3.518 \(\int (a+b \sec (d+e x)) (b^2+2 a b \sec (d+e x)+a^2 \sec ^2(d+e x))^2 \, dx\)

Optimal. Leaf size=184 \[ \frac {a^2 b \left (41 a^2+26 b^2\right ) \tan (d+e x) \sec (d+e x)}{24 e}+\frac {\left (4 a^2+7 b^2\right ) \tan (d+e x) \left (a^2 \sec (d+e x)+a b\right )^2}{12 a e}+\frac {b \tan (d+e x) \left (a^2 \sec (d+e x)+a b\right )^3}{4 a^2 e}+\frac {a \left (4 a^4+50 a^2 b^2+19 b^4\right ) \tan (d+e x)}{6 e}+\frac {b \left (19 a^4+56 a^2 b^2+8 b^4\right ) \tanh ^{-1}(\sin (d+e x))}{8 e}+a b^4 x \]

[Out]

a*b^4*x+1/8*b*(19*a^4+56*a^2*b^2+8*b^4)*arctanh(sin(e*x+d))/e+1/6*a*(4*a^4+50*a^2*b^2+19*b^4)*tan(e*x+d)/e+1/2
4*a^2*b*(41*a^2+26*b^2)*sec(e*x+d)*tan(e*x+d)/e+1/12*(4*a^2+7*b^2)*(a*b+a^2*sec(e*x+d))^2*tan(e*x+d)/a/e+1/4*b
*(a*b+a^2*sec(e*x+d))^3*tan(e*x+d)/a^2/e

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Rubi [A]  time = 0.43, antiderivative size = 184, normalized size of antiderivative = 1.00, number of steps used = 8, number of rules used = 7, integrand size = 39, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.180, Rules used = {4172, 3918, 4056, 4048, 3770, 3767, 8} \[ \frac {a \left (50 a^2 b^2+4 a^4+19 b^4\right ) \tan (d+e x)}{6 e}+\frac {b \left (56 a^2 b^2+19 a^4+8 b^4\right ) \tanh ^{-1}(\sin (d+e x))}{8 e}+\frac {a^2 b \left (41 a^2+26 b^2\right ) \tan (d+e x) \sec (d+e x)}{24 e}+\frac {\left (4 a^2+7 b^2\right ) \tan (d+e x) \left (a^2 \sec (d+e x)+a b\right )^2}{12 a e}+\frac {b \tan (d+e x) \left (a^2 \sec (d+e x)+a b\right )^3}{4 a^2 e}+a b^4 x \]

Antiderivative was successfully verified.

[In]

Int[(a + b*Sec[d + e*x])*(b^2 + 2*a*b*Sec[d + e*x] + a^2*Sec[d + e*x]^2)^2,x]

[Out]

a*b^4*x + (b*(19*a^4 + 56*a^2*b^2 + 8*b^4)*ArcTanh[Sin[d + e*x]])/(8*e) + (a*(4*a^4 + 50*a^2*b^2 + 19*b^4)*Tan
[d + e*x])/(6*e) + (a^2*b*(41*a^2 + 26*b^2)*Sec[d + e*x]*Tan[d + e*x])/(24*e) + ((4*a^2 + 7*b^2)*(a*b + a^2*Se
c[d + e*x])^2*Tan[d + e*x])/(12*a*e) + (b*(a*b + a^2*Sec[d + e*x])^3*Tan[d + e*x])/(4*a^2*e)

Rule 8

Int[a_, x_Symbol] :> Simp[a*x, x] /; FreeQ[a, x]

Rule 3767

Int[csc[(c_.) + (d_.)*(x_)]^(n_), x_Symbol] :> -Dist[d^(-1), Subst[Int[ExpandIntegrand[(1 + x^2)^(n/2 - 1), x]
, x], x, Cot[c + d*x]], x] /; FreeQ[{c, d}, x] && IGtQ[n/2, 0]

Rule 3770

Int[csc[(c_.) + (d_.)*(x_)], x_Symbol] :> -Simp[ArcTanh[Cos[c + d*x]]/d, x] /; FreeQ[{c, d}, x]

Rule 3918

Int[(csc[(e_.) + (f_.)*(x_)]*(b_.) + (a_))^(m_)*(csc[(e_.) + (f_.)*(x_)]*(d_.) + (c_)), x_Symbol] :> -Simp[(b*
d*Cot[e + f*x]*(a + b*Csc[e + f*x])^(m - 1))/(f*m), x] + Dist[1/m, Int[(a + b*Csc[e + f*x])^(m - 2)*Simp[a^2*c
*m + (b^2*d*(m - 1) + 2*a*b*c*m + a^2*d*m)*Csc[e + f*x] + b*(b*c*m + a*d*(2*m - 1))*Csc[e + f*x]^2, x], x], x]
/; FreeQ[{a, b, c, d, e, f}, x] && NeQ[b*c - a*d, 0] && GtQ[m, 1] && NeQ[a^2 - b^2, 0] && IntegerQ[2*m]

Rule 4048

Int[((A_.) + csc[(e_.) + (f_.)*(x_)]*(B_.) + csc[(e_.) + (f_.)*(x_)]^2*(C_.))*(csc[(e_.) + (f_.)*(x_)]*(b_.) +
(a_)), x_Symbol] :> -Simp[(b*C*Csc[e + f*x]*Cot[e + f*x])/(2*f), x] + Dist[1/2, Int[Simp[2*A*a + (2*B*a + b*(
2*A + C))*Csc[e + f*x] + 2*(a*C + B*b)*Csc[e + f*x]^2, x], x], x] /; FreeQ[{a, b, e, f, A, B, C}, x]

Rule 4056

Int[((A_.) + csc[(e_.) + (f_.)*(x_)]*(B_.) + csc[(e_.) + (f_.)*(x_)]^2*(C_.))*(csc[(e_.) + (f_.)*(x_)]*(b_.) +
(a_))^(m_.), x_Symbol] :> -Simp[(C*Cot[e + f*x]*(a + b*Csc[e + f*x])^m)/(f*(m + 1)), x] + Dist[1/(m + 1), Int
[(a + b*Csc[e + f*x])^(m - 1)*Simp[a*A*(m + 1) + ((A*b + a*B)*(m + 1) + b*C*m)*Csc[e + f*x] + (b*B*(m + 1) + a
*C*m)*Csc[e + f*x]^2, x], x], x] /; FreeQ[{a, b, e, f, A, B, C}, x] && NeQ[a^2 - b^2, 0] && IGtQ[2*m, 0]

Rule 4172

Int[((A_) + (B_.)*sec[(d_.) + (e_.)*(x_)])*((a_) + (b_.)*sec[(d_.) + (e_.)*(x_)] + (c_.)*sec[(d_.) + (e_.)*(x_
)]^2)^(n_), x_Symbol] :> Dist[1/(4^n*c^n), Int[(A + B*Sec[d + e*x])*(b + 2*c*Sec[d + e*x])^(2*n), x], x] /; Fr
eeQ[{a, b, c, d, e, A, B}, x] && EqQ[b^2 - 4*a*c, 0] && IntegerQ[n]

Rubi steps

\begin {align*} \int (a+b \sec (d+e x)) \left (b^2+2 a b \sec (d+e x)+a^2 \sec ^2(d+e x)\right )^2 \, dx &=\frac {\int \left (2 a b+2 a^2 \sec (d+e x)\right )^4 (a+b \sec (d+e x)) \, dx}{16 a^4}\\ &=\frac {b \left (a b+a^2 \sec (d+e x)\right )^3 \tan (d+e x)}{4 a^2 e}+\frac {\int \left (2 a b+2 a^2 \sec (d+e x)\right )^2 \left (16 a^3 b^2+4 a^2 b \left (11 a^2+4 b^2\right ) \sec (d+e x)+4 a^3 \left (4 a^2+7 b^2\right ) \sec ^2(d+e x)\right ) \, dx}{64 a^4}\\ &=\frac {\left (4 a^2+7 b^2\right ) \left (a b+a^2 \sec (d+e x)\right )^2 \tan (d+e x)}{12 a e}+\frac {b \left (a b+a^2 \sec (d+e x)\right )^3 \tan (d+e x)}{4 a^2 e}+\frac {\int \left (2 a b+2 a^2 \sec (d+e x)\right ) \left (96 a^4 b^3+8 a^3 \left (8 a^4+59 a^2 b^2+12 b^4\right ) \sec (d+e x)+8 a^4 b \left (41 a^2+26 b^2\right ) \sec ^2(d+e x)\right ) \, dx}{192 a^4}\\ &=\frac {a^2 b \left (41 a^2+26 b^2\right ) \sec (d+e x) \tan (d+e x)}{24 e}+\frac {\left (4 a^2+7 b^2\right ) \left (a b+a^2 \sec (d+e x)\right )^2 \tan (d+e x)}{12 a e}+\frac {b \left (a b+a^2 \sec (d+e x)\right )^3 \tan (d+e x)}{4 a^2 e}+\frac {\int \left (384 a^5 b^4+48 a^4 b \left (19 a^4+56 a^2 b^2+8 b^4\right ) \sec (d+e x)+64 a^5 \left (4 a^4+50 a^2 b^2+19 b^4\right ) \sec ^2(d+e x)\right ) \, dx}{384 a^4}\\ &=a b^4 x+\frac {a^2 b \left (41 a^2+26 b^2\right ) \sec (d+e x) \tan (d+e x)}{24 e}+\frac {\left (4 a^2+7 b^2\right ) \left (a b+a^2 \sec (d+e x)\right )^2 \tan (d+e x)}{12 a e}+\frac {b \left (a b+a^2 \sec (d+e x)\right )^3 \tan (d+e x)}{4 a^2 e}+\frac {1}{8} \left (b \left (19 a^4+56 a^2 b^2+8 b^4\right )\right ) \int \sec (d+e x) \, dx+\frac {1}{6} \left (a \left (4 a^4+50 a^2 b^2+19 b^4\right )\right ) \int \sec ^2(d+e x) \, dx\\ &=a b^4 x+\frac {b \left (19 a^4+56 a^2 b^2+8 b^4\right ) \tanh ^{-1}(\sin (d+e x))}{8 e}+\frac {a^2 b \left (41 a^2+26 b^2\right ) \sec (d+e x) \tan (d+e x)}{24 e}+\frac {\left (4 a^2+7 b^2\right ) \left (a b+a^2 \sec (d+e x)\right )^2 \tan (d+e x)}{12 a e}+\frac {b \left (a b+a^2 \sec (d+e x)\right )^3 \tan (d+e x)}{4 a^2 e}-\frac {\left (a \left (4 a^4+50 a^2 b^2+19 b^4\right )\right ) \operatorname {Subst}(\int 1 \, dx,x,-\tan (d+e x))}{6 e}\\ &=a b^4 x+\frac {b \left (19 a^4+56 a^2 b^2+8 b^4\right ) \tanh ^{-1}(\sin (d+e x))}{8 e}+\frac {a \left (4 a^4+50 a^2 b^2+19 b^4\right ) \tan (d+e x)}{6 e}+\frac {a^2 b \left (41 a^2+26 b^2\right ) \sec (d+e x) \tan (d+e x)}{24 e}+\frac {\left (4 a^2+7 b^2\right ) \left (a b+a^2 \sec (d+e x)\right )^2 \tan (d+e x)}{12 a e}+\frac {b \left (a b+a^2 \sec (d+e x)\right )^3 \tan (d+e x)}{4 a^2 e}\\ \end {align*}

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Mathematica [A]  time = 0.87, size = 130, normalized size = 0.71 \[ \frac {3 b \left (19 a^4+56 a^2 b^2+8 b^4\right ) \tanh ^{-1}(\sin (d+e x))+8 a^3 \left (a^2+4 b^2\right ) \tan ^3(d+e x)+3 a \tan (d+e x) \left (2 a^3 b \sec ^3(d+e x)+a b \left (19 a^2+24 b^2\right ) \sec (d+e x)+8 \left (a^4+10 a^2 b^2+4 b^4\right )\right )+24 a b^4 e x}{24 e} \]

Antiderivative was successfully verified.

[In]

Integrate[(a + b*Sec[d + e*x])*(b^2 + 2*a*b*Sec[d + e*x] + a^2*Sec[d + e*x]^2)^2,x]

[Out]

(24*a*b^4*e*x + 3*b*(19*a^4 + 56*a^2*b^2 + 8*b^4)*ArcTanh[Sin[d + e*x]] + 3*a*(8*(a^4 + 10*a^2*b^2 + 4*b^4) +
a*b*(19*a^2 + 24*b^2)*Sec[d + e*x] + 2*a^3*b*Sec[d + e*x]^3)*Tan[d + e*x] + 8*a^3*(a^2 + 4*b^2)*Tan[d + e*x]^3
)/(24*e)

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fricas [A]  time = 2.43, size = 198, normalized size = 1.08 \[ \frac {48 \, a b^{4} e x \cos \left (e x + d\right )^{4} + 3 \, {\left (19 \, a^{4} b + 56 \, a^{2} b^{3} + 8 \, b^{5}\right )} \cos \left (e x + d\right )^{4} \log \left (\sin \left (e x + d\right ) + 1\right ) - 3 \, {\left (19 \, a^{4} b + 56 \, a^{2} b^{3} + 8 \, b^{5}\right )} \cos \left (e x + d\right )^{4} \log \left (-\sin \left (e x + d\right ) + 1\right ) + 2 \, {\left (6 \, a^{4} b + 16 \, {\left (a^{5} + 13 \, a^{3} b^{2} + 6 \, a b^{4}\right )} \cos \left (e x + d\right )^{3} + 3 \, {\left (19 \, a^{4} b + 24 \, a^{2} b^{3}\right )} \cos \left (e x + d\right )^{2} + 8 \, {\left (a^{5} + 4 \, a^{3} b^{2}\right )} \cos \left (e x + d\right )\right )} \sin \left (e x + d\right )}{48 \, e \cos \left (e x + d\right )^{4}} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate((a+b*sec(e*x+d))*(b^2+2*a*b*sec(e*x+d)+a^2*sec(e*x+d)^2)^2,x, algorithm="fricas")

[Out]

1/48*(48*a*b^4*e*x*cos(e*x + d)^4 + 3*(19*a^4*b + 56*a^2*b^3 + 8*b^5)*cos(e*x + d)^4*log(sin(e*x + d) + 1) - 3
*(19*a^4*b + 56*a^2*b^3 + 8*b^5)*cos(e*x + d)^4*log(-sin(e*x + d) + 1) + 2*(6*a^4*b + 16*(a^5 + 13*a^3*b^2 + 6
*a*b^4)*cos(e*x + d)^3 + 3*(19*a^4*b + 24*a^2*b^3)*cos(e*x + d)^2 + 8*(a^5 + 4*a^3*b^2)*cos(e*x + d))*sin(e*x
+ d))/(e*cos(e*x + d)^4)

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giac [B]  time = 0.39, size = 470, normalized size = 2.55 \[ \frac {1}{24} \, {\left (24 \, {\left (x e + d\right )} a b^{4} + 3 \, {\left (19 \, a^{4} b + 56 \, a^{2} b^{3} + 8 \, b^{5}\right )} \log \left ({\left | \tan \left (\frac {1}{2} \, x e + \frac {1}{2} \, d\right ) + 1 \right |}\right ) - 3 \, {\left (19 \, a^{4} b + 56 \, a^{2} b^{3} + 8 \, b^{5}\right )} \log \left ({\left | \tan \left (\frac {1}{2} \, x e + \frac {1}{2} \, d\right ) - 1 \right |}\right ) - \frac {2 \, {\left (24 \, a^{5} \tan \left (\frac {1}{2} \, x e + \frac {1}{2} \, d\right )^{7} - 63 \, a^{4} b \tan \left (\frac {1}{2} \, x e + \frac {1}{2} \, d\right )^{7} + 240 \, a^{3} b^{2} \tan \left (\frac {1}{2} \, x e + \frac {1}{2} \, d\right )^{7} - 72 \, a^{2} b^{3} \tan \left (\frac {1}{2} \, x e + \frac {1}{2} \, d\right )^{7} + 96 \, a b^{4} \tan \left (\frac {1}{2} \, x e + \frac {1}{2} \, d\right )^{7} - 40 \, a^{5} \tan \left (\frac {1}{2} \, x e + \frac {1}{2} \, d\right )^{5} + 39 \, a^{4} b \tan \left (\frac {1}{2} \, x e + \frac {1}{2} \, d\right )^{5} - 592 \, a^{3} b^{2} \tan \left (\frac {1}{2} \, x e + \frac {1}{2} \, d\right )^{5} + 72 \, a^{2} b^{3} \tan \left (\frac {1}{2} \, x e + \frac {1}{2} \, d\right )^{5} - 288 \, a b^{4} \tan \left (\frac {1}{2} \, x e + \frac {1}{2} \, d\right )^{5} + 40 \, a^{5} \tan \left (\frac {1}{2} \, x e + \frac {1}{2} \, d\right )^{3} + 39 \, a^{4} b \tan \left (\frac {1}{2} \, x e + \frac {1}{2} \, d\right )^{3} + 592 \, a^{3} b^{2} \tan \left (\frac {1}{2} \, x e + \frac {1}{2} \, d\right )^{3} + 72 \, a^{2} b^{3} \tan \left (\frac {1}{2} \, x e + \frac {1}{2} \, d\right )^{3} + 288 \, a b^{4} \tan \left (\frac {1}{2} \, x e + \frac {1}{2} \, d\right )^{3} - 24 \, a^{5} \tan \left (\frac {1}{2} \, x e + \frac {1}{2} \, d\right ) - 63 \, a^{4} b \tan \left (\frac {1}{2} \, x e + \frac {1}{2} \, d\right ) - 240 \, a^{3} b^{2} \tan \left (\frac {1}{2} \, x e + \frac {1}{2} \, d\right ) - 72 \, a^{2} b^{3} \tan \left (\frac {1}{2} \, x e + \frac {1}{2} \, d\right ) - 96 \, a b^{4} \tan \left (\frac {1}{2} \, x e + \frac {1}{2} \, d\right )\right )}}{{\left (\tan \left (\frac {1}{2} \, x e + \frac {1}{2} \, d\right )^{2} - 1\right )}^{4}}\right )} e^{\left (-1\right )} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate((a+b*sec(e*x+d))*(b^2+2*a*b*sec(e*x+d)+a^2*sec(e*x+d)^2)^2,x, algorithm="giac")

[Out]

1/24*(24*(x*e + d)*a*b^4 + 3*(19*a^4*b + 56*a^2*b^3 + 8*b^5)*log(abs(tan(1/2*x*e + 1/2*d) + 1)) - 3*(19*a^4*b
+ 56*a^2*b^3 + 8*b^5)*log(abs(tan(1/2*x*e + 1/2*d) - 1)) - 2*(24*a^5*tan(1/2*x*e + 1/2*d)^7 - 63*a^4*b*tan(1/2
*x*e + 1/2*d)^7 + 240*a^3*b^2*tan(1/2*x*e + 1/2*d)^7 - 72*a^2*b^3*tan(1/2*x*e + 1/2*d)^7 + 96*a*b^4*tan(1/2*x*
e + 1/2*d)^7 - 40*a^5*tan(1/2*x*e + 1/2*d)^5 + 39*a^4*b*tan(1/2*x*e + 1/2*d)^5 - 592*a^3*b^2*tan(1/2*x*e + 1/2
*d)^5 + 72*a^2*b^3*tan(1/2*x*e + 1/2*d)^5 - 288*a*b^4*tan(1/2*x*e + 1/2*d)^5 + 40*a^5*tan(1/2*x*e + 1/2*d)^3 +
39*a^4*b*tan(1/2*x*e + 1/2*d)^3 + 592*a^3*b^2*tan(1/2*x*e + 1/2*d)^3 + 72*a^2*b^3*tan(1/2*x*e + 1/2*d)^3 + 28
8*a*b^4*tan(1/2*x*e + 1/2*d)^3 - 24*a^5*tan(1/2*x*e + 1/2*d) - 63*a^4*b*tan(1/2*x*e + 1/2*d) - 240*a^3*b^2*tan
(1/2*x*e + 1/2*d) - 72*a^2*b^3*tan(1/2*x*e + 1/2*d) - 96*a*b^4*tan(1/2*x*e + 1/2*d))/(tan(1/2*x*e + 1/2*d)^2 -
1)^4)*e^(-1)

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maple [A]  time = 0.14, size = 246, normalized size = 1.34 \[ a \,b^{4} x +\frac {a \,b^{4} d}{e}+\frac {7 a^{2} b^{3} \ln \left (\sec \left (e x +d \right )+\tan \left (e x +d \right )\right )}{e}+\frac {26 a^{3} b^{2} \tan \left (e x +d \right )}{3 e}+\frac {19 a^{4} b \sec \left (e x +d \right ) \tan \left (e x +d \right )}{8 e}+\frac {19 a^{4} b \ln \left (\sec \left (e x +d \right )+\tan \left (e x +d \right )\right )}{8 e}+\frac {2 a^{5} \tan \left (e x +d \right )}{3 e}+\frac {a^{5} \tan \left (e x +d \right ) \left (\sec ^{2}\left (e x +d \right )\right )}{3 e}+\frac {b^{5} \ln \left (\sec \left (e x +d \right )+\tan \left (e x +d \right )\right )}{e}+\frac {4 a \,b^{4} \tan \left (e x +d \right )}{e}+\frac {3 a^{2} b^{3} \sec \left (e x +d \right ) \tan \left (e x +d \right )}{e}+\frac {4 a^{3} b^{2} \tan \left (e x +d \right ) \left (\sec ^{2}\left (e x +d \right )\right )}{3 e}+\frac {a^{4} b \tan \left (e x +d \right ) \left (\sec ^{3}\left (e x +d \right )\right )}{4 e} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

int((a+b*sec(e*x+d))*(b^2+2*a*b*sec(e*x+d)+a^2*sec(e*x+d)^2)^2,x)

[Out]

a*b^4*x+1/e*a*b^4*d+7/e*a^2*b^3*ln(sec(e*x+d)+tan(e*x+d))+26/3/e*a^3*b^2*tan(e*x+d)+19/8/e*a^4*b*sec(e*x+d)*ta
n(e*x+d)+19/8/e*a^4*b*ln(sec(e*x+d)+tan(e*x+d))+2/3/e*a^5*tan(e*x+d)+1/3/e*a^5*tan(e*x+d)*sec(e*x+d)^2+1/e*b^5
*ln(sec(e*x+d)+tan(e*x+d))+4/e*a*b^4*tan(e*x+d)+3/e*a^2*b^3*sec(e*x+d)*tan(e*x+d)+4/3/e*a^3*b^2*tan(e*x+d)*sec
(e*x+d)^2+1/4/e*a^4*b*tan(e*x+d)*sec(e*x+d)^3

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maxima [A]  time = 0.34, size = 299, normalized size = 1.62 \[ \frac {16 \, {\left (\tan \left (e x + d\right )^{3} + 3 \, \tan \left (e x + d\right )\right )} a^{5} + 64 \, {\left (\tan \left (e x + d\right )^{3} + 3 \, \tan \left (e x + d\right )\right )} a^{3} b^{2} + 48 \, {\left (e x + d\right )} a b^{4} - 3 \, a^{4} b {\left (\frac {2 \, {\left (3 \, \sin \left (e x + d\right )^{3} - 5 \, \sin \left (e x + d\right )\right )}}{\sin \left (e x + d\right )^{4} - 2 \, \sin \left (e x + d\right )^{2} + 1} - 3 \, \log \left (\sin \left (e x + d\right ) + 1\right ) + 3 \, \log \left (\sin \left (e x + d\right ) - 1\right )\right )} - 48 \, a^{4} b {\left (\frac {2 \, \sin \left (e x + d\right )}{\sin \left (e x + d\right )^{2} - 1} - \log \left (\sin \left (e x + d\right ) + 1\right ) + \log \left (\sin \left (e x + d\right ) - 1\right )\right )} - 72 \, a^{2} b^{3} {\left (\frac {2 \, \sin \left (e x + d\right )}{\sin \left (e x + d\right )^{2} - 1} - \log \left (\sin \left (e x + d\right ) + 1\right ) + \log \left (\sin \left (e x + d\right ) - 1\right )\right )} + 192 \, a^{2} b^{3} \log \left (\sec \left (e x + d\right ) + \tan \left (e x + d\right )\right ) + 48 \, b^{5} \log \left (\sec \left (e x + d\right ) + \tan \left (e x + d\right )\right ) + 288 \, a^{3} b^{2} \tan \left (e x + d\right ) + 192 \, a b^{4} \tan \left (e x + d\right )}{48 \, e} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate((a+b*sec(e*x+d))*(b^2+2*a*b*sec(e*x+d)+a^2*sec(e*x+d)^2)^2,x, algorithm="maxima")

[Out]

1/48*(16*(tan(e*x + d)^3 + 3*tan(e*x + d))*a^5 + 64*(tan(e*x + d)^3 + 3*tan(e*x + d))*a^3*b^2 + 48*(e*x + d)*a
*b^4 - 3*a^4*b*(2*(3*sin(e*x + d)^3 - 5*sin(e*x + d))/(sin(e*x + d)^4 - 2*sin(e*x + d)^2 + 1) - 3*log(sin(e*x
+ d) + 1) + 3*log(sin(e*x + d) - 1)) - 48*a^4*b*(2*sin(e*x + d)/(sin(e*x + d)^2 - 1) - log(sin(e*x + d) + 1) +
log(sin(e*x + d) - 1)) - 72*a^2*b^3*(2*sin(e*x + d)/(sin(e*x + d)^2 - 1) - log(sin(e*x + d) + 1) + log(sin(e*
x + d) - 1)) + 192*a^2*b^3*log(sec(e*x + d) + tan(e*x + d)) + 48*b^5*log(sec(e*x + d) + tan(e*x + d)) + 288*a^
3*b^2*tan(e*x + d) + 192*a*b^4*tan(e*x + d))/e

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mupad [B]  time = 3.30, size = 323, normalized size = 1.76 \[ \frac {2\,a^5\,\sin \left (d+e\,x\right )}{3\,e\,\cos \left (d+e\,x\right )}+\frac {a^5\,\sin \left (d+e\,x\right )}{3\,e\,{\cos \left (d+e\,x\right )}^3}+\frac {2\,a\,b^4\,\mathrm {atan}\left (\frac {\sin \left (\frac {d}{2}+\frac {e\,x}{2}\right )}{\cos \left (\frac {d}{2}+\frac {e\,x}{2}\right )}\right )}{e}+\frac {4\,a\,b^4\,\sin \left (d+e\,x\right )}{e\,\cos \left (d+e\,x\right )}+\frac {19\,a^4\,b\,\sin \left (d+e\,x\right )}{8\,e\,{\cos \left (d+e\,x\right )}^2}+\frac {a^4\,b\,\sin \left (d+e\,x\right )}{4\,e\,{\cos \left (d+e\,x\right )}^4}+\frac {26\,a^3\,b^2\,\sin \left (d+e\,x\right )}{3\,e\,\cos \left (d+e\,x\right )}+\frac {3\,a^2\,b^3\,\sin \left (d+e\,x\right )}{e\,{\cos \left (d+e\,x\right )}^2}+\frac {4\,a^3\,b^2\,\sin \left (d+e\,x\right )}{3\,e\,{\cos \left (d+e\,x\right )}^3}-\frac {b^5\,\mathrm {atan}\left (\frac {\sin \left (\frac {d}{2}+\frac {e\,x}{2}\right )\,1{}\mathrm {i}}{\cos \left (\frac {d}{2}+\frac {e\,x}{2}\right )}\right )\,2{}\mathrm {i}}{e}-\frac {a^2\,b^3\,\mathrm {atan}\left (\frac {\sin \left (\frac {d}{2}+\frac {e\,x}{2}\right )\,1{}\mathrm {i}}{\cos \left (\frac {d}{2}+\frac {e\,x}{2}\right )}\right )\,14{}\mathrm {i}}{e}-\frac {a^4\,b\,\mathrm {atan}\left (\frac {\sin \left (\frac {d}{2}+\frac {e\,x}{2}\right )\,1{}\mathrm {i}}{\cos \left (\frac {d}{2}+\frac {e\,x}{2}\right )}\right )\,19{}\mathrm {i}}{4\,e} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

int((a + b/cos(d + e*x))*(b^2 + a^2/cos(d + e*x)^2 + (2*a*b)/cos(d + e*x))^2,x)

[Out]

(2*a^5*sin(d + e*x))/(3*e*cos(d + e*x)) - (b^5*atan((sin(d/2 + (e*x)/2)*1i)/cos(d/2 + (e*x)/2))*2i)/e + (a^5*s
in(d + e*x))/(3*e*cos(d + e*x)^3) - (a^2*b^3*atan((sin(d/2 + (e*x)/2)*1i)/cos(d/2 + (e*x)/2))*14i)/e + (2*a*b^
4*atan(sin(d/2 + (e*x)/2)/cos(d/2 + (e*x)/2)))/e - (a^4*b*atan((sin(d/2 + (e*x)/2)*1i)/cos(d/2 + (e*x)/2))*19i
)/(4*e) + (4*a*b^4*sin(d + e*x))/(e*cos(d + e*x)) + (19*a^4*b*sin(d + e*x))/(8*e*cos(d + e*x)^2) + (a^4*b*sin(
d + e*x))/(4*e*cos(d + e*x)^4) + (26*a^3*b^2*sin(d + e*x))/(3*e*cos(d + e*x)) + (3*a^2*b^3*sin(d + e*x))/(e*co
s(d + e*x)^2) + (4*a^3*b^2*sin(d + e*x))/(3*e*cos(d + e*x)^3)

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sympy [F]  time = 0.00, size = 0, normalized size = 0.00 \[ \int \left (a + b \sec {\left (d + e x \right )}\right ) \left (a \sec {\left (d + e x \right )} + b\right )^{4}\, dx \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate((a+b*sec(e*x+d))*(b**2+2*a*b*sec(e*x+d)+a**2*sec(e*x+d)**2)**2,x)

[Out]

Integral((a + b*sec(d + e*x))*(a*sec(d + e*x) + b)**4, x)

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