3.9.77 \(\int \frac {(c d^2-c e^2 x^2)^{3/2}}{(d+e x)^{11/2}} \, dx\) [877]

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

[Out]

-1/3*(-c*e^2*x^2+c*d^2)^(3/2)/e/(e*x+d)^(9/2)-1/32*c^(3/2)*arctanh(1/2*(-c*e^2*x^2+c*d^2)^(1/2)*2^(1/2)/c^(1/2
)/d^(1/2)/(e*x+d)^(1/2))/d^(3/2)/e*2^(1/2)+1/4*c*(-c*e^2*x^2+c*d^2)^(1/2)/e/(e*x+d)^(5/2)-1/16*c*(-c*e^2*x^2+c
*d^2)^(1/2)/d/e/(e*x+d)^(3/2)

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Rubi [A]
time = 0.06, antiderivative size = 178, 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 = {677, 687, 675, 214} \begin {gather*} -\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}} \end {gather*}

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 214

Int[((a_) + (b_.)*(x_)^2)^(-1), x_Symbol] :> Simp[(Rt[-a/b, 2]/a)*ArcTanh[x/Rt[-a/b, 2]], x] /; FreeQ[{a, b},
x] && NegQ[a/b]

Rule 675

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 677

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 687

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)^{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 ) \text {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*}

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Mathematica [A]
time = 0.47, size = 151, normalized size = 0.85 \begin {gather*} -\frac {\left (c \left (d^2-e^2 x^2\right )\right )^{3/2} \left (2 \sqrt {d} \sqrt {d^2-e^2 x^2} \left (7 d^2-22 d e x+3 e^2 x^2\right )+3 \sqrt {2} (d+e x)^{7/2} \tanh ^{-1}\left (\frac {\sqrt {2} \sqrt {d} \sqrt {d+e x}}{\sqrt {d^2-e^2 x^2}}\right )\right )}{96 d^{3/2} e (d+e x)^{7/2} \left (d^2-e^2 x^2\right )^{3/2}} \end {gather*}

Antiderivative was successfully verified.

[In]

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

[Out]

-1/96*((c*(d^2 - e^2*x^2))^(3/2)*(2*Sqrt[d]*Sqrt[d^2 - e^2*x^2]*(7*d^2 - 22*d*e*x + 3*e^2*x^2) + 3*Sqrt[2]*(d
+ e*x)^(7/2)*ArcTanh[(Sqrt[2]*Sqrt[d]*Sqrt[d + e*x])/Sqrt[d^2 - e^2*x^2]]))/(d^(3/2)*e*(d + e*x)^(7/2)*(d^2 -
e^2*x^2)^(3/2))

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Maple [A]
time = 0.50, size = 241, normalized size = 1.35

method	result	size
default	\(-\frac {\sqrt {c \left (-e^{2} x^{2}+d^{2}\right )}\, c \left (3 \sqrt {2}\, \arctanh \left (\frac {\sqrt {c \left (-e x +d \right )}\, \sqrt {2}}{2 \sqrt {c d}}\right ) c \,e^{3} x^{3}+9 \sqrt {2}\, \arctanh \left (\frac {\sqrt {c \left (-e x +d \right )}\, \sqrt {2}}{2 \sqrt {c d}}\right ) c d \,e^{2} x^{2}+9 \sqrt {2}\, \arctanh \left (\frac {\sqrt {c \left (-e x +d \right )}\, \sqrt {2}}{2 \sqrt {c d}}\right ) c \,d^{2} e x +3 \sqrt {2}\, \arctanh \left (\frac {\sqrt {c \left (-e x +d \right )}\, \sqrt {2}}{2 \sqrt {c d}}\right ) c \,d^{3}+6 e^{2} x^{2} \sqrt {c \left (-e x +d \right )}\, \sqrt {c d}-44 d e x \sqrt {c \left (-e x +d \right )}\, \sqrt {c d}+14 \sqrt {c \left (-e x +d \right )}\, \sqrt {c d}\, d^{2}\right )}{96 \left (e x +d \right )^{\frac {7}{2}} \sqrt {c \left (-e x +d \right )}\, e d \sqrt {c d}}\)	\(241\)

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,method=_RETURNVERBOSE)

[Out]

-1/96*(c*(-e^2*x^2+d^2))^(1/2)*c*(3*2^(1/2)*arctanh(1/2*(c*(-e*x+d))^(1/2)*2^(1/2)/(c*d)^(1/2))*c*e^3*x^3+9*2^
(1/2)*arctanh(1/2*(c*(-e*x+d))^(1/2)*2^(1/2)/(c*d)^(1/2))*c*d*e^2*x^2+9*2^(1/2)*arctanh(1/2*(c*(-e*x+d))^(1/2)
*2^(1/2)/(c*d)^(1/2))*c*d^2*e*x+3*2^(1/2)*arctanh(1/2*(c*(-e*x+d))^(1/2)*2^(1/2)/(c*d)^(1/2))*c*d^3+6*e^2*x^2*
(c*(-e*x+d))^(1/2)*(c*d)^(1/2)-44*d*e*x*(c*(-e*x+d))^(1/2)*(c*d)^(1/2)+14*(c*(-e*x+d))^(1/2)*(c*d)^(1/2)*d^2)/
(e*x+d)^(7/2)/(c*(-e*x+d))^(1/2)/e/d/(c*d)^(1/2)

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Maxima [F]
time = 0.00, size = 0, normalized size = 0.00 \begin {gather*} \text {Failed to integrate} \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)^(11/2),x, algorithm="maxima")

[Out]

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

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Fricas [A]
time = 2.50, size = 433, normalized size = 2.43 \begin {gather*} \left [\frac {3 \, \sqrt {\frac {1}{2}} {\left (c x^{4} e^{4} + 4 \, c d x^{3} e^{3} + 6 \, c d^{2} x^{2} e^{2} + 4 \, c d^{3} x e + c d^{4}\right )} \sqrt {\frac {c}{d}} \log \left (-\frac {c x^{2} e^{2} - 2 \, c d x e - 3 \, c d^{2} + 4 \, \sqrt {\frac {1}{2}} \sqrt {-c x^{2} e^{2} + c d^{2}} \sqrt {x e + d} d \sqrt {\frac {c}{d}}}{x^{2} e^{2} + 2 \, d x e + d^{2}}\right ) - 2 \, {\left (3 \, c x^{2} e^{2} - 22 \, c d x e + 7 \, c d^{2}\right )} \sqrt {-c x^{2} e^{2} + c d^{2}} \sqrt {x e + d}}{96 \, {\left (d x^{4} e^{5} + 4 \, d^{2} x^{3} e^{4} + 6 \, d^{3} x^{2} e^{3} + 4 \, d^{4} x e^{2} + d^{5} e\right )}}, -\frac {3 \, \sqrt {\frac {1}{2}} {\left (c x^{4} e^{4} + 4 \, c d x^{3} e^{3} + 6 \, c d^{2} x^{2} e^{2} + 4 \, c d^{3} x e + c d^{4}\right )} \sqrt {-\frac {c}{d}} \arctan \left (\frac {2 \, \sqrt {\frac {1}{2}} \sqrt {-c x^{2} e^{2} + c d^{2}} \sqrt {x e + d} d \sqrt {-\frac {c}{d}}}{c x^{2} e^{2} - c d^{2}}\right ) + {\left (3 \, c x^{2} e^{2} - 22 \, c d x e + 7 \, c d^{2}\right )} \sqrt {-c x^{2} e^{2} + c d^{2}} \sqrt {x e + d}}{48 \, {\left (d x^{4} e^{5} + 4 \, d^{2} x^{3} e^{4} + 6 \, d^{3} x^{2} e^{3} + 4 \, d^{4} x e^{2} + d^{5} 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)^(11/2),x, algorithm="fricas")

[Out]

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

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Sympy [F]
time = 0.00, size = 0, normalized size = 0.00 \begin {gather*} \int \frac {\left (- c \left (- d + e x\right ) \left (d + e x\right )\right )^{\frac {3}{2}}}{\left (d + e x\right )^{\frac {11}{2}}}\, dx \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)**(11/2),x)

[Out]

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

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Giac [A]
time = 1.39, size = 149, normalized size = 0.84 \begin {gather*} \frac {1}{96} \, {\left (\frac {3 \, \sqrt {2} c^{2} \arctan \left (\frac {\sqrt {2} \sqrt {-{\left (x e + d\right )} c + 2 \, c d}}{2 \, \sqrt {-c d}}\right )}{\sqrt {-c d} d} + \frac {2 \, {\left (12 \, \sqrt {-{\left (x e + d\right )} c + 2 \, c d} c^{4} d^{2} - 16 \, {\left (-{\left (x e + d\right )} c + 2 \, c d\right )}^{\frac {3}{2}} c^{3} d - 3 \, {\left ({\left (x e + d\right )} c - 2 \, c d\right )}^{2} \sqrt {-{\left (x e + d\right )} c + 2 \, c d} c^{2}\right )}}{{\left (x e + d\right )}^{3} c^{3} d}\right )} e^{\left (-1\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)^(11/2),x, algorithm="giac")

[Out]

1/96*(3*sqrt(2)*c^2*arctan(1/2*sqrt(2)*sqrt(-(x*e + d)*c + 2*c*d)/sqrt(-c*d))/(sqrt(-c*d)*d) + 2*(12*sqrt(-(x*
e + d)*c + 2*c*d)*c^4*d^2 - 16*(-(x*e + d)*c + 2*c*d)^(3/2)*c^3*d - 3*((x*e + d)*c - 2*c*d)^2*sqrt(-(x*e + d)*
c + 2*c*d)*c^2)/((x*e + d)^3*c^3*d))*e^(-1)

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Mupad [F]
time = 0.00, size = -1, normalized size = -0.01 \begin {gather*} \int \frac {{\left (c\,d^2-c\,e^2\,x^2\right )}^{3/2}}{{\left (d+e\,x\right )}^{11/2}} \,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)^(11/2),x)

[Out]

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

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