3.209 \(\int \frac{\sec (e+f x) (a+a \sec (e+f x))^3}{(c+d \sec (e+f x))^5} \, dx\)

Optimal. Leaf size=266 \[ \frac{5 a^3 (4 c-3 d) \tanh ^{-1}\left (\frac{\sqrt{c-d} \tan \left (\frac{1}{2} (e+f x)\right )}{\sqrt{c+d}}\right )}{4 f (c-d)^{3/2} (c+d)^{9/2}}+\frac{5 a^3 (4 c-3 d) (c+4 d) \tan (e+f x)}{24 d f (c-d) (c+d)^4 (c+d \sec (e+f x))}-\frac{5 a^3 (4 c-3 d) \tan (e+f x)}{24 d f (c+d)^3 (c+d \sec (e+f x))^2}-\frac{d \tan (e+f x) (a \sec (e+f x)+a)^3}{4 f \left (c^2-d^2\right ) (c+d \sec (e+f x))^4}+\frac{a (4 c-3 d) \tan (e+f x) (a \sec (e+f x)+a)^2}{12 f (c-d) (c+d)^2 (c+d \sec (e+f x))^3} \]

[Out]

(5*a^3*(4*c - 3*d)*ArcTanh[(Sqrt[c - d]*Tan[(e + f*x)/2])/Sqrt[c + d]])/(4*(c - d)^(3/2)*(c + d)^(9/2)*f) - (d
*(a + a*Sec[e + f*x])^3*Tan[e + f*x])/(4*(c^2 - d^2)*f*(c + d*Sec[e + f*x])^4) + (a*(4*c - 3*d)*(a + a*Sec[e +
f*x])^2*Tan[e + f*x])/(12*(c - d)*(c + d)^2*f*(c + d*Sec[e + f*x])^3) - (5*a^3*(4*c - 3*d)*Tan[e + f*x])/(24*
d*(c + d)^3*f*(c + d*Sec[e + f*x])^2) + (5*a^3*(4*c - 3*d)*(c + 4*d)*Tan[e + f*x])/(24*(c - d)*d*(c + d)^4*f*(
c + d*Sec[e + f*x]))

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Rubi [A]  time = 0.33683, antiderivative size = 327, normalized size of antiderivative = 1.23, number of steps used = 7, number of rules used = 5, integrand size = 31, \(\frac{\text{number of rules}}{\text{integrand size}}\) = 0.161, Rules used = {3987, 96, 94, 93, 205} \[ -\frac{5 a^4 (4 c-3 d) \tan (e+f x) \tan ^{-1}\left (\frac{\sqrt{c+d} \sqrt{a \sec (e+f x)+a}}{\sqrt{c-d} \sqrt{a-a \sec (e+f x)}}\right )}{4 f (c-d)^{3/2} (c+d)^{9/2} \sqrt{a-a \sec (e+f x)} \sqrt{a \sec (e+f x)+a}}+\frac{5 a^3 (4 c-3 d) \tan (e+f x)}{8 f (c-d) (c+d)^4 (c+d \sec (e+f x))}+\frac{5 (4 c-3 d) \tan (e+f x) \left (a^3 \sec (e+f x)+a^3\right )}{24 f (c-d) (c+d)^3 (c+d \sec (e+f x))^2}-\frac{d \tan (e+f x) (a \sec (e+f x)+a)^3}{4 f \left (c^2-d^2\right ) (c+d \sec (e+f x))^4}+\frac{a (4 c-3 d) \tan (e+f x) (a \sec (e+f x)+a)^2}{12 f (c-d) (c+d)^2 (c+d \sec (e+f x))^3} \]

Antiderivative was successfully verified.

[In]

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

[Out]

(-5*a^4*(4*c - 3*d)*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])/(4*(c - d)^(3/2)*(c + d)^(9/2)*f*Sqrt[a - a*Sec[e + f*x]]*Sqrt[a + a*Sec[e + f*x]]) - (d*(a + a*Sec[
e + f*x])^3*Tan[e + f*x])/(4*(c^2 - d^2)*f*(c + d*Sec[e + f*x])^4) + (a*(4*c - 3*d)*(a + a*Sec[e + f*x])^2*Tan
[e + f*x])/(12*(c - d)*(c + d)^2*f*(c + d*Sec[e + f*x])^3) + (5*(4*c - 3*d)*(a^3 + a^3*Sec[e + f*x])*Tan[e + f
*x])/(24*(c - d)*(c + d)^3*f*(c + d*Sec[e + f*x])^2) + (5*a^3*(4*c - 3*d)*Tan[e + f*x])/(8*(c - d)*(c + d)^4*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 96

Int[((a_.) + (b_.)*(x_))^(m_)*((c_.) + (d_.)*(x_))^(n_.)*((e_.) + (f_.)*(x_))^(p_.), x_Symbol] :> Simp[(b*(a +
b*x)^(m + 1)*(c + d*x)^(n + 1)*(e + f*x)^(p + 1))/((m + 1)*(b*c - a*d)*(b*e - a*f)), x] + Dist[(a*d*f*(m + 1)
+ b*c*f*(n + 1) + b*d*e*(p + 1))/((m + 1)*(b*c - a*d)*(b*e - a*f)), Int[(a + b*x)^(m + 1)*(c + d*x)^n*(e + f*
x)^p, x], x] /; FreeQ[{a, b, c, d, e, f, m, n, p}, x] && EqQ[Simplify[m + n + p + 3], 0] && (LtQ[m, -1] || Sum
SimplerQ[m, 1])

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))^5} \, 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)^5} \, dx,x,\sec (e+f x)\right )}{f \sqrt{a-a \sec (e+f x)} \sqrt{a+a \sec (e+f x)}}\\ &=-\frac{d (a+a \sec (e+f x))^3 \tan (e+f x)}{4 \left (c^2-d^2\right ) f (c+d \sec (e+f x))^4}-\frac{\left (a^2 (4 c-3 d) \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 )}{4 \left (c^2-d^2\right ) f \sqrt{a-a \sec (e+f x)} \sqrt{a+a \sec (e+f x)}}\\ &=-\frac{d (a+a \sec (e+f x))^3 \tan (e+f x)}{4 \left (c^2-d^2\right ) f (c+d \sec (e+f x))^4}+\frac{a (4 c-3 d) (a+a \sec (e+f x))^2 \tan (e+f x)}{12 (c-d) (c+d)^2 f (c+d \sec (e+f x))^3}-\frac{\left (5 a^3 (4 c-3 d) \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 )}{12 (c+d) \left (c^2-d^2\right ) f \sqrt{a-a \sec (e+f x)} \sqrt{a+a \sec (e+f x)}}\\ &=-\frac{d (a+a \sec (e+f x))^3 \tan (e+f x)}{4 \left (c^2-d^2\right ) f (c+d \sec (e+f x))^4}+\frac{a (4 c-3 d) (a+a \sec (e+f x))^2 \tan (e+f x)}{12 (c-d) (c+d)^2 f (c+d \sec (e+f x))^3}+\frac{5 (4 c-3 d) \left (a^3+a^3 \sec (e+f x)\right ) \tan (e+f x)}{24 (c-d) (c+d)^3 f (c+d \sec (e+f x))^2}-\frac{\left (5 a^4 (4 c-3 d) \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 )}{8 (c+d)^2 \left (c^2-d^2\right ) f \sqrt{a-a \sec (e+f x)} \sqrt{a+a \sec (e+f x)}}\\ &=-\frac{d (a+a \sec (e+f x))^3 \tan (e+f x)}{4 \left (c^2-d^2\right ) f (c+d \sec (e+f x))^4}+\frac{a (4 c-3 d) (a+a \sec (e+f x))^2 \tan (e+f x)}{12 (c-d) (c+d)^2 f (c+d \sec (e+f x))^3}+\frac{5 (4 c-3 d) \left (a^3+a^3 \sec (e+f x)\right ) \tan (e+f x)}{24 (c-d) (c+d)^3 f (c+d \sec (e+f x))^2}+\frac{5 a^3 (4 c-3 d) \tan (e+f x)}{8 (c-d) (c+d)^4 f (c+d \sec (e+f x))}-\frac{\left (5 a^5 (4 c-3 d) \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 )}{8 (c+d)^3 \left (c^2-d^2\right ) f \sqrt{a-a \sec (e+f x)} \sqrt{a+a \sec (e+f x)}}\\ &=-\frac{d (a+a \sec (e+f x))^3 \tan (e+f x)}{4 \left (c^2-d^2\right ) f (c+d \sec (e+f x))^4}+\frac{a (4 c-3 d) (a+a \sec (e+f x))^2 \tan (e+f x)}{12 (c-d) (c+d)^2 f (c+d \sec (e+f x))^3}+\frac{5 (4 c-3 d) \left (a^3+a^3 \sec (e+f x)\right ) \tan (e+f x)}{24 (c-d) (c+d)^3 f (c+d \sec (e+f x))^2}+\frac{5 a^3 (4 c-3 d) \tan (e+f x)}{8 (c-d) (c+d)^4 f (c+d \sec (e+f x))}-\frac{\left (5 a^5 (4 c-3 d) \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 )}{4 (c+d)^3 \left (c^2-d^2\right ) f \sqrt{a-a \sec (e+f x)} \sqrt{a+a \sec (e+f x)}}\\ &=-\frac{5 a^4 (4 c-3 d) \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)}{4 (c-d)^{3/2} (c+d)^{9/2} f \sqrt{a-a \sec (e+f x)} \sqrt{a+a \sec (e+f x)}}-\frac{d (a+a \sec (e+f x))^3 \tan (e+f x)}{4 \left (c^2-d^2\right ) f (c+d \sec (e+f x))^4}+\frac{a (4 c-3 d) (a+a \sec (e+f x))^2 \tan (e+f x)}{12 (c-d) (c+d)^2 f (c+d \sec (e+f x))^3}+\frac{5 (4 c-3 d) \left (a^3+a^3 \sec (e+f x)\right ) \tan (e+f x)}{24 (c-d) (c+d)^3 f (c+d \sec (e+f x))^2}+\frac{5 a^3 (4 c-3 d) \tan (e+f x)}{8 (c-d) (c+d)^4 f (c+d \sec (e+f x))}\\ \end{align*}

Mathematica [A]  time = 9.66236, size = 274, normalized size = 1.03 \[ \frac{a^3 \left (-\frac{\sin (e+f x) \left (37 c^2 d^2 \cos (3 (e+f x))+\left (-577 c^2 d^2-84 c^3 d-296 c^4+984 c d^3+198 d^4\right ) \cos (e+f x)+\left (384 c^2 d^2-470 c^3 d-72 c^4+200 c d^3+48 d^4\right ) \cos (2 (e+f x))+336 c^2 d^2+36 c^3 d \cos (3 (e+f x))-478 c^3 d-88 c^4 \cos (3 (e+f x))-72 c^4+24 c d^3 \cos (3 (e+f x))+28 c d^3+6 d^4 \cos (3 (e+f x))+336 d^4\right )}{(c \cos (e+f x)+d)^4}-\frac{120 (4 c-3 d) \tanh ^{-1}\left (\frac{(d-c) \tan \left (\frac{1}{2} (e+f x)\right )}{\sqrt{c^2-d^2}}\right )}{\sqrt{c^2-d^2}}\right )}{96 f (c-d) (c+d)^4} \]

Antiderivative was successfully verified.

[In]

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

[Out]

(a^3*((-120*(4*c - 3*d)*ArcTanh[((-c + d)*Tan[(e + f*x)/2])/Sqrt[c^2 - d^2]])/Sqrt[c^2 - d^2] - ((-72*c^4 - 47
8*c^3*d + 336*c^2*d^2 + 28*c*d^3 + 336*d^4 + (-296*c^4 - 84*c^3*d - 577*c^2*d^2 + 984*c*d^3 + 198*d^4)*Cos[e +
f*x] + (-72*c^4 - 470*c^3*d + 384*c^2*d^2 + 200*c*d^3 + 48*d^4)*Cos[2*(e + f*x)] - 88*c^4*Cos[3*(e + f*x)] +
36*c^3*d*Cos[3*(e + f*x)] + 37*c^2*d^2*Cos[3*(e + f*x)] + 24*c*d^3*Cos[3*(e + f*x)] + 6*d^4*Cos[3*(e + f*x)])*
Sin[e + f*x])/(d + c*Cos[e + f*x])^4))/(96*(c - d)*(c + d)^4*f)

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Maple [A]  time = 0.156, size = 303, normalized size = 1.1 \begin{align*} 16\,{\frac{{a}^{3}}{f} \left ({\frac{1}{ \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 ) ^{4}} \left ( -{\frac{ \left ( 20\,c-15\,d \right ) \left ({c}^{2}-2\,cd+{d}^{2} \right ) \left ( \tan \left ( 1/2\,fx+e/2 \right ) \right ) ^{7}}{64\,{c}^{4}+256\,{c}^{3}d+384\,{c}^{2}{d}^{2}+256\,c{d}^{3}+64\,{d}^{4}}}+{\frac{ \left ( 55\,c-55\,d \right ) \left ( 4\,c-3\,d \right ) \left ( \tan \left ( 1/2\,fx+e/2 \right ) \right ) ^{5}}{192\,{c}^{3}+576\,{c}^{2}d+576\,{d}^{2}c+192\,{d}^{3}}}-{\frac{ \left ( 292\,c-219\,d \right ) \left ( \tan \left ( 1/2\,fx+e/2 \right ) \right ) ^{3}}{192\,{c}^{2}+384\,cd+192\,{d}^{2}}}+{\frac{ \left ( 44\,c-49\,d \right ) \tan \left ( 1/2\,fx+e/2 \right ) }{ \left ( 64\,c+64\,d \right ) \left ( c-d \right ) }} \right ) }+{\frac{20\,c-15\,d}{ \left ( 64\,{c}^{5}+192\,{c}^{4}d+128\,{c}^{3}{d}^{2}-128\,{c}^{2}{d}^{3}-192\,c{d}^{4}-64\,{d}^{5} \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 ) } \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))^5,x)

[Out]

16/f*a^3*((-5/64*(4*c-3*d)*(c^2-2*c*d+d^2)/(c^4+4*c^3*d+6*c^2*d^2+4*c*d^3+d^4)*tan(1/2*f*x+1/2*e)^7+55/192*(c-
d)*(4*c-3*d)/(c^3+3*c^2*d+3*c*d^2+d^3)*tan(1/2*f*x+1/2*e)^5-73/192*(4*c-3*d)/(c^2+2*c*d+d^2)*tan(1/2*f*x+1/2*e
)^3+1/64*(44*c-49*d)/(c+d)/(c-d)*tan(1/2*f*x+1/2*e))/(tan(1/2*f*x+1/2*e)^2*c-tan(1/2*f*x+1/2*e)^2*d-c-d)^4+5/6
4*(4*c-3*d)/(c^5+3*c^4*d+2*c^3*d^2-2*c^2*d^3-3*c*d^4-d^5)/((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))^5,x, algorithm="maxima")

[Out]

Exception raised: ValueError

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Fricas [B]  time = 0.804176, size = 3699, normalized size = 13.91 \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))^5,x, algorithm="fricas")

[Out]

[1/48*(15*(4*a^3*c*d^4 - 3*a^3*d^5 + (4*a^3*c^5 - 3*a^3*c^4*d)*cos(f*x + e)^4 + 4*(4*a^3*c^4*d - 3*a^3*c^3*d^2
)*cos(f*x + e)^3 + 6*(4*a^3*c^3*d^2 - 3*a^3*c^2*d^3)*cos(f*x + e)^2 + 4*(4*a^3*c^2*d^3 - 3*a^3*c*d^4)*cos(f*x
+ e))*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^5*d + 12*a^3
*c^4*d^2 + 41*a^3*c^3*d^3 - 84*a^3*c^2*d^4 - 43*a^3*c*d^5 + 72*a^3*d^6 + (88*a^3*c^6 - 36*a^3*c^5*d - 125*a^3*
c^4*d^2 + 12*a^3*c^3*d^3 + 31*a^3*c^2*d^4 + 24*a^3*c*d^5 + 6*a^3*d^6)*cos(f*x + e)^3 + (36*a^3*c^6 + 235*a^3*c
^5*d - 228*a^3*c^4*d^2 - 335*a^3*c^3*d^3 + 168*a^3*c^2*d^4 + 100*a^3*c*d^5 + 24*a^3*d^6)*cos(f*x + e)^2 + (8*a
^3*c^6 + 48*a^3*c^5*d + 164*a^3*c^4*d^2 - 276*a^3*c^3*d^3 - 217*a^3*c^2*d^4 + 228*a^3*c*d^5 + 45*a^3*d^6)*cos(
f*x + e))*sin(f*x + e))/((c^11 + 3*c^10*d + c^9*d^2 - 5*c^8*d^3 - 5*c^7*d^4 + c^6*d^5 + 3*c^5*d^6 + c^4*d^7)*f
*cos(f*x + e)^4 + 4*(c^10*d + 3*c^9*d^2 + c^8*d^3 - 5*c^7*d^4 - 5*c^6*d^5 + c^5*d^6 + 3*c^4*d^7 + c^3*d^8)*f*c
os(f*x + e)^3 + 6*(c^9*d^2 + 3*c^8*d^3 + c^7*d^4 - 5*c^6*d^5 - 5*c^5*d^6 + c^4*d^7 + 3*c^3*d^8 + c^2*d^9)*f*co
s(f*x + e)^2 + 4*(c^8*d^3 + 3*c^7*d^4 + c^6*d^5 - 5*c^5*d^6 - 5*c^4*d^7 + c^3*d^8 + 3*c^2*d^9 + c*d^10)*f*cos(
f*x + e) + (c^7*d^4 + 3*c^6*d^5 + c^5*d^6 - 5*c^4*d^7 - 5*c^3*d^8 + c^2*d^9 + 3*c*d^10 + d^11)*f), 1/24*(15*(4
*a^3*c*d^4 - 3*a^3*d^5 + (4*a^3*c^5 - 3*a^3*c^4*d)*cos(f*x + e)^4 + 4*(4*a^3*c^4*d - 3*a^3*c^3*d^2)*cos(f*x +
e)^3 + 6*(4*a^3*c^3*d^2 - 3*a^3*c^2*d^3)*cos(f*x + e)^2 + 4*(4*a^3*c^2*d^3 - 3*a^3*c*d^4)*cos(f*x + e))*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^5*d + 12*a^3*c
^4*d^2 + 41*a^3*c^3*d^3 - 84*a^3*c^2*d^4 - 43*a^3*c*d^5 + 72*a^3*d^6 + (88*a^3*c^6 - 36*a^3*c^5*d - 125*a^3*c^
4*d^2 + 12*a^3*c^3*d^3 + 31*a^3*c^2*d^4 + 24*a^3*c*d^5 + 6*a^3*d^6)*cos(f*x + e)^3 + (36*a^3*c^6 + 235*a^3*c^5
*d - 228*a^3*c^4*d^2 - 335*a^3*c^3*d^3 + 168*a^3*c^2*d^4 + 100*a^3*c*d^5 + 24*a^3*d^6)*cos(f*x + e)^2 + (8*a^3
*c^6 + 48*a^3*c^5*d + 164*a^3*c^4*d^2 - 276*a^3*c^3*d^3 - 217*a^3*c^2*d^4 + 228*a^3*c*d^5 + 45*a^3*d^6)*cos(f*
x + e))*sin(f*x + e))/((c^11 + 3*c^10*d + c^9*d^2 - 5*c^8*d^3 - 5*c^7*d^4 + c^6*d^5 + 3*c^5*d^6 + c^4*d^7)*f*c
os(f*x + e)^4 + 4*(c^10*d + 3*c^9*d^2 + c^8*d^3 - 5*c^7*d^4 - 5*c^6*d^5 + c^5*d^6 + 3*c^4*d^7 + c^3*d^8)*f*cos
(f*x + e)^3 + 6*(c^9*d^2 + 3*c^8*d^3 + c^7*d^4 - 5*c^6*d^5 - 5*c^5*d^6 + c^4*d^7 + 3*c^3*d^8 + c^2*d^9)*f*cos(
f*x + e)^2 + 4*(c^8*d^3 + 3*c^7*d^4 + c^6*d^5 - 5*c^5*d^6 - 5*c^4*d^7 + c^3*d^8 + 3*c^2*d^9 + c*d^10)*f*cos(f*
x + e) + (c^7*d^4 + 3*c^6*d^5 + c^5*d^6 - 5*c^4*d^7 - 5*c^3*d^8 + c^2*d^9 + 3*c*d^10 + d^11)*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^{5} + 5 c^{4} d \sec{\left (e + f x \right )} + 10 c^{3} d^{2} \sec ^{2}{\left (e + f x \right )} + 10 c^{2} d^{3} \sec ^{3}{\left (e + f x \right )} + 5 c d^{4} \sec ^{4}{\left (e + f x \right )} + d^{5} \sec ^{5}{\left (e + f x \right )}}\, dx + \int \frac{3 \sec ^{2}{\left (e + f x \right )}}{c^{5} + 5 c^{4} d \sec{\left (e + f x \right )} + 10 c^{3} d^{2} \sec ^{2}{\left (e + f x \right )} + 10 c^{2} d^{3} \sec ^{3}{\left (e + f x \right )} + 5 c d^{4} \sec ^{4}{\left (e + f x \right )} + d^{5} \sec ^{5}{\left (e + f x \right )}}\, dx + \int \frac{3 \sec ^{3}{\left (e + f x \right )}}{c^{5} + 5 c^{4} d \sec{\left (e + f x \right )} + 10 c^{3} d^{2} \sec ^{2}{\left (e + f x \right )} + 10 c^{2} d^{3} \sec ^{3}{\left (e + f x \right )} + 5 c d^{4} \sec ^{4}{\left (e + f x \right )} + d^{5} \sec ^{5}{\left (e + f x \right )}}\, dx + \int \frac{\sec ^{4}{\left (e + f x \right )}}{c^{5} + 5 c^{4} d \sec{\left (e + f x \right )} + 10 c^{3} d^{2} \sec ^{2}{\left (e + f x \right )} + 10 c^{2} d^{3} \sec ^{3}{\left (e + f x \right )} + 5 c d^{4} \sec ^{4}{\left (e + f x \right )} + d^{5} \sec ^{5}{\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))**5,x)

[Out]

a**3*(Integral(sec(e + f*x)/(c**5 + 5*c**4*d*sec(e + f*x) + 10*c**3*d**2*sec(e + f*x)**2 + 10*c**2*d**3*sec(e
+ f*x)**3 + 5*c*d**4*sec(e + f*x)**4 + d**5*sec(e + f*x)**5), x) + Integral(3*sec(e + f*x)**2/(c**5 + 5*c**4*d
*sec(e + f*x) + 10*c**3*d**2*sec(e + f*x)**2 + 10*c**2*d**3*sec(e + f*x)**3 + 5*c*d**4*sec(e + f*x)**4 + d**5*
sec(e + f*x)**5), x) + Integral(3*sec(e + f*x)**3/(c**5 + 5*c**4*d*sec(e + f*x) + 10*c**3*d**2*sec(e + f*x)**2
+ 10*c**2*d**3*sec(e + f*x)**3 + 5*c*d**4*sec(e + f*x)**4 + d**5*sec(e + f*x)**5), x) + Integral(sec(e + f*x)
**4/(c**5 + 5*c**4*d*sec(e + f*x) + 10*c**3*d**2*sec(e + f*x)**2 + 10*c**2*d**3*sec(e + f*x)**3 + 5*c*d**4*sec
(e + f*x)**4 + d**5*sec(e + f*x)**5), x))

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Giac [B]  time = 1.53497, size = 845, normalized size = 3.18 \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))^5,x, algorithm="giac")

[Out]

1/12*(15*(4*a^3*c - 3*a^3*d)*(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)))/((c^5 + 3*c^4*d + 2*c^3*d^2 - 2*c^2*d^3 - 3*c*d^4 - d^5)*sqrt(
-c^2 + d^2)) - (60*a^3*c^4*tan(1/2*f*x + 1/2*e)^7 - 225*a^3*c^3*d*tan(1/2*f*x + 1/2*e)^7 + 315*a^3*c^2*d^2*tan
(1/2*f*x + 1/2*e)^7 - 195*a^3*c*d^3*tan(1/2*f*x + 1/2*e)^7 + 45*a^3*d^4*tan(1/2*f*x + 1/2*e)^7 - 220*a^3*c^4*t
an(1/2*f*x + 1/2*e)^5 + 385*a^3*c^3*d*tan(1/2*f*x + 1/2*e)^5 + 55*a^3*c^2*d^2*tan(1/2*f*x + 1/2*e)^5 - 385*a^3
*c*d^3*tan(1/2*f*x + 1/2*e)^5 + 165*a^3*d^4*tan(1/2*f*x + 1/2*e)^5 + 292*a^3*c^4*tan(1/2*f*x + 1/2*e)^3 + 73*a
^3*c^3*d*tan(1/2*f*x + 1/2*e)^3 - 511*a^3*c^2*d^2*tan(1/2*f*x + 1/2*e)^3 - 73*a^3*c*d^3*tan(1/2*f*x + 1/2*e)^3
+ 219*a^3*d^4*tan(1/2*f*x + 1/2*e)^3 - 132*a^3*c^4*tan(1/2*f*x + 1/2*e) - 249*a^3*c^3*d*tan(1/2*f*x + 1/2*e)
+ 45*a^3*c^2*d^2*tan(1/2*f*x + 1/2*e) + 309*a^3*c*d^3*tan(1/2*f*x + 1/2*e) + 147*a^3*d^4*tan(1/2*f*x + 1/2*e))
/((c^5 + 3*c^4*d + 2*c^3*d^2 - 2*c^2*d^3 - 3*c*d^4 - d^5)*(c*tan(1/2*f*x + 1/2*e)^2 - d*tan(1/2*f*x + 1/2*e)^2
- c - d)^4))/f