3.83 \(\int \frac{c-c \sec (e+f x)}{(a+a \sec (e+f x))^{5/2}} \, dx\)
Optimal. Leaf size=148 \[ \frac{2 c \tan ^{-1}\left (\frac{\sqrt{a} \tan (e+f x)}{\sqrt{a \sec (e+f x)+a}}\right )}{a^{5/2} f}-\frac{23 c \tan ^{-1}\left (\frac{\sqrt{a} \tan (e+f x)}{\sqrt{2} \sqrt{a \sec (e+f x)+a}}\right )}{8 \sqrt{2} a^{5/2} f}-\frac{7 c \tan (e+f x)}{8 a f (a \sec (e+f x)+a)^{3/2}}-\frac{c \tan (e+f x)}{2 f (a \sec (e+f x)+a)^{5/2}} \]
[Out]
(2*c*ArcTan[(Sqrt[a]*Tan[e + f*x])/Sqrt[a + a*Sec[e + f*x]]])/(a^(5/2)*f) - (23*c*ArcTan[(Sqrt[a]*Tan[e + f*x]
)/(Sqrt[2]*Sqrt[a + a*Sec[e + f*x]])])/(8*Sqrt[2]*a^(5/2)*f) - (c*Tan[e + f*x])/(2*f*(a + a*Sec[e + f*x])^(5/2
)) - (7*c*Tan[e + f*x])/(8*a*f*(a + a*Sec[e + f*x])^(3/2))
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Rubi [A] time = 0.191576, antiderivative size = 181, normalized size of antiderivative =
1.22, number of steps used = 7, number of rules used = 6, integrand size = 26, \(\frac{\text{number of rules}}{\text{integrand size}}\) = 0.231, Rules
used = {3904, 3887, 471, 527, 522, 203} \[ -\frac{7 c \sin (e+f x) \sec ^2\left (\frac{1}{2} (e+f x)\right )}{16 a^2 f \sqrt{a \sec (e+f x)+a}}+\frac{2 c \tan ^{-1}\left (\frac{\sqrt{a} \tan (e+f x)}{\sqrt{a \sec (e+f x)+a}}\right )}{a^{5/2} f}-\frac{23 c \tan ^{-1}\left (\frac{\sqrt{a} \tan (e+f x)}{\sqrt{2} \sqrt{a \sec (e+f x)+a}}\right )}{8 \sqrt{2} a^{5/2} f}-\frac{c \sin (e+f x) \cos (e+f x) \sec ^4\left (\frac{1}{2} (e+f x)\right )}{8 a^2 f \sqrt{a \sec (e+f x)+a}} \]
Antiderivative was successfully verified.
[In]
Int[(c - c*Sec[e + f*x])/(a + a*Sec[e + f*x])^(5/2),x]
[Out]
(2*c*ArcTan[(Sqrt[a]*Tan[e + f*x])/Sqrt[a + a*Sec[e + f*x]]])/(a^(5/2)*f) - (23*c*ArcTan[(Sqrt[a]*Tan[e + f*x]
)/(Sqrt[2]*Sqrt[a + a*Sec[e + f*x]])])/(8*Sqrt[2]*a^(5/2)*f) - (7*c*Sec[(e + f*x)/2]^2*Sin[e + f*x])/(16*a^2*f
*Sqrt[a + a*Sec[e + f*x]]) - (c*Cos[e + f*x]*Sec[(e + f*x)/2]^4*Sin[e + f*x])/(8*a^2*f*Sqrt[a + a*Sec[e + f*x]
])
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 471
Int[((e_.)*(x_))^(m_.)*((a_) + (b_.)*(x_)^(n_))^(p_)*((c_) + (d_.)*(x_)^(n_))^(q_), x_Symbol] :> Simp[(e^(n -
1)*(e*x)^(m - n + 1)*(a + b*x^n)^(p + 1)*(c + d*x^n)^(q + 1))/(n*(b*c - a*d)*(p + 1)), x] - Dist[e^n/(n*(b*c -
a*d)*(p + 1)), Int[(e*x)^(m - n)*(a + b*x^n)^(p + 1)*(c + d*x^n)^q*Simp[c*(m - n + 1) + d*(m + n*(p + q + 1)
+ 1)*x^n, x], x], x] /; FreeQ[{a, b, c, d, e, q}, x] && NeQ[b*c - a*d, 0] && IGtQ[n, 0] && LtQ[p, -1] && GeQ[n
, m - n + 1] && GtQ[m - n + 1, 0] && IntBinomialQ[a, b, c, d, e, m, n, p, q, x]
Rule 527
Int[((a_) + (b_.)*(x_)^(n_))^(p_)*((c_) + (d_.)*(x_)^(n_))^(q_.)*((e_) + (f_.)*(x_)^(n_)), x_Symbol] :> -Simp[
((b*e - a*f)*x*(a + b*x^n)^(p + 1)*(c + d*x^n)^(q + 1))/(a*n*(b*c - a*d)*(p + 1)), x] + Dist[1/(a*n*(b*c - a*d
)*(p + 1)), Int[(a + b*x^n)^(p + 1)*(c + d*x^n)^q*Simp[c*(b*e - a*f) + e*n*(b*c - a*d)*(p + 1) + d*(b*e - a*f)
*(n*(p + q + 2) + 1)*x^n, x], x], x] /; FreeQ[{a, b, c, d, e, f, n, q}, x] && LtQ[p, -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{c-c \sec (e+f x)}{(a+a \sec (e+f x))^{5/2}} \, dx &=-\left ((a c) \int \frac{\tan ^2(e+f x)}{(a+a \sec (e+f x))^{7/2}} \, dx\right )\\ &=\frac{(2 c) \operatorname{Subst}\left (\int \frac{x^2}{\left (1+a x^2\right ) \left (2+a x^2\right )^3} \, dx,x,-\frac{\tan (e+f x)}{\sqrt{a+a \sec (e+f x)}}\right )}{a f}\\ &=-\frac{c \cos (e+f x) \sec ^4\left (\frac{1}{2} (e+f x)\right ) \sin (e+f x)}{8 a^2 f \sqrt{a+a \sec (e+f x)}}-\frac{c \operatorname{Subst}\left (\int \frac{1-3 a x^2}{\left (1+a x^2\right ) \left (2+a x^2\right )^2} \, dx,x,-\frac{\tan (e+f x)}{\sqrt{a+a \sec (e+f x)}}\right )}{2 a^2 f}\\ &=-\frac{7 c \sec ^2\left (\frac{1}{2} (e+f x)\right ) \sin (e+f x)}{16 a^2 f \sqrt{a+a \sec (e+f x)}}-\frac{c \cos (e+f x) \sec ^4\left (\frac{1}{2} (e+f x)\right ) \sin (e+f x)}{8 a^2 f \sqrt{a+a \sec (e+f x)}}-\frac{c \operatorname{Subst}\left (\int \frac{9 a-7 a^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 )}{8 a^3 f}\\ &=-\frac{7 c \sec ^2\left (\frac{1}{2} (e+f x)\right ) \sin (e+f x)}{16 a^2 f \sqrt{a+a \sec (e+f x)}}-\frac{c \cos (e+f x) \sec ^4\left (\frac{1}{2} (e+f x)\right ) \sin (e+f x)}{8 a^2 f \sqrt{a+a \sec (e+f x)}}-\frac{(2 c) \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 )}{a^2 f}+\frac{(23 c) \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 )}{8 a^2 f}\\ &=\frac{2 c \tan ^{-1}\left (\frac{\sqrt{a} \tan (e+f x)}{\sqrt{a+a \sec (e+f x)}}\right )}{a^{5/2} f}-\frac{23 c \tan ^{-1}\left (\frac{\sqrt{a} \tan (e+f x)}{\sqrt{2} \sqrt{a+a \sec (e+f x)}}\right )}{8 \sqrt{2} a^{5/2} f}-\frac{7 c \sec ^2\left (\frac{1}{2} (e+f x)\right ) \sin (e+f x)}{16 a^2 f \sqrt{a+a \sec (e+f x)}}-\frac{c \cos (e+f x) \sec ^4\left (\frac{1}{2} (e+f x)\right ) \sin (e+f x)}{8 a^2 f \sqrt{a+a \sec (e+f x)}}\\ \end{align*}
Mathematica [A] time = 1.45808, size = 134, normalized size = 0.91 \[ -\frac{c \cot \left (\frac{1}{2} (e+f x)\right ) \left ((8 \cos (e+f x)-11 \cos (2 (e+f x))+3) \sec ^4\left (\frac{1}{2} (e+f x)\right )-128 \sqrt{\sec (e+f x)-1} \tan ^{-1}\left (\sqrt{\sec (e+f x)-1}\right )+92 \sqrt{2} \sqrt{\sec (e+f x)-1} \tan ^{-1}\left (\frac{\sqrt{\sec (e+f x)-1}}{\sqrt{2}}\right )\right )}{64 a^2 f \sqrt{a (\sec (e+f x)+1)}} \]
Antiderivative was successfully verified.
[In]
Integrate[(c - c*Sec[e + f*x])/(a + a*Sec[e + f*x])^(5/2),x]
[Out]
-(c*Cot[(e + f*x)/2]*((3 + 8*Cos[e + f*x] - 11*Cos[2*(e + f*x)])*Sec[(e + f*x)/2]^4 - 128*ArcTan[Sqrt[-1 + Sec
[e + f*x]]]*Sqrt[-1 + Sec[e + f*x]] + 92*Sqrt[2]*ArcTan[Sqrt[-1 + Sec[e + f*x]]/Sqrt[2]]*Sqrt[-1 + Sec[e + f*x
]]))/(64*a^2*f*Sqrt[a*(1 + Sec[e + f*x])])
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Maple [B] time = 0.179, size = 543, normalized size = 3.7 \begin{align*} \text{result too large to display} \end{align*}
Verification of antiderivative is not currently implemented for this CAS.
[In]
int((c-c*sec(f*x+e))/(a+a*sec(f*x+e))^(5/2),x)
[Out]
-1/16*c/f/a^3*(1/cos(f*x+e)*a*(1+cos(f*x+e)))^(1/2)*(16*cos(f*x+e)^2*sin(f*x+e)*2^(1/2)*(-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))+23*sin(f*x+e)*c
os(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))+32*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))+46*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))+16*2^(1/2)*sin(f*x+e)*a
rctanh(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)+23*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)-c
os(f*x+e)+1)/sin(f*x+e))-22*cos(f*x+e)^3+8*cos(f*x+e)^2+14*cos(f*x+e))/(1+cos(f*x+e))^2/sin(f*x+e)
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Maxima [F] time = 0., size = 0, normalized size = 0. \begin{align*} -\int \frac{c \sec \left (f x + e\right ) - c}{{\left (a \sec \left (f x + e\right ) + a\right )}^{\frac{5}{2}}}\,{d x} \end{align*}
Verification of antiderivative is not currently implemented for this CAS.
[In]
integrate((c-c*sec(f*x+e))/(a+a*sec(f*x+e))^(5/2),x, algorithm="maxima")
[Out]
-integrate((c*sec(f*x + e) - c)/(a*sec(f*x + e) + a)^(5/2), x)
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Fricas [B] time = 3.9638, size = 1605, normalized size = 10.84 \begin{align*} \text{result too large to display} \end{align*}
Verification of antiderivative is not currently implemented for this CAS.
[In]
integrate((c-c*sec(f*x+e))/(a+a*sec(f*x+e))^(5/2),x, algorithm="fricas")
[Out]
[-1/32*(23*sqrt(2)*(c*cos(f*x + e)^3 + 3*c*cos(f*x + e)^2 + 3*c*cos(f*x + e) + c)*sqrt(-a)*log(-(2*sqrt(2)*sqr
t(-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)) + 32*(c*cos(f*x + e)^3 + 3*c*cos(f*x + e)^2 + 3*c*cos(f*x + e) +
c)*sqrt(-a)*log((2*a*cos(f*x + e)^2 + 2*sqrt(-a)*sqrt((a*cos(f*x + e) + a)/cos(f*x + e))*cos(f*x + e)*sin(f*x
+ e) + a*cos(f*x + e) - a)/(cos(f*x + e) + 1)) + 4*(11*c*cos(f*x + e)^2 + 7*c*cos(f*x + e))*sqrt((a*cos(f*x +
e) + a)/cos(f*x + e))*sin(f*x + e))/(a^3*f*cos(f*x + e)^3 + 3*a^3*f*cos(f*x + e)^2 + 3*a^3*f*cos(f*x + e) + a
^3*f), 1/16*(23*sqrt(2)*(c*cos(f*x + e)^3 + 3*c*cos(f*x + e)^2 + 3*c*cos(f*x + e) + c)*sqrt(a)*arctan(sqrt(2)*
sqrt((a*cos(f*x + e) + a)/cos(f*x + e))*cos(f*x + e)/(sqrt(a)*sin(f*x + e))) - 32*(c*cos(f*x + e)^3 + 3*c*cos(
f*x + e)^2 + 3*c*cos(f*x + e) + c)*sqrt(a)*arctan(sqrt((a*cos(f*x + e) + a)/cos(f*x + e))*cos(f*x + e)/(sqrt(a
)*sin(f*x + e))) - 2*(11*c*cos(f*x + e)^2 + 7*c*cos(f*x + e))*sqrt((a*cos(f*x + e) + a)/cos(f*x + e))*sin(f*x
+ e))/(a^3*f*cos(f*x + e)^3 + 3*a^3*f*cos(f*x + e)^2 + 3*a^3*f*cos(f*x + e) + a^3*f)]
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Sympy [F] time = 0., size = 0, normalized size = 0. \begin{align*} - c \left (\int \frac{\sec{\left (e + f x \right )}}{a^{2} \sqrt{a \sec{\left (e + f x \right )} + a} \sec ^{2}{\left (e + f x \right )} + 2 a^{2} \sqrt{a \sec{\left (e + f x \right )} + a} \sec{\left (e + f x \right )} + a^{2} \sqrt{a \sec{\left (e + f x \right )} + a}}\, dx + \int - \frac{1}{a^{2} \sqrt{a \sec{\left (e + f x \right )} + a} \sec ^{2}{\left (e + f x \right )} + 2 a^{2} \sqrt{a \sec{\left (e + f x \right )} + a} \sec{\left (e + f x \right )} + a^{2} \sqrt{a \sec{\left (e + f x \right )} + a}}\, dx\right ) \end{align*}
Verification of antiderivative is not currently implemented for this CAS.
[In]
integrate((c-c*sec(f*x+e))/(a+a*sec(f*x+e))**(5/2),x)
[Out]
-c*(Integral(sec(e + f*x)/(a**2*sqrt(a*sec(e + f*x) + a)*sec(e + f*x)**2 + 2*a**2*sqrt(a*sec(e + f*x) + a)*sec
(e + f*x) + a**2*sqrt(a*sec(e + f*x) + a)), x) + Integral(-1/(a**2*sqrt(a*sec(e + f*x) + a)*sec(e + f*x)**2 +
2*a**2*sqrt(a*sec(e + f*x) + a)*sec(e + f*x) + a**2*sqrt(a*sec(e + f*x) + a)), x))
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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((c-c*sec(f*x+e))/(a+a*sec(f*x+e))^(5/2),x, algorithm="giac")
[Out]
Timed out