3.65 \(\int \frac{(c-c \sec (e+f x))^4}{\sqrt{a+a \sec (e+f x)}} \, dx\)

Optimal. Leaf size=185 \[ \frac{2 a^2 c^4 \tan ^5(e+f x)}{5 f (a \sec (e+f x)+a)^{5/2}}-\frac{2 a c^4 \tan ^3(e+f x)}{f (a \sec (e+f x)+a)^{3/2}}+\frac{2 c^4 \tan ^{-1}\left (\frac{\sqrt{a} \tan (e+f x)}{\sqrt{a \sec (e+f x)+a}}\right )}{\sqrt{a} f}-\frac{16 \sqrt{2} c^4 \tan ^{-1}\left (\frac{\sqrt{a} \tan (e+f x)}{\sqrt{2} \sqrt{a \sec (e+f x)+a}}\right )}{\sqrt{a} f}+\frac{14 c^4 \tan (e+f x)}{f \sqrt{a \sec (e+f x)+a}} \]

[Out]

(2*c^4*ArcTan[(Sqrt[a]*Tan[e + f*x])/Sqrt[a + a*Sec[e + f*x]]])/(Sqrt[a]*f) - (16*Sqrt[2]*c^4*ArcTan[(Sqrt[a]*
Tan[e + f*x])/(Sqrt[2]*Sqrt[a + a*Sec[e + f*x]])])/(Sqrt[a]*f) + (14*c^4*Tan[e + f*x])/(f*Sqrt[a + a*Sec[e + f
*x]]) - (2*a*c^4*Tan[e + f*x]^3)/(f*(a + a*Sec[e + f*x])^(3/2)) + (2*a^2*c^4*Tan[e + f*x]^5)/(5*f*(a + a*Sec[e
+ f*x])^(5/2))

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Rubi [A]  time = 0.276144, antiderivative size = 185, 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, 479, 582, 522, 203} \[ \frac{2 a^2 c^4 \tan ^5(e+f x)}{5 f (a \sec (e+f x)+a)^{5/2}}-\frac{2 a c^4 \tan ^3(e+f x)}{f (a \sec (e+f x)+a)^{3/2}}+\frac{2 c^4 \tan ^{-1}\left (\frac{\sqrt{a} \tan (e+f x)}{\sqrt{a \sec (e+f x)+a}}\right )}{\sqrt{a} f}-\frac{16 \sqrt{2} c^4 \tan ^{-1}\left (\frac{\sqrt{a} \tan (e+f x)}{\sqrt{2} \sqrt{a \sec (e+f x)+a}}\right )}{\sqrt{a} f}+\frac{14 c^4 \tan (e+f x)}{f \sqrt{a \sec (e+f x)+a}} \]

Antiderivative was successfully verified.

[In]

Int[(c - c*Sec[e + f*x])^4/Sqrt[a + a*Sec[e + f*x]],x]

[Out]

(2*c^4*ArcTan[(Sqrt[a]*Tan[e + f*x])/Sqrt[a + a*Sec[e + f*x]]])/(Sqrt[a]*f) - (16*Sqrt[2]*c^4*ArcTan[(Sqrt[a]*
Tan[e + f*x])/(Sqrt[2]*Sqrt[a + a*Sec[e + f*x]])])/(Sqrt[a]*f) + (14*c^4*Tan[e + f*x])/(f*Sqrt[a + a*Sec[e + f
*x]]) - (2*a*c^4*Tan[e + f*x]^3)/(f*(a + a*Sec[e + f*x])^(3/2)) + (2*a^2*c^4*Tan[e + f*x]^5)/(5*f*(a + a*Sec[e
+ f*x])^(5/2))

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 479

Int[((e_.)*(x_))^(m_.)*((a_) + (b_.)*(x_)^(n_))^(p_)*((c_) + (d_.)*(x_)^(n_))^(q_), x_Symbol] :> Simp[(e^(2*n
- 1)*(e*x)^(m - 2*n + 1)*(a + b*x^n)^(p + 1)*(c + d*x^n)^(q + 1))/(b*d*(m + n*(p + q) + 1)), x] - Dist[e^(2*n)
/(b*d*(m + n*(p + q) + 1)), Int[(e*x)^(m - 2*n)*(a + b*x^n)^p*(c + d*x^n)^q*Simp[a*c*(m - 2*n + 1) + (a*d*(m +
n*(q - 1) + 1) + b*c*(m + n*(p - 1) + 1))*x^n, x], x], x] /; FreeQ[{a, b, c, d, e, p, q}, x] && NeQ[b*c - a*d
, 0] && IGtQ[n, 0] && GtQ[m - n + 1, n] && IntBinomialQ[a, b, c, d, e, m, n, p, q, x]

Rule 582

Int[((g_.)*(x_))^(m_.)*((a_) + (b_.)*(x_)^(n_))^(p_.)*((c_) + (d_.)*(x_)^(n_))^(q_.)*((e_) + (f_.)*(x_)^(n_)),
x_Symbol] :> Simp[(f*g^(n - 1)*(g*x)^(m - n + 1)*(a + b*x^n)^(p + 1)*(c + d*x^n)^(q + 1))/(b*d*(m + n*(p + q
+ 1) + 1)), x] - Dist[g^n/(b*d*(m + n*(p + q + 1) + 1)), Int[(g*x)^(m - n)*(a + b*x^n)^p*(c + d*x^n)^q*Simp[a*
f*c*(m - n + 1) + (a*f*d*(m + n*q + 1) + b*(f*c*(m + n*p + 1) - e*d*(m + n*(p + q + 1) + 1)))*x^n, x], x], x]
/; FreeQ[{a, b, c, d, e, f, g, p, q}, x] && IGtQ[n, 0] && GtQ[m, n - 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))^4}{\sqrt{a+a \sec (e+f x)}} \, dx &=\left (a^4 c^4\right ) \int \frac{\tan ^8(e+f x)}{(a+a \sec (e+f x))^{9/2}} \, dx\\ &=-\frac{\left (2 a^4 c^4\right ) \operatorname{Subst}\left (\int \frac{x^8}{\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 )}{f}\\ &=\frac{2 a^2 c^4 \tan ^5(e+f x)}{5 f (a+a \sec (e+f x))^{5/2}}+\frac{\left (2 a^2 c^4\right ) \operatorname{Subst}\left (\int \frac{x^4 \left (10+15 a x^2\right )}{\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 f}\\ &=-\frac{2 a c^4 \tan ^3(e+f x)}{f (a+a \sec (e+f x))^{3/2}}+\frac{2 a^2 c^4 \tan ^5(e+f x)}{5 f (a+a \sec (e+f x))^{5/2}}-\frac{\left (2 c^4\right ) \operatorname{Subst}\left (\int \frac{x^2 \left (90 a+105 a^2 x^2\right )}{\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 )}{15 f}\\ &=\frac{14 c^4 \tan (e+f x)}{f \sqrt{a+a \sec (e+f x)}}-\frac{2 a c^4 \tan ^3(e+f x)}{f (a+a \sec (e+f x))^{3/2}}+\frac{2 a^2 c^4 \tan ^5(e+f x)}{5 f (a+a \sec (e+f x))^{5/2}}+\frac{\left (2 c^4\right ) \operatorname{Subst}\left (\int \frac{210 a^2+225 a^3 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 )}{15 a^2 f}\\ &=\frac{14 c^4 \tan (e+f x)}{f \sqrt{a+a \sec (e+f x)}}-\frac{2 a c^4 \tan ^3(e+f x)}{f (a+a \sec (e+f x))^{3/2}}+\frac{2 a^2 c^4 \tan ^5(e+f x)}{5 f (a+a \sec (e+f x))^{5/2}}-\frac{\left (2 c^4\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 )}{f}+\frac{\left (32 c^4\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 )}{f}\\ &=\frac{2 c^4 \tan ^{-1}\left (\frac{\sqrt{a} \tan (e+f x)}{\sqrt{a+a \sec (e+f x)}}\right )}{\sqrt{a} f}-\frac{16 \sqrt{2} c^4 \tan ^{-1}\left (\frac{\sqrt{a} \tan (e+f x)}{\sqrt{2} \sqrt{a+a \sec (e+f x)}}\right )}{\sqrt{a} f}+\frac{14 c^4 \tan (e+f x)}{f \sqrt{a+a \sec (e+f x)}}-\frac{2 a c^4 \tan ^3(e+f x)}{f (a+a \sec (e+f x))^{3/2}}+\frac{2 a^2 c^4 \tan ^5(e+f x)}{5 f (a+a \sec (e+f x))^{5/2}}\\ \end{align*}

Mathematica [A]  time = 1.42351, size = 153, normalized size = 0.83 \[ \frac{c^4 \cot \left (\frac{1}{2} (e+f x)\right ) \sec ^3(e+f x) \left (-155 \cos (e+f x)+96 \cos (2 (e+f x))-41 \cos (3 (e+f x))+20 \cos ^3(e+f x) \sqrt{\sec (e+f x)-1} \tan ^{-1}\left (\sqrt{\sec (e+f x)-1}\right )-160 \sqrt{2} \cos ^3(e+f x) \sqrt{\sec (e+f x)-1} \tan ^{-1}\left (\frac{\sqrt{\sec (e+f x)-1}}{\sqrt{2}}\right )+100\right )}{10 f \sqrt{a (\sec (e+f x)+1)}} \]

Antiderivative was successfully verified.

[In]

Integrate[(c - c*Sec[e + f*x])^4/Sqrt[a + a*Sec[e + f*x]],x]

[Out]

(c^4*Cot[(e + f*x)/2]*(100 - 155*Cos[e + f*x] + 96*Cos[2*(e + f*x)] - 41*Cos[3*(e + f*x)] + 20*ArcTan[Sqrt[-1
+ Sec[e + f*x]]]*Cos[e + f*x]^3*Sqrt[-1 + Sec[e + f*x]] - 160*Sqrt[2]*ArcTan[Sqrt[-1 + Sec[e + f*x]]/Sqrt[2]]*
Cos[e + f*x]^3*Sqrt[-1 + Sec[e + f*x]])*Sec[e + f*x]^3)/(10*f*Sqrt[a*(1 + Sec[e + f*x])])

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Maple [B]  time = 0.321, size = 544, 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))^4/(a+a*sec(f*x+e))^(1/2),x)

[Out]

-1/20*c^4/f/a*(1/cos(f*x+e)*a*(1+cos(f*x+e)))^(1/2)*(5*2^(1/2)*sin(f*x+e)*cos(f*x+e)^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))*(-2*cos(f*x+e)/(1+cos(f*x+e)))^(5/2)+80*sin(f*x+e)*co
s(f*x+e)^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))*(-2*cos(f*x+e)/(1+cos
(f*x+e)))^(5/2)+10*2^(1/2)*sin(f*x+e)*cos(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)))^(5/2)+160*sin(f*x+e)*cos(f*x+e)*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))*(-2*cos(f*x+e)/(1+cos(f*x+e)))^(5/2)+5*2^(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))*(-2*cos(f*x+e)/(1+cos(f*x+e)))^(5/2)*sin(f*
x+e)+80*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))*(-2*cos(f*x+e)/(1+cos(f*
x+e)))^(5/2)*sin(f*x+e)+328*cos(f*x+e)^3-384*cos(f*x+e)^2+64*cos(f*x+e)-8)/cos(f*x+e)^2/sin(f*x+e)

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Maxima [F(-2)]  time = 0., size = 0, normalized size = 0. \begin{align*} \text{Exception raised: ValueError} \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate((c-c*sec(f*x+e))^4/(a+a*sec(f*x+e))^(1/2),x, algorithm="maxima")

[Out]

Exception raised: ValueError

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Fricas [A]  time = 5.2609, size = 1420, normalized size = 7.68 \begin{align*} \left [\frac{40 \, \sqrt{2}{\left (a c^{4} \cos \left (f x + e\right )^{3} + a c^{4} \cos \left (f x + e\right )^{2}\right )} \sqrt{-\frac{1}{a}} \log \left (\frac{2 \, \sqrt{2} \sqrt{\frac{a \cos \left (f x + e\right ) + a}{\cos \left (f x + e\right )}} \sqrt{-\frac{1}{a}} \cos \left (f x + e\right ) \sin \left (f x + e\right ) + 3 \, \cos \left (f x + e\right )^{2} + 2 \, \cos \left (f x + e\right ) - 1}{\cos \left (f x + e\right )^{2} + 2 \, \cos \left (f x + e\right ) + 1}\right ) - 5 \,{\left (c^{4} \cos \left (f x + e\right )^{3} + c^{4} \cos \left (f x + e\right )^{2}\right )} \sqrt{-a} \log \left (\frac{2 \, a \cos \left (f x + e\right )^{2} + 2 \, \sqrt{-a} \sqrt{\frac{a \cos \left (f x + e\right ) + a}{\cos \left (f x + e\right )}} \cos \left (f x + e\right ) \sin \left (f x + e\right ) + a \cos \left (f x + e\right ) - a}{\cos \left (f x + e\right ) + 1}\right ) + 2 \,{\left (41 \, c^{4} \cos \left (f x + e\right )^{2} - 7 \, c^{4} \cos \left (f x + e\right ) + c^{4}\right )} \sqrt{\frac{a \cos \left (f x + e\right ) + a}{\cos \left (f x + e\right )}} \sin \left (f x + e\right )}{5 \,{\left (a f \cos \left (f x + e\right )^{3} + a f \cos \left (f x + e\right )^{2}\right )}}, -\frac{2 \,{\left (5 \,{\left (c^{4} \cos \left (f x + e\right )^{3} + c^{4} \cos \left (f x + e\right )^{2}\right )} \sqrt{a} \arctan \left (\frac{\sqrt{\frac{a \cos \left (f x + e\right ) + a}{\cos \left (f x + e\right )}} \cos \left (f x + e\right )}{\sqrt{a} \sin \left (f x + e\right )}\right ) -{\left (41 \, c^{4} \cos \left (f x + e\right )^{2} - 7 \, c^{4} \cos \left (f x + e\right ) + c^{4}\right )} \sqrt{\frac{a \cos \left (f x + e\right ) + a}{\cos \left (f x + e\right )}} \sin \left (f x + e\right ) - \frac{40 \, \sqrt{2}{\left (a c^{4} \cos \left (f x + e\right )^{3} + a c^{4} \cos \left (f x + e\right )^{2}\right )} \arctan \left (\frac{\sqrt{2} \sqrt{\frac{a \cos \left (f x + e\right ) + a}{\cos \left (f x + e\right )}} \cos \left (f x + e\right )}{\sqrt{a} \sin \left (f x + e\right )}\right )}{\sqrt{a}}\right )}}{5 \,{\left (a f \cos \left (f x + e\right )^{3} + a f \cos \left (f x + e\right )^{2}\right )}}\right ] \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate((c-c*sec(f*x+e))^4/(a+a*sec(f*x+e))^(1/2),x, algorithm="fricas")

[Out]

[1/5*(40*sqrt(2)*(a*c^4*cos(f*x + e)^3 + a*c^4*cos(f*x + e)^2)*sqrt(-1/a)*log((2*sqrt(2)*sqrt((a*cos(f*x + e)
+ a)/cos(f*x + e))*sqrt(-1/a)*cos(f*x + e)*sin(f*x + e) + 3*cos(f*x + e)^2 + 2*cos(f*x + e) - 1)/(cos(f*x + e)
^2 + 2*cos(f*x + e) + 1)) - 5*(c^4*cos(f*x + e)^3 + c^4*cos(f*x + e)^2)*sqrt(-a)*log((2*a*cos(f*x + e)^2 + 2*s
qrt(-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)) + 2*(41*c^4*cos(f*x + e)^2 - 7*c^4*cos(f*x + e) + c^4)*sqrt((a*cos(f*x + e) + a)/cos(f*x + e))*sin(f*x +
e))/(a*f*cos(f*x + e)^3 + a*f*cos(f*x + e)^2), -2/5*(5*(c^4*cos(f*x + e)^3 + c^4*cos(f*x + e)^2)*sqrt(a)*arct
an(sqrt((a*cos(f*x + e) + a)/cos(f*x + e))*cos(f*x + e)/(sqrt(a)*sin(f*x + e))) - (41*c^4*cos(f*x + e)^2 - 7*c
^4*cos(f*x + e) + c^4)*sqrt((a*cos(f*x + e) + a)/cos(f*x + e))*sin(f*x + e) - 40*sqrt(2)*(a*c^4*cos(f*x + e)^3
+ a*c^4*cos(f*x + e)^2)*arctan(sqrt(2)*sqrt((a*cos(f*x + e) + a)/cos(f*x + e))*cos(f*x + e)/(sqrt(a)*sin(f*x
+ e)))/sqrt(a))/(a*f*cos(f*x + e)^3 + a*f*cos(f*x + e)^2)]

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Sympy [F]  time = 0., size = 0, normalized size = 0. \begin{align*} c^{4} \left (\int - \frac{4 \sec{\left (e + f x \right )}}{\sqrt{a \sec{\left (e + f x \right )} + a}}\, dx + \int \frac{6 \sec ^{2}{\left (e + f x \right )}}{\sqrt{a \sec{\left (e + f x \right )} + a}}\, dx + \int - \frac{4 \sec ^{3}{\left (e + f x \right )}}{\sqrt{a \sec{\left (e + f x \right )} + a}}\, dx + \int \frac{\sec ^{4}{\left (e + f x \right )}}{\sqrt{a \sec{\left (e + f x \right )} + a}}\, dx + \int \frac{1}{\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))**4/(a+a*sec(f*x+e))**(1/2),x)

[Out]

c**4*(Integral(-4*sec(e + f*x)/sqrt(a*sec(e + f*x) + a), x) + Integral(6*sec(e + f*x)**2/sqrt(a*sec(e + f*x) +
a), x) + Integral(-4*sec(e + f*x)**3/sqrt(a*sec(e + f*x) + a), x) + Integral(sec(e + f*x)**4/sqrt(a*sec(e + f
*x) + a), x) + Integral(1/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))^4/(a+a*sec(f*x+e))^(1/2),x, algorithm="giac")

[Out]

Timed out