3.578 \(\int \frac{(d+e x)^4}{(a+c x^2)^{5/2}} \, dx\)

Optimal. Leaf size=161 \[ -\frac{d e \sqrt{a+c x^2} \left (5 a e^2+2 c d^2\right )}{3 a^2 c^2}-\frac{(d+e x) \left (a e \left (3 a e^2+c d^2\right )-2 c d x \left (2 a e^2+c d^2\right )\right )}{3 a^2 c^2 \sqrt{a+c x^2}}+\frac{e^4 \tanh ^{-1}\left (\frac{\sqrt{c} x}{\sqrt{a+c x^2}}\right )}{c^{5/2}}-\frac{(d+e x)^3 (a e-c d x)}{3 a c \left (a+c x^2\right )^{3/2}} \]

[Out]

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

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Rubi [A]  time = 0.118167, antiderivative size = 161, 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, 819, 641, 217, 206} \[ -\frac{d e \sqrt{a+c x^2} \left (5 a e^2+2 c d^2\right )}{3 a^2 c^2}-\frac{(d+e x) \left (a e \left (3 a e^2+c d^2\right )-2 c d x \left (2 a e^2+c d^2\right )\right )}{3 a^2 c^2 \sqrt{a+c x^2}}+\frac{e^4 \tanh ^{-1}\left (\frac{\sqrt{c} x}{\sqrt{a+c x^2}}\right )}{c^{5/2}}-\frac{(d+e x)^3 (a e-c d x)}{3 a c \left (a+c x^2\right )^{3/2}} \]

Antiderivative was successfully verified.

[In]

Int[(d + e*x)^4/(a + c*x^2)^(5/2),x]

[Out]

-((a*e - c*d*x)*(d + e*x)^3)/(3*a*c*(a + c*x^2)^(3/2)) - ((d + e*x)*(a*e*(c*d^2 + 3*a*e^2) - 2*c*d*(c*d^2 + 2*
a*e^2)*x))/(3*a^2*c^2*Sqrt[a + c*x^2]) - (d*e*(2*c*d^2 + 5*a*e^2)*Sqrt[a + c*x^2])/(3*a^2*c^2) + (e^4*ArcTanh[
(Sqrt[c]*x)/Sqrt[a + c*x^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 819

Int[((d_.) + (e_.)*(x_))^(m_)*((f_.) + (g_.)*(x_))*((a_) + (c_.)*(x_)^2)^(p_.), x_Symbol] :> Simp[((d + e*x)^(
m - 1)*(a + c*x^2)^(p + 1)*(a*(e*f + d*g) - (c*d*f - a*e*g)*x))/(2*a*c*(p + 1)), x] - Dist[1/(2*a*c*(p + 1)),
Int[(d + e*x)^(m - 2)*(a + c*x^2)^(p + 1)*Simp[a*e*(e*f*(m - 1) + d*g*m) - c*d^2*f*(2*p + 3) + e*(a*e*g*m - c*
d*f*(m + 2*p + 2))*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, 1] && (EqQ[d, 0] || (EqQ[m, 2] && EqQ[p, -3] && RationalQ[a, c, d, e, f, g]) ||  !ILtQ[m + 2*p + 3, 0])

Rule 641

Int[((d_) + (e_.)*(x_))*((a_) + (c_.)*(x_)^2)^(p_.), x_Symbol] :> Simp[(e*(a + c*x^2)^(p + 1))/(2*c*(p + 1)),
x] + Dist[d, Int[(a + c*x^2)^p, x], x] /; FreeQ[{a, c, d, e, p}, x] && NeQ[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 )^{5/2}} \, dx &=-\frac{(a e-c d x) (d+e x)^3}{3 a c \left (a+c x^2\right )^{3/2}}+\frac{\int \frac{(d+e x)^2 \left (2 c d^2+3 a e^2-c d e x\right )}{\left (a+c x^2\right )^{3/2}} \, dx}{3 a c}\\ &=-\frac{(a e-c d x) (d+e x)^3}{3 a c \left (a+c x^2\right )^{3/2}}-\frac{(d+e x) \left (a e \left (c d^2+3 a e^2\right )-2 c d \left (c d^2+2 a e^2\right ) x\right )}{3 a^2 c^2 \sqrt{a+c x^2}}+\frac{\int \frac{3 a^2 e^4-c d e \left (2 c d^2+5 a e^2\right ) x}{\sqrt{a+c x^2}} \, dx}{3 a^2 c^2}\\ &=-\frac{(a e-c d x) (d+e x)^3}{3 a c \left (a+c x^2\right )^{3/2}}-\frac{(d+e x) \left (a e \left (c d^2+3 a e^2\right )-2 c d \left (c d^2+2 a e^2\right ) x\right )}{3 a^2 c^2 \sqrt{a+c x^2}}-\frac{d e \left (2 c d^2+5 a e^2\right ) \sqrt{a+c x^2}}{3 a^2 c^2}+\frac{e^4 \int \frac{1}{\sqrt{a+c x^2}} \, dx}{c^2}\\ &=-\frac{(a e-c d x) (d+e x)^3}{3 a c \left (a+c x^2\right )^{3/2}}-\frac{(d+e x) \left (a e \left (c d^2+3 a e^2\right )-2 c d \left (c d^2+2 a e^2\right ) x\right )}{3 a^2 c^2 \sqrt{a+c x^2}}-\frac{d e \left (2 c d^2+5 a e^2\right ) \sqrt{a+c x^2}}{3 a^2 c^2}+\frac{e^4 \operatorname{Subst}\left (\int \frac{1}{1-c x^2} \, dx,x,\frac{x}{\sqrt{a+c x^2}}\right )}{c^2}\\ &=-\frac{(a e-c d x) (d+e x)^3}{3 a c \left (a+c x^2\right )^{3/2}}-\frac{(d+e x) \left (a e \left (c d^2+3 a e^2\right )-2 c d \left (c d^2+2 a e^2\right ) x\right )}{3 a^2 c^2 \sqrt{a+c x^2}}-\frac{d e \left (2 c d^2+5 a e^2\right ) \sqrt{a+c x^2}}{3 a^2 c^2}+\frac{e^4 \tanh ^{-1}\left (\frac{\sqrt{c} x}{\sqrt{a+c x^2}}\right )}{c^{5/2}}\\ \end{align*}

Mathematica [A]  time = 0.212186, size = 130, normalized size = 0.81 \[ \frac{-4 a^2 c e \left (d^3+3 d e^2 x^2+e^3 x^3\right )-a^3 e^3 (8 d+3 e x)+3 a c^2 d^2 x \left (d^2+2 e^2 x^2\right )+2 c^3 d^4 x^3}{3 a^2 c^2 \left (a+c x^2\right )^{3/2}}+\frac{e^4 \log \left (\sqrt{c} \sqrt{a+c x^2}+c x\right )}{c^{5/2}} \]

Antiderivative was successfully verified.

[In]

Integrate[(d + e*x)^4/(a + c*x^2)^(5/2),x]

[Out]

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

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Maple [A]  time = 0.053, size = 202, normalized size = 1.3 \begin{align*} -{\frac{{e}^{4}{x}^{3}}{3\,c} \left ( c{x}^{2}+a \right ) ^{-{\frac{3}{2}}}}-{\frac{{e}^{4}x}{{c}^{2}}{\frac{1}{\sqrt{c{x}^{2}+a}}}}+{{e}^{4}\ln \left ( x\sqrt{c}+\sqrt{c{x}^{2}+a} \right ){c}^{-{\frac{5}{2}}}}-4\,{\frac{d{e}^{3}{x}^{2}}{c \left ( c{x}^{2}+a \right ) ^{3/2}}}-{\frac{8\,d{e}^{3}a}{3\,{c}^{2}} \left ( c{x}^{2}+a \right ) ^{-{\frac{3}{2}}}}-2\,{\frac{{d}^{2}{e}^{2}x}{c \left ( c{x}^{2}+a \right ) ^{3/2}}}+2\,{\frac{{d}^{2}{e}^{2}x}{ac\sqrt{c{x}^{2}+a}}}-{\frac{4\,{d}^{3}e}{3\,c} \left ( c{x}^{2}+a \right ) ^{-{\frac{3}{2}}}}+{\frac{{d}^{4}x}{3\,a} \left ( c{x}^{2}+a \right ) ^{-{\frac{3}{2}}}}+{\frac{2\,{d}^{4}x}{3\,{a}^{2}}{\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)^(5/2),x)

[Out]

-1/3*e^4*x^3/c/(c*x^2+a)^(3/2)-e^4/c^2*x/(c*x^2+a)^(1/2)+e^4/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)^(3/2)-8/3*d*e^3*a/c^2/(c*x^2+a)^(3/2)-2*d^2*e^2/c*x/(c*x^2+a)^(3/2)+2*d^2*e^2/a/c*x/(c*x^2+a)^(1/
2)-4/3*d^3*e/c/(c*x^2+a)^(3/2)+1/3*d^4*x/a/(c*x^2+a)^(3/2)+2/3*d^4/a^2*x/(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)^(5/2),x, algorithm="maxima")

[Out]

Exception raised: ValueError

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Fricas [A]  time = 2.3249, size = 825, normalized size = 5.12 \begin{align*} \left [\frac{3 \,{\left (a^{2} c^{2} e^{4} x^{4} + 2 \, a^{3} c e^{4} x^{2} + a^{4} e^{4}\right )} \sqrt{c} \log \left (-2 \, c x^{2} - 2 \, \sqrt{c x^{2} + a} \sqrt{c} x - a\right ) - 2 \,{\left (12 \, a^{2} c^{2} d e^{3} x^{2} + 4 \, a^{2} c^{2} d^{3} e + 8 \, a^{3} c d e^{3} - 2 \,{\left (c^{4} d^{4} + 3 \, a c^{3} d^{2} e^{2} - 2 \, a^{2} c^{2} e^{4}\right )} x^{3} - 3 \,{\left (a c^{3} d^{4} - a^{3} c e^{4}\right )} x\right )} \sqrt{c x^{2} + a}}{6 \,{\left (a^{2} c^{5} x^{4} + 2 \, a^{3} c^{4} x^{2} + a^{4} c^{3}\right )}}, -\frac{3 \,{\left (a^{2} c^{2} e^{4} x^{4} + 2 \, a^{3} c e^{4} x^{2} + a^{4} e^{4}\right )} \sqrt{-c} \arctan \left (\frac{\sqrt{-c} x}{\sqrt{c x^{2} + a}}\right ) +{\left (12 \, a^{2} c^{2} d e^{3} x^{2} + 4 \, a^{2} c^{2} d^{3} e + 8 \, a^{3} c d e^{3} - 2 \,{\left (c^{4} d^{4} + 3 \, a c^{3} d^{2} e^{2} - 2 \, a^{2} c^{2} e^{4}\right )} x^{3} - 3 \,{\left (a c^{3} d^{4} - a^{3} c e^{4}\right )} x\right )} \sqrt{c x^{2} + a}}{3 \,{\left (a^{2} c^{5} x^{4} + 2 \, a^{3} c^{4} x^{2} + a^{4} 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)^(5/2),x, algorithm="fricas")

[Out]

[1/6*(3*(a^2*c^2*e^4*x^4 + 2*a^3*c*e^4*x^2 + a^4*e^4)*sqrt(c)*log(-2*c*x^2 - 2*sqrt(c*x^2 + a)*sqrt(c)*x - a)
- 2*(12*a^2*c^2*d*e^3*x^2 + 4*a^2*c^2*d^3*e + 8*a^3*c*d*e^3 - 2*(c^4*d^4 + 3*a*c^3*d^2*e^2 - 2*a^2*c^2*e^4)*x^
3 - 3*(a*c^3*d^4 - a^3*c*e^4)*x)*sqrt(c*x^2 + a))/(a^2*c^5*x^4 + 2*a^3*c^4*x^2 + a^4*c^3), -1/3*(3*(a^2*c^2*e^
4*x^4 + 2*a^3*c*e^4*x^2 + a^4*e^4)*sqrt(-c)*arctan(sqrt(-c)*x/sqrt(c*x^2 + a)) + (12*a^2*c^2*d*e^3*x^2 + 4*a^2
*c^2*d^3*e + 8*a^3*c*d*e^3 - 2*(c^4*d^4 + 3*a*c^3*d^2*e^2 - 2*a^2*c^2*e^4)*x^3 - 3*(a*c^3*d^4 - a^3*c*e^4)*x)*
sqrt(c*x^2 + a))/(a^2*c^5*x^4 + 2*a^3*c^4*x^2 + a^4*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{5}{2}}}\, dx \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate((e*x+d)**4/(c*x**2+a)**(5/2),x)

[Out]

Integral((d + e*x)**4/(a + c*x**2)**(5/2), x)

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Giac [A]  time = 1.32629, size = 203, normalized size = 1.26 \begin{align*} -\frac{{\left (2 \, x{\left (\frac{6 \, d e^{3}}{c} - \frac{{\left (c^{5} d^{4} + 3 \, a c^{4} d^{2} e^{2} - 2 \, a^{2} c^{3} e^{4}\right )} x}{a^{2} c^{4}}\right )} - \frac{3 \,{\left (a c^{4} d^{4} - a^{3} c^{2} e^{4}\right )}}{a^{2} c^{4}}\right )} x + \frac{4 \,{\left (a^{2} c^{3} d^{3} e + 2 \, a^{3} c^{2} d e^{3}\right )}}{a^{2} c^{4}}}{3 \,{\left (c x^{2} + a\right )}^{\frac{3}{2}}} - \frac{e^{4} \log \left ({\left | -\sqrt{c} x + \sqrt{c x^{2} + a} \right |}\right )}{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)^(5/2),x, algorithm="giac")

[Out]

-1/3*((2*x*(6*d*e^3/c - (c^5*d^4 + 3*a*c^4*d^2*e^2 - 2*a^2*c^3*e^4)*x/(a^2*c^4)) - 3*(a*c^4*d^4 - a^3*c^2*e^4)
/(a^2*c^4))*x + 4*(a^2*c^3*d^3*e + 2*a^3*c^2*d*e^3)/(a^2*c^4))/(c*x^2 + a)^(3/2) - e^4*log(abs(-sqrt(c)*x + sq
rt(c*x^2 + a)))/c^(5/2)