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

Optimal. Leaf size=162 \[ \frac {3 e^2 \left (4 c d^2-a e^2\right ) \tanh ^{-1}\left (\frac {\sqrt {c} x}{\sqrt {a+c x^2}}\right )}{2 c^{5/2}}-\frac {e \sqrt {a+c x^2} \left (e x \left (2 c d^2-3 a e^2\right )+4 d \left (c d^2-4 a e^2\right )\right )}{2 a c^2}-\frac {(d+e x)^3 (a e-c d x)}{a c \sqrt {a+c x^2}}-\frac {d e \sqrt {a+c x^2} (d+e x)^2}{a c} \]

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Rubi [A]  time = 0.15, antiderivative size = 162, normalized size of antiderivative = 1.00, number of steps used = 5, number of rules used = 5, integrand size = 19, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.263, Rules used = {739, 833, 780, 217, 206} \begin {gather*} -\frac {e \sqrt {a+c x^2} \left (e x \left (2 c d^2-3 a e^2\right )+4 d \left (c d^2-4 a e^2\right )\right )}{2 a c^2}+\frac {3 e^2 \left (4 c d^2-a e^2\right ) \tanh ^{-1}\left (\frac {\sqrt {c} x}{\sqrt {a+c x^2}}\right )}{2 c^{5/2}}-\frac {(d+e x)^3 (a e-c d x)}{a c \sqrt {a+c x^2}}-\frac {d e \sqrt {a+c x^2} (d+e x)^2}{a c} \end {gather*}

Antiderivative was successfully verified.

[In]

Int[(d + e*x)^4/(a + c*x^2)^(3/2),x]

[Out]

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

Rule 206

Int[((a_) + (b_.)*(x_)^2)^(-1), x_Symbol] :> Simp[(1*ArcTanh[(Rt[-b, 2]*x)/Rt[a, 2]])/(Rt[a, 2]*Rt[-b, 2]), x]
/; FreeQ[{a, b}, x] && NegQ[a/b] && (GtQ[a, 0] || LtQ[b, 0])

Rule 217

Int[1/Sqrt[(a_) + (b_.)*(x_)^2], x_Symbol] :> Subst[Int[1/(1 - b*x^2), x], x, x/Sqrt[a + b*x^2]] /; FreeQ[{a,
b}, x] &&  !GtQ[a, 0]

Rule 739

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[1/((p + 1)*(-2*a*c)), Int[(d + e*x)^(m - 2)*Simp[a*e^2*(m - 1) -
c*d^2*(2*p + 3) - d*c*e*(m + 2*p + 2)*x, 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, 1] && IntQuadraticQ[a, 0, c, d, e, m, p, x]

Rule 780

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

Rule 833

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

Rubi steps

\begin {align*} \int \frac {(d+e x)^4}{\left (a+c x^2\right )^{3/2}} \, dx &=-\frac {(a e-c d x) (d+e x)^3}{a c \sqrt {a+c x^2}}+\frac {\int \frac {(d+e x)^2 \left (3 a e^2-3 c d e x\right )}{\sqrt {a+c x^2}} \, dx}{a c}\\ &=-\frac {(a e-c d x) (d+e x)^3}{a c \sqrt {a+c x^2}}-\frac {d e (d+e x)^2 \sqrt {a+c x^2}}{a c}+\frac {\int \frac {(d+e x) \left (15 a c d e^2-3 c e \left (2 c d^2-3 a e^2\right ) x\right )}{\sqrt {a+c x^2}} \, dx}{3 a c^2}\\ &=-\frac {(a e-c d x) (d+e x)^3}{a c \sqrt {a+c x^2}}-\frac {d e (d+e x)^2 \sqrt {a+c x^2}}{a c}-\frac {e \left (4 d \left (c d^2-4 a e^2\right )+e \left (2 c d^2-3 a e^2\right ) x\right ) \sqrt {a+c x^2}}{2 a c^2}+\frac {\left (3 e^2 \left (4 c d^2-a e^2\right )\right ) \int \frac {1}{\sqrt {a+c x^2}} \, dx}{2 c^2}\\ &=-\frac {(a e-c d x) (d+e x)^3}{a c \sqrt {a+c x^2}}-\frac {d e (d+e x)^2 \sqrt {a+c x^2}}{a c}-\frac {e \left (4 d \left (c d^2-4 a e^2\right )+e \left (2 c d^2-3 a e^2\right ) x\right ) \sqrt {a+c x^2}}{2 a c^2}+\frac {\left (3 e^2 \left (4 c d^2-a e^2\right )\right ) \operatorname {Subst}\left (\int \frac {1}{1-c x^2} \, dx,x,\frac {x}{\sqrt {a+c x^2}}\right )}{2 c^2}\\ &=-\frac {(a e-c d x) (d+e x)^3}{a c \sqrt {a+c x^2}}-\frac {d e (d+e x)^2 \sqrt {a+c x^2}}{a c}-\frac {e \left (4 d \left (c d^2-4 a e^2\right )+e \left (2 c d^2-3 a e^2\right ) x\right ) \sqrt {a+c x^2}}{2 a c^2}+\frac {3 e^2 \left (4 c d^2-a e^2\right ) \tanh ^{-1}\left (\frac {\sqrt {c} x}{\sqrt {a+c x^2}}\right )}{2 c^{5/2}}\\ \end {align*}

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Mathematica [A]  time = 0.14, size = 127, normalized size = 0.78 \begin {gather*} \frac {a^2 e^3 (16 d+3 e x)+a c e \left (-8 d^3-12 d^2 e x+8 d e^2 x^2+e^3 x^3\right )+2 c^2 d^4 x}{2 a c^2 \sqrt {a+c x^2}}+\frac {3 \left (4 c d^2 e^2-a e^4\right ) \log \left (\sqrt {c} \sqrt {a+c x^2}+c x\right )}{2 c^{5/2}} \end {gather*}

Antiderivative was successfully verified.

[In]

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

[Out]

(2*c^2*d^4*x + a^2*e^3*(16*d + 3*e*x) + a*c*e*(-8*d^3 - 12*d^2*e*x + 8*d*e^2*x^2 + e^3*x^3))/(2*a*c^2*Sqrt[a +
c*x^2]) + (3*(4*c*d^2*e^2 - a*e^4)*Log[c*x + Sqrt[c]*Sqrt[a + c*x^2]])/(2*c^(5/2))

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IntegrateAlgebraic [A]  time = 0.52, size = 135, normalized size = 0.83 \begin {gather*} \frac {16 a^2 d e^3+3 a^2 e^4 x-8 a c d^3 e-12 a c d^2 e^2 x+8 a c d e^3 x^2+a c e^4 x^3+2 c^2 d^4 x}{2 a c^2 \sqrt {a+c x^2}}-\frac {3 \left (4 c d^2 e^2-a e^4\right ) \log \left (\sqrt {a+c x^2}-\sqrt {c} x\right )}{2 c^{5/2}} \end {gather*}

Antiderivative was successfully verified.

[In]

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

[Out]

(-8*a*c*d^3*e + 16*a^2*d*e^3 + 2*c^2*d^4*x - 12*a*c*d^2*e^2*x + 3*a^2*e^4*x + 8*a*c*d*e^3*x^2 + a*c*e^4*x^3)/(
2*a*c^2*Sqrt[a + c*x^2]) - (3*(4*c*d^2*e^2 - a*e^4)*Log[-(Sqrt[c]*x) + Sqrt[a + c*x^2]])/(2*c^(5/2))

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fricas [A]  time = 0.44, size = 368, normalized size = 2.27 \begin {gather*} \left [\frac {3 \, {\left (4 \, a^{2} c d^{2} e^{2} - a^{3} e^{4} + {\left (4 \, a c^{2} d^{2} e^{2} - a^{2} c e^{4}\right )} x^{2}\right )} \sqrt {c} \log \left (-2 \, c x^{2} - 2 \, \sqrt {c x^{2} + a} \sqrt {c} x - a\right ) + 2 \, {\left (a c^{2} e^{4} x^{3} + 8 \, a c^{2} d e^{3} x^{2} - 8 \, a c^{2} d^{3} e + 16 \, a^{2} c d e^{3} + {\left (2 \, c^{3} d^{4} - 12 \, a c^{2} d^{2} e^{2} + 3 \, a^{2} c e^{4}\right )} x\right )} \sqrt {c x^{2} + a}}{4 \, {\left (a c^{4} x^{2} + a^{2} c^{3}\right )}}, -\frac {3 \, {\left (4 \, a^{2} c d^{2} e^{2} - a^{3} e^{4} + {\left (4 \, a c^{2} d^{2} e^{2} - a^{2} c e^{4}\right )} x^{2}\right )} \sqrt {-c} \arctan \left (\frac {\sqrt {-c} x}{\sqrt {c x^{2} + a}}\right ) - {\left (a c^{2} e^{4} x^{3} + 8 \, a c^{2} d e^{3} x^{2} - 8 \, a c^{2} d^{3} e + 16 \, a^{2} c d e^{3} + {\left (2 \, c^{3} d^{4} - 12 \, a c^{2} d^{2} e^{2} + 3 \, a^{2} c e^{4}\right )} x\right )} \sqrt {c x^{2} + a}}{2 \, {\left (a c^{4} x^{2} + a^{2} 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)^(3/2),x, algorithm="fricas")

[Out]

[1/4*(3*(4*a^2*c*d^2*e^2 - a^3*e^4 + (4*a*c^2*d^2*e^2 - a^2*c*e^4)*x^2)*sqrt(c)*log(-2*c*x^2 - 2*sqrt(c*x^2 +
a)*sqrt(c)*x - a) + 2*(a*c^2*e^4*x^3 + 8*a*c^2*d*e^3*x^2 - 8*a*c^2*d^3*e + 16*a^2*c*d*e^3 + (2*c^3*d^4 - 12*a*
c^2*d^2*e^2 + 3*a^2*c*e^4)*x)*sqrt(c*x^2 + a))/(a*c^4*x^2 + a^2*c^3), -1/2*(3*(4*a^2*c*d^2*e^2 - a^3*e^4 + (4*
a*c^2*d^2*e^2 - a^2*c*e^4)*x^2)*sqrt(-c)*arctan(sqrt(-c)*x/sqrt(c*x^2 + a)) - (a*c^2*e^4*x^3 + 8*a*c^2*d*e^3*x
^2 - 8*a*c^2*d^3*e + 16*a^2*c*d*e^3 + (2*c^3*d^4 - 12*a*c^2*d^2*e^2 + 3*a^2*c*e^4)*x)*sqrt(c*x^2 + a))/(a*c^4*
x^2 + a^2*c^3)]

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giac [A]  time = 0.28, size = 138, normalized size = 0.85 \begin {gather*} \frac {{\left (x {\left (\frac {x e^{4}}{c} + \frac {8 \, d e^{3}}{c}\right )} + \frac {2 \, c^{4} d^{4} - 12 \, a c^{3} d^{2} e^{2} + 3 \, a^{2} c^{2} e^{4}}{a c^{4}}\right )} x - \frac {8 \, {\left (a c^{3} d^{3} e - 2 \, a^{2} c^{2} d e^{3}\right )}}{a c^{4}}}{2 \, \sqrt {c x^{2} + a}} - \frac {3 \, {\left (4 \, c d^{2} e^{2} - a e^{4}\right )} \log \left ({\left | -\sqrt {c} x + \sqrt {c x^{2} + a} \right |}\right )}{2 \, c^{\frac {5}{2}}} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

1/2*((x*(x*e^4/c + 8*d*e^3/c) + (2*c^4*d^4 - 12*a*c^3*d^2*e^2 + 3*a^2*c^2*e^4)/(a*c^4))*x - 8*(a*c^3*d^3*e - 2
*a^2*c^2*d*e^3)/(a*c^4))/sqrt(c*x^2 + a) - 3/2*(4*c*d^2*e^2 - a*e^4)*log(abs(-sqrt(c)*x + sqrt(c*x^2 + a)))/c^
(5/2)

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maple [A]  time = 0.05, size = 189, normalized size = 1.17 \begin {gather*} \frac {e^{4} x^{3}}{2 \sqrt {c \,x^{2}+a}\, c}+\frac {4 d \,e^{3} x^{2}}{\sqrt {c \,x^{2}+a}\, c}+\frac {3 a \,e^{4} x}{2 \sqrt {c \,x^{2}+a}\, c^{2}}+\frac {d^{4} x}{\sqrt {c \,x^{2}+a}\, a}-\frac {6 d^{2} e^{2} x}{\sqrt {c \,x^{2}+a}\, c}-\frac {3 a \,e^{4} \ln \left (\sqrt {c}\, x +\sqrt {c \,x^{2}+a}\right )}{2 c^{\frac {5}{2}}}+\frac {6 d^{2} e^{2} \ln \left (\sqrt {c}\, x +\sqrt {c \,x^{2}+a}\right )}{c^{\frac {3}{2}}}+\frac {8 a d \,e^{3}}{\sqrt {c \,x^{2}+a}\, c^{2}}-\frac {4 d^{3} e}{\sqrt {c \,x^{2}+a}\, c} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

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

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maxima [A]  time = 1.36, size = 174, normalized size = 1.07 \begin {gather*} \frac {e^{4} x^{3}}{2 \, \sqrt {c x^{2} + a} c} + \frac {4 \, d e^{3} x^{2}}{\sqrt {c x^{2} + a} c} + \frac {d^{4} x}{\sqrt {c x^{2} + a} a} - \frac {6 \, d^{2} e^{2} x}{\sqrt {c x^{2} + a} c} + \frac {3 \, a e^{4} x}{2 \, \sqrt {c x^{2} + a} c^{2}} + \frac {6 \, d^{2} e^{2} \operatorname {arsinh}\left (\frac {c x}{\sqrt {a c}}\right )}{c^{\frac {3}{2}}} - \frac {3 \, a e^{4} \operatorname {arsinh}\left (\frac {c x}{\sqrt {a c}}\right )}{2 \, c^{\frac {5}{2}}} - \frac {4 \, d^{3} e}{\sqrt {c x^{2} + a} c} + \frac {8 \, a d e^{3}}{\sqrt {c x^{2} + a} c^{2}} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

1/2*e^4*x^3/(sqrt(c*x^2 + a)*c) + 4*d*e^3*x^2/(sqrt(c*x^2 + a)*c) + d^4*x/(sqrt(c*x^2 + a)*a) - 6*d^2*e^2*x/(s
qrt(c*x^2 + a)*c) + 3/2*a*e^4*x/(sqrt(c*x^2 + a)*c^2) + 6*d^2*e^2*arcsinh(c*x/sqrt(a*c))/c^(3/2) - 3/2*a*e^4*a
rcsinh(c*x/sqrt(a*c))/c^(5/2) - 4*d^3*e/(sqrt(c*x^2 + a)*c) + 8*a*d*e^3/(sqrt(c*x^2 + a)*c^2)

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mupad [F]  time = 0.00, size = -1, normalized size = -0.01 \begin {gather*} \int \frac {{\left (d+e\,x\right )}^4}{{\left (c\,x^2+a\right )}^{3/2}} \,d x \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

int((d + e*x)^4/(a + c*x^2)^(3/2),x)

[Out]

int((d + e*x)^4/(a + c*x^2)^(3/2), x)

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

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

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

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