### 3.561 $$\int \frac{(d+e x)^4}{\sqrt{a+c x^2}} \, dx$$

Optimal. Leaf size=161 $\frac{\left (3 a^2 e^4-24 a c d^2 e^2+8 c^2 d^4\right ) \tanh ^{-1}\left (\frac{\sqrt{c} x}{\sqrt{a+c x^2}}\right )}{8 c^{5/2}}+\frac{e \sqrt{a+c x^2} \left (e x \left (26 c d^2-9 a e^2\right )+4 d \left (19 c d^2-16 a e^2\right )\right )}{24 c^2}+\frac{e \sqrt{a+c x^2} (d+e x)^3}{4 c}+\frac{7 d e \sqrt{a+c x^2} (d+e x)^2}{12 c}$

[Out]

(7*d*e*(d + e*x)^2*Sqrt[a + c*x^2])/(12*c) + (e*(d + e*x)^3*Sqrt[a + c*x^2])/(4*c) + (e*(4*d*(19*c*d^2 - 16*a*
e^2) + e*(26*c*d^2 - 9*a*e^2)*x)*Sqrt[a + c*x^2])/(24*c^2) + ((8*c^2*d^4 - 24*a*c*d^2*e^2 + 3*a^2*e^4)*ArcTanh
[(Sqrt[c]*x)/Sqrt[a + c*x^2]])/(8*c^(5/2))

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Rubi [A]  time = 0.161434, 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 = {743, 833, 780, 217, 206} $\frac{\left (3 a^2 e^4-24 a c d^2 e^2+8 c^2 d^4\right ) \tanh ^{-1}\left (\frac{\sqrt{c} x}{\sqrt{a+c x^2}}\right )}{8 c^{5/2}}+\frac{e \sqrt{a+c x^2} \left (e x \left (26 c d^2-9 a e^2\right )+4 d \left (19 c d^2-16 a e^2\right )\right )}{24 c^2}+\frac{e \sqrt{a+c x^2} (d+e x)^3}{4 c}+\frac{7 d e \sqrt{a+c x^2} (d+e x)^2}{12 c}$

Antiderivative was successfully veriﬁed.

[In]

Int[(d + e*x)^4/Sqrt[a + c*x^2],x]

[Out]

(7*d*e*(d + e*x)^2*Sqrt[a + c*x^2])/(12*c) + (e*(d + e*x)^3*Sqrt[a + c*x^2])/(4*c) + (e*(4*d*(19*c*d^2 - 16*a*
e^2) + e*(26*c*d^2 - 9*a*e^2)*x)*Sqrt[a + c*x^2])/(24*c^2) + ((8*c^2*d^4 - 24*a*c*d^2*e^2 + 3*a^2*e^4)*ArcTanh
[(Sqrt[c]*x)/Sqrt[a + c*x^2]])/(8*c^(5/2))

Rule 743

Int[((d_) + (e_.)*(x_))^(m_)*((a_) + (c_.)*(x_)^2)^(p_), x_Symbol] :> Simp[(e*(d + e*x)^(m - 1)*(a + c*x^2)^(p
+ 1))/(c*(m + 2*p + 1)), x] + Dist[1/(c*(m + 2*p + 1)), Int[(d + e*x)^(m - 2)*Simp[c*d^2*(m + 2*p + 1) - a*e^
2*(m - 1) + 2*c*d*e*(m + p)*x, x]*(a + c*x^2)^p, x], x] /; FreeQ[{a, c, d, e, m, p}, x] && NeQ[c*d^2 + a*e^2,
0] && If[RationalQ[m], GtQ[m, 1], SumSimplerQ[m, -2]] && NeQ[m + 2*p + 1, 0] && 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}{\sqrt{a+c x^2}} \, dx &=\frac{e (d+e x)^3 \sqrt{a+c x^2}}{4 c}+\frac{\int \frac{(d+e x)^2 \left (4 c d^2-3 a e^2+7 c d e x\right )}{\sqrt{a+c x^2}} \, dx}{4 c}\\ &=\frac{7 d e (d+e x)^2 \sqrt{a+c x^2}}{12 c}+\frac{e (d+e x)^3 \sqrt{a+c x^2}}{4 c}+\frac{\int \frac{(d+e x) \left (c d \left (12 c d^2-23 a e^2\right )+c e \left (26 c d^2-9 a e^2\right ) x\right )}{\sqrt{a+c x^2}} \, dx}{12 c^2}\\ &=\frac{7 d e (d+e x)^2 \sqrt{a+c x^2}}{12 c}+\frac{e (d+e x)^3 \sqrt{a+c x^2}}{4 c}+\frac{e \left (4 d \left (19 c d^2-16 a e^2\right )+e \left (26 c d^2-9 a e^2\right ) x\right ) \sqrt{a+c x^2}}{24 c^2}+\frac{\left (8 c^2 d^4-24 a c d^2 e^2+3 a^2 e^4\right ) \int \frac{1}{\sqrt{a+c x^2}} \, dx}{8 c^2}\\ &=\frac{7 d e (d+e x)^2 \sqrt{a+c x^2}}{12 c}+\frac{e (d+e x)^3 \sqrt{a+c x^2}}{4 c}+\frac{e \left (4 d \left (19 c d^2-16 a e^2\right )+e \left (26 c d^2-9 a e^2\right ) x\right ) \sqrt{a+c x^2}}{24 c^2}+\frac{\left (8 c^2 d^4-24 a c d^2 e^2+3 a^2 e^4\right ) \operatorname{Subst}\left (\int \frac{1}{1-c x^2} \, dx,x,\frac{x}{\sqrt{a+c x^2}}\right )}{8 c^2}\\ &=\frac{7 d e (d+e x)^2 \sqrt{a+c x^2}}{12 c}+\frac{e (d+e x)^3 \sqrt{a+c x^2}}{4 c}+\frac{e \left (4 d \left (19 c d^2-16 a e^2\right )+e \left (26 c d^2-9 a e^2\right ) x\right ) \sqrt{a+c x^2}}{24 c^2}+\frac{\left (8 c^2 d^4-24 a c d^2 e^2+3 a^2 e^4\right ) \tanh ^{-1}\left (\frac{\sqrt{c} x}{\sqrt{a+c x^2}}\right )}{8 c^{5/2}}\\ \end{align*}

Mathematica [A]  time = 0.107137, size = 126, normalized size = 0.78 $\frac{3 \left (3 a^2 e^4-24 a c d^2 e^2+8 c^2 d^4\right ) \log \left (\sqrt{c} \sqrt{a+c x^2}+c x\right )+\sqrt{c} e \sqrt{a+c x^2} \left (c \left (72 d^2 e x+96 d^3+32 d e^2 x^2+6 e^3 x^3\right )-a e^2 (64 d+9 e x)\right )}{24 c^{5/2}}$

Antiderivative was successfully veriﬁed.

[In]

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

[Out]

(Sqrt[c]*e*Sqrt[a + c*x^2]*(-(a*e^2*(64*d + 9*e*x)) + c*(96*d^3 + 72*d^2*e*x + 32*d*e^2*x^2 + 6*e^3*x^3)) + 3*
(8*c^2*d^4 - 24*a*c*d^2*e^2 + 3*a^2*e^4)*Log[c*x + Sqrt[c]*Sqrt[a + c*x^2]])/(24*c^(5/2))

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Maple [A]  time = 0.049, size = 198, normalized size = 1.2 \begin{align*}{\frac{{e}^{4}{x}^{3}}{4\,c}\sqrt{c{x}^{2}+a}}-{\frac{3\,{e}^{4}ax}{8\,{c}^{2}}\sqrt{c{x}^{2}+a}}+{\frac{3\,{a}^{2}{e}^{4}}{8}\ln \left ( x\sqrt{c}+\sqrt{c{x}^{2}+a} \right ){c}^{-{\frac{5}{2}}}}+{\frac{4\,d{e}^{3}{x}^{2}}{3\,c}\sqrt{c{x}^{2}+a}}-{\frac{8\,d{e}^{3}a}{3\,{c}^{2}}\sqrt{c{x}^{2}+a}}+3\,{\frac{{d}^{2}{e}^{2}x\sqrt{c{x}^{2}+a}}{c}}-3\,{\frac{{d}^{2}{e}^{2}a\ln \left ( x\sqrt{c}+\sqrt{c{x}^{2}+a} \right ) }{{c}^{3/2}}}+4\,{\frac{{d}^{3}e\sqrt{c{x}^{2}+a}}{c}}+{{d}^{4}\ln \left ( x\sqrt{c}+\sqrt{c{x}^{2}+a} \right ){\frac{1}{\sqrt{c}}}} \end{align*}

Veriﬁcation of antiderivative is not currently implemented for this CAS.

[In]

int((e*x+d)^4/(c*x^2+a)^(1/2),x)

[Out]

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

[Out]

Exception raised: ValueError

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Fricas [A]  time = 1.94591, size = 603, normalized size = 3.75 \begin{align*} \left [\frac{3 \,{\left (8 \, c^{2} d^{4} - 24 \, a c d^{2} e^{2} + 3 \, a^{2} e^{4}\right )} \sqrt{c} \log \left (-2 \, c x^{2} - 2 \, \sqrt{c x^{2} + a} \sqrt{c} x - a\right ) + 2 \,{\left (6 \, c^{2} e^{4} x^{3} + 32 \, c^{2} d e^{3} x^{2} + 96 \, c^{2} d^{3} e - 64 \, a c d e^{3} + 9 \,{\left (8 \, c^{2} d^{2} e^{2} - a c e^{4}\right )} x\right )} \sqrt{c x^{2} + a}}{48 \, c^{3}}, -\frac{3 \,{\left (8 \, c^{2} d^{4} - 24 \, a c d^{2} e^{2} + 3 \, a^{2} e^{4}\right )} \sqrt{-c} \arctan \left (\frac{\sqrt{-c} x}{\sqrt{c x^{2} + a}}\right ) -{\left (6 \, c^{2} e^{4} x^{3} + 32 \, c^{2} d e^{3} x^{2} + 96 \, c^{2} d^{3} e - 64 \, a c d e^{3} + 9 \,{\left (8 \, c^{2} d^{2} e^{2} - a c e^{4}\right )} x\right )} \sqrt{c x^{2} + a}}{24 \, c^{3}}\right ] \end{align*}

Veriﬁcation of antiderivative is not currently implemented for this CAS.

[In]

integrate((e*x+d)^4/(c*x^2+a)^(1/2),x, algorithm="fricas")

[Out]

[1/48*(3*(8*c^2*d^4 - 24*a*c*d^2*e^2 + 3*a^2*e^4)*sqrt(c)*log(-2*c*x^2 - 2*sqrt(c*x^2 + a)*sqrt(c)*x - a) + 2*
(6*c^2*e^4*x^3 + 32*c^2*d*e^3*x^2 + 96*c^2*d^3*e - 64*a*c*d*e^3 + 9*(8*c^2*d^2*e^2 - a*c*e^4)*x)*sqrt(c*x^2 +
a))/c^3, -1/24*(3*(8*c^2*d^4 - 24*a*c*d^2*e^2 + 3*a^2*e^4)*sqrt(-c)*arctan(sqrt(-c)*x/sqrt(c*x^2 + a)) - (6*c^
2*e^4*x^3 + 32*c^2*d*e^3*x^2 + 96*c^2*d^3*e - 64*a*c*d*e^3 + 9*(8*c^2*d^2*e^2 - a*c*e^4)*x)*sqrt(c*x^2 + a))/c
^3]

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Sympy [A]  time = 8.7505, size = 330, normalized size = 2.05 \begin{align*} - \frac{3 a^{\frac{3}{2}} e^{4} x}{8 c^{2} \sqrt{1 + \frac{c x^{2}}{a}}} + \frac{3 \sqrt{a} d^{2} e^{2} x \sqrt{1 + \frac{c x^{2}}{a}}}{c} - \frac{\sqrt{a} e^{4} x^{3}}{8 c \sqrt{1 + \frac{c x^{2}}{a}}} + \frac{3 a^{2} e^{4} \operatorname{asinh}{\left (\frac{\sqrt{c} x}{\sqrt{a}} \right )}}{8 c^{\frac{5}{2}}} - \frac{3 a d^{2} e^{2} \operatorname{asinh}{\left (\frac{\sqrt{c} x}{\sqrt{a}} \right )}}{c^{\frac{3}{2}}} + d^{4} \left (\begin{cases} \frac{\sqrt{- \frac{a}{c}} \operatorname{asin}{\left (x \sqrt{- \frac{c}{a}} \right )}}{\sqrt{a}} & \text{for}\: a > 0 \wedge c < 0 \\\frac{\sqrt{\frac{a}{c}} \operatorname{asinh}{\left (x \sqrt{\frac{c}{a}} \right )}}{\sqrt{a}} & \text{for}\: a > 0 \wedge c > 0 \\\frac{\sqrt{- \frac{a}{c}} \operatorname{acosh}{\left (x \sqrt{- \frac{c}{a}} \right )}}{\sqrt{- a}} & \text{for}\: c > 0 \wedge a < 0 \end{cases}\right ) + 4 d^{3} e \left (\begin{cases} \frac{x^{2}}{2 \sqrt{a}} & \text{for}\: c = 0 \\\frac{\sqrt{a + c x^{2}}}{c} & \text{otherwise} \end{cases}\right ) + 4 d e^{3} \left (\begin{cases} - \frac{2 a \sqrt{a + c x^{2}}}{3 c^{2}} + \frac{x^{2} \sqrt{a + c x^{2}}}{3 c} & \text{for}\: c \neq 0 \\\frac{x^{4}}{4 \sqrt{a}} & \text{otherwise} \end{cases}\right ) + \frac{e^{4} x^{5}}{4 \sqrt{a} \sqrt{1 + \frac{c x^{2}}{a}}} \end{align*}

Veriﬁcation of antiderivative is not currently implemented for this CAS.

[In]

integrate((e*x+d)**4/(c*x**2+a)**(1/2),x)

[Out]

-3*a**(3/2)*e**4*x/(8*c**2*sqrt(1 + c*x**2/a)) + 3*sqrt(a)*d**2*e**2*x*sqrt(1 + c*x**2/a)/c - sqrt(a)*e**4*x**
3/(8*c*sqrt(1 + c*x**2/a)) + 3*a**2*e**4*asinh(sqrt(c)*x/sqrt(a))/(8*c**(5/2)) - 3*a*d**2*e**2*asinh(sqrt(c)*x
/sqrt(a))/c**(3/2) + d**4*Piecewise((sqrt(-a/c)*asin(x*sqrt(-c/a))/sqrt(a), (a > 0) & (c < 0)), (sqrt(a/c)*asi
nh(x*sqrt(c/a))/sqrt(a), (a > 0) & (c > 0)), (sqrt(-a/c)*acosh(x*sqrt(-c/a))/sqrt(-a), (c > 0) & (a < 0))) + 4
*d**3*e*Piecewise((x**2/(2*sqrt(a)), Eq(c, 0)), (sqrt(a + c*x**2)/c, True)) + 4*d*e**3*Piecewise((-2*a*sqrt(a
+ c*x**2)/(3*c**2) + x**2*sqrt(a + c*x**2)/(3*c), Ne(c, 0)), (x**4/(4*sqrt(a)), True)) + e**4*x**5/(4*sqrt(a)*
sqrt(1 + c*x**2/a))

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Giac [A]  time = 1.32206, size = 180, normalized size = 1.12 \begin{align*} \frac{1}{24} \, \sqrt{c x^{2} + a}{\left ({\left (2 \, x{\left (\frac{3 \, x e^{4}}{c} + \frac{16 \, d e^{3}}{c}\right )} + \frac{9 \,{\left (8 \, c^{3} d^{2} e^{2} - a c^{2} e^{4}\right )}}{c^{4}}\right )} x + \frac{32 \,{\left (3 \, c^{3} d^{3} e - 2 \, a c^{2} d e^{3}\right )}}{c^{4}}\right )} - \frac{{\left (8 \, c^{2} d^{4} - 24 \, a c d^{2} e^{2} + 3 \, a^{2} e^{4}\right )} \log \left ({\left | -\sqrt{c} x + \sqrt{c x^{2} + a} \right |}\right )}{8 \, 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)^(1/2),x, algorithm="giac")

[Out]

1/24*sqrt(c*x^2 + a)*((2*x*(3*x*e^4/c + 16*d*e^3/c) + 9*(8*c^3*d^2*e^2 - a*c^2*e^4)/c^4)*x + 32*(3*c^3*d^3*e -
2*a*c^2*d*e^3)/c^4) - 1/8*(8*c^2*d^4 - 24*a*c*d^2*e^2 + 3*a^2*e^4)*log(abs(-sqrt(c)*x + sqrt(c*x^2 + a)))/c^(
5/2)