3.8.33 \(\int \frac {1}{(d+e x)^{3/2} (c d^2-c e^2 x^2)^{3/2}} \, dx\)

Optimal. Leaf size=191 \[ -\frac {15 \tanh ^{-1}\left (\frac {\sqrt {c d^2-c e^2 x^2}}{\sqrt {2} \sqrt {c} \sqrt {d} \sqrt {d+e x}}\right )}{32 \sqrt {2} c^{3/2} d^{7/2} e}-\frac {5}{16 c d^2 e \sqrt {d+e x} \sqrt {c d^2-c e^2 x^2}}-\frac {1}{4 c d e (d+e x)^{3/2} \sqrt {c d^2-c e^2 x^2}}+\frac {15 \sqrt {d+e x}}{32 c d^3 e \sqrt {c d^2-c e^2 x^2}} \]

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Rubi [A]  time = 0.11, antiderivative size = 191, normalized size of antiderivative = 1.00, 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 = {673, 667, 661, 208} \begin {gather*} -\frac {15 \tanh ^{-1}\left (\frac {\sqrt {c d^2-c e^2 x^2}}{\sqrt {2} \sqrt {c} \sqrt {d} \sqrt {d+e x}}\right )}{32 \sqrt {2} c^{3/2} d^{7/2} e}+\frac {15 \sqrt {d+e x}}{32 c d^3 e \sqrt {c d^2-c e^2 x^2}}-\frac {5}{16 c d^2 e \sqrt {d+e x} \sqrt {c d^2-c e^2 x^2}}-\frac {1}{4 c d e (d+e x)^{3/2} \sqrt {c d^2-c e^2 x^2}} \end {gather*}

Antiderivative was successfully verified.

[In]

Int[1/((d + e*x)^(3/2)*(c*d^2 - c*e^2*x^2)^(3/2)),x]

[Out]

-1/(4*c*d*e*(d + e*x)^(3/2)*Sqrt[c*d^2 - c*e^2*x^2]) - 5/(16*c*d^2*e*Sqrt[d + e*x]*Sqrt[c*d^2 - c*e^2*x^2]) +
(15*Sqrt[d + e*x])/(32*c*d^3*e*Sqrt[c*d^2 - c*e^2*x^2]) - (15*ArcTanh[Sqrt[c*d^2 - c*e^2*x^2]/(Sqrt[2]*Sqrt[c]
*Sqrt[d]*Sqrt[d + e*x])])/(32*Sqrt[2]*c^(3/2)*d^(7/2)*e)

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]

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 667

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

Rubi steps

\begin {align*} \int \frac {1}{(d+e x)^{3/2} \left (c d^2-c e^2 x^2\right )^{3/2}} \, dx &=-\frac {1}{4 c d e (d+e x)^{3/2} \sqrt {c d^2-c e^2 x^2}}+\frac {5 \int \frac {1}{\sqrt {d+e x} \left (c d^2-c e^2 x^2\right )^{3/2}} \, dx}{8 d}\\ &=-\frac {1}{4 c d e (d+e x)^{3/2} \sqrt {c d^2-c e^2 x^2}}-\frac {5}{16 c d^2 e \sqrt {d+e x} \sqrt {c d^2-c e^2 x^2}}+\frac {15 \int \frac {\sqrt {d+e x}}{\left (c d^2-c e^2 x^2\right )^{3/2}} \, dx}{32 d^2}\\ &=-\frac {1}{4 c d e (d+e x)^{3/2} \sqrt {c d^2-c e^2 x^2}}-\frac {5}{16 c d^2 e \sqrt {d+e x} \sqrt {c d^2-c e^2 x^2}}+\frac {15 \sqrt {d+e x}}{32 c d^3 e \sqrt {c d^2-c e^2 x^2}}+\frac {15 \int \frac {1}{\sqrt {d+e x} \sqrt {c d^2-c e^2 x^2}} \, dx}{64 c d^3}\\ &=-\frac {1}{4 c d e (d+e x)^{3/2} \sqrt {c d^2-c e^2 x^2}}-\frac {5}{16 c d^2 e \sqrt {d+e x} \sqrt {c d^2-c e^2 x^2}}+\frac {15 \sqrt {d+e x}}{32 c d^3 e \sqrt {c d^2-c e^2 x^2}}+\frac {(15 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 )}{32 c d^3}\\ &=-\frac {1}{4 c d e (d+e x)^{3/2} \sqrt {c d^2-c e^2 x^2}}-\frac {5}{16 c d^2 e \sqrt {d+e x} \sqrt {c d^2-c e^2 x^2}}+\frac {15 \sqrt {d+e x}}{32 c d^3 e \sqrt {c d^2-c e^2 x^2}}-\frac {15 \tanh ^{-1}\left (\frac {\sqrt {c d^2-c e^2 x^2}}{\sqrt {2} \sqrt {c} \sqrt {d} \sqrt {d+e x}}\right )}{32 \sqrt {2} c^{3/2} d^{7/2} e}\\ \end {align*}

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Mathematica [A]  time = 0.11, size = 143, normalized size = 0.75 \begin {gather*} \frac {2 \sqrt {d} \sqrt {d+e x} \left (-3 d^2+20 d e x+15 e^2 x^2\right )-15 \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 )}{64 c d^{7/2} e (d+e x)^2 \sqrt {c \left (d^2-e^2 x^2\right )}} \end {gather*}

Antiderivative was successfully verified.

[In]

Integrate[1/((d + e*x)^(3/2)*(c*d^2 - c*e^2*x^2)^(3/2)),x]

[Out]

(2*Sqrt[d]*Sqrt[d + e*x]*(-3*d^2 + 20*d*e*x + 15*e^2*x^2) - 15*Sqrt[2]*(d + e*x)^2*Sqrt[d^2 - e^2*x^2]*ArcTanh
[Sqrt[d^2 - e^2*x^2]/(Sqrt[2]*Sqrt[d]*Sqrt[d + e*x])])/(64*c*d^(7/2)*e*(d + e*x)^2*Sqrt[c*(d^2 - e^2*x^2)])

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IntegrateAlgebraic [A]  time = 1.93, size = 159, normalized size = 0.83 \begin {gather*} \frac {15 \tanh ^{-1}\left (\frac {\sqrt {2} \sqrt {d} \sqrt {2 c d (d+e x)-c (d+e x)^2}}{\sqrt {c} (e x-d) \sqrt {d+e x}}\right )}{32 \sqrt {2} c^{3/2} d^{7/2} e}+\frac {\left (-8 d^2-10 d (d+e x)+15 (d+e x)^2\right ) \sqrt {2 c d (d+e x)-c (d+e x)^2}}{32 c^2 d^3 e (d-e x) (d+e x)^{5/2}} \end {gather*}

Antiderivative was successfully verified.

[In]

IntegrateAlgebraic[1/((d + e*x)^(3/2)*(c*d^2 - c*e^2*x^2)^(3/2)),x]

[Out]

((-8*d^2 - 10*d*(d + e*x) + 15*(d + e*x)^2)*Sqrt[2*c*d*(d + e*x) - c*(d + e*x)^2])/(32*c^2*d^3*e*(d - e*x)*(d
+ e*x)^(5/2)) + (15*ArcTanh[(Sqrt[2]*Sqrt[d]*Sqrt[2*c*d*(d + e*x) - c*(d + e*x)^2])/(Sqrt[c]*(-d + e*x)*Sqrt[d
+ e*x])])/(32*Sqrt[2]*c^(3/2)*d^(7/2)*e)

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fricas [A]  time = 0.41, size = 412, normalized size = 2.16 \begin {gather*} \left [\frac {15 \, \sqrt {2} {\left (e^{4} x^{4} + 2 \, d e^{3} x^{3} - 2 \, d^{3} e x - d^{4}\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 (15 \, d e^{2} x^{2} + 20 \, d^{2} e x - 3 \, d^{3}\right )} \sqrt {e x + d}}{128 \, {\left (c^{2} d^{4} e^{5} x^{4} + 2 \, c^{2} d^{5} e^{4} x^{3} - 2 \, c^{2} d^{7} e^{2} x - c^{2} d^{8} e\right )}}, -\frac {15 \, \sqrt {2} {\left (e^{4} x^{4} + 2 \, d e^{3} x^{3} - 2 \, d^{3} e x - d^{4}\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 (15 \, d e^{2} x^{2} + 20 \, d^{2} e x - 3 \, d^{3}\right )} \sqrt {e x + d}}{64 \, {\left (c^{2} d^{4} e^{5} x^{4} + 2 \, c^{2} d^{5} e^{4} x^{3} - 2 \, c^{2} d^{7} e^{2} x - c^{2} d^{8} e\right )}}\right ] \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate(1/(e*x+d)^(3/2)/(-c*e^2*x^2+c*d^2)^(3/2),x, algorithm="fricas")

[Out]

[1/128*(15*sqrt(2)*(e^4*x^4 + 2*d*e^3*x^3 - 2*d^3*e*x - d^4)*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)*(15*d*e^2*x^2 + 20*d^2*e*x - 3*d^3)*sqrt(e*x + d))/(c^2*d^4*e^5*x^4 + 2*c^2*d^5*e^4*x^3 - 2*c^2*d^7*e^2
*x - c^2*d^8*e), -1/64*(15*sqrt(2)*(e^4*x^4 + 2*d*e^3*x^3 - 2*d^3*e*x - d^4)*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)*(15*d*e^2*x^2 + 2
0*d^2*e*x - 3*d^3)*sqrt(e*x + d))/(c^2*d^4*e^5*x^4 + 2*c^2*d^5*e^4*x^3 - 2*c^2*d^7*e^2*x - c^2*d^8*e)]

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giac [F]  time = 0.00, size = 0, normalized size = 0.00 \begin {gather*} \mathit {sage}_{0} x \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate(1/(e*x+d)^(3/2)/(-c*e^2*x^2+c*d^2)^(3/2),x, algorithm="giac")

[Out]

sage0*x

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maple [A]  time = 0.07, size = 217, normalized size = 1.14 \begin {gather*} \frac {\sqrt {-\left (e^{2} x^{2}-d^{2}\right ) c}\, \left (15 \sqrt {-\left (e x -d \right ) c}\, \sqrt {2}\, e^{2} x^{2} \arctanh \left (\frac {\sqrt {-\left (e x -d \right ) c}\, \sqrt {2}}{2 \sqrt {c d}}\right )+30 \sqrt {-\left (e x -d \right ) c}\, \sqrt {2}\, d e x \arctanh \left (\frac {\sqrt {-\left (e x -d \right ) c}\, \sqrt {2}}{2 \sqrt {c d}}\right )-30 \sqrt {c d}\, e^{2} x^{2}+15 \sqrt {-\left (e x -d \right ) c}\, \sqrt {2}\, d^{2} \arctanh \left (\frac {\sqrt {-\left (e x -d \right ) c}\, \sqrt {2}}{2 \sqrt {c d}}\right )-40 \sqrt {c d}\, d e x +6 \sqrt {c d}\, d^{2}\right )}{64 \left (e x +d \right )^{\frac {5}{2}} \left (e x -d \right ) \sqrt {c d}\, c^{2} d^{3} e} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

int(1/(e*x+d)^(3/2)/(-c*e^2*x^2+c*d^2)^(3/2),x)

[Out]

1/64/(e*x+d)^(5/2)*(-(e^2*x^2-d^2)*c)^(1/2)/c^2*(15*(-(e*x-d)*c)^(1/2)*2^(1/2)*arctanh(1/2*(-(e*x-d)*c)^(1/2)*
2^(1/2)/(c*d)^(1/2))*x^2*e^2+30*(-(e*x-d)*c)^(1/2)*2^(1/2)*arctanh(1/2*(-(e*x-d)*c)^(1/2)*2^(1/2)/(c*d)^(1/2))
*x*d*e+15*(-(e*x-d)*c)^(1/2)*2^(1/2)*arctanh(1/2*(-(e*x-d)*c)^(1/2)*2^(1/2)/(c*d)^(1/2))*d^2-30*(c*d)^(1/2)*x^
2*e^2-40*(c*d)^(1/2)*x*d*e+6*(c*d)^(1/2)*d^2)/(e*x-d)/e/d^3/(c*d)^(1/2)

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maxima [F]  time = 0.00, size = 0, normalized size = 0.00 \begin {gather*} \int \frac {1}{{\left (-c e^{2} x^{2} + c d^{2}\right )}^{\frac {3}{2}} {\left (e x + d\right )}^{\frac {3}{2}}}\,{d x} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate(1/(e*x+d)^(3/2)/(-c*e^2*x^2+c*d^2)^(3/2),x, algorithm="maxima")

[Out]

integrate(1/((-c*e^2*x^2 + c*d^2)^(3/2)*(e*x + d)^(3/2)), x)

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mupad [F]  time = 0.00, size = -1, normalized size = -0.01 \begin {gather*} \int \frac {1}{{\left (c\,d^2-c\,e^2\,x^2\right )}^{3/2}\,{\left (d+e\,x\right )}^{3/2}} \,d x \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

int(1/((c*d^2 - c*e^2*x^2)^(3/2)*(d + e*x)^(3/2)),x)

[Out]

int(1/((c*d^2 - c*e^2*x^2)^(3/2)*(d + e*x)^(3/2)), x)

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sympy [F]  time = 0.00, size = 0, normalized size = 0.00 \begin {gather*} \int \frac {1}{\left (- c \left (- d + e x\right ) \left (d + e x\right )\right )^{\frac {3}{2}} \left (d + e x\right )^{\frac {3}{2}}}\, dx \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate(1/(e*x+d)**(3/2)/(-c*e**2*x**2+c*d**2)**(3/2),x)

[Out]

Integral(1/((-c*(-d + e*x)*(d + e*x))**(3/2)*(d + e*x)**(3/2)), x)

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