### 3.875 $$\int \frac{(c d^2-c e^2 x^2)^{3/2}}{(d+e x)^{7/2}} \, dx$$

Optimal. Leaf size=133 $\frac{3 \sqrt{2} c^{3/2} \sqrt{d} \tanh ^{-1}\left (\frac{\sqrt{c d^2-c e^2 x^2}}{\sqrt{2} \sqrt{c} \sqrt{d} \sqrt{d+e x}}\right )}{e}-\frac{3 c \sqrt{c d^2-c e^2 x^2}}{e \sqrt{d+e x}}-\frac{\left (c d^2-c e^2 x^2\right )^{3/2}}{e (d+e x)^{5/2}}$

[Out]

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

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Rubi [A]  time = 0.0789978, antiderivative size = 133, normalized size of antiderivative = 1., number of steps used = 4, number of rules used = 4, integrand size = 29, $$\frac{\text{number of rules}}{\text{integrand size}}$$ = 0.138, Rules used = {663, 665, 661, 208} $\frac{3 \sqrt{2} c^{3/2} \sqrt{d} \tanh ^{-1}\left (\frac{\sqrt{c d^2-c e^2 x^2}}{\sqrt{2} \sqrt{c} \sqrt{d} \sqrt{d+e x}}\right )}{e}-\frac{3 c \sqrt{c d^2-c e^2 x^2}}{e \sqrt{d+e x}}-\frac{\left (c d^2-c e^2 x^2\right )^{3/2}}{e (d+e x)^{5/2}}$

Antiderivative was successfully veriﬁed.

[In]

Int[(c*d^2 - c*e^2*x^2)^(3/2)/(d + e*x)^(7/2),x]

[Out]

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

Rule 663

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

Rule 665

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

Rule 661

Int[1/(Sqrt[(d_) + (e_.)*(x_)]*Sqrt[(a_) + (c_.)*(x_)^2]), x_Symbol] :> Dist[2*e, Subst[Int[1/(2*c*d + e^2*x^2
), x], x, Sqrt[a + c*x^2]/Sqrt[d + e*x]], x] /; FreeQ[{a, c, d, e}, x] && EqQ[c*d^2 + a*e^2, 0]

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{\left (c d^2-c e^2 x^2\right )^{3/2}}{(d+e x)^{7/2}} \, dx &=-\frac{\left (c d^2-c e^2 x^2\right )^{3/2}}{e (d+e x)^{5/2}}-\frac{1}{2} (3 c) \int \frac{\sqrt{c d^2-c e^2 x^2}}{(d+e x)^{3/2}} \, dx\\ &=-\frac{3 c \sqrt{c d^2-c e^2 x^2}}{e \sqrt{d+e x}}-\frac{\left (c d^2-c e^2 x^2\right )^{3/2}}{e (d+e x)^{5/2}}-\left (3 c^2 d\right ) \int \frac{1}{\sqrt{d+e x} \sqrt{c d^2-c e^2 x^2}} \, dx\\ &=-\frac{3 c \sqrt{c d^2-c e^2 x^2}}{e \sqrt{d+e x}}-\frac{\left (c d^2-c e^2 x^2\right )^{3/2}}{e (d+e x)^{5/2}}-\left (6 c^2 d e\right ) \operatorname{Subst}\left (\int \frac{1}{-2 c d e^2+e^2 x^2} \, dx,x,\frac{\sqrt{c d^2-c e^2 x^2}}{\sqrt{d+e x}}\right )\\ &=-\frac{3 c \sqrt{c d^2-c e^2 x^2}}{e \sqrt{d+e x}}-\frac{\left (c d^2-c e^2 x^2\right )^{3/2}}{e (d+e x)^{5/2}}+\frac{3 \sqrt{2} c^{3/2} \sqrt{d} \tanh ^{-1}\left (\frac{\sqrt{c d^2-c e^2 x^2}}{\sqrt{2} \sqrt{c} \sqrt{d} \sqrt{d+e x}}\right )}{e}\\ \end{align*}

Mathematica [A]  time = 0.203967, size = 107, normalized size = 0.8 $\frac{c \sqrt{c \left (d^2-e^2 x^2\right )} \left (\frac{3 \sqrt{2} \sqrt{d} \tanh ^{-1}\left (\frac{\sqrt{d^2-e^2 x^2}}{\sqrt{2} \sqrt{d} \sqrt{d+e x}}\right )}{\sqrt{d^2-e^2 x^2}}-\frac{2 (2 d+e x)}{(d+e x)^{3/2}}\right )}{e}$

Antiderivative was successfully veriﬁed.

[In]

Integrate[(c*d^2 - c*e^2*x^2)^(3/2)/(d + e*x)^(7/2),x]

[Out]

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

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Maple [A]  time = 0.171, size = 154, normalized size = 1.2 \begin{align*}{\frac{c}{e}\sqrt{-c \left ({e}^{2}{x}^{2}-{d}^{2} \right ) } \left ( 3\,\sqrt{2}{\it Artanh} \left ( 1/2\,{\frac{\sqrt{- \left ( ex-d \right ) c}\sqrt{2}}{\sqrt{cd}}} \right ) xcde+3\,\sqrt{2}{\it Artanh} \left ( 1/2\,{\frac{\sqrt{- \left ( ex-d \right ) c}\sqrt{2}}{\sqrt{cd}}} \right ) c{d}^{2}-2\,xe\sqrt{- \left ( ex-d \right ) c}\sqrt{cd}-4\,\sqrt{- \left ( ex-d \right ) c}\sqrt{cd}d \right ) \left ( ex+d \right ) ^{-{\frac{3}{2}}}{\frac{1}{\sqrt{- \left ( ex-d \right ) c}}}{\frac{1}{\sqrt{cd}}}} \end{align*}

Veriﬁcation of antiderivative is not currently implemented for this CAS.

[In]

int((-c*e^2*x^2+c*d^2)^(3/2)/(e*x+d)^(7/2),x)

[Out]

(-c*(e^2*x^2-d^2))^(1/2)*c*(3*2^(1/2)*arctanh(1/2*(-(e*x-d)*c)^(1/2)*2^(1/2)/(c*d)^(1/2))*x*c*d*e+3*2^(1/2)*ar
ctanh(1/2*(-(e*x-d)*c)^(1/2)*2^(1/2)/(c*d)^(1/2))*c*d^2-2*x*e*(-(e*x-d)*c)^(1/2)*(c*d)^(1/2)-4*(-(e*x-d)*c)^(1
/2)*(c*d)^(1/2)*d)/(e*x+d)^(3/2)/(-(e*x-d)*c)^(1/2)/e/(c*d)^(1/2)

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Maxima [F]  time = 0., size = 0, normalized size = 0. \begin{align*} \int \frac{{\left (-c e^{2} x^{2} + c d^{2}\right )}^{\frac{3}{2}}}{{\left (e x + d\right )}^{\frac{7}{2}}}\,{d x} \end{align*}

Veriﬁcation of antiderivative is not currently implemented for this CAS.

[In]

integrate((-c*e^2*x^2+c*d^2)^(3/2)/(e*x+d)^(7/2),x, algorithm="maxima")

[Out]

integrate((-c*e^2*x^2 + c*d^2)^(3/2)/(e*x + d)^(7/2), x)

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Fricas [A]  time = 2.21309, size = 689, normalized size = 5.18 \begin{align*} \left [\frac{3 \, \sqrt{2}{\left (c e^{2} x^{2} + 2 \, c d e x + c d^{2}\right )} \sqrt{c d} \log \left (-\frac{c e^{2} x^{2} - 2 \, c d e x - 3 \, c d^{2} - 2 \, \sqrt{2} \sqrt{-c e^{2} x^{2} + c d^{2}} \sqrt{c d} \sqrt{e x + d}}{e^{2} x^{2} + 2 \, d e x + d^{2}}\right ) - 4 \, \sqrt{-c e^{2} x^{2} + c d^{2}}{\left (c e x + 2 \, c d\right )} \sqrt{e x + d}}{2 \,{\left (e^{3} x^{2} + 2 \, d e^{2} x + d^{2} e\right )}}, \frac{3 \, \sqrt{2}{\left (c e^{2} x^{2} + 2 \, c d e x + c d^{2}\right )} \sqrt{-c d} \arctan \left (\frac{\sqrt{2} \sqrt{-c e^{2} x^{2} + c d^{2}} \sqrt{-c d} \sqrt{e x + d}}{c e^{2} x^{2} - c d^{2}}\right ) - 2 \, \sqrt{-c e^{2} x^{2} + c d^{2}}{\left (c e x + 2 \, c d\right )} \sqrt{e x + d}}{e^{3} x^{2} + 2 \, d e^{2} x + d^{2} e}\right ] \end{align*}

Veriﬁcation of antiderivative is not currently implemented for this CAS.

[In]

integrate((-c*e^2*x^2+c*d^2)^(3/2)/(e*x+d)^(7/2),x, algorithm="fricas")

[Out]

[1/2*(3*sqrt(2)*(c*e^2*x^2 + 2*c*d*e*x + c*d^2)*sqrt(c*d)*log(-(c*e^2*x^2 - 2*c*d*e*x - 3*c*d^2 - 2*sqrt(2)*sq
rt(-c*e^2*x^2 + c*d^2)*sqrt(c*d)*sqrt(e*x + d))/(e^2*x^2 + 2*d*e*x + d^2)) - 4*sqrt(-c*e^2*x^2 + c*d^2)*(c*e*x
+ 2*c*d)*sqrt(e*x + d))/(e^3*x^2 + 2*d*e^2*x + d^2*e), (3*sqrt(2)*(c*e^2*x^2 + 2*c*d*e*x + c*d^2)*sqrt(-c*d)*
arctan(sqrt(2)*sqrt(-c*e^2*x^2 + c*d^2)*sqrt(-c*d)*sqrt(e*x + d)/(c*e^2*x^2 - c*d^2)) - 2*sqrt(-c*e^2*x^2 + c*
d^2)*(c*e*x + 2*c*d)*sqrt(e*x + d))/(e^3*x^2 + 2*d*e^2*x + d^2*e)]

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Sympy [F]  time = 0., size = 0, normalized size = 0. \begin{align*} \int \frac{\left (- c \left (- d + e x\right ) \left (d + e x\right )\right )^{\frac{3}{2}}}{\left (d + e x\right )^{\frac{7}{2}}}\, dx \end{align*}

Veriﬁcation of antiderivative is not currently implemented for this CAS.

[In]

integrate((-c*e**2*x**2+c*d**2)**(3/2)/(e*x+d)**(7/2),x)

[Out]

Integral((-c*(-d + e*x)*(d + e*x))**(3/2)/(d + e*x)**(7/2), x)

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Giac [F]  time = 0., size = 0, normalized size = 0. \begin{align*} \int \frac{{\left (-c e^{2} x^{2} + c d^{2}\right )}^{\frac{3}{2}}}{{\left (e x + d\right )}^{\frac{7}{2}}}\,{d x} \end{align*}

Veriﬁcation of antiderivative is not currently implemented for this CAS.

[In]

integrate((-c*e^2*x^2+c*d^2)^(3/2)/(e*x+d)^(7/2),x, algorithm="giac")

[Out]

integrate((-c*e^2*x^2 + c*d^2)^(3/2)/(e*x + d)^(7/2), x)