3.877 \(\int \frac{(c d^2-c e^2 x^2)^{3/2}}{(d+e x)^{11/2}} \, dx\)

Optimal. Leaf size=178 \[ -\frac{c^{3/2} \tanh ^{-1}\left (\frac{\sqrt{c d^2-c e^2 x^2}}{\sqrt{2} \sqrt{c} \sqrt{d} \sqrt{d+e x}}\right )}{16 \sqrt{2} d^{3/2} e}-\frac{c \sqrt{c d^2-c e^2 x^2}}{16 d e (d+e x)^{3/2}}+\frac{c \sqrt{c d^2-c e^2 x^2}}{4 e (d+e x)^{5/2}}-\frac{\left (c d^2-c e^2 x^2\right )^{3/2}}{3 e (d+e x)^{9/2}} \]

[Out]

(c*Sqrt[c*d^2 - c*e^2*x^2])/(4*e*(d + e*x)^(5/2)) - (c*Sqrt[c*d^2 - c*e^2*x^2])/(16*d*e*(d + e*x)^(3/2)) - (c*
d^2 - c*e^2*x^2)^(3/2)/(3*e*(d + e*x)^(9/2)) - (c^(3/2)*ArcTanh[Sqrt[c*d^2 - c*e^2*x^2]/(Sqrt[2]*Sqrt[c]*Sqrt[
d]*Sqrt[d + e*x])])/(16*Sqrt[2]*d^(3/2)*e)

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Rubi [A]  time = 0.0967206, antiderivative size = 178, normalized size of antiderivative = 1., number of steps used = 5, number of rules used = 4, integrand size = 29, \(\frac{\text{number of rules}}{\text{integrand size}}\) = 0.138, Rules used = {663, 673, 661, 208} \[ -\frac{c^{3/2} \tanh ^{-1}\left (\frac{\sqrt{c d^2-c e^2 x^2}}{\sqrt{2} \sqrt{c} \sqrt{d} \sqrt{d+e x}}\right )}{16 \sqrt{2} d^{3/2} e}-\frac{c \sqrt{c d^2-c e^2 x^2}}{16 d e (d+e x)^{3/2}}+\frac{c \sqrt{c d^2-c e^2 x^2}}{4 e (d+e x)^{5/2}}-\frac{\left (c d^2-c e^2 x^2\right )^{3/2}}{3 e (d+e x)^{9/2}} \]

Antiderivative was successfully verified.

[In]

Int[(c*d^2 - c*e^2*x^2)^(3/2)/(d + e*x)^(11/2),x]

[Out]

(c*Sqrt[c*d^2 - c*e^2*x^2])/(4*e*(d + e*x)^(5/2)) - (c*Sqrt[c*d^2 - c*e^2*x^2])/(16*d*e*(d + e*x)^(3/2)) - (c*
d^2 - c*e^2*x^2)^(3/2)/(3*e*(d + e*x)^(9/2)) - (c^(3/2)*ArcTanh[Sqrt[c*d^2 - c*e^2*x^2]/(Sqrt[2]*Sqrt[c]*Sqrt[
d]*Sqrt[d + e*x])])/(16*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{\left (c d^2-c e^2 x^2\right )^{3/2}}{(d+e x)^{11/2}} \, dx &=-\frac{\left (c d^2-c e^2 x^2\right )^{3/2}}{3 e (d+e x)^{9/2}}-\frac{1}{2} c \int \frac{\sqrt{c d^2-c e^2 x^2}}{(d+e x)^{7/2}} \, dx\\ &=\frac{c \sqrt{c d^2-c e^2 x^2}}{4 e (d+e x)^{5/2}}-\frac{\left (c d^2-c e^2 x^2\right )^{3/2}}{3 e (d+e x)^{9/2}}+\frac{1}{8} c^2 \int \frac{1}{(d+e x)^{3/2} \sqrt{c d^2-c e^2 x^2}} \, dx\\ &=\frac{c \sqrt{c d^2-c e^2 x^2}}{4 e (d+e x)^{5/2}}-\frac{c \sqrt{c d^2-c e^2 x^2}}{16 d e (d+e x)^{3/2}}-\frac{\left (c d^2-c e^2 x^2\right )^{3/2}}{3 e (d+e x)^{9/2}}+\frac{c^2 \int \frac{1}{\sqrt{d+e x} \sqrt{c d^2-c e^2 x^2}} \, dx}{32 d}\\ &=\frac{c \sqrt{c d^2-c e^2 x^2}}{4 e (d+e x)^{5/2}}-\frac{c \sqrt{c d^2-c e^2 x^2}}{16 d e (d+e x)^{3/2}}-\frac{\left (c d^2-c e^2 x^2\right )^{3/2}}{3 e (d+e x)^{9/2}}+\frac{\left (c^2 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 )}{16 d}\\ &=\frac{c \sqrt{c d^2-c e^2 x^2}}{4 e (d+e x)^{5/2}}-\frac{c \sqrt{c d^2-c e^2 x^2}}{16 d e (d+e x)^{3/2}}-\frac{\left (c d^2-c e^2 x^2\right )^{3/2}}{3 e (d+e x)^{9/2}}-\frac{c^{3/2} \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} d^{3/2} e}\\ \end{align*}

Mathematica [A]  time = 0.247028, size = 134, normalized size = 0.75 \[ \frac{\left (c \left (d^2-e^2 x^2\right )\right )^{3/2} \left (-\frac{2 \sqrt{d} \left (7 d^2-22 d e x+3 e^2 x^2\right )}{(d-e x) (d+e x)^{9/2}}-\frac{3 \sqrt{2} \tanh ^{-1}\left (\frac{\sqrt{d^2-e^2 x^2}}{\sqrt{2} \sqrt{d} \sqrt{d+e x}}\right )}{\left (d^2-e^2 x^2\right )^{3/2}}\right )}{96 d^{3/2} e} \]

Antiderivative was successfully verified.

[In]

Integrate[(c*d^2 - c*e^2*x^2)^(3/2)/(d + e*x)^(11/2),x]

[Out]

((c*(d^2 - e^2*x^2))^(3/2)*((-2*Sqrt[d]*(7*d^2 - 22*d*e*x + 3*e^2*x^2))/((d - e*x)*(d + e*x)^(9/2)) - (3*Sqrt[
2]*ArcTanh[Sqrt[d^2 - e^2*x^2]/(Sqrt[2]*Sqrt[d]*Sqrt[d + e*x])])/(d^2 - e^2*x^2)^(3/2)))/(96*d^(3/2)*e)

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Maple [A]  time = 0.174, size = 259, normalized size = 1.5 \begin{align*} -{\frac{c}{96\,de}\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}^{3}c{e}^{3}+9\,\sqrt{2}{\it Artanh} \left ( 1/2\,{\frac{\sqrt{- \left ( ex-d \right ) c}\sqrt{2}}{\sqrt{cd}}} \right ){x}^{2}cd{e}^{2}+9\,\sqrt{2}{\it Artanh} \left ( 1/2\,{\frac{\sqrt{- \left ( ex-d \right ) c}\sqrt{2}}{\sqrt{cd}}} \right ) xc{d}^{2}e+3\,\sqrt{2}{\it Artanh} \left ( 1/2\,{\frac{\sqrt{- \left ( ex-d \right ) c}\sqrt{2}}{\sqrt{cd}}} \right ) c{d}^{3}+6\,{x}^{2}{e}^{2}\sqrt{- \left ( ex-d \right ) c}\sqrt{cd}-44\,xde\sqrt{- \left ( ex-d \right ) c}\sqrt{cd}+14\,\sqrt{- \left ( ex-d \right ) c}\sqrt{cd}{d}^{2} \right ) \left ( ex+d \right ) ^{-{\frac{7}{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)^(11/2),x)

[Out]

-1/96*(-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^3*c*e^3+9*2^
(1/2)*arctanh(1/2*(-(e*x-d)*c)^(1/2)*2^(1/2)/(c*d)^(1/2))*x^2*c*d*e^2+9*2^(1/2)*arctanh(1/2*(-(e*x-d)*c)^(1/2)
*2^(1/2)/(c*d)^(1/2))*x*c*d^2*e+3*2^(1/2)*arctanh(1/2*(-(e*x-d)*c)^(1/2)*2^(1/2)/(c*d)^(1/2))*c*d^3+6*x^2*e^2*
(-(e*x-d)*c)^(1/2)*(c*d)^(1/2)-44*x*d*e*(-(e*x-d)*c)^(1/2)*(c*d)^(1/2)+14*(-(e*x-d)*c)^(1/2)*(c*d)^(1/2)*d^2)/
(e*x+d)^(7/2)/(-(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{{\left (-c e^{2} x^{2} + c d^{2}\right )}^{\frac{3}{2}}}{{\left (e x + d\right )}^{\frac{11}{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)^(11/2),x, algorithm="maxima")

[Out]

integrate((-c*e^2*x^2 + c*d^2)^(3/2)/(e*x + d)^(11/2), x)

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Fricas [A]  time = 2.25512, size = 965, normalized size = 5.42 \begin{align*} \left [\frac{3 \, \sqrt{\frac{1}{2}}{\left (c e^{4} x^{4} + 4 \, c d e^{3} x^{3} + 6 \, c d^{2} e^{2} x^{2} + 4 \, c d^{3} e x + c d^{4}\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 \,{\left (3 \, c e^{2} x^{2} - 22 \, c d e x + 7 \, c d^{2}\right )} \sqrt{-c e^{2} x^{2} + c d^{2}} \sqrt{e x + d}}{96 \,{\left (d e^{5} x^{4} + 4 \, d^{2} e^{4} x^{3} + 6 \, d^{3} e^{3} x^{2} + 4 \, d^{4} e^{2} x + d^{5} e\right )}}, -\frac{3 \, \sqrt{\frac{1}{2}}{\left (c e^{4} x^{4} + 4 \, c d e^{3} x^{3} + 6 \, c d^{2} e^{2} x^{2} + 4 \, c d^{3} e x + c d^{4}\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 ) +{\left (3 \, c e^{2} x^{2} - 22 \, c d e x + 7 \, c d^{2}\right )} \sqrt{-c e^{2} x^{2} + c d^{2}} \sqrt{e x + d}}{48 \,{\left (d e^{5} x^{4} + 4 \, d^{2} e^{4} x^{3} + 6 \, d^{3} e^{3} x^{2} + 4 \, d^{4} e^{2} x + d^{5} 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)^(3/2)/(e*x+d)^(11/2),x, algorithm="fricas")

[Out]

[1/96*(3*sqrt(1/2)*(c*e^4*x^4 + 4*c*d*e^3*x^3 + 6*c*d^2*e^2*x^2 + 4*c*d^3*e*x + c*d^4)*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*(3*c*e^2*x^2 - 22*c*d*e*x + 7*c*d^2)*sqrt(-c*e^2*x^2 + c*d^2)*sqrt(e*x + d))/(d*e^5*x^4 + 4*d^2*e^
4*x^3 + 6*d^3*e^3*x^2 + 4*d^4*e^2*x + d^5*e), -1/48*(3*sqrt(1/2)*(c*e^4*x^4 + 4*c*d*e^3*x^3 + 6*c*d^2*e^2*x^2
+ 4*c*d^3*e*x + c*d^4)*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)) + (3*c*e^2*x^2 - 22*c*d*e*x + 7*c*d^2)*sqrt(-c*e^2*x^2 + c*d^2)*sqrt(e*x + d))/(d*e^5*x^4 + 4*
d^2*e^4*x^3 + 6*d^3*e^3*x^2 + 4*d^4*e^2*x + d^5*e)]

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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((-c*e**2*x**2+c*d**2)**(3/2)/(e*x+d)**(11/2),x)

[Out]

Timed out

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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{11}{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)^(11/2),x, algorithm="giac")

[Out]

integrate((-c*e^2*x^2 + c*d^2)^(3/2)/(e*x + d)^(11/2), x)