3.208 \(\int \frac{\sec (e+f x) (a+a \sec (e+f x))^3}{(c+d \sec (e+f x))^4} \, dx\)

Optimal. Leaf size=178 \[ \frac{5 a^3 \tanh ^{-1}\left (\frac{\sqrt{c-d} \tan \left (\frac{1}{2} (e+f x)\right )}{\sqrt{c+d}}\right )}{f \sqrt{c-d} (c+d)^{7/2}}+\frac{5 a^3 (c+4 d) \tan (e+f x)}{6 d f (c+d)^3 (c+d \sec (e+f x))}-\frac{5 a^3 (c-d) \tan (e+f x)}{6 d f (c+d)^2 (c+d \sec (e+f x))^2}+\frac{a \tan (e+f x) (a \sec (e+f x)+a)^2}{3 f (c+d) (c+d \sec (e+f x))^3} \]

[Out]

(5*a^3*ArcTanh[(Sqrt[c - d]*Tan[(e + f*x)/2])/Sqrt[c + d]])/(Sqrt[c - d]*(c + d)^(7/2)*f) + (a*(a + a*Sec[e +
f*x])^2*Tan[e + f*x])/(3*(c + d)*f*(c + d*Sec[e + f*x])^3) - (5*a^3*(c - d)*Tan[e + f*x])/(6*d*(c + d)^2*f*(c
+ d*Sec[e + f*x])^2) + (5*a^3*(c + 4*d)*Tan[e + f*x])/(6*d*(c + d)^3*f*(c + d*Sec[e + f*x]))

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Rubi [A]  time = 0.223182, antiderivative size = 227, normalized size of antiderivative = 1.28, number of steps used = 6, number of rules used = 4, integrand size = 31, \(\frac{\text{number of rules}}{\text{integrand size}}\) = 0.129, Rules used = {3987, 94, 93, 205} \[ -\frac{5 a^4 \tan (e+f x) \tan ^{-1}\left (\frac{\sqrt{c+d} \sqrt{a \sec (e+f x)+a}}{\sqrt{c-d} \sqrt{a-a \sec (e+f x)}}\right )}{f \sqrt{c-d} (c+d)^{7/2} \sqrt{a-a \sec (e+f x)} \sqrt{a \sec (e+f x)+a}}+\frac{5 a^3 \tan (e+f x)}{2 f (c+d)^3 (c+d \sec (e+f x))}+\frac{5 \tan (e+f x) \left (a^3 \sec (e+f x)+a^3\right )}{6 f (c+d)^2 (c+d \sec (e+f x))^2}+\frac{a \tan (e+f x) (a \sec (e+f x)+a)^2}{3 f (c+d) (c+d \sec (e+f x))^3} \]

Antiderivative was successfully verified.

[In]

Int[(Sec[e + f*x]*(a + a*Sec[e + f*x])^3)/(c + d*Sec[e + f*x])^4,x]

[Out]

(-5*a^4*ArcTan[(Sqrt[c + d]*Sqrt[a + a*Sec[e + f*x]])/(Sqrt[c - d]*Sqrt[a - a*Sec[e + f*x]])]*Tan[e + f*x])/(S
qrt[c - d]*(c + d)^(7/2)*f*Sqrt[a - a*Sec[e + f*x]]*Sqrt[a + a*Sec[e + f*x]]) + (a*(a + a*Sec[e + f*x])^2*Tan[
e + f*x])/(3*(c + d)*f*(c + d*Sec[e + f*x])^3) + (5*(a^3 + a^3*Sec[e + f*x])*Tan[e + f*x])/(6*(c + d)^2*f*(c +
d*Sec[e + f*x])^2) + (5*a^3*Tan[e + f*x])/(2*(c + d)^3*f*(c + d*Sec[e + f*x]))

Rule 3987

Int[(csc[(e_.) + (f_.)*(x_)]*(g_.))^(p_.)*(csc[(e_.) + (f_.)*(x_)]*(b_.) + (a_))^(m_)*(csc[(e_.) + (f_.)*(x_)]
*(d_.) + (c_))^(n_), x_Symbol] :> Dist[(a^2*g*Cot[e + f*x])/(f*Sqrt[a + b*Csc[e + f*x]]*Sqrt[a - b*Csc[e + f*x
]]), Subst[Int[((g*x)^(p - 1)*(a + b*x)^(m - 1/2)*(c + d*x)^n)/Sqrt[a - b*x], x], x, Csc[e + f*x]], x] /; Free
Q[{a, b, c, d, e, f, g, m, n, p}, x] && NeQ[b*c - a*d, 0] && EqQ[a^2 - b^2, 0] && NeQ[c^2 - d^2, 0] && (EqQ[p,
1] || IntegerQ[m - 1/2])

Rule 94

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

Rule 93

Int[(((a_.) + (b_.)*(x_))^(m_)*((c_.) + (d_.)*(x_))^(n_))/((e_.) + (f_.)*(x_)), x_Symbol] :> With[{q = Denomin
ator[m]}, Dist[q, Subst[Int[x^(q*(m + 1) - 1)/(b*e - a*f - (d*e - c*f)*x^q), x], x, (a + b*x)^(1/q)/(c + d*x)^
(1/q)], x]] /; FreeQ[{a, b, c, d, e, f}, x] && EqQ[m + n + 1, 0] && RationalQ[n] && LtQ[-1, m, 0] && SimplerQ[
a + b*x, c + d*x]

Rule 205

Int[((a_) + (b_.)*(x_)^2)^(-1), x_Symbol] :> Simp[(Rt[a/b, 2]*ArcTan[x/Rt[a/b, 2]])/a, x] /; FreeQ[{a, b}, x]
&& PosQ[a/b]

Rubi steps

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

Mathematica [C]  time = 3.51964, size = 398, normalized size = 2.24 \[ \frac{a^3 \sec ^6\left (\frac{1}{2} (e+f x)\right ) \sec (e+f x) (\sec (e+f x)+1)^3 (c \cos (e+f x)+d) \left (\frac{c \sec (e) \left (-3 \left (30 c^2 d^2-3 c^3 d+6 c^4+18 c d^3+4 d^4\right ) \sin (2 e+f x)+c \left (3 \left (38 c^2 d+3 c^3+12 c d^2+2 d^3\right ) \sin (e+2 f x)+3 \left (-6 c^2 d+3 c^3-6 c d^2-2 d^3\right ) \sin (3 e+2 f x)+c \left (22 c^2+9 c d+2 d^2\right ) \sin (2 e+3 f x)\right )+6 \left (30 c^2 d^2+6 c^3 d+8 c^4+9 c d^3+2 d^4\right ) \sin (f x)\right )-2 d \left (50 c^2 d^2+27 c^3 d+66 c^4+18 c d^3+4 d^4\right ) \tan (e)}{c^3}-\frac{120 i (\cos (e)-i \sin (e)) (c \cos (e+f x)+d)^3 \tan ^{-1}\left (\frac{(\sin (e)+i \cos (e)) \left (\tan \left (\frac{f x}{2}\right ) (c \cos (e)-d)+c \sin (e)\right )}{\sqrt{c^2-d^2} \sqrt{(\cos (e)-i \sin (e))^2}}\right )}{\sqrt{c^2-d^2} \sqrt{(\cos (e)-i \sin (e))^2}}\right )}{192 f (c+d)^3 (c+d \sec (e+f x))^4} \]

Antiderivative was successfully verified.

[In]

Integrate[(Sec[e + f*x]*(a + a*Sec[e + f*x])^3)/(c + d*Sec[e + f*x])^4,x]

[Out]

(a^3*(d + c*Cos[e + f*x])*Sec[(e + f*x)/2]^6*Sec[e + f*x]*(1 + Sec[e + f*x])^3*(((-120*I)*ArcTan[((I*Cos[e] +
Sin[e])*(c*Sin[e] + (-d + c*Cos[e])*Tan[(f*x)/2]))/(Sqrt[c^2 - d^2]*Sqrt[(Cos[e] - I*Sin[e])^2])]*(d + c*Cos[e
+ f*x])^3*(Cos[e] - I*Sin[e]))/(Sqrt[c^2 - d^2]*Sqrt[(Cos[e] - I*Sin[e])^2]) + (c*Sec[e]*(6*(8*c^4 + 6*c^3*d
+ 30*c^2*d^2 + 9*c*d^3 + 2*d^4)*Sin[f*x] - 3*(6*c^4 - 3*c^3*d + 30*c^2*d^2 + 18*c*d^3 + 4*d^4)*Sin[2*e + f*x]
+ c*(3*(3*c^3 + 38*c^2*d + 12*c*d^2 + 2*d^3)*Sin[e + 2*f*x] + 3*(3*c^3 - 6*c^2*d - 6*c*d^2 - 2*d^3)*Sin[3*e +
2*f*x] + c*(22*c^2 + 9*c*d + 2*d^2)*Sin[2*e + 3*f*x])) - 2*d*(66*c^4 + 27*c^3*d + 50*c^2*d^2 + 18*c*d^3 + 4*d^
4)*Tan[e])/c^3))/(192*(c + d)^3*f*(c + d*Sec[e + f*x])^4)

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Maple [A]  time = 0.128, size = 227, normalized size = 1.3 \begin{align*} 16\,{\frac{{a}^{3}}{f} \left ( -1/6\,{\frac{\tan \left ( 1/2\,fx+e/2 \right ) }{ \left ( c+d \right ) \left ( \left ( \tan \left ( 1/2\,fx+e/2 \right ) \right ) ^{2}c- \left ( \tan \left ( 1/2\,fx+e/2 \right ) \right ) ^{2}d-c-d \right ) ^{3}}}-5/6\,{\frac{1}{c+d} \left ( -1/4\,{\frac{\tan \left ( 1/2\,fx+e/2 \right ) }{ \left ( c+d \right ) \left ( \left ( \tan \left ( 1/2\,fx+e/2 \right ) \right ) ^{2}c- \left ( \tan \left ( 1/2\,fx+e/2 \right ) \right ) ^{2}d-c-d \right ) ^{2}}}-3/4\,{\frac{1}{c+d} \left ( -1/2\,{\frac{\tan \left ( 1/2\,fx+e/2 \right ) }{ \left ( c+d \right ) \left ( \left ( \tan \left ( 1/2\,fx+e/2 \right ) \right ) ^{2}c- \left ( \tan \left ( 1/2\,fx+e/2 \right ) \right ) ^{2}d-c-d \right ) }}+1/2\,{\frac{1}{ \left ( c+d \right ) \sqrt{ \left ( c+d \right ) \left ( c-d \right ) }}{\it Artanh} \left ({\frac{\tan \left ( 1/2\,fx+e/2 \right ) \left ( c-d \right ) }{\sqrt{ \left ( c+d \right ) \left ( c-d \right ) }}} \right ) } \right ) } \right ) } \right ) } \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

int(sec(f*x+e)*(a+a*sec(f*x+e))^3/(c+d*sec(f*x+e))^4,x)

[Out]

16/f*a^3*(-1/6*tan(1/2*f*x+1/2*e)/(c+d)/(tan(1/2*f*x+1/2*e)^2*c-tan(1/2*f*x+1/2*e)^2*d-c-d)^3-5/6/(c+d)*(-1/4*
tan(1/2*f*x+1/2*e)/(c+d)/(tan(1/2*f*x+1/2*e)^2*c-tan(1/2*f*x+1/2*e)^2*d-c-d)^2-3/4/(c+d)*(-1/2*tan(1/2*f*x+1/2
*e)/(c+d)/(tan(1/2*f*x+1/2*e)^2*c-tan(1/2*f*x+1/2*e)^2*d-c-d)+1/2/(c+d)/((c+d)*(c-d))^(1/2)*arctanh(tan(1/2*f*
x+1/2*e)*(c-d)/((c+d)*(c-d))^(1/2)))))

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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(sec(f*x+e)*(a+a*sec(f*x+e))^3/(c+d*sec(f*x+e))^4,x, algorithm="maxima")

[Out]

Exception raised: ValueError

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Fricas [B]  time = 0.642665, size = 2178, normalized size = 12.24 \begin{align*} \text{result too large to display} \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate(sec(f*x+e)*(a+a*sec(f*x+e))^3/(c+d*sec(f*x+e))^4,x, algorithm="fricas")

[Out]

[1/12*(15*(a^3*c^3*cos(f*x + e)^3 + 3*a^3*c^2*d*cos(f*x + e)^2 + 3*a^3*c*d^2*cos(f*x + e) + a^3*d^3)*sqrt(c^2
- d^2)*log((2*c*d*cos(f*x + e) - (c^2 - 2*d^2)*cos(f*x + e)^2 + 2*sqrt(c^2 - d^2)*(d*cos(f*x + e) + c)*sin(f*x
+ e) + 2*c^2 - d^2)/(c^2*cos(f*x + e)^2 + 2*c*d*cos(f*x + e) + d^2)) + 2*(2*a^3*c^4 + 9*a^3*c^3*d + 20*a^3*c^
2*d^2 - 9*a^3*c*d^3 - 22*a^3*d^4 + (22*a^3*c^4 + 9*a^3*c^3*d - 20*a^3*c^2*d^2 - 9*a^3*c*d^3 - 2*a^3*d^4)*cos(f
*x + e)^2 + 3*(3*a^3*c^4 + 16*a^3*c^3*d - 16*a^3*c*d^3 - 3*a^3*d^4)*cos(f*x + e))*sin(f*x + e))/((c^8 + 3*c^7*
d + 2*c^6*d^2 - 2*c^5*d^3 - 3*c^4*d^4 - c^3*d^5)*f*cos(f*x + e)^3 + 3*(c^7*d + 3*c^6*d^2 + 2*c^5*d^3 - 2*c^4*d
^4 - 3*c^3*d^5 - c^2*d^6)*f*cos(f*x + e)^2 + 3*(c^6*d^2 + 3*c^5*d^3 + 2*c^4*d^4 - 2*c^3*d^5 - 3*c^2*d^6 - c*d^
7)*f*cos(f*x + e) + (c^5*d^3 + 3*c^4*d^4 + 2*c^3*d^5 - 2*c^2*d^6 - 3*c*d^7 - d^8)*f), 1/6*(15*(a^3*c^3*cos(f*x
+ e)^3 + 3*a^3*c^2*d*cos(f*x + e)^2 + 3*a^3*c*d^2*cos(f*x + e) + a^3*d^3)*sqrt(-c^2 + d^2)*arctan(-sqrt(-c^2
+ d^2)*(d*cos(f*x + e) + c)/((c^2 - d^2)*sin(f*x + e))) + (2*a^3*c^4 + 9*a^3*c^3*d + 20*a^3*c^2*d^2 - 9*a^3*c*
d^3 - 22*a^3*d^4 + (22*a^3*c^4 + 9*a^3*c^3*d - 20*a^3*c^2*d^2 - 9*a^3*c*d^3 - 2*a^3*d^4)*cos(f*x + e)^2 + 3*(3
*a^3*c^4 + 16*a^3*c^3*d - 16*a^3*c*d^3 - 3*a^3*d^4)*cos(f*x + e))*sin(f*x + e))/((c^8 + 3*c^7*d + 2*c^6*d^2 -
2*c^5*d^3 - 3*c^4*d^4 - c^3*d^5)*f*cos(f*x + e)^3 + 3*(c^7*d + 3*c^6*d^2 + 2*c^5*d^3 - 2*c^4*d^4 - 3*c^3*d^5 -
c^2*d^6)*f*cos(f*x + e)^2 + 3*(c^6*d^2 + 3*c^5*d^3 + 2*c^4*d^4 - 2*c^3*d^5 - 3*c^2*d^6 - c*d^7)*f*cos(f*x + e
) + (c^5*d^3 + 3*c^4*d^4 + 2*c^3*d^5 - 2*c^2*d^6 - 3*c*d^7 - d^8)*f)]

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

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate(sec(f*x+e)*(a+a*sec(f*x+e))**3/(c+d*sec(f*x+e))**4,x)

[Out]

a**3*(Integral(sec(e + f*x)/(c**4 + 4*c**3*d*sec(e + f*x) + 6*c**2*d**2*sec(e + f*x)**2 + 4*c*d**3*sec(e + f*x
)**3 + d**4*sec(e + f*x)**4), x) + Integral(3*sec(e + f*x)**2/(c**4 + 4*c**3*d*sec(e + f*x) + 6*c**2*d**2*sec(
e + f*x)**2 + 4*c*d**3*sec(e + f*x)**3 + d**4*sec(e + f*x)**4), x) + Integral(3*sec(e + f*x)**3/(c**4 + 4*c**3
*d*sec(e + f*x) + 6*c**2*d**2*sec(e + f*x)**2 + 4*c*d**3*sec(e + f*x)**3 + d**4*sec(e + f*x)**4), x) + Integra
l(sec(e + f*x)**4/(c**4 + 4*c**3*d*sec(e + f*x) + 6*c**2*d**2*sec(e + f*x)**2 + 4*c*d**3*sec(e + f*x)**3 + d**
4*sec(e + f*x)**4), x))

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Giac [A]  time = 1.45993, size = 432, normalized size = 2.43 \begin{align*} -\frac{\frac{15 \,{\left (\pi \left \lfloor \frac{f x + e}{2 \, \pi } + \frac{1}{2} \right \rfloor \mathrm{sgn}\left (2 \, c - 2 \, d\right ) + \arctan \left (\frac{c \tan \left (\frac{1}{2} \, f x + \frac{1}{2} \, e\right ) - d \tan \left (\frac{1}{2} \, f x + \frac{1}{2} \, e\right )}{\sqrt{-c^{2} + d^{2}}}\right )\right )} a^{3}}{{\left (c^{3} + 3 \, c^{2} d + 3 \, c d^{2} + d^{3}\right )} \sqrt{-c^{2} + d^{2}}} + \frac{15 \, a^{3} c^{2} \tan \left (\frac{1}{2} \, f x + \frac{1}{2} \, e\right )^{5} - 30 \, a^{3} c d \tan \left (\frac{1}{2} \, f x + \frac{1}{2} \, e\right )^{5} + 15 \, a^{3} d^{2} \tan \left (\frac{1}{2} \, f x + \frac{1}{2} \, e\right )^{5} - 40 \, a^{3} c^{2} \tan \left (\frac{1}{2} \, f x + \frac{1}{2} \, e\right )^{3} + 40 \, a^{3} d^{2} \tan \left (\frac{1}{2} \, f x + \frac{1}{2} \, e\right )^{3} + 33 \, a^{3} c^{2} \tan \left (\frac{1}{2} \, f x + \frac{1}{2} \, e\right ) + 66 \, a^{3} c d \tan \left (\frac{1}{2} \, f x + \frac{1}{2} \, e\right ) + 33 \, a^{3} d^{2} \tan \left (\frac{1}{2} \, f x + \frac{1}{2} \, e\right )}{{\left (c^{3} + 3 \, c^{2} d + 3 \, c d^{2} + d^{3}\right )}{\left (c \tan \left (\frac{1}{2} \, f x + \frac{1}{2} \, e\right )^{2} - d \tan \left (\frac{1}{2} \, f x + \frac{1}{2} \, e\right )^{2} - c - d\right )}^{3}}}{3 \, f} \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate(sec(f*x+e)*(a+a*sec(f*x+e))^3/(c+d*sec(f*x+e))^4,x, algorithm="giac")

[Out]

-1/3*(15*(pi*floor(1/2*(f*x + e)/pi + 1/2)*sgn(2*c - 2*d) + arctan((c*tan(1/2*f*x + 1/2*e) - d*tan(1/2*f*x + 1
/2*e))/sqrt(-c^2 + d^2)))*a^3/((c^3 + 3*c^2*d + 3*c*d^2 + d^3)*sqrt(-c^2 + d^2)) + (15*a^3*c^2*tan(1/2*f*x + 1
/2*e)^5 - 30*a^3*c*d*tan(1/2*f*x + 1/2*e)^5 + 15*a^3*d^2*tan(1/2*f*x + 1/2*e)^5 - 40*a^3*c^2*tan(1/2*f*x + 1/2
*e)^3 + 40*a^3*d^2*tan(1/2*f*x + 1/2*e)^3 + 33*a^3*c^2*tan(1/2*f*x + 1/2*e) + 66*a^3*c*d*tan(1/2*f*x + 1/2*e)
+ 33*a^3*d^2*tan(1/2*f*x + 1/2*e))/((c^3 + 3*c^2*d + 3*c*d^2 + d^3)*(c*tan(1/2*f*x + 1/2*e)^2 - d*tan(1/2*f*x
+ 1/2*e)^2 - c - d)^3))/f