3.3.18 \(\int \frac {\sec (e+f x) (c+d \sec (e+f x))^4}{(a+a \sec (e+f x))^2} \, dx\) [218]

Optimal. Leaf size=193 \[ \frac {d^2 \left (12 c^2-16 c d+7 d^2\right ) \tanh ^{-1}(\sin (e+f x))}{2 a^2 f}+\frac {(c-d) (c+8 d) (c+d \sec (e+f x))^2 \tan (e+f x)}{3 f \left (a^2+a^2 \sec (e+f x)\right )}+\frac {(c-d) (c+d \sec (e+f x))^3 \tan (e+f x)}{3 f (a+a \sec (e+f x))^2}-\frac {d \left (4 \left (c^3+8 c^2 d-20 c d^2+8 d^3\right )+d \left (2 c^2+16 c d-21 d^2\right ) \sec (e+f x)\right ) \tan (e+f x)}{6 a^2 f} \]

[Out]

1/2*d^2*(12*c^2-16*c*d+7*d^2)*arctanh(sin(f*x+e))/a^2/f+1/3*(c-d)*(c+8*d)*(c+d*sec(f*x+e))^2*tan(f*x+e)/f/(a^2
+a^2*sec(f*x+e))+1/3*(c-d)*(c+d*sec(f*x+e))^3*tan(f*x+e)/f/(a+a*sec(f*x+e))^2-1/6*d*(4*c^3+32*c^2*d-80*c*d^2+3
2*d^3+d*(2*c^2+16*c*d-21*d^2)*sec(f*x+e))*tan(f*x+e)/a^2/f

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Rubi [A]
time = 0.21, antiderivative size = 249, normalized size of antiderivative = 1.29, number of steps used = 7, number of rules used = 7, integrand size = 31, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.226, Rules used = {4072, 100, 155, 152, 65, 223, 209} \begin {gather*} -\frac {d \tan (e+f x) \left (d \left (2 c^2+16 c d-21 d^2\right ) \sec (e+f x)+4 \left (c^3+8 c^2 d-20 c d^2+8 d^3\right )\right )}{6 a^2 f}+\frac {(c-d) (c+8 d) \tan (e+f x) (c+d \sec (e+f x))^2}{3 f \left (a^2 \sec (e+f x)+a^2\right )}+\frac {d^2 \left (12 c^2-16 c d+7 d^2\right ) \tan (e+f x) \text {ArcTan}\left (\frac {\sqrt {a-a \sec (e+f x)}}{\sqrt {a (\sec (e+f x)+1)}}\right )}{a f \sqrt {a-a \sec (e+f x)} \sqrt {a \sec (e+f x)+a}}+\frac {(c-d) \tan (e+f x) (c+d \sec (e+f x))^3}{3 f (a \sec (e+f x)+a)^2} \end {gather*}

Antiderivative was successfully verified.

[In]

Int[(Sec[e + f*x]*(c + d*Sec[e + f*x])^4)/(a + a*Sec[e + f*x])^2,x]

[Out]

(d^2*(12*c^2 - 16*c*d + 7*d^2)*ArcTan[Sqrt[a - a*Sec[e + f*x]]/Sqrt[a*(1 + Sec[e + f*x])]]*Tan[e + f*x])/(a*f*
Sqrt[a - a*Sec[e + f*x]]*Sqrt[a + a*Sec[e + f*x]]) + ((c - d)*(c + 8*d)*(c + d*Sec[e + f*x])^2*Tan[e + f*x])/(
3*f*(a^2 + a^2*Sec[e + f*x])) + ((c - d)*(c + d*Sec[e + f*x])^3*Tan[e + f*x])/(3*f*(a + a*Sec[e + f*x])^2) - (
d*(4*(c^3 + 8*c^2*d - 20*c*d^2 + 8*d^3) + d*(2*c^2 + 16*c*d - 21*d^2)*Sec[e + f*x])*Tan[e + f*x])/(6*a^2*f)

Rule 65

Int[((a_.) + (b_.)*(x_))^(m_)*((c_.) + (d_.)*(x_))^(n_), x_Symbol] :> With[{p = Denominator[m]}, Dist[p/b, Sub
st[Int[x^(p*(m + 1) - 1)*(c - a*(d/b) + d*(x^p/b))^n, x], x, (a + b*x)^(1/p)], x]] /; FreeQ[{a, b, c, d}, x] &
& NeQ[b*c - a*d, 0] && LtQ[-1, m, 0] && LeQ[-1, n, 0] && LeQ[Denominator[n], Denominator[m]] && IntLinearQ[a,
b, c, d, m, n, x]

Rule 100

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

Rule 152

Int[((a_.) + (b_.)*(x_))^(m_.)*((c_.) + (d_.)*(x_))^(n_.)*((e_) + (f_.)*(x_))*((g_.) + (h_.)*(x_)), x_Symbol]
:> Simp[(-(a*d*f*h*(n + 2) + b*c*f*h*(m + 2) - b*d*(f*g + e*h)*(m + n + 3) - b*d*f*h*(m + n + 2)*x))*(a + b*x)
^(m + 1)*((c + d*x)^(n + 1)/(b^2*d^2*(m + n + 2)*(m + n + 3))), x] + Dist[(a^2*d^2*f*h*(n + 1)*(n + 2) + a*b*d
*(n + 1)*(2*c*f*h*(m + 1) - d*(f*g + e*h)*(m + n + 3)) + b^2*(c^2*f*h*(m + 1)*(m + 2) - c*d*(f*g + e*h)*(m + 1
)*(m + n + 3) + d^2*e*g*(m + n + 2)*(m + n + 3)))/(b^2*d^2*(m + n + 2)*(m + n + 3)), Int[(a + b*x)^m*(c + d*x)
^n, x], x] /; FreeQ[{a, b, c, d, e, f, g, h, m, n}, x] && NeQ[m + n + 2, 0] && NeQ[m + n + 3, 0]

Rule 155

Int[((a_.) + (b_.)*(x_))^(m_)*((c_.) + (d_.)*(x_))^(n_)*((e_.) + (f_.)*(x_))^(p_)*((g_.) + (h_.)*(x_)), x_Symb
ol] :> Simp[(b*g - a*h)*(a + b*x)^(m + 1)*(c + d*x)^n*((e + f*x)^(p + 1)/(b*(b*e - a*f)*(m + 1))), x] - Dist[1
/(b*(b*e - a*f)*(m + 1)), Int[(a + b*x)^(m + 1)*(c + d*x)^(n - 1)*(e + f*x)^p*Simp[b*c*(f*g - e*h)*(m + 1) + (
b*g - a*h)*(d*e*n + c*f*(p + 1)) + d*(b*(f*g - e*h)*(m + 1) + f*(b*g - a*h)*(n + p + 1))*x, x], x], x] /; Free
Q[{a, b, c, d, e, f, g, h, p}, x] && LtQ[m, -1] && GtQ[n, 0] && IntegersQ[2*m, 2*n, 2*p]

Rule 209

Int[((a_) + (b_.)*(x_)^2)^(-1), x_Symbol] :> Simp[(1/(Rt[a, 2]*Rt[b, 2]))*ArcTan[Rt[b, 2]*(x/Rt[a, 2])], x] /;
FreeQ[{a, b}, x] && PosQ[a/b] && (GtQ[a, 0] || GtQ[b, 0])

Rule 223

Int[1/Sqrt[(a_) + (b_.)*(x_)^2], x_Symbol] :> Subst[Int[1/(1 - b*x^2), x], x, x/Sqrt[a + b*x^2]] /; FreeQ[{a,
b}, x] &&  !GtQ[a, 0]

Rule 4072

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])

Rubi steps

\begin {align*} \int \frac {\sec (e+f x) (c+d \sec (e+f x))^4}{(a+a \sec (e+f x))^2} \, dx &=-\frac {\left (a^2 \tan (e+f x)\right ) \text {Subst}\left (\int \frac {(c+d x)^4}{\sqrt {a-a x} (a+a x)^{5/2}} \, dx,x,\sec (e+f x)\right )}{f \sqrt {a-a \sec (e+f x)} \sqrt {a+a \sec (e+f x)}}\\ &=\frac {(c-d) (c+d \sec (e+f x))^3 \tan (e+f x)}{3 f (a+a \sec (e+f x))^2}+\frac {\tan (e+f x) \text {Subst}\left (\int \frac {(c+d x)^2 \left (-a^2 \left (c^2+5 c d-3 d^2\right )+a^2 (2 c-5 d) d x\right )}{\sqrt {a-a x} (a+a x)^{3/2}} \, dx,x,\sec (e+f x)\right )}{3 a f \sqrt {a-a \sec (e+f x)} \sqrt {a+a \sec (e+f x)}}\\ &=\frac {(c-d) (c+8 d) (c+d \sec (e+f x))^2 \tan (e+f x)}{3 f \left (a^2+a^2 \sec (e+f x)\right )}+\frac {(c-d) (c+d \sec (e+f x))^3 \tan (e+f x)}{3 f (a+a \sec (e+f x))^2}+\frac {\tan (e+f x) \text {Subst}\left (\int \frac {(c+d x) \left (-a^4 (19 c-16 d) d^2+a^4 d \left (2 c^2+16 c d-21 d^2\right ) x\right )}{\sqrt {a-a x} \sqrt {a+a x}} \, dx,x,\sec (e+f x)\right )}{3 a^4 f \sqrt {a-a \sec (e+f x)} \sqrt {a+a \sec (e+f x)}}\\ &=\frac {(c-d) (c+8 d) (c+d \sec (e+f x))^2 \tan (e+f x)}{3 f \left (a^2+a^2 \sec (e+f x)\right )}+\frac {(c-d) (c+d \sec (e+f x))^3 \tan (e+f x)}{3 f (a+a \sec (e+f x))^2}-\frac {d \left (4 \left (c^3+8 c^2 d-20 c d^2+8 d^3\right )+d \left (2 c^2+16 c d-21 d^2\right ) \sec (e+f x)\right ) \tan (e+f x)}{6 a^2 f}-\frac {\left (d^2 \left (12 c^2-16 c d+7 d^2\right ) \tan (e+f x)\right ) \text {Subst}\left (\int \frac {1}{\sqrt {a-a x} \sqrt {a+a x}} \, dx,x,\sec (e+f x)\right )}{2 f \sqrt {a-a \sec (e+f x)} \sqrt {a+a \sec (e+f x)}}\\ &=\frac {(c-d) (c+8 d) (c+d \sec (e+f x))^2 \tan (e+f x)}{3 f \left (a^2+a^2 \sec (e+f x)\right )}+\frac {(c-d) (c+d \sec (e+f x))^3 \tan (e+f x)}{3 f (a+a \sec (e+f x))^2}-\frac {d \left (4 \left (c^3+8 c^2 d-20 c d^2+8 d^3\right )+d \left (2 c^2+16 c d-21 d^2\right ) \sec (e+f x)\right ) \tan (e+f x)}{6 a^2 f}+\frac {\left (d^2 \left (12 c^2-16 c d+7 d^2\right ) \tan (e+f x)\right ) \text {Subst}\left (\int \frac {1}{\sqrt {2 a-x^2}} \, dx,x,\sqrt {a-a \sec (e+f x)}\right )}{a f \sqrt {a-a \sec (e+f x)} \sqrt {a+a \sec (e+f x)}}\\ &=\frac {(c-d) (c+8 d) (c+d \sec (e+f x))^2 \tan (e+f x)}{3 f \left (a^2+a^2 \sec (e+f x)\right )}+\frac {(c-d) (c+d \sec (e+f x))^3 \tan (e+f x)}{3 f (a+a \sec (e+f x))^2}-\frac {d \left (4 \left (c^3+8 c^2 d-20 c d^2+8 d^3\right )+d \left (2 c^2+16 c d-21 d^2\right ) \sec (e+f x)\right ) \tan (e+f x)}{6 a^2 f}+\frac {\left (d^2 \left (12 c^2-16 c d+7 d^2\right ) \tan (e+f x)\right ) \text {Subst}\left (\int \frac {1}{1+x^2} \, dx,x,\frac {\sqrt {a-a \sec (e+f x)}}{\sqrt {a+a \sec (e+f x)}}\right )}{a f \sqrt {a-a \sec (e+f x)} \sqrt {a+a \sec (e+f x)}}\\ &=\frac {d^2 \left (12 c^2-16 c d+7 d^2\right ) \tan ^{-1}\left (\frac {\sqrt {a-a \sec (e+f x)}}{\sqrt {a+a \sec (e+f x)}}\right ) \tan (e+f x)}{a f \sqrt {a-a \sec (e+f x)} \sqrt {a+a \sec (e+f x)}}+\frac {(c-d) (c+8 d) (c+d \sec (e+f x))^2 \tan (e+f x)}{3 f \left (a^2+a^2 \sec (e+f x)\right )}+\frac {(c-d) (c+d \sec (e+f x))^3 \tan (e+f x)}{3 f (a+a \sec (e+f x))^2}-\frac {d \left (4 \left (c^3+8 c^2 d-20 c d^2+8 d^3\right )+d \left (2 c^2+16 c d-21 d^2\right ) \sec (e+f x)\right ) \tan (e+f x)}{6 a^2 f}\\ \end {align*}

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Mathematica [A]
time = 2.88, size = 310, normalized size = 1.61 \begin {gather*} \frac {-24 d^2 \left (12 c^2-16 c d+7 d^2\right ) \cos ^4\left (\frac {1}{2} (e+f x)\right ) \left (\log \left (\cos \left (\frac {1}{2} (e+f x)\right )-\sin \left (\frac {1}{2} (e+f x)\right )\right )-\log \left (\cos \left (\frac {1}{2} (e+f x)\right )+\sin \left (\frac {1}{2} (e+f x)\right )\right )\right )+2 \cos \left (\frac {1}{2} (e+f x)\right ) \left (2 c^4+16 c^3 d-60 c^2 d^2+112 c d^3-37 d^4+6 \left (c^4+2 c^3 d-12 c^2 d^2+28 c d^3-10 d^4\right ) \cos (e+f x)+\left (2 c^4+16 c^3 d-60 c^2 d^2+112 c d^3-43 d^4\right ) \cos (2 (e+f x))+2 c^4 \cos (3 (e+f x))+4 c^3 d \cos (3 (e+f x))-24 c^2 d^2 \cos (3 (e+f x))+40 c d^3 \cos (3 (e+f x))-16 d^4 \cos (3 (e+f x))\right ) \sec ^2(e+f x) \sin \left (\frac {1}{2} (e+f x)\right )}{12 a^2 f (1+\cos (e+f x))^2} \end {gather*}

Antiderivative was successfully verified.

[In]

Integrate[(Sec[e + f*x]*(c + d*Sec[e + f*x])^4)/(a + a*Sec[e + f*x])^2,x]

[Out]

(-24*d^2*(12*c^2 - 16*c*d + 7*d^2)*Cos[(e + f*x)/2]^4*(Log[Cos[(e + f*x)/2] - Sin[(e + f*x)/2]] - Log[Cos[(e +
f*x)/2] + Sin[(e + f*x)/2]]) + 2*Cos[(e + f*x)/2]*(2*c^4 + 16*c^3*d - 60*c^2*d^2 + 112*c*d^3 - 37*d^4 + 6*(c^
4 + 2*c^3*d - 12*c^2*d^2 + 28*c*d^3 - 10*d^4)*Cos[e + f*x] + (2*c^4 + 16*c^3*d - 60*c^2*d^2 + 112*c*d^3 - 43*d
^4)*Cos[2*(e + f*x)] + 2*c^4*Cos[3*(e + f*x)] + 4*c^3*d*Cos[3*(e + f*x)] - 24*c^2*d^2*Cos[3*(e + f*x)] + 40*c*
d^3*Cos[3*(e + f*x)] - 16*d^4*Cos[3*(e + f*x)])*Sec[e + f*x]^2*Sin[(e + f*x)/2])/(12*a^2*f*(1 + Cos[e + f*x])^
2)

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Maple [A]
time = 0.24, size = 317, normalized size = 1.64 Too large to display

Verification of antiderivative is not currently implemented for this CAS.

[In]

int(sec(f*x+e)*(c+d*sec(f*x+e))^4/(a+a*sec(f*x+e))^2,x,method=_RETURNVERBOSE)

[Out]

1/2/f/a^2*(-1/3*c^4*tan(1/2*f*x+1/2*e)^3+4/3*c^3*d*tan(1/2*f*x+1/2*e)^3-2*c^2*d^2*tan(1/2*f*x+1/2*e)^3+4/3*c*d
^3*tan(1/2*f*x+1/2*e)^3-1/3*d^4*tan(1/2*f*x+1/2*e)^3+c^4*tan(1/2*f*x+1/2*e)+4*c^3*d*tan(1/2*f*x+1/2*e)-18*c^2*
d^2*tan(1/2*f*x+1/2*e)+20*c*d^3*tan(1/2*f*x+1/2*e)-7*d^4*tan(1/2*f*x+1/2*e)+d^2*(12*c^2-16*c*d+7*d^2)*ln(tan(1
/2*f*x+1/2*e)+1)-d^3*(8*c-5*d)/(tan(1/2*f*x+1/2*e)+1)-d^4/(tan(1/2*f*x+1/2*e)+1)^2-d^3*(8*c-5*d)/(tan(1/2*f*x+
1/2*e)-1)+d^4/(tan(1/2*f*x+1/2*e)-1)^2-d^2*(12*c^2-16*c*d+7*d^2)*ln(tan(1/2*f*x+1/2*e)-1))

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Maxima [B] Leaf count of result is larger than twice the leaf count of optimal. 580 vs. \(2 (193) = 386\).
time = 0.29, size = 580, normalized size = 3.01 \begin {gather*} -\frac {d^{4} {\left (\frac {6 \, {\left (\frac {3 \, \sin \left (f x + e\right )}{\cos \left (f x + e\right ) + 1} - \frac {5 \, \sin \left (f x + e\right )^{3}}{{\left (\cos \left (f x + e\right ) + 1\right )}^{3}}\right )}}{a^{2} - \frac {2 \, a^{2} \sin \left (f x + e\right )^{2}}{{\left (\cos \left (f x + e\right ) + 1\right )}^{2}} + \frac {a^{2} \sin \left (f x + e\right )^{4}}{{\left (\cos \left (f x + e\right ) + 1\right )}^{4}}} + \frac {\frac {21 \, \sin \left (f x + e\right )}{\cos \left (f x + e\right ) + 1} + \frac {\sin \left (f x + e\right )^{3}}{{\left (\cos \left (f x + e\right ) + 1\right )}^{3}}}{a^{2}} - \frac {21 \, \log \left (\frac {\sin \left (f x + e\right )}{\cos \left (f x + e\right ) + 1} + 1\right )}{a^{2}} + \frac {21 \, \log \left (\frac {\sin \left (f x + e\right )}{\cos \left (f x + e\right ) + 1} - 1\right )}{a^{2}}\right )} - 4 \, c d^{3} {\left (\frac {\frac {15 \, \sin \left (f x + e\right )}{\cos \left (f x + e\right ) + 1} + \frac {\sin \left (f x + e\right )^{3}}{{\left (\cos \left (f x + e\right ) + 1\right )}^{3}}}{a^{2}} - \frac {12 \, \log \left (\frac {\sin \left (f x + e\right )}{\cos \left (f x + e\right ) + 1} + 1\right )}{a^{2}} + \frac {12 \, \log \left (\frac {\sin \left (f x + e\right )}{\cos \left (f x + e\right ) + 1} - 1\right )}{a^{2}} + \frac {12 \, \sin \left (f x + e\right )}{{\left (a^{2} - \frac {a^{2} \sin \left (f x + e\right )^{2}}{{\left (\cos \left (f x + e\right ) + 1\right )}^{2}}\right )} {\left (\cos \left (f x + e\right ) + 1\right )}}\right )} + 6 \, c^{2} d^{2} {\left (\frac {\frac {9 \, \sin \left (f x + e\right )}{\cos \left (f x + e\right ) + 1} + \frac {\sin \left (f x + e\right )^{3}}{{\left (\cos \left (f x + e\right ) + 1\right )}^{3}}}{a^{2}} - \frac {6 \, \log \left (\frac {\sin \left (f x + e\right )}{\cos \left (f x + e\right ) + 1} + 1\right )}{a^{2}} + \frac {6 \, \log \left (\frac {\sin \left (f x + e\right )}{\cos \left (f x + e\right ) + 1} - 1\right )}{a^{2}}\right )} - \frac {4 \, c^{3} d {\left (\frac {3 \, \sin \left (f x + e\right )}{\cos \left (f x + e\right ) + 1} + \frac {\sin \left (f x + e\right )^{3}}{{\left (\cos \left (f x + e\right ) + 1\right )}^{3}}\right )}}{a^{2}} - \frac {c^{4} {\left (\frac {3 \, \sin \left (f x + e\right )}{\cos \left (f x + e\right ) + 1} - \frac {\sin \left (f x + e\right )^{3}}{{\left (\cos \left (f x + e\right ) + 1\right )}^{3}}\right )}}{a^{2}}}{6 \, f} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

-1/6*(d^4*(6*(3*sin(f*x + e)/(cos(f*x + e) + 1) - 5*sin(f*x + e)^3/(cos(f*x + e) + 1)^3)/(a^2 - 2*a^2*sin(f*x
+ e)^2/(cos(f*x + e) + 1)^2 + a^2*sin(f*x + e)^4/(cos(f*x + e) + 1)^4) + (21*sin(f*x + e)/(cos(f*x + e) + 1) +
sin(f*x + e)^3/(cos(f*x + e) + 1)^3)/a^2 - 21*log(sin(f*x + e)/(cos(f*x + e) + 1) + 1)/a^2 + 21*log(sin(f*x +
e)/(cos(f*x + e) + 1) - 1)/a^2) - 4*c*d^3*((15*sin(f*x + e)/(cos(f*x + e) + 1) + sin(f*x + e)^3/(cos(f*x + e)
+ 1)^3)/a^2 - 12*log(sin(f*x + e)/(cos(f*x + e) + 1) + 1)/a^2 + 12*log(sin(f*x + e)/(cos(f*x + e) + 1) - 1)/a
^2 + 12*sin(f*x + e)/((a^2 - a^2*sin(f*x + e)^2/(cos(f*x + e) + 1)^2)*(cos(f*x + e) + 1))) + 6*c^2*d^2*((9*sin
(f*x + e)/(cos(f*x + e) + 1) + sin(f*x + e)^3/(cos(f*x + e) + 1)^3)/a^2 - 6*log(sin(f*x + e)/(cos(f*x + e) + 1
) + 1)/a^2 + 6*log(sin(f*x + e)/(cos(f*x + e) + 1) - 1)/a^2) - 4*c^3*d*(3*sin(f*x + e)/(cos(f*x + e) + 1) + si
n(f*x + e)^3/(cos(f*x + e) + 1)^3)/a^2 - c^4*(3*sin(f*x + e)/(cos(f*x + e) + 1) - sin(f*x + e)^3/(cos(f*x + e)
+ 1)^3)/a^2)/f

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Fricas [A]
time = 2.32, size = 376, normalized size = 1.95 \begin {gather*} \frac {3 \, {\left ({\left (12 \, c^{2} d^{2} - 16 \, c d^{3} + 7 \, d^{4}\right )} \cos \left (f x + e\right )^{4} + 2 \, {\left (12 \, c^{2} d^{2} - 16 \, c d^{3} + 7 \, d^{4}\right )} \cos \left (f x + e\right )^{3} + {\left (12 \, c^{2} d^{2} - 16 \, c d^{3} + 7 \, d^{4}\right )} \cos \left (f x + e\right )^{2}\right )} \log \left (\sin \left (f x + e\right ) + 1\right ) - 3 \, {\left ({\left (12 \, c^{2} d^{2} - 16 \, c d^{3} + 7 \, d^{4}\right )} \cos \left (f x + e\right )^{4} + 2 \, {\left (12 \, c^{2} d^{2} - 16 \, c d^{3} + 7 \, d^{4}\right )} \cos \left (f x + e\right )^{3} + {\left (12 \, c^{2} d^{2} - 16 \, c d^{3} + 7 \, d^{4}\right )} \cos \left (f x + e\right )^{2}\right )} \log \left (-\sin \left (f x + e\right ) + 1\right ) + 2 \, {\left (3 \, d^{4} + 4 \, {\left (c^{4} + 2 \, c^{3} d - 12 \, c^{2} d^{2} + 20 \, c d^{3} - 8 \, d^{4}\right )} \cos \left (f x + e\right )^{3} + {\left (2 \, c^{4} + 16 \, c^{3} d - 60 \, c^{2} d^{2} + 112 \, c d^{3} - 43 \, d^{4}\right )} \cos \left (f x + e\right )^{2} + 6 \, {\left (4 \, c d^{3} - d^{4}\right )} \cos \left (f x + e\right )\right )} \sin \left (f x + e\right )}{12 \, {\left (a^{2} f \cos \left (f x + e\right )^{4} + 2 \, a^{2} f \cos \left (f x + e\right )^{3} + a^{2} f \cos \left (f x + e\right )^{2}\right )}} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

1/12*(3*((12*c^2*d^2 - 16*c*d^3 + 7*d^4)*cos(f*x + e)^4 + 2*(12*c^2*d^2 - 16*c*d^3 + 7*d^4)*cos(f*x + e)^3 + (
12*c^2*d^2 - 16*c*d^3 + 7*d^4)*cos(f*x + e)^2)*log(sin(f*x + e) + 1) - 3*((12*c^2*d^2 - 16*c*d^3 + 7*d^4)*cos(
f*x + e)^4 + 2*(12*c^2*d^2 - 16*c*d^3 + 7*d^4)*cos(f*x + e)^3 + (12*c^2*d^2 - 16*c*d^3 + 7*d^4)*cos(f*x + e)^2
)*log(-sin(f*x + e) + 1) + 2*(3*d^4 + 4*(c^4 + 2*c^3*d - 12*c^2*d^2 + 20*c*d^3 - 8*d^4)*cos(f*x + e)^3 + (2*c^
4 + 16*c^3*d - 60*c^2*d^2 + 112*c*d^3 - 43*d^4)*cos(f*x + e)^2 + 6*(4*c*d^3 - d^4)*cos(f*x + e))*sin(f*x + e))
/(a^2*f*cos(f*x + e)^4 + 2*a^2*f*cos(f*x + e)^3 + a^2*f*cos(f*x + e)^2)

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Sympy [F]
time = 0.00, size = 0, normalized size = 0.00 \begin {gather*} \frac {\int \frac {c^{4} \sec {\left (e + f x \right )}}{\sec ^{2}{\left (e + f x \right )} + 2 \sec {\left (e + f x \right )} + 1}\, dx + \int \frac {d^{4} \sec ^{5}{\left (e + f x \right )}}{\sec ^{2}{\left (e + f x \right )} + 2 \sec {\left (e + f x \right )} + 1}\, dx + \int \frac {4 c d^{3} \sec ^{4}{\left (e + f x \right )}}{\sec ^{2}{\left (e + f x \right )} + 2 \sec {\left (e + f x \right )} + 1}\, dx + \int \frac {6 c^{2} d^{2} \sec ^{3}{\left (e + f x \right )}}{\sec ^{2}{\left (e + f x \right )} + 2 \sec {\left (e + f x \right )} + 1}\, dx + \int \frac {4 c^{3} d \sec ^{2}{\left (e + f x \right )}}{\sec ^{2}{\left (e + f x \right )} + 2 \sec {\left (e + f x \right )} + 1}\, dx}{a^{2}} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate(sec(f*x+e)*(c+d*sec(f*x+e))**4/(a+a*sec(f*x+e))**2,x)

[Out]

(Integral(c**4*sec(e + f*x)/(sec(e + f*x)**2 + 2*sec(e + f*x) + 1), x) + Integral(d**4*sec(e + f*x)**5/(sec(e
+ f*x)**2 + 2*sec(e + f*x) + 1), x) + Integral(4*c*d**3*sec(e + f*x)**4/(sec(e + f*x)**2 + 2*sec(e + f*x) + 1)
, x) + Integral(6*c**2*d**2*sec(e + f*x)**3/(sec(e + f*x)**2 + 2*sec(e + f*x) + 1), x) + Integral(4*c**3*d*sec
(e + f*x)**2/(sec(e + f*x)**2 + 2*sec(e + f*x) + 1), x))/a**2

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Giac [A]
time = 0.56, size = 359, normalized size = 1.86 \begin {gather*} \frac {\frac {3 \, {\left (12 \, c^{2} d^{2} - 16 \, c d^{3} + 7 \, d^{4}\right )} \log \left ({\left | \tan \left (\frac {1}{2} \, f x + \frac {1}{2} \, e\right ) + 1 \right |}\right )}{a^{2}} - \frac {3 \, {\left (12 \, c^{2} d^{2} - 16 \, c d^{3} + 7 \, d^{4}\right )} \log \left ({\left | \tan \left (\frac {1}{2} \, f x + \frac {1}{2} \, e\right ) - 1 \right |}\right )}{a^{2}} - \frac {6 \, {\left (8 \, c d^{3} \tan \left (\frac {1}{2} \, f x + \frac {1}{2} \, e\right )^{3} - 5 \, d^{4} \tan \left (\frac {1}{2} \, f x + \frac {1}{2} \, e\right )^{3} - 8 \, c d^{3} \tan \left (\frac {1}{2} \, f x + \frac {1}{2} \, e\right ) + 3 \, d^{4} \tan \left (\frac {1}{2} \, f x + \frac {1}{2} \, e\right )\right )}}{{\left (\tan \left (\frac {1}{2} \, f x + \frac {1}{2} \, e\right )^{2} - 1\right )}^{2} a^{2}} - \frac {a^{4} c^{4} \tan \left (\frac {1}{2} \, f x + \frac {1}{2} \, e\right )^{3} - 4 \, a^{4} c^{3} d \tan \left (\frac {1}{2} \, f x + \frac {1}{2} \, e\right )^{3} + 6 \, a^{4} c^{2} d^{2} \tan \left (\frac {1}{2} \, f x + \frac {1}{2} \, e\right )^{3} - 4 \, a^{4} c d^{3} \tan \left (\frac {1}{2} \, f x + \frac {1}{2} \, e\right )^{3} + a^{4} d^{4} \tan \left (\frac {1}{2} \, f x + \frac {1}{2} \, e\right )^{3} - 3 \, a^{4} c^{4} \tan \left (\frac {1}{2} \, f x + \frac {1}{2} \, e\right ) - 12 \, a^{4} c^{3} d \tan \left (\frac {1}{2} \, f x + \frac {1}{2} \, e\right ) + 54 \, a^{4} c^{2} d^{2} \tan \left (\frac {1}{2} \, f x + \frac {1}{2} \, e\right ) - 60 \, a^{4} c d^{3} \tan \left (\frac {1}{2} \, f x + \frac {1}{2} \, e\right ) + 21 \, a^{4} d^{4} \tan \left (\frac {1}{2} \, f x + \frac {1}{2} \, e\right )}{a^{6}}}{6 \, f} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate(sec(f*x+e)*(c+d*sec(f*x+e))^4/(a+a*sec(f*x+e))^2,x, algorithm="giac")

[Out]

1/6*(3*(12*c^2*d^2 - 16*c*d^3 + 7*d^4)*log(abs(tan(1/2*f*x + 1/2*e) + 1))/a^2 - 3*(12*c^2*d^2 - 16*c*d^3 + 7*d
^4)*log(abs(tan(1/2*f*x + 1/2*e) - 1))/a^2 - 6*(8*c*d^3*tan(1/2*f*x + 1/2*e)^3 - 5*d^4*tan(1/2*f*x + 1/2*e)^3
- 8*c*d^3*tan(1/2*f*x + 1/2*e) + 3*d^4*tan(1/2*f*x + 1/2*e))/((tan(1/2*f*x + 1/2*e)^2 - 1)^2*a^2) - (a^4*c^4*t
an(1/2*f*x + 1/2*e)^3 - 4*a^4*c^3*d*tan(1/2*f*x + 1/2*e)^3 + 6*a^4*c^2*d^2*tan(1/2*f*x + 1/2*e)^3 - 4*a^4*c*d^
3*tan(1/2*f*x + 1/2*e)^3 + a^4*d^4*tan(1/2*f*x + 1/2*e)^3 - 3*a^4*c^4*tan(1/2*f*x + 1/2*e) - 12*a^4*c^3*d*tan(
1/2*f*x + 1/2*e) + 54*a^4*c^2*d^2*tan(1/2*f*x + 1/2*e) - 60*a^4*c*d^3*tan(1/2*f*x + 1/2*e) + 21*a^4*d^4*tan(1/
2*f*x + 1/2*e))/a^6)/f

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Mupad [B]
time = 1.91, size = 193, normalized size = 1.00 \begin {gather*} \frac {\mathrm {tan}\left (\frac {e}{2}+\frac {f\,x}{2}\right )\,\left (8\,c\,d^3-3\,d^4\right )-{\mathrm {tan}\left (\frac {e}{2}+\frac {f\,x}{2}\right )}^3\,\left (8\,c\,d^3-5\,d^4\right )}{f\,\left (a^2\,{\mathrm {tan}\left (\frac {e}{2}+\frac {f\,x}{2}\right )}^4-2\,a^2\,{\mathrm {tan}\left (\frac {e}{2}+\frac {f\,x}{2}\right )}^2+a^2\right )}-\frac {\mathrm {tan}\left (\frac {e}{2}+\frac {f\,x}{2}\right )\,\left (\frac {3\,{\left (c-d\right )}^4}{2\,a^2}-\frac {2\,\left (c+d\right )\,{\left (c-d\right )}^3}{a^2}\right )}{f}-\frac {{\mathrm {tan}\left (\frac {e}{2}+\frac {f\,x}{2}\right )}^3\,{\left (c-d\right )}^4}{6\,a^2\,f}+\frac {d^2\,\mathrm {atanh}\left (\mathrm {tan}\left (\frac {e}{2}+\frac {f\,x}{2}\right )\right )\,\left (12\,c^2-16\,c\,d+7\,d^2\right )}{a^2\,f} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

int((c + d/cos(e + f*x))^4/(cos(e + f*x)*(a + a/cos(e + f*x))^2),x)

[Out]

(tan(e/2 + (f*x)/2)*(8*c*d^3 - 3*d^4) - tan(e/2 + (f*x)/2)^3*(8*c*d^3 - 5*d^4))/(f*(a^2*tan(e/2 + (f*x)/2)^4 -
2*a^2*tan(e/2 + (f*x)/2)^2 + a^2)) - (tan(e/2 + (f*x)/2)*((3*(c - d)^4)/(2*a^2) - (2*(c + d)*(c - d)^3)/a^2))
/f - (tan(e/2 + (f*x)/2)^3*(c - d)^4)/(6*a^2*f) + (d^2*atanh(tan(e/2 + (f*x)/2))*(12*c^2 - 16*c*d + 7*d^2))/(a
^2*f)

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