3.11 \(\int \sec ^3(c+d x) (a+a \sec (c+d x))^2 \, dx\)

Optimal. Leaf size=96 \[ \frac {2 a^2 \tan ^3(c+d x)}{3 d}+\frac {2 a^2 \tan (c+d x)}{d}+\frac {7 a^2 \tanh ^{-1}(\sin (c+d x))}{8 d}+\frac {a^2 \tan (c+d x) \sec ^3(c+d x)}{4 d}+\frac {7 a^2 \tan (c+d x) \sec (c+d x)}{8 d} \]

[Out]

7/8*a^2*arctanh(sin(d*x+c))/d+2*a^2*tan(d*x+c)/d+7/8*a^2*sec(d*x+c)*tan(d*x+c)/d+1/4*a^2*sec(d*x+c)^3*tan(d*x+
c)/d+2/3*a^2*tan(d*x+c)^3/d

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Rubi [A]  time = 0.08, antiderivative size = 96, normalized size of antiderivative = 1.00, number of steps used = 6, number of rules used = 5, integrand size = 21, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.238, Rules used = {3788, 3767, 4046, 3768, 3770} \[ \frac {2 a^2 \tan ^3(c+d x)}{3 d}+\frac {2 a^2 \tan (c+d x)}{d}+\frac {7 a^2 \tanh ^{-1}(\sin (c+d x))}{8 d}+\frac {a^2 \tan (c+d x) \sec ^3(c+d x)}{4 d}+\frac {7 a^2 \tan (c+d x) \sec (c+d x)}{8 d} \]

Antiderivative was successfully verified.

[In]

Int[Sec[c + d*x]^3*(a + a*Sec[c + d*x])^2,x]

[Out]

(7*a^2*ArcTanh[Sin[c + d*x]])/(8*d) + (2*a^2*Tan[c + d*x])/d + (7*a^2*Sec[c + d*x]*Tan[c + d*x])/(8*d) + (a^2*
Sec[c + d*x]^3*Tan[c + d*x])/(4*d) + (2*a^2*Tan[c + d*x]^3)/(3*d)

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 3768

Int[(csc[(c_.) + (d_.)*(x_)]*(b_.))^(n_), x_Symbol] :> -Simp[(b*Cos[c + d*x]*(b*Csc[c + d*x])^(n - 1))/(d*(n -
1)), x] + Dist[(b^2*(n - 2))/(n - 1), Int[(b*Csc[c + d*x])^(n - 2), x], x] /; FreeQ[{b, c, d}, x] && GtQ[n, 1
] && IntegerQ[2*n]

Rule 3770

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

Rule 3788

Int[(csc[(e_.) + (f_.)*(x_)]*(d_.))^(n_.)*(csc[(e_.) + (f_.)*(x_)]*(b_.) + (a_))^2, x_Symbol] :> Dist[(2*a*b)/
d, Int[(d*Csc[e + f*x])^(n + 1), x], x] + Int[(d*Csc[e + f*x])^n*(a^2 + b^2*Csc[e + f*x]^2), x] /; FreeQ[{a, b
, d, e, f, n}, x]

Rule 4046

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

Rubi steps

\begin {align*} \int \sec ^3(c+d x) (a+a \sec (c+d x))^2 \, dx &=\left (2 a^2\right ) \int \sec ^4(c+d x) \, dx+\int \sec ^3(c+d x) \left (a^2+a^2 \sec ^2(c+d x)\right ) \, dx\\ &=\frac {a^2 \sec ^3(c+d x) \tan (c+d x)}{4 d}+\frac {1}{4} \left (7 a^2\right ) \int \sec ^3(c+d x) \, dx-\frac {\left (2 a^2\right ) \operatorname {Subst}\left (\int \left (1+x^2\right ) \, dx,x,-\tan (c+d x)\right )}{d}\\ &=\frac {2 a^2 \tan (c+d x)}{d}+\frac {7 a^2 \sec (c+d x) \tan (c+d x)}{8 d}+\frac {a^2 \sec ^3(c+d x) \tan (c+d x)}{4 d}+\frac {2 a^2 \tan ^3(c+d x)}{3 d}+\frac {1}{8} \left (7 a^2\right ) \int \sec (c+d x) \, dx\\ &=\frac {7 a^2 \tanh ^{-1}(\sin (c+d x))}{8 d}+\frac {2 a^2 \tan (c+d x)}{d}+\frac {7 a^2 \sec (c+d x) \tan (c+d x)}{8 d}+\frac {a^2 \sec ^3(c+d x) \tan (c+d x)}{4 d}+\frac {2 a^2 \tan ^3(c+d x)}{3 d}\\ \end {align*}

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Mathematica [B]  time = 6.46, size = 877, normalized size = 9.14 \[ -\frac {7 \cos ^2(c+d x) \log \left (\cos \left (\frac {c}{2}+\frac {d x}{2}\right )-\sin \left (\frac {c}{2}+\frac {d x}{2}\right )\right ) (\sec (c+d x) a+a)^2 \sec ^4\left (\frac {c}{2}+\frac {d x}{2}\right )}{32 d}+\frac {7 \cos ^2(c+d x) \log \left (\cos \left (\frac {c}{2}+\frac {d x}{2}\right )+\sin \left (\frac {c}{2}+\frac {d x}{2}\right )\right ) (\sec (c+d x) a+a)^2 \sec ^4\left (\frac {c}{2}+\frac {d x}{2}\right )}{32 d}+\frac {\cos ^2(c+d x) (\sec (c+d x) a+a)^2 \sin \left (\frac {d x}{2}\right ) \sec ^4\left (\frac {c}{2}+\frac {d x}{2}\right )}{3 d \left (\cos \left (\frac {c}{2}\right )-\sin \left (\frac {c}{2}\right )\right ) \left (\cos \left (\frac {c}{2}+\frac {d x}{2}\right )-\sin \left (\frac {c}{2}+\frac {d x}{2}\right )\right )}+\frac {\cos ^2(c+d x) (\sec (c+d x) a+a)^2 \sin \left (\frac {d x}{2}\right ) \sec ^4\left (\frac {c}{2}+\frac {d x}{2}\right )}{3 d \left (\cos \left (\frac {c}{2}\right )+\sin \left (\frac {c}{2}\right )\right ) \left (\cos \left (\frac {c}{2}+\frac {d x}{2}\right )+\sin \left (\frac {c}{2}+\frac {d x}{2}\right )\right )}+\frac {\cos ^2(c+d x) (\sec (c+d x) a+a)^2 \left (29 \cos \left (\frac {c}{2}\right )-13 \sin \left (\frac {c}{2}\right )\right ) \sec ^4\left (\frac {c}{2}+\frac {d x}{2}\right )}{192 d \left (\cos \left (\frac {c}{2}\right )-\sin \left (\frac {c}{2}\right )\right ) \left (\cos \left (\frac {c}{2}+\frac {d x}{2}\right )-\sin \left (\frac {c}{2}+\frac {d x}{2}\right )\right )^2}+\frac {\cos ^2(c+d x) (\sec (c+d x) a+a)^2 \left (-29 \cos \left (\frac {c}{2}\right )-13 \sin \left (\frac {c}{2}\right )\right ) \sec ^4\left (\frac {c}{2}+\frac {d x}{2}\right )}{192 d \left (\cos \left (\frac {c}{2}\right )+\sin \left (\frac {c}{2}\right )\right ) \left (\cos \left (\frac {c}{2}+\frac {d x}{2}\right )+\sin \left (\frac {c}{2}+\frac {d x}{2}\right )\right )^2}+\frac {\cos ^2(c+d x) (\sec (c+d x) a+a)^2 \sin \left (\frac {d x}{2}\right ) \sec ^4\left (\frac {c}{2}+\frac {d x}{2}\right )}{12 d \left (\cos \left (\frac {c}{2}\right )-\sin \left (\frac {c}{2}\right )\right ) \left (\cos \left (\frac {c}{2}+\frac {d x}{2}\right )-\sin \left (\frac {c}{2}+\frac {d x}{2}\right )\right )^3}+\frac {\cos ^2(c+d x) (\sec (c+d x) a+a)^2 \sin \left (\frac {d x}{2}\right ) \sec ^4\left (\frac {c}{2}+\frac {d x}{2}\right )}{12 d \left (\cos \left (\frac {c}{2}\right )+\sin \left (\frac {c}{2}\right )\right ) \left (\cos \left (\frac {c}{2}+\frac {d x}{2}\right )+\sin \left (\frac {c}{2}+\frac {d x}{2}\right )\right )^3}+\frac {\cos ^2(c+d x) (\sec (c+d x) a+a)^2 \sec ^4\left (\frac {c}{2}+\frac {d x}{2}\right )}{64 d \left (\cos \left (\frac {c}{2}+\frac {d x}{2}\right )-\sin \left (\frac {c}{2}+\frac {d x}{2}\right )\right )^4}-\frac {\cos ^2(c+d x) (\sec (c+d x) a+a)^2 \sec ^4\left (\frac {c}{2}+\frac {d x}{2}\right )}{64 d \left (\cos \left (\frac {c}{2}+\frac {d x}{2}\right )+\sin \left (\frac {c}{2}+\frac {d x}{2}\right )\right )^4} \]

Warning: Unable to verify antiderivative.

[In]

Integrate[Sec[c + d*x]^3*(a + a*Sec[c + d*x])^2,x]

[Out]

(-7*Cos[c + d*x]^2*Log[Cos[c/2 + (d*x)/2] - Sin[c/2 + (d*x)/2]]*Sec[c/2 + (d*x)/2]^4*(a + a*Sec[c + d*x])^2)/(
32*d) + (7*Cos[c + d*x]^2*Log[Cos[c/2 + (d*x)/2] + Sin[c/2 + (d*x)/2]]*Sec[c/2 + (d*x)/2]^4*(a + a*Sec[c + d*x
])^2)/(32*d) + (Cos[c + d*x]^2*Sec[c/2 + (d*x)/2]^4*(a + a*Sec[c + d*x])^2)/(64*d*(Cos[c/2 + (d*x)/2] - Sin[c/
2 + (d*x)/2])^4) + (Cos[c + d*x]^2*Sec[c/2 + (d*x)/2]^4*(a + a*Sec[c + d*x])^2*Sin[(d*x)/2])/(12*d*(Cos[c/2] -
Sin[c/2])*(Cos[c/2 + (d*x)/2] - Sin[c/2 + (d*x)/2])^3) + (Cos[c + d*x]^2*Sec[c/2 + (d*x)/2]^4*(a + a*Sec[c +
d*x])^2*(29*Cos[c/2] - 13*Sin[c/2]))/(192*d*(Cos[c/2] - Sin[c/2])*(Cos[c/2 + (d*x)/2] - Sin[c/2 + (d*x)/2])^2)
+ (Cos[c + d*x]^2*Sec[c/2 + (d*x)/2]^4*(a + a*Sec[c + d*x])^2*Sin[(d*x)/2])/(3*d*(Cos[c/2] - Sin[c/2])*(Cos[c
/2 + (d*x)/2] - Sin[c/2 + (d*x)/2])) - (Cos[c + d*x]^2*Sec[c/2 + (d*x)/2]^4*(a + a*Sec[c + d*x])^2)/(64*d*(Cos
[c/2 + (d*x)/2] + Sin[c/2 + (d*x)/2])^4) + (Cos[c + d*x]^2*Sec[c/2 + (d*x)/2]^4*(a + a*Sec[c + d*x])^2*Sin[(d*
x)/2])/(12*d*(Cos[c/2] + Sin[c/2])*(Cos[c/2 + (d*x)/2] + Sin[c/2 + (d*x)/2])^3) + (Cos[c + d*x]^2*Sec[c/2 + (d
*x)/2]^4*(a + a*Sec[c + d*x])^2*(-29*Cos[c/2] - 13*Sin[c/2]))/(192*d*(Cos[c/2] + Sin[c/2])*(Cos[c/2 + (d*x)/2]
+ Sin[c/2 + (d*x)/2])^2) + (Cos[c + d*x]^2*Sec[c/2 + (d*x)/2]^4*(a + a*Sec[c + d*x])^2*Sin[(d*x)/2])/(3*d*(Co
s[c/2] + Sin[c/2])*(Cos[c/2 + (d*x)/2] + Sin[c/2 + (d*x)/2]))

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fricas [A]  time = 0.80, size = 111, normalized size = 1.16 \[ \frac {21 \, a^{2} \cos \left (d x + c\right )^{4} \log \left (\sin \left (d x + c\right ) + 1\right ) - 21 \, a^{2} \cos \left (d x + c\right )^{4} \log \left (-\sin \left (d x + c\right ) + 1\right ) + 2 \, {\left (32 \, a^{2} \cos \left (d x + c\right )^{3} + 21 \, a^{2} \cos \left (d x + c\right )^{2} + 16 \, a^{2} \cos \left (d x + c\right ) + 6 \, a^{2}\right )} \sin \left (d x + c\right )}{48 \, d \cos \left (d x + c\right )^{4}} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate(sec(d*x+c)^3*(a+a*sec(d*x+c))^2,x, algorithm="fricas")

[Out]

1/48*(21*a^2*cos(d*x + c)^4*log(sin(d*x + c) + 1) - 21*a^2*cos(d*x + c)^4*log(-sin(d*x + c) + 1) + 2*(32*a^2*c
os(d*x + c)^3 + 21*a^2*cos(d*x + c)^2 + 16*a^2*cos(d*x + c) + 6*a^2)*sin(d*x + c))/(d*cos(d*x + c)^4)

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giac [A]  time = 0.48, size = 122, normalized size = 1.27 \[ \frac {21 \, a^{2} \log \left ({\left | \tan \left (\frac {1}{2} \, d x + \frac {1}{2} \, c\right ) + 1 \right |}\right ) - 21 \, a^{2} \log \left ({\left | \tan \left (\frac {1}{2} \, d x + \frac {1}{2} \, c\right ) - 1 \right |}\right ) - \frac {2 \, {\left (21 \, a^{2} \tan \left (\frac {1}{2} \, d x + \frac {1}{2} \, c\right )^{7} - 77 \, a^{2} \tan \left (\frac {1}{2} \, d x + \frac {1}{2} \, c\right )^{5} + 83 \, a^{2} \tan \left (\frac {1}{2} \, d x + \frac {1}{2} \, c\right )^{3} - 75 \, a^{2} \tan \left (\frac {1}{2} \, d x + \frac {1}{2} \, c\right )\right )}}{{\left (\tan \left (\frac {1}{2} \, d x + \frac {1}{2} \, c\right )^{2} - 1\right )}^{4}}}{24 \, d} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate(sec(d*x+c)^3*(a+a*sec(d*x+c))^2,x, algorithm="giac")

[Out]

1/24*(21*a^2*log(abs(tan(1/2*d*x + 1/2*c) + 1)) - 21*a^2*log(abs(tan(1/2*d*x + 1/2*c) - 1)) - 2*(21*a^2*tan(1/
2*d*x + 1/2*c)^7 - 77*a^2*tan(1/2*d*x + 1/2*c)^5 + 83*a^2*tan(1/2*d*x + 1/2*c)^3 - 75*a^2*tan(1/2*d*x + 1/2*c)
)/(tan(1/2*d*x + 1/2*c)^2 - 1)^4)/d

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maple [A]  time = 0.87, size = 102, normalized size = 1.06 \[ \frac {7 a^{2} \sec \left (d x +c \right ) \tan \left (d x +c \right )}{8 d}+\frac {7 a^{2} \ln \left (\sec \left (d x +c \right )+\tan \left (d x +c \right )\right )}{8 d}+\frac {4 a^{2} \tan \left (d x +c \right )}{3 d}+\frac {2 a^{2} \left (\sec ^{2}\left (d x +c \right )\right ) \tan \left (d x +c \right )}{3 d}+\frac {a^{2} \left (\sec ^{3}\left (d x +c \right )\right ) \tan \left (d x +c \right )}{4 d} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

int(sec(d*x+c)^3*(a+a*sec(d*x+c))^2,x)

[Out]

7/8*a^2*sec(d*x+c)*tan(d*x+c)/d+7/8/d*a^2*ln(sec(d*x+c)+tan(d*x+c))+4/3*a^2*tan(d*x+c)/d+2/3*a^2*sec(d*x+c)^2*
tan(d*x+c)/d+1/4*a^2*sec(d*x+c)^3*tan(d*x+c)/d

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maxima [A]  time = 0.77, size = 145, normalized size = 1.51 \[ \frac {32 \, {\left (\tan \left (d x + c\right )^{3} + 3 \, \tan \left (d x + c\right )\right )} a^{2} - 3 \, a^{2} {\left (\frac {2 \, {\left (3 \, \sin \left (d x + c\right )^{3} - 5 \, \sin \left (d x + c\right )\right )}}{\sin \left (d x + c\right )^{4} - 2 \, \sin \left (d x + c\right )^{2} + 1} - 3 \, \log \left (\sin \left (d x + c\right ) + 1\right ) + 3 \, \log \left (\sin \left (d x + c\right ) - 1\right )\right )} - 12 \, a^{2} {\left (\frac {2 \, \sin \left (d x + c\right )}{\sin \left (d x + c\right )^{2} - 1} - \log \left (\sin \left (d x + c\right ) + 1\right ) + \log \left (\sin \left (d x + c\right ) - 1\right )\right )}}{48 \, d} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate(sec(d*x+c)^3*(a+a*sec(d*x+c))^2,x, algorithm="maxima")

[Out]

1/48*(32*(tan(d*x + c)^3 + 3*tan(d*x + c))*a^2 - 3*a^2*(2*(3*sin(d*x + c)^3 - 5*sin(d*x + c))/(sin(d*x + c)^4
- 2*sin(d*x + c)^2 + 1) - 3*log(sin(d*x + c) + 1) + 3*log(sin(d*x + c) - 1)) - 12*a^2*(2*sin(d*x + c)/(sin(d*x
+ c)^2 - 1) - log(sin(d*x + c) + 1) + log(sin(d*x + c) - 1)))/d

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mupad [B]  time = 3.96, size = 141, normalized size = 1.47 \[ \frac {7\,a^2\,\mathrm {atanh}\left (\mathrm {tan}\left (\frac {c}{2}+\frac {d\,x}{2}\right )\right )}{4\,d}-\frac {\frac {7\,a^2\,{\mathrm {tan}\left (\frac {c}{2}+\frac {d\,x}{2}\right )}^7}{4}-\frac {77\,a^2\,{\mathrm {tan}\left (\frac {c}{2}+\frac {d\,x}{2}\right )}^5}{12}+\frac {83\,a^2\,{\mathrm {tan}\left (\frac {c}{2}+\frac {d\,x}{2}\right )}^3}{12}-\frac {25\,a^2\,\mathrm {tan}\left (\frac {c}{2}+\frac {d\,x}{2}\right )}{4}}{d\,\left ({\mathrm {tan}\left (\frac {c}{2}+\frac {d\,x}{2}\right )}^8-4\,{\mathrm {tan}\left (\frac {c}{2}+\frac {d\,x}{2}\right )}^6+6\,{\mathrm {tan}\left (\frac {c}{2}+\frac {d\,x}{2}\right )}^4-4\,{\mathrm {tan}\left (\frac {c}{2}+\frac {d\,x}{2}\right )}^2+1\right )} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

int((a + a/cos(c + d*x))^2/cos(c + d*x)^3,x)

[Out]

(7*a^2*atanh(tan(c/2 + (d*x)/2)))/(4*d) - ((83*a^2*tan(c/2 + (d*x)/2)^3)/12 - (77*a^2*tan(c/2 + (d*x)/2)^5)/12
+ (7*a^2*tan(c/2 + (d*x)/2)^7)/4 - (25*a^2*tan(c/2 + (d*x)/2))/4)/(d*(6*tan(c/2 + (d*x)/2)^4 - 4*tan(c/2 + (d
*x)/2)^2 - 4*tan(c/2 + (d*x)/2)^6 + tan(c/2 + (d*x)/2)^8 + 1))

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sympy [F]  time = 0.00, size = 0, normalized size = 0.00 \[ a^{2} \left (\int \sec ^{3}{\left (c + d x \right )}\, dx + \int 2 \sec ^{4}{\left (c + d x \right )}\, dx + \int \sec ^{5}{\left (c + d x \right )}\, dx\right ) \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate(sec(d*x+c)**3*(a+a*sec(d*x+c))**2,x)

[Out]

a**2*(Integral(sec(c + d*x)**3, x) + Integral(2*sec(c + d*x)**4, x) + Integral(sec(c + d*x)**5, x))

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