3.71 \(\int \frac{1}{\sqrt{a+a \sec (e+f x)} (c-c \sec (e+f x))^3} \, dx\)
Optimal. Leaf size=196 \[ \frac{\cot ^5(e+f x) (a \sec (e+f x)+a)^{5/2}}{5 a^3 c^3 f}-\frac{\cot ^3(e+f x) (a \sec (e+f x)+a)^{3/2}}{2 a^2 c^3 f}+\frac{7 \cot (e+f x) \sqrt{a \sec (e+f x)+a}}{4 a c^3 f}+\frac{2 \tan ^{-1}\left (\frac{\sqrt{a} \tan (e+f x)}{\sqrt{a \sec (e+f x)+a}}\right )}{\sqrt{a} c^3 f}-\frac{\tan ^{-1}\left (\frac{\sqrt{a} \tan (e+f x)}{\sqrt{2} \sqrt{a \sec (e+f x)+a}}\right )}{4 \sqrt{2} \sqrt{a} c^3 f} \]
[Out]
(2*ArcTan[(Sqrt[a]*Tan[e + f*x])/Sqrt[a + a*Sec[e + f*x]]])/(Sqrt[a]*c^3*f) - ArcTan[(Sqrt[a]*Tan[e + f*x])/(S
qrt[2]*Sqrt[a + a*Sec[e + f*x]])]/(4*Sqrt[2]*Sqrt[a]*c^3*f) + (7*Cot[e + f*x]*Sqrt[a + a*Sec[e + f*x]])/(4*a*c
^3*f) - (Cot[e + f*x]^3*(a + a*Sec[e + f*x])^(3/2))/(2*a^2*c^3*f) + (Cot[e + f*x]^5*(a + a*Sec[e + f*x])^(5/2)
)/(5*a^3*c^3*f)
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Rubi [A] time = 0.284238, antiderivative size = 196, normalized size of antiderivative = 1.,
number of steps used = 8, number of rules used = 6, integrand size = 28, \(\frac{\text{number of rules}}{\text{integrand size}}\) = 0.214, Rules used =
{3904, 3887, 480, 583, 522, 203} \[ \frac{\cot ^5(e+f x) (a \sec (e+f x)+a)^{5/2}}{5 a^3 c^3 f}-\frac{\cot ^3(e+f x) (a \sec (e+f x)+a)^{3/2}}{2 a^2 c^3 f}+\frac{7 \cot (e+f x) \sqrt{a \sec (e+f x)+a}}{4 a c^3 f}+\frac{2 \tan ^{-1}\left (\frac{\sqrt{a} \tan (e+f x)}{\sqrt{a \sec (e+f x)+a}}\right )}{\sqrt{a} c^3 f}-\frac{\tan ^{-1}\left (\frac{\sqrt{a} \tan (e+f x)}{\sqrt{2} \sqrt{a \sec (e+f x)+a}}\right )}{4 \sqrt{2} \sqrt{a} c^3 f} \]
Antiderivative was successfully verified.
[In]
Int[1/(Sqrt[a + a*Sec[e + f*x]]*(c - c*Sec[e + f*x])^3),x]
[Out]
(2*ArcTan[(Sqrt[a]*Tan[e + f*x])/Sqrt[a + a*Sec[e + f*x]]])/(Sqrt[a]*c^3*f) - ArcTan[(Sqrt[a]*Tan[e + f*x])/(S
qrt[2]*Sqrt[a + a*Sec[e + f*x]])]/(4*Sqrt[2]*Sqrt[a]*c^3*f) + (7*Cot[e + f*x]*Sqrt[a + a*Sec[e + f*x]])/(4*a*c
^3*f) - (Cot[e + f*x]^3*(a + a*Sec[e + f*x])^(3/2))/(2*a^2*c^3*f) + (Cot[e + f*x]^5*(a + a*Sec[e + f*x])^(5/2)
)/(5*a^3*c^3*f)
Rule 3904
Int[(csc[(e_.) + (f_.)*(x_)]*(b_.) + (a_))^(m_.)*(csc[(e_.) + (f_.)*(x_)]*(d_.) + (c_))^(n_.), x_Symbol] :> Di
st[(-(a*c))^m, Int[Cot[e + f*x]^(2*m)*(c + d*Csc[e + f*x])^(n - m), x], x] /; FreeQ[{a, b, c, d, e, f, n}, x]
&& EqQ[b*c + a*d, 0] && EqQ[a^2 - b^2, 0] && IntegerQ[m] && RationalQ[n] && !(IntegerQ[n] && GtQ[m - n, 0])
Rule 3887
Int[cot[(c_.) + (d_.)*(x_)]^(m_.)*(csc[(c_.) + (d_.)*(x_)]*(b_.) + (a_))^(n_.), x_Symbol] :> Dist[(-2*a^(m/2 +
n + 1/2))/d, Subst[Int[(x^m*(2 + a*x^2)^(m/2 + n - 1/2))/(1 + a*x^2), x], x, Cot[c + d*x]/Sqrt[a + b*Csc[c +
d*x]]], x] /; FreeQ[{a, b, c, d}, x] && EqQ[a^2 - b^2, 0] && IntegerQ[m/2] && IntegerQ[n - 1/2]
Rule 480
Int[((e_.)*(x_))^(m_)*((a_) + (b_.)*(x_)^(n_))^(p_)*((c_) + (d_.)*(x_)^(n_))^(q_), x_Symbol] :> Simp[((e*x)^(m
+ 1)*(a + b*x^n)^(p + 1)*(c + d*x^n)^(q + 1))/(a*c*e*(m + 1)), x] - Dist[1/(a*c*e^n*(m + 1)), Int[(e*x)^(m +
n)*(a + b*x^n)^p*(c + d*x^n)^q*Simp[(b*c + a*d)*(m + n + 1) + n*(b*c*p + a*d*q) + b*d*(m + n*(p + q + 2) + 1)*
x^n, x], x], x] /; FreeQ[{a, b, c, d, e, p, q}, x] && NeQ[b*c - a*d, 0] && IGtQ[n, 0] && LtQ[m, -1] && IntBino
mialQ[a, b, c, d, e, m, n, p, q, x]
Rule 583
Int[((g_.)*(x_))^(m_)*((a_) + (b_.)*(x_)^(n_))^(p_.)*((c_) + (d_.)*(x_)^(n_))^(q_.)*((e_) + (f_.)*(x_)^(n_)),
x_Symbol] :> Simp[(e*(g*x)^(m + 1)*(a + b*x^n)^(p + 1)*(c + d*x^n)^(q + 1))/(a*c*g*(m + 1)), x] + Dist[1/(a*c*
g^n*(m + 1)), Int[(g*x)^(m + n)*(a + b*x^n)^p*(c + d*x^n)^q*Simp[a*f*c*(m + 1) - e*(b*c + a*d)*(m + n + 1) - e
*n*(b*c*p + a*d*q) - b*e*d*(m + n*(p + q + 2) + 1)*x^n, x], x], x] /; FreeQ[{a, b, c, d, e, f, g, p, q}, x] &&
IGtQ[n, 0] && LtQ[m, -1]
Rule 522
Int[((e_) + (f_.)*(x_)^(n_))/(((a_) + (b_.)*(x_)^(n_))*((c_) + (d_.)*(x_)^(n_))), x_Symbol] :> Dist[(b*e - a*f
)/(b*c - a*d), Int[1/(a + b*x^n), x], x] - Dist[(d*e - c*f)/(b*c - a*d), Int[1/(c + d*x^n), x], x] /; FreeQ[{a
, b, c, d, e, f, n}, x]
Rule 203
Int[((a_) + (b_.)*(x_)^2)^(-1), x_Symbol] :> Simp[(1*ArcTan[(Rt[b, 2]*x)/Rt[a, 2]])/(Rt[a, 2]*Rt[b, 2]), x] /;
FreeQ[{a, b}, x] && PosQ[a/b] && (GtQ[a, 0] || GtQ[b, 0])
Rubi steps
\begin{align*} \int \frac{1}{\sqrt{a+a \sec (e+f x)} (c-c \sec (e+f x))^3} \, dx &=-\frac{\int \cot ^6(e+f x) (a+a \sec (e+f x))^{5/2} \, dx}{a^3 c^3}\\ &=\frac{2 \operatorname{Subst}\left (\int \frac{1}{x^6 \left (1+a x^2\right ) \left (2+a x^2\right )} \, dx,x,-\frac{\tan (e+f x)}{\sqrt{a+a \sec (e+f x)}}\right )}{a^3 c^3 f}\\ &=\frac{\cot ^5(e+f x) (a+a \sec (e+f x))^{5/2}}{5 a^3 c^3 f}+\frac{\operatorname{Subst}\left (\int \frac{-15 a-5 a^2 x^2}{x^4 \left (1+a x^2\right ) \left (2+a x^2\right )} \, dx,x,-\frac{\tan (e+f x)}{\sqrt{a+a \sec (e+f x)}}\right )}{5 a^3 c^3 f}\\ &=-\frac{\cot ^3(e+f x) (a+a \sec (e+f x))^{3/2}}{2 a^2 c^3 f}+\frac{\cot ^5(e+f x) (a+a \sec (e+f x))^{5/2}}{5 a^3 c^3 f}-\frac{\operatorname{Subst}\left (\int \frac{-105 a^2-45 a^3 x^2}{x^2 \left (1+a x^2\right ) \left (2+a x^2\right )} \, dx,x,-\frac{\tan (e+f x)}{\sqrt{a+a \sec (e+f x)}}\right )}{30 a^3 c^3 f}\\ &=\frac{7 \cot (e+f x) \sqrt{a+a \sec (e+f x)}}{4 a c^3 f}-\frac{\cot ^3(e+f x) (a+a \sec (e+f x))^{3/2}}{2 a^2 c^3 f}+\frac{\cot ^5(e+f x) (a+a \sec (e+f x))^{5/2}}{5 a^3 c^3 f}+\frac{\operatorname{Subst}\left (\int \frac{-225 a^3-105 a^4 x^2}{\left (1+a x^2\right ) \left (2+a x^2\right )} \, dx,x,-\frac{\tan (e+f x)}{\sqrt{a+a \sec (e+f x)}}\right )}{60 a^3 c^3 f}\\ &=\frac{7 \cot (e+f x) \sqrt{a+a \sec (e+f x)}}{4 a c^3 f}-\frac{\cot ^3(e+f x) (a+a \sec (e+f x))^{3/2}}{2 a^2 c^3 f}+\frac{\cot ^5(e+f x) (a+a \sec (e+f x))^{5/2}}{5 a^3 c^3 f}+\frac{\operatorname{Subst}\left (\int \frac{1}{2+a x^2} \, dx,x,-\frac{\tan (e+f x)}{\sqrt{a+a \sec (e+f x)}}\right )}{4 c^3 f}-\frac{2 \operatorname{Subst}\left (\int \frac{1}{1+a x^2} \, dx,x,-\frac{\tan (e+f x)}{\sqrt{a+a \sec (e+f x)}}\right )}{c^3 f}\\ &=\frac{2 \tan ^{-1}\left (\frac{\sqrt{a} \tan (e+f x)}{\sqrt{a+a \sec (e+f x)}}\right )}{\sqrt{a} c^3 f}-\frac{\tan ^{-1}\left (\frac{\sqrt{a} \tan (e+f x)}{\sqrt{2} \sqrt{a+a \sec (e+f x)}}\right )}{4 \sqrt{2} \sqrt{a} c^3 f}+\frac{7 \cot (e+f x) \sqrt{a+a \sec (e+f x)}}{4 a c^3 f}-\frac{\cot ^3(e+f x) (a+a \sec (e+f x))^{3/2}}{2 a^2 c^3 f}+\frac{\cot ^5(e+f x) (a+a \sec (e+f x))^{5/2}}{5 a^3 c^3 f}\\ \end{align*}
Mathematica [C] time = 23.5009, size = 5602, normalized size = 28.58 \[ \text{Result too large to show} \]
Warning: Unable to verify antiderivative.
[In]
Integrate[1/(Sqrt[a + a*Sec[e + f*x]]*(c - c*Sec[e + f*x])^3),x]
[Out]
Result too large to show
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Maple [B] time = 0.359, size = 545, normalized size = 2.8 \begin{align*} \text{result too large to display} \end{align*}
Verification of antiderivative is not currently implemented for this CAS.
[In]
int(1/(c-c*sec(f*x+e))^3/(a+a*sec(f*x+e))^(1/2),x)
[Out]
-1/40/c^3/f/a*(1/cos(f*x+e)*a*(1+cos(f*x+e)))^(1/2)*(1+cos(f*x+e))^2*(40*cos(f*x+e)^2*sin(f*x+e)*2^(1/2)*(-2*c
os(f*x+e)/(1+cos(f*x+e)))^(1/2)*arctanh(1/2*2^(1/2)*(-2*cos(f*x+e)/(1+cos(f*x+e)))^(1/2)*sin(f*x+e)/cos(f*x+e)
)+5*sin(f*x+e)*cos(f*x+e)^2*(-2*cos(f*x+e)/(1+cos(f*x+e)))^(1/2)*ln(((-2*cos(f*x+e)/(1+cos(f*x+e)))^(1/2)*sin(
f*x+e)-cos(f*x+e)+1)/sin(f*x+e))-80*2^(1/2)*cos(f*x+e)*sin(f*x+e)*(-2*cos(f*x+e)/(1+cos(f*x+e)))^(1/2)*arctanh
(1/2*2^(1/2)*(-2*cos(f*x+e)/(1+cos(f*x+e)))^(1/2)*sin(f*x+e)/cos(f*x+e))-10*sin(f*x+e)*cos(f*x+e)*(-2*cos(f*x+
e)/(1+cos(f*x+e)))^(1/2)*ln(((-2*cos(f*x+e)/(1+cos(f*x+e)))^(1/2)*sin(f*x+e)-cos(f*x+e)+1)/sin(f*x+e))+40*2^(1
/2)*sin(f*x+e)*arctanh(1/2*2^(1/2)*(-2*cos(f*x+e)/(1+cos(f*x+e)))^(1/2)*sin(f*x+e)/cos(f*x+e))*(-2*cos(f*x+e)/
(1+cos(f*x+e)))^(1/2)+5*sin(f*x+e)*(-2*cos(f*x+e)/(1+cos(f*x+e)))^(1/2)*ln(((-2*cos(f*x+e)/(1+cos(f*x+e)))^(1/
2)*sin(f*x+e)-cos(f*x+e)+1)/sin(f*x+e))-98*cos(f*x+e)^3+160*cos(f*x+e)^2-70*cos(f*x+e))/sin(f*x+e)^5
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Maxima [F] time = 0., size = 0, normalized size = 0. \begin{align*} -\int \frac{1}{\sqrt{a \sec \left (f x + e\right ) + a}{\left (c \sec \left (f x + e\right ) - c\right )}^{3}}\,{d x} \end{align*}
Verification of antiderivative is not currently implemented for this CAS.
[In]
integrate(1/(c-c*sec(f*x+e))^3/(a+a*sec(f*x+e))^(1/2),x, algorithm="maxima")
[Out]
-integrate(1/(sqrt(a*sec(f*x + e) + a)*(c*sec(f*x + e) - c)^3), x)
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Fricas [A] time = 2.62276, size = 1627, normalized size = 8.3 \begin{align*} \text{result too large to display} \end{align*}
Verification of antiderivative is not currently implemented for this CAS.
[In]
integrate(1/(c-c*sec(f*x+e))^3/(a+a*sec(f*x+e))^(1/2),x, algorithm="fricas")
[Out]
[-1/80*(5*sqrt(2)*(cos(f*x + e)^2 - 2*cos(f*x + e) + 1)*sqrt(-a)*log(-(2*sqrt(2)*sqrt(-a)*sqrt((a*cos(f*x + e)
+ a)/cos(f*x + e))*cos(f*x + e)*sin(f*x + e) - 3*a*cos(f*x + e)^2 - 2*a*cos(f*x + e) + a)/(cos(f*x + e)^2 + 2
*cos(f*x + e) + 1))*sin(f*x + e) + 40*(cos(f*x + e)^2 - 2*cos(f*x + e) + 1)*sqrt(-a)*log(-(8*a*cos(f*x + e)^3
+ 4*(2*cos(f*x + e)^2 - cos(f*x + e))*sqrt(-a)*sqrt((a*cos(f*x + e) + a)/cos(f*x + e))*sin(f*x + e) - 7*a*cos(
f*x + e) + a)/(cos(f*x + e) + 1))*sin(f*x + e) - 4*(49*cos(f*x + e)^3 - 80*cos(f*x + e)^2 + 35*cos(f*x + e))*s
qrt((a*cos(f*x + e) + a)/cos(f*x + e)))/((a*c^3*f*cos(f*x + e)^2 - 2*a*c^3*f*cos(f*x + e) + a*c^3*f)*sin(f*x +
e)), 1/40*(5*sqrt(2)*(cos(f*x + e)^2 - 2*cos(f*x + e) + 1)*sqrt(a)*arctan(sqrt(2)*sqrt((a*cos(f*x + e) + a)/c
os(f*x + e))*cos(f*x + e)/(sqrt(a)*sin(f*x + e)))*sin(f*x + e) + 40*(cos(f*x + e)^2 - 2*cos(f*x + e) + 1)*sqrt
(a)*arctan(2*sqrt(a)*sqrt((a*cos(f*x + e) + a)/cos(f*x + e))*cos(f*x + e)*sin(f*x + e)/(2*a*cos(f*x + e)^2 + a
*cos(f*x + e) - a))*sin(f*x + e) + 2*(49*cos(f*x + e)^3 - 80*cos(f*x + e)^2 + 35*cos(f*x + e))*sqrt((a*cos(f*x
+ e) + a)/cos(f*x + e)))/((a*c^3*f*cos(f*x + e)^2 - 2*a*c^3*f*cos(f*x + e) + a*c^3*f)*sin(f*x + e))]
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Sympy [F] time = 0., size = 0, normalized size = 0. \begin{align*} - \frac{\int \frac{1}{\sqrt{a \sec{\left (e + f x \right )} + a} \sec ^{3}{\left (e + f x \right )} - 3 \sqrt{a \sec{\left (e + f x \right )} + a} \sec ^{2}{\left (e + f x \right )} + 3 \sqrt{a \sec{\left (e + f x \right )} + a} \sec{\left (e + f x \right )} - \sqrt{a \sec{\left (e + f x \right )} + a}}\, dx}{c^{3}} \end{align*}
Verification of antiderivative is not currently implemented for this CAS.
[In]
integrate(1/(c-c*sec(f*x+e))**3/(a+a*sec(f*x+e))**(1/2),x)
[Out]
-Integral(1/(sqrt(a*sec(e + f*x) + a)*sec(e + f*x)**3 - 3*sqrt(a*sec(e + f*x) + a)*sec(e + f*x)**2 + 3*sqrt(a*
sec(e + f*x) + a)*sec(e + f*x) - sqrt(a*sec(e + f*x) + a)), x)/c**3
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Giac [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(1/(c-c*sec(f*x+e))^3/(a+a*sec(f*x+e))^(1/2),x, algorithm="giac")
[Out]
Timed out