2 + 5*c*d**2)/(3*a*d*e**2 + 5*c*d**3) + x)/32 + (-4*a**4*e**3 - 24*a**3*c*d**2*e - 12*a**3*c*e**3*x**2 + x
**5*(9*a*c**3*d*e**2 + 15*c**4*d**3) + x**3*(24*a**2*c**2*d*e**2 + 40*a*c**3*d**3) + x*(-9*a**3*c*d*e**2 + 33*
a**2*c**2*d**3))/(48*a**6*c**2 + 144*a**5*c**3*x**2 + 144*a**4*c**4*x**4 + 48*a**3*c**5*x**6)

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Giac [A]
time = 0.98, size = 154, normalized size = 0.99 \begin {gather*} \frac {{\left (5 \, c d^{3} + 3 \, a d e^{2}\right )} \arctan \left (\frac {c x}{\sqrt {a c}}\right )}{16 \, \sqrt {a c} a^{3} c} + \frac {15 \, c^{4} d^{3} x^{5} + 9 \, a c^{3} d x^{5} e^{2} + 40 \, a c^{3} d^{3} x^{3} + 24 \, a^{2} c^{2} d x^{3} e^{2} + 33 \, a^{2} c^{2} d^{3} x - 12 \, a^{3} c x^{2} e^{3} - 9 \, a^{3} c d x e^{2} - 24 \, a^{3} c d^{2} e - 4 \, a^{4} e^{3}}{48 \, {\left (c x^{2} + a\right )}^{3} a^{3} c^{2}} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

1/16*(5*c*d^3 + 3*a*d*e^2)*arctan(c*x/sqrt(a*c))/(sqrt(a*c)*a^3*c) + 1/48*(15*c^4*d^3*x^5 + 9*a*c^3*d*x^5*e^2
+ 40*a*c^3*d^3*x^3 + 24*a^2*c^2*d*x^3*e^2 + 33*a^2*c^2*d^3*x - 12*a^3*c*x^2*e^3 - 9*a^3*c*d*x*e^2 - 24*a^3*c*d
^2*e - 4*a^4*e^3)/((c*x^2 + a)^3*a^3*c^2)

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Mupad [B]
time = 0.13, size = 163, normalized size = 1.04 \begin {gather*} \frac {d\,\mathrm {atan}\left (\frac {\sqrt {c}\,x}{\sqrt {a}}\right )\,\left (5\,c\,d^2+3\,a\,e^2\right )}{16\,a^{7/2}\,c^{3/2}}-\frac {\frac {e^3\,x^2}{4\,c}+\frac {e\,\left (6\,c\,d^2+a\,e^2\right )}{12\,c^2}-\frac {d\,x^3\,\left (5\,c\,d^2+3\,a\,e^2\right )}{6\,a^2}+\frac {d\,x\,\left (3\,a\,e^2-11\,c\,d^2\right )}{16\,a\,c}-\frac {c\,d\,x^5\,\left (5\,c\,d^2+3\,a\,e^2\right )}{16\,a^3}}{a^3+3\,a^2\,c\,x^2+3\,a\,c^2\,x^4+c^3\,x^6} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

int((d + e*x)^3/(a + c*x^2)^4,x)

[Out]

(d*atan((c^(1/2)*x)/a^(1/2))*(3*a*e^2 + 5*c*d^2))/(16*a^(7/2)*c^(3/2)) - ((e^3*x^2)/(4*c) + (e*(a*e^2 + 6*c*d^
2))/(12*c^2) - (d*x^3*(3*a*e^2 + 5*c*d^2))/(6*a^2) + (d*x*(3*a*e^2 - 11*c*d^2))/(16*a*c) - (c*d*x^5*(3*a*e^2 +
5*c*d^2))/(16*a^3))/(a^3 + c^3*x^6 + 3*a^2*c*x^2 + 3*a*c^2*x^4)

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