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

Optimal. Leaf size=133 \[ \frac{d^3 \tanh ^{-1}(\sin (e+f x))}{a^3 f}+\frac{(c-d) \tan (e+f x) \left (\left (2 c^2+11 c d+29 d^2\right ) \sec (e+f x)+2 \left (2 c^2+8 c d+11 d^2\right )\right )}{15 a f (a \sec (e+f x)+a)^2}+\frac{(c-d) \tan (e+f x) (c+d \sec (e+f x))^2}{5 f (a \sec (e+f x)+a)^3} \]

[Out]

(d^3*ArcTanh[Sin[e + f*x]])/(a^3*f) + ((c - d)*(c + d*Sec[e + f*x])^2*Tan[e + f*x])/(5*f*(a + a*Sec[e + f*x])^
3) + ((c - d)*(2*(2*c^2 + 8*c*d + 11*d^2) + (2*c^2 + 11*c*d + 29*d^2)*Sec[e + f*x])*Tan[e + f*x])/(15*a*f*(a +
a*Sec[e + f*x])^2)

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Rubi [A]  time = 0.200819, antiderivative size = 193, normalized size of antiderivative = 1.45, number of steps used = 6, number of rules used = 6, integrand size = 31, \(\frac{\text{number of rules}}{\text{integrand size}}\) = 0.194, Rules used = {3987, 98, 145, 63, 217, 203} \[ \frac{2 d^3 \tan (e+f x) \tan ^{-1}\left (\frac{\sqrt{a-a \sec (e+f x)}}{\sqrt{a (\sec (e+f x)+1)}}\right )}{a^2 f \sqrt{a-a \sec (e+f x)} \sqrt{a \sec (e+f x)+a}}+\frac{(c-d) \tan (e+f x) \left (\left (2 c^2+11 c d+29 d^2\right ) \sec (e+f x)+2 \left (2 c^2+8 c d+11 d^2\right )\right )}{15 a f (a \sec (e+f x)+a)^2}+\frac{(c-d) \tan (e+f x) (c+d \sec (e+f x))^2}{5 f (a \sec (e+f x)+a)^3} \]

Antiderivative was successfully verified.

[In]

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

[Out]

(2*d^3*ArcTan[Sqrt[a - a*Sec[e + f*x]]/Sqrt[a*(1 + Sec[e + f*x])]]*Tan[e + f*x])/(a^2*f*Sqrt[a - a*Sec[e + f*x
]]*Sqrt[a + a*Sec[e + f*x]]) + ((c - d)*(c + d*Sec[e + f*x])^2*Tan[e + f*x])/(5*f*(a + a*Sec[e + f*x])^3) + ((
c - d)*(2*(2*c^2 + 8*c*d + 11*d^2) + (2*c^2 + 11*c*d + 29*d^2)*Sec[e + f*x])*Tan[e + f*x])/(15*a*f*(a + a*Sec[
e + f*x])^2)

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 98

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 145

Int[((a_.) + (b_.)*(x_))^(m_)*((c_.) + (d_.)*(x_))^(n_.)*((e_) + (f_.)*(x_))*((g_.) + (h_.)*(x_)), x_Symbol] :
> Simp[((b^3*c*e*g*(m + 2) - a^3*d*f*h*(n + 2) - a^2*b*(c*f*h*m - d*(f*g + e*h)*(m + n + 3)) - a*b^2*(c*(f*g +
e*h) + d*e*g*(2*m + n + 4)) + b*(a^2*d*f*h*(m - n) - a*b*(2*c*f*h*(m + 1) - d*(f*g + e*h)*(n + 1)) + b^2*(c*(
f*g + e*h)*(m + 1) - d*e*g*(m + n + 2)))*x)*(a + b*x)^(m + 1)*(c + d*x)^(n + 1))/(b^2*(b*c - a*d)^2*(m + 1)*(m
+ 2)), x] + Dist[(f*h)/b^2 - (d*(m + n + 3)*(a^2*d*f*h*(m - n) - a*b*(2*c*f*h*(m + 1) - d*(f*g + e*h)*(n + 1)
) + b^2*(c*(f*g + e*h)*(m + 1) - d*e*g*(m + n + 2))))/(b^2*(b*c - a*d)^2*(m + 1)*(m + 2)), Int[(a + b*x)^(m +
2)*(c + d*x)^n, x], x] /; FreeQ[{a, b, c, d, e, f, g, h, m, n}, x] && (LtQ[m, -2] || (EqQ[m + n + 3, 0] &&  !L
tQ[n, -2]))

Rule 63

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 217

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 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{\sec (e+f x) (c+d \sec (e+f x))^3}{(a+a \sec (e+f x))^3} \, dx &=-\frac{\left (a^2 \tan (e+f x)\right ) \operatorname{Subst}\left (\int \frac{(c+d x)^3}{\sqrt{a-a x} (a+a x)^{7/2}} \, 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))^2 \tan (e+f x)}{5 f (a+a \sec (e+f x))^3}+\frac{\tan (e+f x) \operatorname{Subst}\left (\int \frac{(c+d x) \left (-a^2 \left (2 c^2+5 c d-2 d^2\right )-5 a^2 d^2 x\right )}{\sqrt{a-a x} (a+a x)^{5/2}} \, dx,x,\sec (e+f x)\right )}{5 a f \sqrt{a-a \sec (e+f x)} \sqrt{a+a \sec (e+f x)}}\\ &=\frac{(c-d) (c+d \sec (e+f x))^2 \tan (e+f x)}{5 f (a+a \sec (e+f x))^3}+\frac{(c-d) \left (2 \left (2 c^2+8 c d+11 d^2\right )+\left (2 c^2+11 c d+29 d^2\right ) \sec (e+f x)\right ) \tan (e+f x)}{15 a f (a+a \sec (e+f x))^2}-\frac{\left (d^3 \tan (e+f x)\right ) \operatorname{Subst}\left (\int \frac{1}{\sqrt{a-a x} \sqrt{a+a x}} \, dx,x,\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+d \sec (e+f x))^2 \tan (e+f x)}{5 f (a+a \sec (e+f x))^3}+\frac{(c-d) \left (2 \left (2 c^2+8 c d+11 d^2\right )+\left (2 c^2+11 c d+29 d^2\right ) \sec (e+f x)\right ) \tan (e+f x)}{15 a f (a+a \sec (e+f x))^2}+\frac{\left (2 d^3 \tan (e+f x)\right ) \operatorname{Subst}\left (\int \frac{1}{\sqrt{2 a-x^2}} \, dx,x,\sqrt{a-a \sec (e+f x)}\right )}{a^2 f \sqrt{a-a \sec (e+f x)} \sqrt{a+a \sec (e+f x)}}\\ &=\frac{(c-d) (c+d \sec (e+f x))^2 \tan (e+f x)}{5 f (a+a \sec (e+f x))^3}+\frac{(c-d) \left (2 \left (2 c^2+8 c d+11 d^2\right )+\left (2 c^2+11 c d+29 d^2\right ) \sec (e+f x)\right ) \tan (e+f x)}{15 a f (a+a \sec (e+f x))^2}+\frac{\left (2 d^3 \tan (e+f x)\right ) \operatorname{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^2 f \sqrt{a-a \sec (e+f x)} \sqrt{a+a \sec (e+f x)}}\\ &=\frac{2 d^3 \tan ^{-1}\left (\frac{\sqrt{a-a \sec (e+f x)}}{\sqrt{a+a \sec (e+f x)}}\right ) \tan (e+f x)}{a^2 f \sqrt{a-a \sec (e+f x)} \sqrt{a+a \sec (e+f x)}}+\frac{(c-d) (c+d \sec (e+f x))^2 \tan (e+f x)}{5 f (a+a \sec (e+f x))^3}+\frac{(c-d) \left (2 \left (2 c^2+8 c d+11 d^2\right )+\left (2 c^2+11 c d+29 d^2\right ) \sec (e+f x)\right ) \tan (e+f x)}{15 a f (a+a \sec (e+f x))^2}\\ \end{align*}

Mathematica [B]  time = 1.48399, size = 295, normalized size = 2.22 \[ \frac{(c-d) \sec \left (\frac{e}{2}\right ) \cos \left (\frac{1}{2} (e+f x)\right ) \left (-15 \left (2 c^2+5 c d+5 d^2\right ) \sin \left (e+\frac{f x}{2}\right )+5 \left (8 c^2+17 c d+29 d^2\right ) \sin \left (\frac{f x}{2}\right )+20 c^2 \sin \left (e+\frac{3 f x}{2}\right )-15 c^2 \sin \left (2 e+\frac{3 f x}{2}\right )+7 c^2 \sin \left (2 e+\frac{5 f x}{2}\right )+65 c d \sin \left (e+\frac{3 f x}{2}\right )-15 c d \sin \left (2 e+\frac{3 f x}{2}\right )+16 c d \sin \left (2 e+\frac{5 f x}{2}\right )+95 d^2 \sin \left (e+\frac{3 f x}{2}\right )-15 d^2 \sin \left (2 e+\frac{3 f x}{2}\right )+22 d^2 \sin \left (2 e+\frac{5 f x}{2}\right )\right )-240 d^3 \cos ^6\left (\frac{1}{2} (e+f x)\right ) \left (\log \left (\cos \left (\frac{1}{2} (e+f x)\right )-\sin \left (\frac{1}{2} (e+f x)\right )\right )-\log \left (\sin \left (\frac{1}{2} (e+f x)\right )+\cos \left (\frac{1}{2} (e+f x)\right )\right )\right )}{30 a^3 f (\cos (e+f x)+1)^3} \]

Antiderivative was successfully verified.

[In]

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

[Out]

(-240*d^3*Cos[(e + f*x)/2]^6*(Log[Cos[(e + f*x)/2] - Sin[(e + f*x)/2]] - Log[Cos[(e + f*x)/2] + Sin[(e + f*x)/
2]]) + (c - d)*Cos[(e + f*x)/2]*Sec[e/2]*(5*(8*c^2 + 17*c*d + 29*d^2)*Sin[(f*x)/2] - 15*(2*c^2 + 5*c*d + 5*d^2
)*Sin[e + (f*x)/2] + 20*c^2*Sin[e + (3*f*x)/2] + 65*c*d*Sin[e + (3*f*x)/2] + 95*d^2*Sin[e + (3*f*x)/2] - 15*c^
2*Sin[2*e + (3*f*x)/2] - 15*c*d*Sin[2*e + (3*f*x)/2] - 15*d^2*Sin[2*e + (3*f*x)/2] + 7*c^2*Sin[2*e + (5*f*x)/2
] + 16*c*d*Sin[2*e + (5*f*x)/2] + 22*d^2*Sin[2*e + (5*f*x)/2]))/(30*a^3*f*(1 + Cos[e + f*x])^3)

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Maple [B]  time = 0.071, size = 286, normalized size = 2.2 \begin{align*} -{\frac{3\,{c}^{2}d}{20\,f{a}^{3}} \left ( \tan \left ({\frac{fx}{2}}+{\frac{e}{2}} \right ) \right ) ^{5}}+{\frac{3\,{d}^{2}c}{20\,f{a}^{3}} \left ( \tan \left ({\frac{fx}{2}}+{\frac{e}{2}} \right ) \right ) ^{5}}+{\frac{{d}^{2}c}{2\,f{a}^{3}} \left ( \tan \left ({\frac{fx}{2}}+{\frac{e}{2}} \right ) \right ) ^{3}}+{\frac{3\,{c}^{2}d}{4\,f{a}^{3}}\tan \left ({\frac{fx}{2}}+{\frac{e}{2}} \right ) }+{\frac{3\,{d}^{2}c}{4\,f{a}^{3}}\tan \left ({\frac{fx}{2}}+{\frac{e}{2}} \right ) }-{\frac{{c}^{3}}{6\,f{a}^{3}} \left ( \tan \left ({\frac{fx}{2}}+{\frac{e}{2}} \right ) \right ) ^{3}}-{\frac{{d}^{3}}{3\,f{a}^{3}} \left ( \tan \left ({\frac{fx}{2}}+{\frac{e}{2}} \right ) \right ) ^{3}}+{\frac{{c}^{3}}{4\,f{a}^{3}}\tan \left ({\frac{fx}{2}}+{\frac{e}{2}} \right ) }-{\frac{7\,{d}^{3}}{4\,f{a}^{3}}\tan \left ({\frac{fx}{2}}+{\frac{e}{2}} \right ) }+{\frac{{d}^{3}}{f{a}^{3}}\ln \left ( \tan \left ({\frac{fx}{2}}+{\frac{e}{2}} \right ) +1 \right ) }-{\frac{{d}^{3}}{f{a}^{3}}\ln \left ( \tan \left ({\frac{fx}{2}}+{\frac{e}{2}} \right ) -1 \right ) }+{\frac{{c}^{3}}{20\,f{a}^{3}} \left ( \tan \left ({\frac{fx}{2}}+{\frac{e}{2}} \right ) \right ) ^{5}}-{\frac{{d}^{3}}{20\,f{a}^{3}} \left ( \tan \left ({\frac{fx}{2}}+{\frac{e}{2}} \right ) \right ) ^{5}} \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

-3/20/f/a^3*tan(1/2*f*x+1/2*e)^5*c^2*d+3/20/f/a^3*tan(1/2*f*x+1/2*e)^5*c*d^2+1/2/f/a^3*tan(1/2*f*x+1/2*e)^3*c*
d^2+3/4/f/a^3*tan(1/2*f*x+1/2*e)*c^2*d+3/4/f/a^3*tan(1/2*f*x+1/2*e)*c*d^2-1/6/f/a^3*tan(1/2*f*x+1/2*e)^3*c^3-1
/3/f/a^3*tan(1/2*f*x+1/2*e)^3*d^3+1/4/f/a^3*tan(1/2*f*x+1/2*e)*c^3-7/4/f/a^3*tan(1/2*f*x+1/2*e)*d^3+1/f/a^3*d^
3*ln(tan(1/2*f*x+1/2*e)+1)-1/f/a^3*d^3*ln(tan(1/2*f*x+1/2*e)-1)+1/20/f/a^3*tan(1/2*f*x+1/2*e)^5*c^3-1/20/f/a^3
*tan(1/2*f*x+1/2*e)^5*d^3

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Maxima [B]  time = 1.02522, size = 414, normalized size = 3.11 \begin{align*} -\frac{d^{3}{\left (\frac{\frac{105 \, \sin \left (f x + e\right )}{\cos \left (f x + e\right ) + 1} + \frac{20 \, \sin \left (f x + e\right )^{3}}{{\left (\cos \left (f x + e\right ) + 1\right )}^{3}} + \frac{3 \, \sin \left (f x + e\right )^{5}}{{\left (\cos \left (f x + e\right ) + 1\right )}^{5}}}{a^{3}} - \frac{60 \, \log \left (\frac{\sin \left (f x + e\right )}{\cos \left (f x + e\right ) + 1} + 1\right )}{a^{3}} + \frac{60 \, \log \left (\frac{\sin \left (f x + e\right )}{\cos \left (f x + e\right ) + 1} - 1\right )}{a^{3}}\right )} - \frac{3 \, c d^{2}{\left (\frac{15 \, \sin \left (f x + e\right )}{\cos \left (f x + e\right ) + 1} + \frac{10 \, \sin \left (f x + e\right )^{3}}{{\left (\cos \left (f x + e\right ) + 1\right )}^{3}} + \frac{3 \, \sin \left (f x + e\right )^{5}}{{\left (\cos \left (f x + e\right ) + 1\right )}^{5}}\right )}}{a^{3}} - \frac{c^{3}{\left (\frac{15 \, \sin \left (f x + e\right )}{\cos \left (f x + e\right ) + 1} - \frac{10 \, \sin \left (f x + e\right )^{3}}{{\left (\cos \left (f x + e\right ) + 1\right )}^{3}} + \frac{3 \, \sin \left (f x + e\right )^{5}}{{\left (\cos \left (f x + e\right ) + 1\right )}^{5}}\right )}}{a^{3}} - \frac{9 \, c^{2} d{\left (\frac{5 \, \sin \left (f x + e\right )}{\cos \left (f x + e\right ) + 1} - \frac{\sin \left (f x + e\right )^{5}}{{\left (\cos \left (f x + e\right ) + 1\right )}^{5}}\right )}}{a^{3}}}{60 \, f} \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

-1/60*(d^3*((105*sin(f*x + e)/(cos(f*x + e) + 1) + 20*sin(f*x + e)^3/(cos(f*x + e) + 1)^3 + 3*sin(f*x + e)^5/(
cos(f*x + e) + 1)^5)/a^3 - 60*log(sin(f*x + e)/(cos(f*x + e) + 1) + 1)/a^3 + 60*log(sin(f*x + e)/(cos(f*x + e)
+ 1) - 1)/a^3) - 3*c*d^2*(15*sin(f*x + e)/(cos(f*x + e) + 1) + 10*sin(f*x + e)^3/(cos(f*x + e) + 1)^3 + 3*sin
(f*x + e)^5/(cos(f*x + e) + 1)^5)/a^3 - c^3*(15*sin(f*x + e)/(cos(f*x + e) + 1) - 10*sin(f*x + e)^3/(cos(f*x +
e) + 1)^3 + 3*sin(f*x + e)^5/(cos(f*x + e) + 1)^5)/a^3 - 9*c^2*d*(5*sin(f*x + e)/(cos(f*x + e) + 1) - sin(f*x
+ e)^5/(cos(f*x + e) + 1)^5)/a^3)/f

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Fricas [A]  time = 0.507776, size = 601, normalized size = 4.52 \begin{align*} \frac{15 \,{\left (d^{3} \cos \left (f x + e\right )^{3} + 3 \, d^{3} \cos \left (f x + e\right )^{2} + 3 \, d^{3} \cos \left (f x + e\right ) + d^{3}\right )} \log \left (\sin \left (f x + e\right ) + 1\right ) - 15 \,{\left (d^{3} \cos \left (f x + e\right )^{3} + 3 \, d^{3} \cos \left (f x + e\right )^{2} + 3 \, d^{3} \cos \left (f x + e\right ) + d^{3}\right )} \log \left (-\sin \left (f x + e\right ) + 1\right ) + 2 \,{\left (2 \, c^{3} + 9 \, c^{2} d + 21 \, c d^{2} - 32 \, d^{3} +{\left (7 \, c^{3} + 9 \, c^{2} d + 6 \, c d^{2} - 22 \, d^{3}\right )} \cos \left (f x + e\right )^{2} + 3 \,{\left (2 \, c^{3} + 9 \, c^{2} d + 6 \, c d^{2} - 17 \, d^{3}\right )} \cos \left (f x + e\right )\right )} \sin \left (f x + e\right )}{30 \,{\left (a^{3} f \cos \left (f x + e\right )^{3} + 3 \, a^{3} f \cos \left (f x + e\right )^{2} + 3 \, a^{3} f \cos \left (f x + e\right ) + a^{3} f\right )}} \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

1/30*(15*(d^3*cos(f*x + e)^3 + 3*d^3*cos(f*x + e)^2 + 3*d^3*cos(f*x + e) + d^3)*log(sin(f*x + e) + 1) - 15*(d^
3*cos(f*x + e)^3 + 3*d^3*cos(f*x + e)^2 + 3*d^3*cos(f*x + e) + d^3)*log(-sin(f*x + e) + 1) + 2*(2*c^3 + 9*c^2*
d + 21*c*d^2 - 32*d^3 + (7*c^3 + 9*c^2*d + 6*c*d^2 - 22*d^3)*cos(f*x + e)^2 + 3*(2*c^3 + 9*c^2*d + 6*c*d^2 - 1
7*d^3)*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*} \frac{\int \frac{c^{3} \sec{\left (e + f x \right )}}{\sec ^{3}{\left (e + f x \right )} + 3 \sec ^{2}{\left (e + f x \right )} + 3 \sec{\left (e + f x \right )} + 1}\, dx + \int \frac{d^{3} \sec ^{4}{\left (e + f x \right )}}{\sec ^{3}{\left (e + f x \right )} + 3 \sec ^{2}{\left (e + f x \right )} + 3 \sec{\left (e + f x \right )} + 1}\, dx + \int \frac{3 c d^{2} \sec ^{3}{\left (e + f x \right )}}{\sec ^{3}{\left (e + f x \right )} + 3 \sec ^{2}{\left (e + f x \right )} + 3 \sec{\left (e + f x \right )} + 1}\, dx + \int \frac{3 c^{2} d \sec ^{2}{\left (e + f x \right )}}{\sec ^{3}{\left (e + f x \right )} + 3 \sec ^{2}{\left (e + f x \right )} + 3 \sec{\left (e + f x \right )} + 1}\, dx}{a^{3}} \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

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

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Giac [B]  time = 1.34942, size = 367, normalized size = 2.76 \begin{align*} \frac{\frac{60 \, d^{3} \log \left ({\left | \tan \left (\frac{1}{2} \, f x + \frac{1}{2} \, e\right ) + 1 \right |}\right )}{a^{3}} - \frac{60 \, d^{3} \log \left ({\left | \tan \left (\frac{1}{2} \, f x + \frac{1}{2} \, e\right ) - 1 \right |}\right )}{a^{3}} + \frac{3 \, a^{12} c^{3} \tan \left (\frac{1}{2} \, f x + \frac{1}{2} \, e\right )^{5} - 9 \, a^{12} c^{2} d \tan \left (\frac{1}{2} \, f x + \frac{1}{2} \, e\right )^{5} + 9 \, a^{12} c d^{2} \tan \left (\frac{1}{2} \, f x + \frac{1}{2} \, e\right )^{5} - 3 \, a^{12} d^{3} \tan \left (\frac{1}{2} \, f x + \frac{1}{2} \, e\right )^{5} - 10 \, a^{12} c^{3} \tan \left (\frac{1}{2} \, f x + \frac{1}{2} \, e\right )^{3} + 30 \, a^{12} c d^{2} \tan \left (\frac{1}{2} \, f x + \frac{1}{2} \, e\right )^{3} - 20 \, a^{12} d^{3} \tan \left (\frac{1}{2} \, f x + \frac{1}{2} \, e\right )^{3} + 15 \, a^{12} c^{3} \tan \left (\frac{1}{2} \, f x + \frac{1}{2} \, e\right ) + 45 \, a^{12} c^{2} d \tan \left (\frac{1}{2} \, f x + \frac{1}{2} \, e\right ) + 45 \, a^{12} c d^{2} \tan \left (\frac{1}{2} \, f x + \frac{1}{2} \, e\right ) - 105 \, a^{12} d^{3} \tan \left (\frac{1}{2} \, f x + \frac{1}{2} \, e\right )}{a^{15}}}{60 \, f} \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

1/60*(60*d^3*log(abs(tan(1/2*f*x + 1/2*e) + 1))/a^3 - 60*d^3*log(abs(tan(1/2*f*x + 1/2*e) - 1))/a^3 + (3*a^12*
c^3*tan(1/2*f*x + 1/2*e)^5 - 9*a^12*c^2*d*tan(1/2*f*x + 1/2*e)^5 + 9*a^12*c*d^2*tan(1/2*f*x + 1/2*e)^5 - 3*a^1
2*d^3*tan(1/2*f*x + 1/2*e)^5 - 10*a^12*c^3*tan(1/2*f*x + 1/2*e)^3 + 30*a^12*c*d^2*tan(1/2*f*x + 1/2*e)^3 - 20*
a^12*d^3*tan(1/2*f*x + 1/2*e)^3 + 15*a^12*c^3*tan(1/2*f*x + 1/2*e) + 45*a^12*c^2*d*tan(1/2*f*x + 1/2*e) + 45*a
^12*c*d^2*tan(1/2*f*x + 1/2*e) - 105*a^12*d^3*tan(1/2*f*x + 1/2*e))/a^15)/f