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

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

[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))

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Rubi [A]  time = 0.147554, antiderivative size = 162, normalized size of antiderivative = 1., 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} \[ -\frac{e \sqrt{a+c x^2} \left (e x \left (2 c d^2-3 a e^2\right )+4 d \left (c d^2-4 a e^2\right )\right )}{2 a c^2}+\frac{3 e^2 \left (4 c d^2-a e^2\right ) \tanh ^{-1}\left (\frac{\sqrt{c} x}{\sqrt{a+c x^2}}\right )}{2 c^{5/2}}-\frac{(d+e x)^3 (a e-c d x)}{a c \sqrt{a+c x^2}}-\frac{d e \sqrt{a+c x^2} (d+e x)^2}{a c} \]

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 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 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])

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 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 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])

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*}

Mathematica [A]  time = 0.141832, size = 127, normalized size = 0.78 \[ \frac{a^2 e^3 (16 d+3 e x)+a c e \left (-12 d^2 e x-8 d^3+8 d e^2 x^2+e^3 x^3\right )+2 c^2 d^4 x}{2 a c^2 \sqrt{a+c x^2}}+\frac{3 \left (4 c d^2 e^2-a e^4\right ) \log \left (\sqrt{c} \sqrt{a+c x^2}+c x\right )}{2 c^{5/2}} \]

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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Maple [A]  time = 0.049, size = 189, normalized size = 1.2 \begin{align*}{\frac{{e}^{4}{x}^{3}}{2\,c}{\frac{1}{\sqrt{c{x}^{2}+a}}}}+{\frac{3\,a{e}^{4}x}{2\,{c}^{2}}{\frac{1}{\sqrt{c{x}^{2}+a}}}}-{\frac{3\,a{e}^{4}}{2}\ln \left ( x\sqrt{c}+\sqrt{c{x}^{2}+a} \right ){c}^{-{\frac{5}{2}}}}+4\,{\frac{d{e}^{3}{x}^{2}}{c\sqrt{c{x}^{2}+a}}}+8\,{\frac{d{e}^{3}a}{{c}^{2}\sqrt{c{x}^{2}+a}}}-6\,{\frac{{d}^{2}{e}^{2}x}{c\sqrt{c{x}^{2}+a}}}+6\,{\frac{{d}^{2}{e}^{2}\ln \left ( x\sqrt{c}+\sqrt{c{x}^{2}+a} \right ) }{{c}^{3/2}}}-4\,{\frac{{d}^{3}e}{c\sqrt{c{x}^{2}+a}}}+{\frac{{d}^{4}x}{a}{\frac{1}{\sqrt{c{x}^{2}+a}}}} \end{align*}

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(x*c^(1/2)+(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
(x*c^(1/2)+(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 [F(-2)]  time = 0., size = 0, normalized size = 0. \begin{align*} \text{Exception raised: ValueError} \end{align*}

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]

Exception raised: ValueError

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Fricas [A]  time = 1.96434, size = 768, normalized size = 4.74 \begin{align*} \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{align*}

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

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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Giac [A]  time = 1.34381, size = 186, normalized size = 1.15 \begin{align*} \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{align*}

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)