3.8.16 \(\int \frac {(c d^2-c e^2 x^2)^{3/2}}{(d+e x)^{7/2}} \, dx\)

Optimal. Leaf size=133 \[ \frac {3 \sqrt {2} c^{3/2} \sqrt {d} \tanh ^{-1}\left (\frac {\sqrt {c d^2-c e^2 x^2}}{\sqrt {2} \sqrt {c} \sqrt {d} \sqrt {d+e x}}\right )}{e}-\frac {3 c \sqrt {c d^2-c e^2 x^2}}{e \sqrt {d+e x}}-\frac {\left (c d^2-c e^2 x^2\right )^{3/2}}{e (d+e x)^{5/2}} \]

________________________________________________________________________________________

Rubi [A]  time = 0.08, antiderivative size = 133, normalized size of antiderivative = 1.00, number of steps used = 4, number of rules used = 4, integrand size = 29, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.138, Rules used = {663, 665, 661, 208} \begin {gather*} \frac {3 \sqrt {2} c^{3/2} \sqrt {d} \tanh ^{-1}\left (\frac {\sqrt {c d^2-c e^2 x^2}}{\sqrt {2} \sqrt {c} \sqrt {d} \sqrt {d+e x}}\right )}{e}-\frac {3 c \sqrt {c d^2-c e^2 x^2}}{e \sqrt {d+e x}}-\frac {\left (c d^2-c e^2 x^2\right )^{3/2}}{e (d+e x)^{5/2}} \end {gather*}

Antiderivative was successfully verified.

[In]

Int[(c*d^2 - c*e^2*x^2)^(3/2)/(d + e*x)^(7/2),x]

[Out]

(-3*c*Sqrt[c*d^2 - c*e^2*x^2])/(e*Sqrt[d + e*x]) - (c*d^2 - c*e^2*x^2)^(3/2)/(e*(d + e*x)^(5/2)) + (3*Sqrt[2]*
c^(3/2)*Sqrt[d]*ArcTanh[Sqrt[c*d^2 - c*e^2*x^2]/(Sqrt[2]*Sqrt[c]*Sqrt[d]*Sqrt[d + e*x])])/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 665

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 + 2*p + 1)), x] - Dist[(2*c*d*p)/(e^2*(m + 2*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] && GtQ[p, 0] && (LeQ[-2, m, 0] || EqQ[m + p + 1, 0]) && NeQ[
m + 2*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)^{7/2}} \, dx &=-\frac {\left (c d^2-c e^2 x^2\right )^{3/2}}{e (d+e x)^{5/2}}-\frac {1}{2} (3 c) \int \frac {\sqrt {c d^2-c e^2 x^2}}{(d+e x)^{3/2}} \, dx\\ &=-\frac {3 c \sqrt {c d^2-c e^2 x^2}}{e \sqrt {d+e x}}-\frac {\left (c d^2-c e^2 x^2\right )^{3/2}}{e (d+e x)^{5/2}}-\left (3 c^2 d\right ) \int \frac {1}{\sqrt {d+e x} \sqrt {c d^2-c e^2 x^2}} \, dx\\ &=-\frac {3 c \sqrt {c d^2-c e^2 x^2}}{e \sqrt {d+e x}}-\frac {\left (c d^2-c e^2 x^2\right )^{3/2}}{e (d+e x)^{5/2}}-\left (6 c^2 d 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 )\\ &=-\frac {3 c \sqrt {c d^2-c e^2 x^2}}{e \sqrt {d+e x}}-\frac {\left (c d^2-c e^2 x^2\right )^{3/2}}{e (d+e x)^{5/2}}+\frac {3 \sqrt {2} c^{3/2} \sqrt {d} \tanh ^{-1}\left (\frac {\sqrt {c d^2-c e^2 x^2}}{\sqrt {2} \sqrt {c} \sqrt {d} \sqrt {d+e x}}\right )}{e}\\ \end {align*}

________________________________________________________________________________________

Mathematica [A]  time = 0.19, size = 107, normalized size = 0.80 \begin {gather*} \frac {c \sqrt {c \left (d^2-e^2 x^2\right )} \left (\frac {3 \sqrt {2} \sqrt {d} \tanh ^{-1}\left (\frac {\sqrt {d^2-e^2 x^2}}{\sqrt {2} \sqrt {d} \sqrt {d+e x}}\right )}{\sqrt {d^2-e^2 x^2}}-\frac {2 (2 d+e x)}{(d+e x)^{3/2}}\right )}{e} \end {gather*}

Antiderivative was successfully verified.

[In]

Integrate[(c*d^2 - c*e^2*x^2)^(3/2)/(d + e*x)^(7/2),x]

[Out]

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

________________________________________________________________________________________

IntegrateAlgebraic [A]  time = 1.10, size = 126, normalized size = 0.95 \begin {gather*} -\frac {3 \sqrt {2} c^{3/2} \sqrt {d} \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 )}{e}-\frac {2 c (2 d+e x) \sqrt {2 c d (d+e x)-c (d+e x)^2}}{e (d+e x)^{3/2}} \end {gather*}

Antiderivative was successfully verified.

[In]

IntegrateAlgebraic[(c*d^2 - c*e^2*x^2)^(3/2)/(d + e*x)^(7/2),x]

[Out]

(-2*c*(2*d + e*x)*Sqrt[2*c*d*(d + e*x) - c*(d + e*x)^2])/(e*(d + e*x)^(3/2)) - (3*Sqrt[2]*c^(3/2)*Sqrt[d]*ArcT
anh[(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])])/e

________________________________________________________________________________________

fricas [A]  time = 0.43, size = 309, normalized size = 2.32 \begin {gather*} \left [\frac {3 \, \sqrt {2} {\left (c e^{2} x^{2} + 2 \, c d e x + c d^{2}\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 (c e x + 2 \, c d\right )} \sqrt {e x + d}}{2 \, {\left (e^{3} x^{2} + 2 \, d e^{2} x + d^{2} e\right )}}, \frac {3 \, \sqrt {2} {\left (c e^{2} x^{2} + 2 \, c d e x + c d^{2}\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 (c e x + 2 \, c d\right )} \sqrt {e x + d}}{e^{3} x^{2} + 2 \, d e^{2} x + d^{2} e}\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)^(7/2),x, algorithm="fricas")

[Out]

[1/2*(3*sqrt(2)*(c*e^2*x^2 + 2*c*d*e*x + c*d^2)*sqrt(c*d)*log(-(c*e^2*x^2 - 2*c*d*e*x - 3*c*d^2 - 2*sqrt(2)*sq
rt(-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)*(c*e*x
+ 2*c*d)*sqrt(e*x + d))/(e^3*x^2 + 2*d*e^2*x + d^2*e), (3*sqrt(2)*(c*e^2*x^2 + 2*c*d*e*x + c*d^2)*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)*(c*e*x + 2*c*d)*sqrt(e*x + d))/(e^3*x^2 + 2*d*e^2*x + d^2*e)]

________________________________________________________________________________________

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)^(7/2),x, algorithm="giac")

[Out]

Timed out

________________________________________________________________________________________

maple [A]  time = 0.07, size = 154, normalized size = 1.16 \begin {gather*} \frac {\sqrt {-\left (e^{2} x^{2}-d^{2}\right ) c}\, \left (3 \sqrt {2}\, c d e x \arctanh \left (\frac {\sqrt {-\left (e x -d \right ) c}\, \sqrt {2}}{2 \sqrt {c d}}\right )+3 \sqrt {2}\, c \,d^{2} \arctanh \left (\frac {\sqrt {-\left (e x -d \right ) c}\, \sqrt {2}}{2 \sqrt {c d}}\right )-2 \sqrt {c d}\, \sqrt {-\left (e x -d \right ) c}\, e x -4 \sqrt {-\left (e x -d \right ) c}\, \sqrt {c d}\, d \right ) c}{\left (e x +d \right )^{\frac {3}{2}} \sqrt {-\left (e x -d \right ) c}\, \sqrt {c d}\, 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)^(7/2),x)

[Out]

(-(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*c*d*e+3*2^(1/2)*ar
ctanh(1/2*(-(e*x-d)*c)^(1/2)*2^(1/2)/(c*d)^(1/2))*c*d^2-2*(c*d)^(1/2)*(-(e*x-d)*c)^(1/2)*e*x-4*(-(e*x-d)*c)^(1
/2)*(c*d)^(1/2)*d)/(e*x+d)^(3/2)/(-(e*x-d)*c)^(1/2)/e/(c*d)^(1/2)

________________________________________________________________________________________

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 {7}{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)^(7/2),x, algorithm="maxima")

[Out]

integrate((-c*e^2*x^2 + c*d^2)^(3/2)/(e*x + d)^(7/2), x)

________________________________________________________________________________________

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 )}^{7/2}} \,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)^(7/2),x)

[Out]

int((c*d^2 - c*e^2*x^2)^(3/2)/(d + e*x)^(7/2), x)

________________________________________________________________________________________

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 {7}{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)**(7/2),x)

[Out]

Integral((-c*(-d + e*x)*(d + e*x))**(3/2)/(d + e*x)**(7/2), x)

________________________________________________________________________________________