3.81 \(\int \frac{(c-c \sec (e+f x))^3}{(a+a \sec (e+f x))^{5/2}} \, dx\)

Optimal. Leaf size=191 \[ -\frac{c^3 \sin (e+f x) \sec ^2\left (\frac{1}{2} (e+f x)\right )}{4 a^2 f \sqrt{a \sec (e+f x)+a}}+\frac{2 c^3 \tan ^{-1}\left (\frac{\sqrt{a} \tan (e+f x)}{\sqrt{a \sec (e+f x)+a}}\right )}{a^{5/2} f}-\frac{7 c^3 \tan ^{-1}\left (\frac{\sqrt{a} \tan (e+f x)}{\sqrt{2} \sqrt{a \sec (e+f x)+a}}\right )}{2 \sqrt{2} a^{5/2} f}+\frac{c^3 \sin ^2(e+f x) \tan (e+f x) \sec ^4\left (\frac{1}{2} (e+f x)\right )}{4 a f (a \sec (e+f x)+a)^{3/2}} \]

[Out]

(2*c^3*ArcTan[(Sqrt[a]*Tan[e + f*x])/Sqrt[a + a*Sec[e + f*x]]])/(a^(5/2)*f) - (7*c^3*ArcTan[(Sqrt[a]*Tan[e + f
*x])/(Sqrt[2]*Sqrt[a + a*Sec[e + f*x]])])/(2*Sqrt[2]*a^(5/2)*f) - (c^3*Sec[(e + f*x)/2]^2*Sin[e + f*x])/(4*a^2
*f*Sqrt[a + a*Sec[e + f*x]]) + (c^3*Sec[(e + f*x)/2]^4*Sin[e + f*x]^2*Tan[e + f*x])/(4*a*f*(a + a*Sec[e + f*x]
)^(3/2))

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Rubi [A]  time = 0.252055, antiderivative size = 191, normalized size of antiderivative = 1., number of steps used = 7, number of rules used = 6, integrand size = 28, \(\frac{\text{number of rules}}{\text{integrand size}}\) = 0.214, Rules used = {3904, 3887, 470, 578, 522, 203} \[ -\frac{c^3 \sin (e+f x) \sec ^2\left (\frac{1}{2} (e+f x)\right )}{4 a^2 f \sqrt{a \sec (e+f x)+a}}+\frac{2 c^3 \tan ^{-1}\left (\frac{\sqrt{a} \tan (e+f x)}{\sqrt{a \sec (e+f x)+a}}\right )}{a^{5/2} f}-\frac{7 c^3 \tan ^{-1}\left (\frac{\sqrt{a} \tan (e+f x)}{\sqrt{2} \sqrt{a \sec (e+f x)+a}}\right )}{2 \sqrt{2} a^{5/2} f}+\frac{c^3 \sin ^2(e+f x) \tan (e+f x) \sec ^4\left (\frac{1}{2} (e+f x)\right )}{4 a f (a \sec (e+f x)+a)^{3/2}} \]

Antiderivative was successfully verified.

[In]

Int[(c - c*Sec[e + f*x])^3/(a + a*Sec[e + f*x])^(5/2),x]

[Out]

(2*c^3*ArcTan[(Sqrt[a]*Tan[e + f*x])/Sqrt[a + a*Sec[e + f*x]]])/(a^(5/2)*f) - (7*c^3*ArcTan[(Sqrt[a]*Tan[e + f
*x])/(Sqrt[2]*Sqrt[a + a*Sec[e + f*x]])])/(2*Sqrt[2]*a^(5/2)*f) - (c^3*Sec[(e + f*x)/2]^2*Sin[e + f*x])/(4*a^2
*f*Sqrt[a + a*Sec[e + f*x]]) + (c^3*Sec[(e + f*x)/2]^4*Sin[e + f*x]^2*Tan[e + f*x])/(4*a*f*(a + a*Sec[e + f*x]
)^(3/2))

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 470

Int[((e_.)*(x_))^(m_.)*((a_) + (b_.)*(x_)^(n_))^(p_)*((c_) + (d_.)*(x_)^(n_))^(q_), x_Symbol] :> -Simp[(a*e^(2
*n - 1)*(e*x)^(m - 2*n + 1)*(a + b*x^n)^(p + 1)*(c + d*x^n)^(q + 1))/(b*n*(b*c - a*d)*(p + 1)), x] + Dist[e^(2
*n)/(b*n*(b*c - a*d)*(p + 1)), Int[(e*x)^(m - 2*n)*(a + b*x^n)^(p + 1)*(c + d*x^n)^q*Simp[a*c*(m - 2*n + 1) +
(a*d*(m - n + n*q + 1) + b*c*n*(p + 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] && GtQ[m - n + 1, n] && IntBinomialQ[a, b, c, d, e, m, n, p, q, x]

Rule 578

Int[((g_.)*(x_))^(m_.)*((a_) + (b_.)*(x_)^(n_))^(p_)*((c_) + (d_.)*(x_)^(n_))^(q_)*((e_) + (f_.)*(x_)^(n_)), x
_Symbol] :> Simp[(g^(n - 1)*(b*e - a*f)*(g*x)^(m - n + 1)*(a + b*x^n)^(p + 1)*(c + d*x^n)^(q + 1))/(b*n*(b*c -
a*d)*(p + 1)), x] - Dist[g^n/(b*n*(b*c - a*d)*(p + 1)), Int[(g*x)^(m - n)*(a + b*x^n)^(p + 1)*(c + d*x^n)^q*S
imp[c*(b*e - a*f)*(m - n + 1) + (d*(b*e - a*f)*(m + n*q + 1) - b*n*(c*f - d*e)*(p + 1))*x^n, x], x], x] /; Fre
eQ[{a, b, c, d, e, f, g, q}, x] && IGtQ[n, 0] && LtQ[p, -1] && GtQ[m - n + 1, 0]

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))^3}{(a+a \sec (e+f x))^{5/2}} \, dx &=-\left (\left (a^3 c^3\right ) \int \frac{\tan ^6(e+f x)}{(a+a \sec (e+f x))^{11/2}} \, dx\right )\\ &=\frac{\left (2 a c^3\right ) \operatorname{Subst}\left (\int \frac{x^6}{\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 )}{f}\\ &=\frac{c^3 \sec ^4\left (\frac{1}{2} (e+f x)\right ) \sin ^2(e+f x) \tan (e+f x)}{4 a f (a+a \sec (e+f x))^{3/2}}+\frac{c^3 \operatorname{Subst}\left (\int \frac{x^2 \left (6+2 a x^2\right )}{\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 f}\\ &=-\frac{c^3 \sec ^2\left (\frac{1}{2} (e+f x)\right ) \sin (e+f x)}{4 a^2 f \sqrt{a+a \sec (e+f x)}}+\frac{c^3 \sec ^4\left (\frac{1}{2} (e+f x)\right ) \sin ^2(e+f x) \tan (e+f x)}{4 a f (a+a \sec (e+f x))^{3/2}}-\frac{c^3 \operatorname{Subst}\left (\int \frac{2 a-6 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 )}{4 a^3 f}\\ &=-\frac{c^3 \sec ^2\left (\frac{1}{2} (e+f x)\right ) \sin (e+f x)}{4 a^2 f \sqrt{a+a \sec (e+f x)}}+\frac{c^3 \sec ^4\left (\frac{1}{2} (e+f x)\right ) \sin ^2(e+f x) \tan (e+f x)}{4 a f (a+a \sec (e+f x))^{3/2}}-\frac{\left (2 c^3\right ) \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{\left (7 c^3\right ) \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 )}{2 a^2 f}\\ &=\frac{2 c^3 \tan ^{-1}\left (\frac{\sqrt{a} \tan (e+f x)}{\sqrt{a+a \sec (e+f x)}}\right )}{a^{5/2} f}-\frac{7 c^3 \tan ^{-1}\left (\frac{\sqrt{a} \tan (e+f x)}{\sqrt{2} \sqrt{a+a \sec (e+f x)}}\right )}{2 \sqrt{2} a^{5/2} f}-\frac{c^3 \sec ^2\left (\frac{1}{2} (e+f x)\right ) \sin (e+f x)}{4 a^2 f \sqrt{a+a \sec (e+f x)}}+\frac{c^3 \sec ^4\left (\frac{1}{2} (e+f x)\right ) \sin ^2(e+f x) \tan (e+f x)}{4 a f (a+a \sec (e+f x))^{3/2}}\\ \end{align*}

Mathematica [A]  time = 1.53731, size = 136, normalized size = 0.71 \[ -\frac{c^3 \cot \left (\frac{1}{2} (e+f x)\right ) \left ((8 \cos (e+f x)-3 \cos (2 (e+f x))-5) \sec ^4\left (\frac{1}{2} (e+f x)\right )-32 \sqrt{\sec (e+f x)-1} \tan ^{-1}\left (\sqrt{\sec (e+f x)-1}\right )+28 \sqrt{2} \sqrt{\sec (e+f x)-1} \tan ^{-1}\left (\frac{\sqrt{\sec (e+f x)-1}}{\sqrt{2}}\right )\right )}{16 a^2 f \sqrt{a (\sec (e+f x)+1)}} \]

Antiderivative was successfully verified.

[In]

Integrate[(c - c*Sec[e + f*x])^3/(a + a*Sec[e + f*x])^(5/2),x]

[Out]

-(c^3*Cot[(e + f*x)/2]*((-5 + 8*Cos[e + f*x] - 3*Cos[2*(e + f*x)])*Sec[(e + f*x)/2]^4 - 32*ArcTan[Sqrt[-1 + Se
c[e + f*x]]]*Sqrt[-1 + Sec[e + f*x]] + 28*Sqrt[2]*ArcTan[Sqrt[-1 + Sec[e + f*x]]/Sqrt[2]]*Sqrt[-1 + Sec[e + f*
x]]))/(16*a^2*f*Sqrt[a*(1 + Sec[e + f*x])])

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Maple [B]  time = 0.252, size = 553, normalized size = 2.9 \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))^3/(a+a*sec(f*x+e))^(5/2),x)

[Out]

-1/4*c^3/f/a^3*(1/cos(f*x+e)*a*(1+cos(f*x+e)))^(1/2)*(-1+cos(f*x+e))*(-4*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)
)-7*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))-8*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))-14*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))-4*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)+6*cos(f*x+e)^3-7*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))-8*cos(f*x+e)^2+2*cos(f*x+e))/(1+cos(f*x+e))/sin(f*x+e)^3

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Maxima [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))^3/(a+a*sec(f*x+e))^(5/2),x, algorithm="maxima")

[Out]

Timed out

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Fricas [A]  time = 9.10625, size = 1643, normalized size = 8.6 \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))^3/(a+a*sec(f*x+e))^(5/2),x, algorithm="fricas")

[Out]

[-1/8*(7*sqrt(2)*(c^3*cos(f*x + e)^3 + 3*c^3*cos(f*x + e)^2 + 3*c^3*cos(f*x + e) + c^3)*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)) + 8*(c^3*cos(f*x + e)^3 + 3*c^3*cos(f*x + e)^2 + 3*c^3*cos
(f*x + e) + c^3)*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*(3*c^3*cos(f*x + e)^2 - c^3*cos(f*x + e))*sqr
t((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*co
s(f*x + e) + a^3*f), 1/4*(7*sqrt(2)*(c^3*cos(f*x + e)^3 + 3*c^3*cos(f*x + e)^2 + 3*c^3*cos(f*x + e) + c^3)*sqr
t(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))) - 8*(c^3*cos(
f*x + e)^3 + 3*c^3*cos(f*x + e)^2 + 3*c^3*cos(f*x + e) + c^3)*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*(3*c^3*cos(f*x + e)^2 - c^3*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^{3} \left (\int \frac{3 \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{3 \sec ^{2}{\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{\sec ^{3}{\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))**3/(a+a*sec(f*x+e))**(5/2),x)

[Out]

-c**3*(Integral(3*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(-3*sec(e + f*x)**2/(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) + Integ
ral(sec(e + f*x)**3/(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))^3/(a+a*sec(f*x+e))^(5/2),x, algorithm="giac")

[Out]

Timed out