### 3.639 $$\int \frac{(d+e x)^{5/2}}{(a-c x^2)^3} \, dx$$

Optimal. Leaf size=279 $-\frac{3 \sqrt{\sqrt{c} d-\sqrt{a} e} \left (2 \sqrt{a} \sqrt{c} d e-a e^2+4 c d^2\right ) \tanh ^{-1}\left (\frac{\sqrt [4]{c} \sqrt{d+e x}}{\sqrt{\sqrt{c} d-\sqrt{a} e}}\right )}{32 a^{5/2} c^{7/4}}+\frac{3 \sqrt{\sqrt{a} e+\sqrt{c} d} \left (-2 \sqrt{a} \sqrt{c} d e-a e^2+4 c d^2\right ) \tanh ^{-1}\left (\frac{\sqrt [4]{c} \sqrt{d+e x}}{\sqrt{\sqrt{a} e+\sqrt{c} d}}\right )}{32 a^{5/2} c^{7/4}}+\frac{3 \sqrt{d+e x} \left (x \left (2 c d^2-a e^2\right )+a d e\right )}{16 a^2 c \left (a-c x^2\right )}+\frac{(d+e x)^{3/2} (a e+c d x)}{4 a c \left (a-c x^2\right )^2}$

[Out]

((a*e + c*d*x)*(d + e*x)^(3/2))/(4*a*c*(a - c*x^2)^2) + (3*Sqrt[d + e*x]*(a*d*e + (2*c*d^2 - a*e^2)*x))/(16*a^
2*c*(a - c*x^2)) - (3*Sqrt[Sqrt[c]*d - Sqrt[a]*e]*(4*c*d^2 + 2*Sqrt[a]*Sqrt[c]*d*e - a*e^2)*ArcTanh[(c^(1/4)*S
qrt[d + e*x])/Sqrt[Sqrt[c]*d - Sqrt[a]*e]])/(32*a^(5/2)*c^(7/4)) + (3*Sqrt[Sqrt[c]*d + Sqrt[a]*e]*(4*c*d^2 - 2
*Sqrt[a]*Sqrt[c]*d*e - a*e^2)*ArcTanh[(c^(1/4)*Sqrt[d + e*x])/Sqrt[Sqrt[c]*d + Sqrt[a]*e]])/(32*a^(5/2)*c^(7/4
))

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Rubi [A]  time = 0.458589, antiderivative size = 279, normalized size of antiderivative = 1., number of steps used = 6, number of rules used = 5, integrand size = 20, $$\frac{\text{number of rules}}{\text{integrand size}}$$ = 0.25, Rules used = {739, 821, 827, 1166, 208} $-\frac{3 \sqrt{\sqrt{c} d-\sqrt{a} e} \left (2 \sqrt{a} \sqrt{c} d e-a e^2+4 c d^2\right ) \tanh ^{-1}\left (\frac{\sqrt [4]{c} \sqrt{d+e x}}{\sqrt{\sqrt{c} d-\sqrt{a} e}}\right )}{32 a^{5/2} c^{7/4}}+\frac{3 \sqrt{\sqrt{a} e+\sqrt{c} d} \left (-2 \sqrt{a} \sqrt{c} d e-a e^2+4 c d^2\right ) \tanh ^{-1}\left (\frac{\sqrt [4]{c} \sqrt{d+e x}}{\sqrt{\sqrt{a} e+\sqrt{c} d}}\right )}{32 a^{5/2} c^{7/4}}+\frac{3 \sqrt{d+e x} \left (x \left (2 c d^2-a e^2\right )+a d e\right )}{16 a^2 c \left (a-c x^2\right )}+\frac{(d+e x)^{3/2} (a e+c d x)}{4 a c \left (a-c x^2\right )^2}$

Antiderivative was successfully veriﬁed.

[In]

Int[(d + e*x)^(5/2)/(a - c*x^2)^3,x]

[Out]

((a*e + c*d*x)*(d + e*x)^(3/2))/(4*a*c*(a - c*x^2)^2) + (3*Sqrt[d + e*x]*(a*d*e + (2*c*d^2 - a*e^2)*x))/(16*a^
2*c*(a - c*x^2)) - (3*Sqrt[Sqrt[c]*d - Sqrt[a]*e]*(4*c*d^2 + 2*Sqrt[a]*Sqrt[c]*d*e - a*e^2)*ArcTanh[(c^(1/4)*S
qrt[d + e*x])/Sqrt[Sqrt[c]*d - Sqrt[a]*e]])/(32*a^(5/2)*c^(7/4)) + (3*Sqrt[Sqrt[c]*d + Sqrt[a]*e]*(4*c*d^2 - 2
*Sqrt[a]*Sqrt[c]*d*e - a*e^2)*ArcTanh[(c^(1/4)*Sqrt[d + e*x])/Sqrt[Sqrt[c]*d + Sqrt[a]*e]])/(32*a^(5/2)*c^(7/4
))

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 821

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])

Rule 827

Int[((f_.) + (g_.)*(x_))/(Sqrt[(d_.) + (e_.)*(x_)]*((a_) + (c_.)*(x_)^2)), x_Symbol] :> Dist[2, Subst[Int[(e*f
- d*g + g*x^2)/(c*d^2 + a*e^2 - 2*c*d*x^2 + c*x^4), x], x, Sqrt[d + e*x]], x] /; FreeQ[{a, c, d, e, f, g}, x]
&& NeQ[c*d^2 + a*e^2, 0]

Rule 1166

Int[((d_) + (e_.)*(x_)^2)/((a_) + (b_.)*(x_)^2 + (c_.)*(x_)^4), x_Symbol] :> With[{q = Rt[b^2 - 4*a*c, 2]}, Di
st[e/2 + (2*c*d - b*e)/(2*q), Int[1/(b/2 - q/2 + c*x^2), x], x] + Dist[e/2 - (2*c*d - b*e)/(2*q), Int[1/(b/2 +
q/2 + c*x^2), x], x]] /; FreeQ[{a, b, c, d, e}, x] && NeQ[b^2 - 4*a*c, 0] && NeQ[c*d^2 - a*e^2, 0] && PosQ[b^
2 - 4*a*c]

Rule 208

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

Rubi steps

\begin{align*} \int \frac{(d+e x)^{5/2}}{\left (a-c x^2\right )^3} \, dx &=\frac{(a e+c d x) (d+e x)^{3/2}}{4 a c \left (a-c x^2\right )^2}-\frac{\int \frac{\sqrt{d+e x} \left (-\frac{3}{2} \left (2 c d^2-a e^2\right )-\frac{3}{2} c d e x\right )}{\left (a-c x^2\right )^2} \, dx}{4 a c}\\ &=\frac{(a e+c d x) (d+e x)^{3/2}}{4 a c \left (a-c x^2\right )^2}+\frac{3 \sqrt{d+e x} \left (a d e+\left (2 c d^2-a e^2\right ) x\right )}{16 a^2 c \left (a-c x^2\right )}+\frac{\int \frac{\frac{3}{4} c d \left (4 c d^2-3 a e^2\right )+\frac{3}{4} c e \left (2 c d^2-a e^2\right ) x}{\sqrt{d+e x} \left (a-c x^2\right )} \, dx}{8 a^2 c^2}\\ &=\frac{(a e+c d x) (d+e x)^{3/2}}{4 a c \left (a-c x^2\right )^2}+\frac{3 \sqrt{d+e x} \left (a d e+\left (2 c d^2-a e^2\right ) x\right )}{16 a^2 c \left (a-c x^2\right )}+\frac{\operatorname{Subst}\left (\int \frac{\frac{3}{4} c d e \left (4 c d^2-3 a e^2\right )-\frac{3}{4} c d e \left (2 c d^2-a e^2\right )+\frac{3}{4} c e \left (2 c d^2-a e^2\right ) x^2}{-c d^2+a e^2+2 c d x^2-c x^4} \, dx,x,\sqrt{d+e x}\right )}{4 a^2 c^2}\\ &=\frac{(a e+c d x) (d+e x)^{3/2}}{4 a c \left (a-c x^2\right )^2}+\frac{3 \sqrt{d+e x} \left (a d e+\left (2 c d^2-a e^2\right ) x\right )}{16 a^2 c \left (a-c x^2\right )}+\frac{\left (3 \left (\sqrt{c} d+\sqrt{a} e\right ) \left (4 c d^2-2 \sqrt{a} \sqrt{c} d e-a e^2\right )\right ) \operatorname{Subst}\left (\int \frac{1}{c d+\sqrt{a} \sqrt{c} e-c x^2} \, dx,x,\sqrt{d+e x}\right )}{32 a^{5/2} c}-\frac{\left (3 \left (\sqrt{c} d-\sqrt{a} e\right ) \left (4 c d^2+2 \sqrt{a} \sqrt{c} d e-a e^2\right )\right ) \operatorname{Subst}\left (\int \frac{1}{c d-\sqrt{a} \sqrt{c} e-c x^2} \, dx,x,\sqrt{d+e x}\right )}{32 a^{5/2} c}\\ &=\frac{(a e+c d x) (d+e x)^{3/2}}{4 a c \left (a-c x^2\right )^2}+\frac{3 \sqrt{d+e x} \left (a d e+\left (2 c d^2-a e^2\right ) x\right )}{16 a^2 c \left (a-c x^2\right )}-\frac{3 \sqrt{\sqrt{c} d-\sqrt{a} e} \left (4 c d^2+2 \sqrt{a} \sqrt{c} d e-a e^2\right ) \tanh ^{-1}\left (\frac{\sqrt [4]{c} \sqrt{d+e x}}{\sqrt{\sqrt{c} d-\sqrt{a} e}}\right )}{32 a^{5/2} c^{7/4}}+\frac{3 \sqrt{\sqrt{c} d+\sqrt{a} e} \left (4 c d^2-2 \sqrt{a} \sqrt{c} d e-a e^2\right ) \tanh ^{-1}\left (\frac{\sqrt [4]{c} \sqrt{d+e x}}{\sqrt{\sqrt{c} d+\sqrt{a} e}}\right )}{32 a^{5/2} c^{7/4}}\\ \end{align*}

Mathematica [A]  time = 0.520435, size = 260, normalized size = 0.93 $\frac{\frac{2 \sqrt{a} c^{3/4} \sqrt{d+e x} \left (a^2 e (7 d+e x)+a c x \left (10 d^2+d e x+3 e^2 x^2\right )-6 c^2 d^2 x^3\right )}{\left (a-c x^2\right )^2}-3 \sqrt{\sqrt{c} d-\sqrt{a} e} \left (2 \sqrt{a} \sqrt{c} d e-a e^2+4 c d^2\right ) \tanh ^{-1}\left (\frac{\sqrt [4]{c} \sqrt{d+e x}}{\sqrt{\sqrt{c} d-\sqrt{a} e}}\right )+3 \sqrt{\sqrt{a} e+\sqrt{c} d} \left (-2 \sqrt{a} \sqrt{c} d e-a e^2+4 c d^2\right ) \tanh ^{-1}\left (\frac{\sqrt [4]{c} \sqrt{d+e x}}{\sqrt{\sqrt{a} e+\sqrt{c} d}}\right )}{32 a^{5/2} c^{7/4}}$

Antiderivative was successfully veriﬁed.

[In]

Integrate[(d + e*x)^(5/2)/(a - c*x^2)^3,x]

[Out]

((2*Sqrt[a]*c^(3/4)*Sqrt[d + e*x]*(-6*c^2*d^2*x^3 + a^2*e*(7*d + e*x) + a*c*x*(10*d^2 + d*e*x + 3*e^2*x^2)))/(
a - c*x^2)^2 - 3*Sqrt[Sqrt[c]*d - Sqrt[a]*e]*(4*c*d^2 + 2*Sqrt[a]*Sqrt[c]*d*e - a*e^2)*ArcTanh[(c^(1/4)*Sqrt[d
+ e*x])/Sqrt[Sqrt[c]*d - Sqrt[a]*e]] + 3*Sqrt[Sqrt[c]*d + Sqrt[a]*e]*(4*c*d^2 - 2*Sqrt[a]*Sqrt[c]*d*e - a*e^2
)*ArcTanh[(c^(1/4)*Sqrt[d + e*x])/Sqrt[Sqrt[c]*d + Sqrt[a]*e]])/(32*a^(5/2)*c^(7/4))

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Maple [B]  time = 0.226, size = 792, normalized size = 2.8 \begin{align*} \text{result too large to display} \end{align*}

Veriﬁcation of antiderivative is not currently implemented for this CAS.

[In]

int((e*x+d)^(5/2)/(-c*x^2+a)^3,x)

[Out]

3/16*e^3/(c*e^2*x^2-a*e^2)^2/a*(e*x+d)^(7/2)-3/8*e/(c*e^2*x^2-a*e^2)^2/a^2*(e*x+d)^(7/2)*c*d^2-1/2*e^3/(c*e^2*
x^2-a*e^2)^2*d/a*(e*x+d)^(5/2)+9/8*e/(c*e^2*x^2-a*e^2)^2*d^3/a^2*(e*x+d)^(5/2)*c+1/16*e^5/(c*e^2*x^2-a*e^2)^2/
c*(e*x+d)^(3/2)+17/16*e^3/(c*e^2*x^2-a*e^2)^2/a*(e*x+d)^(3/2)*d^2-9/8*e/(c*e^2*x^2-a*e^2)^2/a^2*c*(e*x+d)^(3/2
)*d^4+3/8*e^5/(c*e^2*x^2-a*e^2)^2*d/c*(e*x+d)^(1/2)-3/4*e^3/(c*e^2*x^2-a*e^2)^2*d^3/a*(e*x+d)^(1/2)+3/8*e/(c*e
^2*x^2-a*e^2)^2*d^5/a^2*c*(e*x+d)^(1/2)-9/32*e^3/a/(a*c*e^2)^(1/2)/((-c*d+(a*c*e^2)^(1/2))*c)^(1/2)*arctan((e*
x+d)^(1/2)*c/((-c*d+(a*c*e^2)^(1/2))*c)^(1/2))*d+3/8*e/a^2*c/(a*c*e^2)^(1/2)/((-c*d+(a*c*e^2)^(1/2))*c)^(1/2)*
arctan((e*x+d)^(1/2)*c/((-c*d+(a*c*e^2)^(1/2))*c)^(1/2))*d^3+3/32*e^3/a/c/((-c*d+(a*c*e^2)^(1/2))*c)^(1/2)*arc
tan((e*x+d)^(1/2)*c/((-c*d+(a*c*e^2)^(1/2))*c)^(1/2))-3/16*e/a^2/((-c*d+(a*c*e^2)^(1/2))*c)^(1/2)*arctan((e*x+
d)^(1/2)*c/((-c*d+(a*c*e^2)^(1/2))*c)^(1/2))*d^2-9/32*e^3/a/(a*c*e^2)^(1/2)/((c*d+(a*c*e^2)^(1/2))*c)^(1/2)*ar
ctanh((e*x+d)^(1/2)*c/((c*d+(a*c*e^2)^(1/2))*c)^(1/2))*d+3/8*e/a^2*c/(a*c*e^2)^(1/2)/((c*d+(a*c*e^2)^(1/2))*c)
^(1/2)*arctanh((e*x+d)^(1/2)*c/((c*d+(a*c*e^2)^(1/2))*c)^(1/2))*d^3-3/32*e^3/a/c/((c*d+(a*c*e^2)^(1/2))*c)^(1/
2)*arctanh((e*x+d)^(1/2)*c/((c*d+(a*c*e^2)^(1/2))*c)^(1/2))+3/16*e/a^2/((c*d+(a*c*e^2)^(1/2))*c)^(1/2)*arctanh
((e*x+d)^(1/2)*c/((c*d+(a*c*e^2)^(1/2))*c)^(1/2))*d^2

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Maxima [F]  time = 0., size = 0, normalized size = 0. \begin{align*} -\int \frac{{\left (e x + d\right )}^{\frac{5}{2}}}{{\left (c x^{2} - a\right )}^{3}}\,{d x} \end{align*}

Veriﬁcation of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

-integrate((e*x + d)^(5/2)/(c*x^2 - a)^3, x)

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Fricas [B]  time = 2.56309, size = 2178, normalized size = 7.81 \begin{align*} \text{result too large to display} \end{align*}

Veriﬁcation of antiderivative is not currently implemented for this CAS.

[In]

integrate((e*x+d)^(5/2)/(-c*x^2+a)^3,x, algorithm="fricas")

[Out]

1/64*(3*(a^2*c^3*x^4 - 2*a^3*c^2*x^2 + a^4*c)*sqrt((a^5*c^3*sqrt(e^10/(a^5*c^7)) + 16*c^2*d^5 - 20*a*c*d^3*e^2
+ 5*a^2*d*e^4)/(a^5*c^3))*log(27*(16*c^2*d^4*e^5 - 12*a*c*d^2*e^7 + a^2*e^9)*sqrt(e*x + d) + 27*(2*a^3*c^2*d*
e^6 + (4*a^5*c^6*d^2 - a^6*c^5*e^2)*sqrt(e^10/(a^5*c^7)))*sqrt((a^5*c^3*sqrt(e^10/(a^5*c^7)) + 16*c^2*d^5 - 20
*a*c*d^3*e^2 + 5*a^2*d*e^4)/(a^5*c^3))) - 3*(a^2*c^3*x^4 - 2*a^3*c^2*x^2 + a^4*c)*sqrt((a^5*c^3*sqrt(e^10/(a^5
*c^7)) + 16*c^2*d^5 - 20*a*c*d^3*e^2 + 5*a^2*d*e^4)/(a^5*c^3))*log(27*(16*c^2*d^4*e^5 - 12*a*c*d^2*e^7 + a^2*e
^9)*sqrt(e*x + d) - 27*(2*a^3*c^2*d*e^6 + (4*a^5*c^6*d^2 - a^6*c^5*e^2)*sqrt(e^10/(a^5*c^7)))*sqrt((a^5*c^3*sq
rt(e^10/(a^5*c^7)) + 16*c^2*d^5 - 20*a*c*d^3*e^2 + 5*a^2*d*e^4)/(a^5*c^3))) + 3*(a^2*c^3*x^4 - 2*a^3*c^2*x^2 +
a^4*c)*sqrt(-(a^5*c^3*sqrt(e^10/(a^5*c^7)) - 16*c^2*d^5 + 20*a*c*d^3*e^2 - 5*a^2*d*e^4)/(a^5*c^3))*log(27*(16
*c^2*d^4*e^5 - 12*a*c*d^2*e^7 + a^2*e^9)*sqrt(e*x + d) + 27*(2*a^3*c^2*d*e^6 - (4*a^5*c^6*d^2 - a^6*c^5*e^2)*s
qrt(e^10/(a^5*c^7)))*sqrt(-(a^5*c^3*sqrt(e^10/(a^5*c^7)) - 16*c^2*d^5 + 20*a*c*d^3*e^2 - 5*a^2*d*e^4)/(a^5*c^3
))) - 3*(a^2*c^3*x^4 - 2*a^3*c^2*x^2 + a^4*c)*sqrt(-(a^5*c^3*sqrt(e^10/(a^5*c^7)) - 16*c^2*d^5 + 20*a*c*d^3*e^
2 - 5*a^2*d*e^4)/(a^5*c^3))*log(27*(16*c^2*d^4*e^5 - 12*a*c*d^2*e^7 + a^2*e^9)*sqrt(e*x + d) - 27*(2*a^3*c^2*d
*e^6 - (4*a^5*c^6*d^2 - a^6*c^5*e^2)*sqrt(e^10/(a^5*c^7)))*sqrt(-(a^5*c^3*sqrt(e^10/(a^5*c^7)) - 16*c^2*d^5 +
20*a*c*d^3*e^2 - 5*a^2*d*e^4)/(a^5*c^3))) + 4*(a*c*d*e*x^2 + 7*a^2*d*e - 3*(2*c^2*d^2 - a*c*e^2)*x^3 + (10*a*c
*d^2 + a^2*e^2)*x)*sqrt(e*x + d))/(a^2*c^3*x^4 - 2*a^3*c^2*x^2 + a^4*c)

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Sympy [F(-1)]  time = 0., size = 0, normalized size = 0. \begin{align*} \text{Timed out} \end{align*}

Veriﬁcation of antiderivative is not currently implemented for this CAS.

[In]

integrate((e*x+d)**(5/2)/(-c*x**2+a)**3,x)

[Out]

Timed out

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Giac [F(-1)]  time = 0., size = 0, normalized size = 0. \begin{align*} \text{Timed out} \end{align*}

Veriﬁcation of antiderivative is not currently implemented for this CAS.

[In]

integrate((e*x+d)^(5/2)/(-c*x^2+a)^3,x, algorithm="giac")

[Out]

Timed out