### 3.885 $$\int \frac{1}{(d+e x)^{5/2} \sqrt{c d^2-c e^2 x^2}} \, dx$$

Optimal. Leaf size=150 $-\frac{3 \sqrt{c d^2-c e^2 x^2}}{16 c d^2 e (d+e x)^{3/2}}-\frac{\sqrt{c d^2-c e^2 x^2}}{4 c d e (d+e x)^{5/2}}-\frac{3 \tanh ^{-1}\left (\frac{\sqrt{c d^2-c e^2 x^2}}{\sqrt{2} \sqrt{c} \sqrt{d} \sqrt{d+e x}}\right )}{16 \sqrt{2} \sqrt{c} d^{5/2} e}$

[Out]

-Sqrt[c*d^2 - c*e^2*x^2]/(4*c*d*e*(d + e*x)^(5/2)) - (3*Sqrt[c*d^2 - c*e^2*x^2])/(16*c*d^2*e*(d + e*x)^(3/2))
- (3*ArcTanh[Sqrt[c*d^2 - c*e^2*x^2]/(Sqrt[2]*Sqrt[c]*Sqrt[d]*Sqrt[d + e*x])])/(16*Sqrt[2]*Sqrt[c]*d^(5/2)*e)

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Rubi [A]  time = 0.0720232, antiderivative size = 150, normalized size of antiderivative = 1., number of steps used = 4, number of rules used = 3, integrand size = 29, $$\frac{\text{number of rules}}{\text{integrand size}}$$ = 0.103, Rules used = {673, 661, 208} $-\frac{3 \sqrt{c d^2-c e^2 x^2}}{16 c d^2 e (d+e x)^{3/2}}-\frac{\sqrt{c d^2-c e^2 x^2}}{4 c d e (d+e x)^{5/2}}-\frac{3 \tanh ^{-1}\left (\frac{\sqrt{c d^2-c e^2 x^2}}{\sqrt{2} \sqrt{c} \sqrt{d} \sqrt{d+e x}}\right )}{16 \sqrt{2} \sqrt{c} d^{5/2} e}$

Antiderivative was successfully veriﬁed.

[In]

Int[1/((d + e*x)^(5/2)*Sqrt[c*d^2 - c*e^2*x^2]),x]

[Out]

-Sqrt[c*d^2 - c*e^2*x^2]/(4*c*d*e*(d + e*x)^(5/2)) - (3*Sqrt[c*d^2 - c*e^2*x^2])/(16*c*d^2*e*(d + e*x)^(3/2))
- (3*ArcTanh[Sqrt[c*d^2 - c*e^2*x^2]/(Sqrt[2]*Sqrt[c]*Sqrt[d]*Sqrt[d + e*x])])/(16*Sqrt[2]*Sqrt[c]*d^(5/2)*e)

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

Mathematica [A]  time = 0.110124, size = 140, normalized size = 0.93 $\frac{2 \sqrt{d} \sqrt{d+e x} \left (-7 d^2+4 d e x+3 e^2 x^2\right )-3 \sqrt{2} (d+e x)^2 \sqrt{d^2-e^2 x^2} \tanh ^{-1}\left (\frac{\sqrt{d^2-e^2 x^2}}{\sqrt{2} \sqrt{d} \sqrt{d+e x}}\right )}{32 d^{5/2} e (d+e x)^2 \sqrt{c \left (d^2-e^2 x^2\right )}}$

Antiderivative was successfully veriﬁed.

[In]

Integrate[1/((d + e*x)^(5/2)*Sqrt[c*d^2 - c*e^2*x^2]),x]

[Out]

(2*Sqrt[d]*Sqrt[d + e*x]*(-7*d^2 + 4*d*e*x + 3*e^2*x^2) - 3*Sqrt[2]*(d + e*x)^2*Sqrt[d^2 - e^2*x^2]*ArcTanh[Sq
rt[d^2 - e^2*x^2]/(Sqrt[2]*Sqrt[d]*Sqrt[d + e*x])])/(32*d^(5/2)*e*(d + e*x)^2*Sqrt[c*(d^2 - e^2*x^2)])

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Maple [A]  time = 0.165, size = 195, normalized size = 1.3 \begin{align*} -{\frac{1}{32\,{d}^{2}ec}\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 ){x}^{2}c{e}^{2}+6\,\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}+6\,xe\sqrt{- \left ( ex-d \right ) c}\sqrt{cd}+14\,\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(1/(e*x+d)^(5/2)/(-c*e^2*x^2+c*d^2)^(1/2),x)

[Out]

-1/32/(e*x+d)^(5/2)*(-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^2*c*e^2+6*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)*arctanh(1/2*(-(e*x-d
)*c)^(1/2)*2^(1/2)/(c*d)^(1/2))*c*d^2+6*x*e*(-(e*x-d)*c)^(1/2)*(c*d)^(1/2)+14*(-(e*x-d)*c)^(1/2)*(c*d)^(1/2)*d
)/(-(e*x-d)*c)^(1/2)/e/d^2/(c*d)^(1/2)

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Maxima [F]  time = 0., size = 0, normalized size = 0. \begin{align*} \int \frac{1}{\sqrt{-c e^{2} x^{2} + c d^{2}}{\left (e x + d\right )}^{\frac{5}{2}}}\,{d x} \end{align*}

Veriﬁcation of antiderivative is not currently implemented for this CAS.

[In]

integrate(1/(e*x+d)^(5/2)/(-c*e^2*x^2+c*d^2)^(1/2),x, algorithm="maxima")

[Out]

integrate(1/(sqrt(-c*e^2*x^2 + c*d^2)*(e*x + d)^(5/2)), x)

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Fricas [A]  time = 2.10015, size = 811, normalized size = 5.41 \begin{align*} \left [\frac{3 \, \sqrt{2}{\left (e^{3} x^{3} + 3 \, d e^{2} x^{2} + 3 \, d^{2} e x + d^{3}\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 (3 \, d e x + 7 \, d^{2}\right )} \sqrt{e x + d}}{64 \,{\left (c d^{3} e^{4} x^{3} + 3 \, c d^{4} e^{3} x^{2} + 3 \, c d^{5} e^{2} x + c d^{6} e\right )}}, -\frac{3 \, \sqrt{2}{\left (e^{3} x^{3} + 3 \, d e^{2} x^{2} + 3 \, d^{2} e x + d^{3}\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 (3 \, d e x + 7 \, d^{2}\right )} \sqrt{e x + d}}{32 \,{\left (c d^{3} e^{4} x^{3} + 3 \, c d^{4} e^{3} x^{2} + 3 \, c d^{5} e^{2} x + c d^{6} e\right )}}\right ] \end{align*}

Veriﬁcation of antiderivative is not currently implemented for this CAS.

[In]

integrate(1/(e*x+d)^(5/2)/(-c*e^2*x^2+c*d^2)^(1/2),x, algorithm="fricas")

[Out]

[1/64*(3*sqrt(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 + 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)) - 4*sqrt(-c*e^2*x^2 + c*
d^2)*(3*d*e*x + 7*d^2)*sqrt(e*x + d))/(c*d^3*e^4*x^3 + 3*c*d^4*e^3*x^2 + 3*c*d^5*e^2*x + c*d^6*e), -1/32*(3*sq
rt(2)*(e^3*x^3 + 3*d*e^2*x^2 + 3*d^2*e*x + d^3)*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)*(3*d*e*x + 7*d^2)*sqrt(e*x + d))/(c*d^3*e^4*x^
3 + 3*c*d^4*e^3*x^2 + 3*c*d^5*e^2*x + c*d^6*e)]

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Sympy [F]  time = 0., size = 0, normalized size = 0. \begin{align*} \int \frac{1}{\sqrt{- c \left (- d + e x\right ) \left (d + e x\right )} \left (d + e x\right )^{\frac{5}{2}}}\, dx \end{align*}

Veriﬁcation of antiderivative is not currently implemented for this CAS.

[In]

integrate(1/(e*x+d)**(5/2)/(-c*e**2*x**2+c*d**2)**(1/2),x)

[Out]

Integral(1/(sqrt(-c*(-d + e*x)*(d + e*x))*(d + e*x)**(5/2)), x)

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Giac [F]  time = 0., size = 0, normalized size = 0. \begin{align*} \mathit{sage}_{0} x \end{align*}

Veriﬁcation of antiderivative is not currently implemented for this CAS.

[In]

integrate(1/(e*x+d)^(5/2)/(-c*e^2*x^2+c*d^2)^(1/2),x, algorithm="giac")

[Out]

sage0*x