### 3.868 $$\int \frac{\sqrt{c d^2-c e^2 x^2}}{(d+e x)^{7/2}} \, dx$$

Optimal. Leaf size=141 $\frac{\sqrt{c d^2-c e^2 x^2}}{8 d e (d+e x)^{3/2}}-\frac{\sqrt{c d^2-c e^2 x^2}}{2 e (d+e x)^{5/2}}+\frac{\sqrt{c} \tanh ^{-1}\left (\frac{\sqrt{c d^2-c e^2 x^2}}{\sqrt{2} \sqrt{c} \sqrt{d} \sqrt{d+e x}}\right )}{8 \sqrt{2} d^{3/2} e}$

[Out]

-Sqrt[c*d^2 - c*e^2*x^2]/(2*e*(d + e*x)^(5/2)) + Sqrt[c*d^2 - c*e^2*x^2]/(8*d*e*(d + e*x)^(3/2)) + (Sqrt[c]*Ar
cTanh[Sqrt[c*d^2 - c*e^2*x^2]/(Sqrt[2]*Sqrt[c]*Sqrt[d]*Sqrt[d + e*x])])/(8*Sqrt[2]*d^(3/2)*e)

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Rubi [A]  time = 0.0776742, antiderivative size = 141, 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, 673, 661, 208} $\frac{\sqrt{c d^2-c e^2 x^2}}{8 d e (d+e x)^{3/2}}-\frac{\sqrt{c d^2-c e^2 x^2}}{2 e (d+e x)^{5/2}}+\frac{\sqrt{c} \tanh ^{-1}\left (\frac{\sqrt{c d^2-c e^2 x^2}}{\sqrt{2} \sqrt{c} \sqrt{d} \sqrt{d+e x}}\right )}{8 \sqrt{2} d^{3/2} e}$

Antiderivative was successfully veriﬁed.

[In]

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

[Out]

-Sqrt[c*d^2 - c*e^2*x^2]/(2*e*(d + e*x)^(5/2)) + Sqrt[c*d^2 - c*e^2*x^2]/(8*d*e*(d + e*x)^(3/2)) + (Sqrt[c]*Ar
cTanh[Sqrt[c*d^2 - c*e^2*x^2]/(Sqrt[2]*Sqrt[c]*Sqrt[d]*Sqrt[d + e*x])])/(8*Sqrt[2]*d^(3/2)*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 673

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

Mathematica [A]  time = 0.144759, size = 111, normalized size = 0.79 $\frac{\sqrt{c \left (d^2-e^2 x^2\right )} \left (\frac{\sqrt{2} \tanh ^{-1}\left (\frac{\sqrt{d^2-e^2 x^2}}{\sqrt{2} \sqrt{d} \sqrt{d+e x}}\right )}{d^{3/2} \sqrt{d^2-e^2 x^2}}+\frac{2 e x-6 d}{d (d+e x)^{5/2}}\right )}{16 e}$

Antiderivative was successfully veriﬁed.

[In]

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

[Out]

(Sqrt[c*(d^2 - e^2*x^2)]*((-6*d + 2*e*x)/(d*(d + e*x)^(5/2)) + (Sqrt[2]*ArcTanh[Sqrt[d^2 - e^2*x^2]/(Sqrt[2]*S
qrt[d]*Sqrt[d + e*x])])/(d^(3/2)*Sqrt[d^2 - e^2*x^2])))/(16*e)

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Maple [A]  time = 0.169, size = 190, normalized size = 1.4 \begin{align*}{\frac{1}{16\,de}\sqrt{-c \left ({e}^{2}{x}^{2}-{d}^{2} \right ) } \left ( \sqrt{2}{\it Artanh} \left ({\frac{\sqrt{2}}{2}\sqrt{- \left ( ex-d \right ) c}{\frac{1}{\sqrt{cd}}}} \right ){x}^{2}c{e}^{2}+2\,\sqrt{2}{\it Artanh} \left ( 1/2\,{\frac{\sqrt{- \left ( ex-d \right ) c}\sqrt{2}}{\sqrt{cd}}} \right ) xcde+\sqrt{2}{\it Artanh} \left ({\frac{\sqrt{2}}{2}\sqrt{- \left ( ex-d \right ) c}{\frac{1}{\sqrt{cd}}}} \right ) c{d}^{2}+2\,xe\sqrt{- \left ( ex-d \right ) c}\sqrt{cd}-6\,\sqrt{- \left ( ex-d \right ) c}\sqrt{cd}d \right ) \left ( ex+d \right ) ^{-{\frac{5}{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)^(1/2)/(e*x+d)^(7/2),x)

[Out]

1/16*(-c*(e^2*x^2-d^2))^(1/2)*(2^(1/2)*arctanh(1/2*(-(e*x-d)*c)^(1/2)*2^(1/2)/(c*d)^(1/2))*x^2*c*e^2+2*2^(1/2)
*arctanh(1/2*(-(e*x-d)*c)^(1/2)*2^(1/2)/(c*d)^(1/2))*x*c*d*e+2^(1/2)*arctanh(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)-6*(-(e*x-d)*c)^(1/2)*(c*d)^(1/2)*d)/(e*x+d)^(5/2)/(-(e*x
-d)*c)^(1/2)/e/d/(c*d)^(1/2)

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Maxima [F]  time = 0., size = 0, normalized size = 0. \begin{align*} \int \frac{\sqrt{-c e^{2} x^{2} + c d^{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)^(1/2)/(e*x+d)^(7/2),x, algorithm="maxima")

[Out]

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

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Fricas [A]  time = 2.38676, size = 776, normalized size = 5.5 \begin{align*} \left [\frac{\sqrt{\frac{1}{2}}{\left (e^{3} x^{3} + 3 \, d e^{2} x^{2} + 3 \, d^{2} e x + d^{3}\right )} \sqrt{\frac{c}{d}} \log \left (-\frac{c e^{2} x^{2} - 2 \, c d e x - 3 \, c d^{2} - 4 \, \sqrt{\frac{1}{2}} \sqrt{-c e^{2} x^{2} + c d^{2}} \sqrt{e x + d} d \sqrt{\frac{c}{d}}}{e^{2} x^{2} + 2 \, d e x + d^{2}}\right ) + 2 \, \sqrt{-c e^{2} x^{2} + c d^{2}} \sqrt{e x + d}{\left (e x - 3 \, d\right )}}{16 \,{\left (d e^{4} x^{3} + 3 \, d^{2} e^{3} x^{2} + 3 \, d^{3} e^{2} x + d^{4} e\right )}}, \frac{\sqrt{\frac{1}{2}}{\left (e^{3} x^{3} + 3 \, d e^{2} x^{2} + 3 \, d^{2} e x + d^{3}\right )} \sqrt{-\frac{c}{d}} \arctan \left (\frac{2 \, \sqrt{\frac{1}{2}} \sqrt{-c e^{2} x^{2} + c d^{2}} \sqrt{e x + d} d \sqrt{-\frac{c}{d}}}{c e^{2} x^{2} - c d^{2}}\right ) + \sqrt{-c e^{2} x^{2} + c d^{2}} \sqrt{e x + d}{\left (e x - 3 \, d\right )}}{8 \,{\left (d e^{4} x^{3} + 3 \, d^{2} e^{3} x^{2} + 3 \, d^{3} e^{2} x + d^{4} e\right )}}\right ] \end{align*}

Veriﬁcation of antiderivative is not currently implemented for this CAS.

[In]

integrate((-c*e^2*x^2+c*d^2)^(1/2)/(e*x+d)^(7/2),x, algorithm="fricas")

[Out]

[1/16*(sqrt(1/2)*(e^3*x^3 + 3*d*e^2*x^2 + 3*d^2*e*x + d^3)*sqrt(c/d)*log(-(c*e^2*x^2 - 2*c*d*e*x - 3*c*d^2 - 4
*sqrt(1/2)*sqrt(-c*e^2*x^2 + c*d^2)*sqrt(e*x + d)*d*sqrt(c/d))/(e^2*x^2 + 2*d*e*x + d^2)) + 2*sqrt(-c*e^2*x^2
+ c*d^2)*sqrt(e*x + d)*(e*x - 3*d))/(d*e^4*x^3 + 3*d^2*e^3*x^2 + 3*d^3*e^2*x + d^4*e), 1/8*(sqrt(1/2)*(e^3*x^3
+ 3*d*e^2*x^2 + 3*d^2*e*x + d^3)*sqrt(-c/d)*arctan(2*sqrt(1/2)*sqrt(-c*e^2*x^2 + c*d^2)*sqrt(e*x + d)*d*sqrt(
-c/d)/(c*e^2*x^2 - c*d^2)) + sqrt(-c*e^2*x^2 + c*d^2)*sqrt(e*x + d)*(e*x - 3*d))/(d*e^4*x^3 + 3*d^2*e^3*x^2 +
3*d^3*e^2*x + d^4*e)]

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Sympy [F]  time = 0., size = 0, normalized size = 0. \begin{align*} \int \frac{\sqrt{- c \left (- d + e x\right ) \left (d + e x\right )}}{\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)**(1/2)/(e*x+d)**(7/2),x)

[Out]

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

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Giac [F]  time = 0., size = 0, normalized size = 0. \begin{align*} \int \frac{\sqrt{-c e^{2} x^{2} + c d^{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)^(1/2)/(e*x+d)^(7/2),x, algorithm="giac")

[Out]

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