3.8.19 \(\int \frac {(c d^2-c e^2 x^2)^{3/2}}{(d+e x)^{13/2}} \, dx\)

Optimal. Leaf size=217 \[ -\frac {3 c^{3/2} \tanh ^{-1}\left (\frac {\sqrt {c d^2-c e^2 x^2}}{\sqrt {2} \sqrt {c} \sqrt {d} \sqrt {d+e x}}\right )}{256 \sqrt {2} d^{5/2} e}-\frac {3 c \sqrt {c d^2-c e^2 x^2}}{256 d^2 e (d+e x)^{3/2}}-\frac {c \sqrt {c d^2-c e^2 x^2}}{64 d e (d+e x)^{5/2}}+\frac {c \sqrt {c d^2-c e^2 x^2}}{8 e (d+e x)^{7/2}}-\frac {\left (c d^2-c e^2 x^2\right )^{3/2}}{4 e (d+e x)^{11/2}} \]

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Rubi [A]  time = 0.13, antiderivative size = 217, normalized size of antiderivative = 1.00, number of steps used = 6, number of rules used = 4, integrand size = 29, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.138, Rules used = {663, 673, 661, 208} \begin {gather*} -\frac {3 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 )}{256 \sqrt {2} d^{5/2} e}-\frac {3 c \sqrt {c d^2-c e^2 x^2}}{256 d^2 e (d+e x)^{3/2}}-\frac {c \sqrt {c d^2-c e^2 x^2}}{64 d e (d+e x)^{5/2}}+\frac {c \sqrt {c d^2-c e^2 x^2}}{8 e (d+e x)^{7/2}}-\frac {\left (c d^2-c e^2 x^2\right )^{3/2}}{4 e (d+e x)^{11/2}} \end {gather*}

Antiderivative was successfully verified.

[In]

Int[(c*d^2 - c*e^2*x^2)^(3/2)/(d + e*x)^(13/2),x]

[Out]

(c*Sqrt[c*d^2 - c*e^2*x^2])/(8*e*(d + e*x)^(7/2)) - (c*Sqrt[c*d^2 - c*e^2*x^2])/(64*d*e*(d + e*x)^(5/2)) - (3*
c*Sqrt[c*d^2 - c*e^2*x^2])/(256*d^2*e*(d + e*x)^(3/2)) - (c*d^2 - c*e^2*x^2)^(3/2)/(4*e*(d + e*x)^(11/2)) - (3
*c^(3/2)*ArcTanh[Sqrt[c*d^2 - c*e^2*x^2]/(Sqrt[2]*Sqrt[c]*Sqrt[d]*Sqrt[d + e*x])])/(256*Sqrt[2]*d^(5/2)*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 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]

Rubi steps

\begin {align*} \int \frac {\left (c d^2-c e^2 x^2\right )^{3/2}}{(d+e x)^{13/2}} \, dx &=-\frac {\left (c d^2-c e^2 x^2\right )^{3/2}}{4 e (d+e x)^{11/2}}-\frac {1}{8} (3 c) \int \frac {\sqrt {c d^2-c e^2 x^2}}{(d+e x)^{9/2}} \, dx\\ &=\frac {c \sqrt {c d^2-c e^2 x^2}}{8 e (d+e x)^{7/2}}-\frac {\left (c d^2-c e^2 x^2\right )^{3/2}}{4 e (d+e x)^{11/2}}+\frac {1}{16} c^2 \int \frac {1}{(d+e x)^{5/2} \sqrt {c d^2-c e^2 x^2}} \, dx\\ &=\frac {c \sqrt {c d^2-c e^2 x^2}}{8 e (d+e x)^{7/2}}-\frac {c \sqrt {c d^2-c e^2 x^2}}{64 d e (d+e x)^{5/2}}-\frac {\left (c d^2-c e^2 x^2\right )^{3/2}}{4 e (d+e x)^{11/2}}+\frac {\left (3 c^2\right ) \int \frac {1}{(d+e x)^{3/2} \sqrt {c d^2-c e^2 x^2}} \, dx}{128 d}\\ &=\frac {c \sqrt {c d^2-c e^2 x^2}}{8 e (d+e x)^{7/2}}-\frac {c \sqrt {c d^2-c e^2 x^2}}{64 d e (d+e x)^{5/2}}-\frac {3 c \sqrt {c d^2-c e^2 x^2}}{256 d^2 e (d+e x)^{3/2}}-\frac {\left (c d^2-c e^2 x^2\right )^{3/2}}{4 e (d+e x)^{11/2}}+\frac {\left (3 c^2\right ) \int \frac {1}{\sqrt {d+e x} \sqrt {c d^2-c e^2 x^2}} \, dx}{512 d^2}\\ &=\frac {c \sqrt {c d^2-c e^2 x^2}}{8 e (d+e x)^{7/2}}-\frac {c \sqrt {c d^2-c e^2 x^2}}{64 d e (d+e x)^{5/2}}-\frac {3 c \sqrt {c d^2-c e^2 x^2}}{256 d^2 e (d+e x)^{3/2}}-\frac {\left (c d^2-c e^2 x^2\right )^{3/2}}{4 e (d+e x)^{11/2}}+\frac {\left (3 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 )}{256 d^2}\\ &=\frac {c \sqrt {c d^2-c e^2 x^2}}{8 e (d+e x)^{7/2}}-\frac {c \sqrt {c d^2-c e^2 x^2}}{64 d e (d+e x)^{5/2}}-\frac {3 c \sqrt {c d^2-c e^2 x^2}}{256 d^2 e (d+e x)^{3/2}}-\frac {\left (c d^2-c e^2 x^2\right )^{3/2}}{4 e (d+e x)^{11/2}}-\frac {3 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 )}{256 \sqrt {2} d^{5/2} e}\\ \end {align*}

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Mathematica [A]  time = 0.26, size = 145, normalized size = 0.67 \begin {gather*} \frac {\left (c \left (d^2-e^2 x^2\right )\right )^{3/2} \left (-\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}}-\frac {2 \sqrt {d} \left (39 d^3-79 d^2 e x+13 d e^2 x^2+3 e^3 x^3\right )}{(d-e x) (d+e x)^{11/2}}\right )}{512 d^{5/2} e} \end {gather*}

Antiderivative was successfully verified.

[In]

Integrate[(c*d^2 - c*e^2*x^2)^(3/2)/(d + e*x)^(13/2),x]

[Out]

((c*(d^2 - e^2*x^2))^(3/2)*((-2*Sqrt[d]*(39*d^3 - 79*d^2*e*x + 13*d*e^2*x^2 + 3*e^3*x^3))/((d - e*x)*(d + e*x)
^(11/2)) - (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)))/(5
12*d^(5/2)*e)

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IntegrateAlgebraic [A]  time = 2.10, size = 161, normalized size = 0.74 \begin {gather*} \frac {3 c^{3/2} \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 )}{256 \sqrt {2} d^{5/2} e}-\frac {c \left (128 d^3-96 d^2 (d+e x)+4 d (d+e x)^2+3 (d+e x)^3\right ) \sqrt {2 c d (d+e x)-c (d+e x)^2}}{256 d^2 e (d+e x)^{9/2}} \end {gather*}

Antiderivative was successfully verified.

[In]

IntegrateAlgebraic[(c*d^2 - c*e^2*x^2)^(3/2)/(d + e*x)^(13/2),x]

[Out]

-1/256*(c*Sqrt[2*c*d*(d + e*x) - c*(d + e*x)^2]*(128*d^3 - 96*d^2*(d + e*x) + 4*d*(d + e*x)^2 + 3*(d + e*x)^3)
)/(d^2*e*(d + e*x)^(9/2)) + (3*c^(3/2)*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])])/(256*Sqrt[2]*d^(5/2)*e)

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fricas [A]  time = 0.44, size = 518, normalized size = 2.39 \begin {gather*} \left [\frac {3 \, \sqrt {\frac {1}{2}} {\left (c e^{5} x^{5} + 5 \, c d e^{4} x^{4} + 10 \, c d^{2} e^{3} x^{3} + 10 \, c d^{3} e^{2} x^{2} + 5 \, c d^{4} e x + c d^{5}\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^{3} x^{3} + 13 \, c d e^{2} x^{2} - 79 \, c d^{2} e x + 39 \, c d^{3}\right )} \sqrt {-c e^{2} x^{2} + c d^{2}} \sqrt {e x + d}}{512 \, {\left (d^{2} e^{6} x^{5} + 5 \, d^{3} e^{5} x^{4} + 10 \, d^{4} e^{4} x^{3} + 10 \, d^{5} e^{3} x^{2} + 5 \, d^{6} e^{2} x + d^{7} e\right )}}, -\frac {3 \, \sqrt {\frac {1}{2}} {\left (c e^{5} x^{5} + 5 \, c d e^{4} x^{4} + 10 \, c d^{2} e^{3} x^{3} + 10 \, c d^{3} e^{2} x^{2} + 5 \, c d^{4} e x + c d^{5}\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^{3} x^{3} + 13 \, c d e^{2} x^{2} - 79 \, c d^{2} e x + 39 \, c d^{3}\right )} \sqrt {-c e^{2} x^{2} + c d^{2}} \sqrt {e x + d}}{256 \, {\left (d^{2} e^{6} x^{5} + 5 \, d^{3} e^{5} x^{4} + 10 \, d^{4} e^{4} x^{3} + 10 \, d^{5} e^{3} x^{2} + 5 \, d^{6} e^{2} x + d^{7} e\right )}}\right ] \end {gather*}

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)^(13/2),x, algorithm="fricas")

[Out]

[1/512*(3*sqrt(1/2)*(c*e^5*x^5 + 5*c*d*e^4*x^4 + 10*c*d^2*e^3*x^3 + 10*c*d^3*e^2*x^2 + 5*c*d^4*e*x + c*d^5)*sq
rt(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^3*x^3 + 13*c*d*e^2*x^2 - 79*c*d^2*e*x + 39*c*d^3)*sqrt(-c*e^2*x^2 + c*
d^2)*sqrt(e*x + d))/(d^2*e^6*x^5 + 5*d^3*e^5*x^4 + 10*d^4*e^4*x^3 + 10*d^5*e^3*x^2 + 5*d^6*e^2*x + d^7*e), -1/
256*(3*sqrt(1/2)*(c*e^5*x^5 + 5*c*d*e^4*x^4 + 10*c*d^2*e^3*x^3 + 10*c*d^3*e^2*x^2 + 5*c*d^4*e*x + c*d^5)*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^3*x
^3 + 13*c*d*e^2*x^2 - 79*c*d^2*e*x + 39*c*d^3)*sqrt(-c*e^2*x^2 + c*d^2)*sqrt(e*x + d))/(d^2*e^6*x^5 + 5*d^3*e^
5*x^4 + 10*d^4*e^4*x^3 + 10*d^5*e^3*x^2 + 5*d^6*e^2*x + d^7*e)]

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giac [F(-1)]  time = 0.00, size = 0, normalized size = 0.00 \begin {gather*} \text {Timed out} \end {gather*}

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)^(13/2),x, algorithm="giac")

[Out]

Timed out

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maple [A]  time = 0.08, size = 325, normalized size = 1.50 \begin {gather*} -\frac {\sqrt {-\left (e^{2} x^{2}-d^{2}\right ) c}\, \left (3 \sqrt {2}\, c \,e^{4} x^{4} \arctanh \left (\frac {\sqrt {-\left (e x -d \right ) c}\, \sqrt {2}}{2 \sqrt {c d}}\right )+12 \sqrt {2}\, c d \,e^{3} x^{3} \arctanh \left (\frac {\sqrt {-\left (e x -d \right ) c}\, \sqrt {2}}{2 \sqrt {c d}}\right )+18 \sqrt {2}\, c \,d^{2} e^{2} x^{2} \arctanh \left (\frac {\sqrt {-\left (e x -d \right ) c}\, \sqrt {2}}{2 \sqrt {c d}}\right )+12 \sqrt {2}\, c \,d^{3} e x \arctanh \left (\frac {\sqrt {-\left (e x -d \right ) c}\, \sqrt {2}}{2 \sqrt {c d}}\right )+3 \sqrt {2}\, c \,d^{4} \arctanh \left (\frac {\sqrt {-\left (e x -d \right ) c}\, \sqrt {2}}{2 \sqrt {c d}}\right )+6 \sqrt {c d}\, \sqrt {-\left (e x -d \right ) c}\, e^{3} x^{3}+26 \sqrt {c d}\, \sqrt {-\left (e x -d \right ) c}\, d \,e^{2} x^{2}-158 \sqrt {c d}\, \sqrt {-\left (e x -d \right ) c}\, d^{2} e x +78 \sqrt {-\left (e x -d \right ) c}\, \sqrt {c d}\, d^{3}\right ) c}{512 \left (e x +d \right )^{\frac {9}{2}} \sqrt {-\left (e x -d \right ) c}\, \sqrt {c d}\, d^{2} e} \end {gather*}

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)^(13/2),x)

[Out]

-1/512*(-(e^2*x^2-d^2)*c)^(1/2)*c*(3*2^(1/2)*arctanh(1/2*(-(e*x-d)*c)^(1/2)*2^(1/2)/(c*d)^(1/2))*x^4*c*e^4+12*
2^(1/2)*arctanh(1/2*(-(e*x-d)*c)^(1/2)*2^(1/2)/(c*d)^(1/2))*x^3*c*d*e^3+18*2^(1/2)*arctanh(1/2*(-(e*x-d)*c)^(1
/2)*2^(1/2)/(c*d)^(1/2))*x^2*c*d^2*e^2+12*2^(1/2)*arctanh(1/2*(-(e*x-d)*c)^(1/2)*2^(1/2)/(c*d)^(1/2))*x*c*d^3*
e+6*x^3*e^3*(c*d)^(1/2)*(-(e*x-d)*c)^(1/2)+3*2^(1/2)*arctanh(1/2*(-(e*x-d)*c)^(1/2)*2^(1/2)/(c*d)^(1/2))*c*d^4
+26*x^2*d*e^2*(c*d)^(1/2)*(-(e*x-d)*c)^(1/2)-158*x*d^2*e*(c*d)^(1/2)*(-(e*x-d)*c)^(1/2)+78*(-(e*x-d)*c)^(1/2)*
(c*d)^(1/2)*d^3)/(e*x+d)^(9/2)/(-(e*x-d)*c)^(1/2)/e/d^2/(c*d)^(1/2)

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maxima [F]  time = 0.00, size = 0, normalized size = 0.00 \begin {gather*} \int \frac {{\left (-c e^{2} x^{2} + c d^{2}\right )}^{\frac {3}{2}}}{{\left (e x + d\right )}^{\frac {13}{2}}}\,{d x} \end {gather*}

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)^(13/2),x, algorithm="maxima")

[Out]

integrate((-c*e^2*x^2 + c*d^2)^(3/2)/(e*x + d)^(13/2), x)

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mupad [F]  time = 0.00, size = -1, normalized size = -0.00 \begin {gather*} \int \frac {{\left (c\,d^2-c\,e^2\,x^2\right )}^{3/2}}{{\left (d+e\,x\right )}^{13/2}} \,d x \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

int((c*d^2 - c*e^2*x^2)^(3/2)/(d + e*x)^(13/2),x)

[Out]

int((c*d^2 - c*e^2*x^2)^(3/2)/(d + e*x)^(13/2), x)

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sympy [F(-1)]  time = 0.00, size = 0, normalized size = 0.00 \begin {gather*} \text {Timed out} \end {gather*}

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)**(13/2),x)

[Out]

Timed out

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