3.44 \(\int \sec ^3(c+d x) (a+a \sec (c+d x)) (A+B \sec (c+d x)) \, dx\)

Optimal. Leaf size=106 \[ \frac{a (A+B) \tan ^3(c+d x)}{3 d}+\frac{a (A+B) \tan (c+d x)}{d}+\frac{a (4 A+3 B) \tanh ^{-1}(\sin (c+d x))}{8 d}+\frac{a (4 A+3 B) \tan (c+d x) \sec (c+d x)}{8 d}+\frac{a B \tan (c+d x) \sec ^3(c+d x)}{4 d} \]

[Out]

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

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Rubi [A]  time = 0.123469, antiderivative size = 106, normalized size of antiderivative = 1., number of steps used = 6, number of rules used = 5, integrand size = 29, \(\frac{\text{number of rules}}{\text{integrand size}}\) = 0.172, Rules used = {3997, 3787, 3768, 3770, 3767} \[ \frac{a (A+B) \tan ^3(c+d x)}{3 d}+\frac{a (A+B) \tan (c+d x)}{d}+\frac{a (4 A+3 B) \tanh ^{-1}(\sin (c+d x))}{8 d}+\frac{a (4 A+3 B) \tan (c+d x) \sec (c+d x)}{8 d}+\frac{a B \tan (c+d x) \sec ^3(c+d x)}{4 d} \]

Antiderivative was successfully verified.

[In]

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

[Out]

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

Rule 3997

Int[(csc[(e_.) + (f_.)*(x_)]*(d_.))^(n_.)*(csc[(e_.) + (f_.)*(x_)]*(b_.) + (a_))*(csc[(e_.) + (f_.)*(x_)]*(B_.
) + (A_)), x_Symbol] :> -Simp[(b*B*Cot[e + f*x]*(d*Csc[e + f*x])^n)/(f*(n + 1)), x] + Dist[1/(n + 1), Int[(d*C
sc[e + f*x])^n*Simp[A*a*(n + 1) + B*b*n + (A*b + B*a)*(n + 1)*Csc[e + f*x], x], x], x] /; FreeQ[{a, b, d, e, f
, A, B}, x] && NeQ[A*b - a*B, 0] &&  !LeQ[n, -1]

Rule 3787

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

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 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]

Rubi steps

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

Mathematica [A]  time = 0.38896, size = 77, normalized size = 0.73 \[ \frac{a \left (3 (4 A+3 B) \tanh ^{-1}(\sin (c+d x))+\tan (c+d x) \sec (c+d x) \left (8 (A+B) (\cos (2 (c+d x))+2) \sec (c+d x)+12 A+6 B \sec ^2(c+d x)+9 B\right )\right )}{24 d} \]

Antiderivative was successfully verified.

[In]

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

[Out]

(a*(3*(4*A + 3*B)*ArcTanh[Sin[c + d*x]] + Sec[c + d*x]*(12*A + 9*B + 8*(A + B)*(2 + Cos[2*(c + d*x)])*Sec[c +
d*x] + 6*B*Sec[c + d*x]^2)*Tan[c + d*x]))/(24*d)

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Maple [A]  time = 0.041, size = 171, normalized size = 1.6 \begin{align*}{\frac{Aa\tan \left ( dx+c \right ) \sec \left ( dx+c \right ) }{2\,d}}+{\frac{Aa\ln \left ( \sec \left ( dx+c \right ) +\tan \left ( dx+c \right ) \right ) }{2\,d}}+{\frac{2\,Ba\tan \left ( dx+c \right ) }{3\,d}}+{\frac{Ba \left ( \sec \left ( dx+c \right ) \right ) ^{2}\tan \left ( dx+c \right ) }{3\,d}}+{\frac{2\,Aa\tan \left ( dx+c \right ) }{3\,d}}+{\frac{Aa\tan \left ( dx+c \right ) \left ( \sec \left ( dx+c \right ) \right ) ^{2}}{3\,d}}+{\frac{Ba \left ( \sec \left ( dx+c \right ) \right ) ^{3}\tan \left ( dx+c \right ) }{4\,d}}+{\frac{3\,Ba\sec \left ( dx+c \right ) \tan \left ( dx+c \right ) }{8\,d}}+{\frac{3\,Ba\ln \left ( \sec \left ( dx+c \right ) +\tan \left ( dx+c \right ) \right ) }{8\,d}} \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

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

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Maxima [A]  time = 0.982083, size = 220, normalized size = 2.08 \begin{align*} \frac{16 \,{\left (\tan \left (d x + c\right )^{3} + 3 \, \tan \left (d x + c\right )\right )} A a + 16 \,{\left (\tan \left (d x + c\right )^{3} + 3 \, \tan \left (d x + c\right )\right )} B a - 3 \, B a{\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 a{\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} \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

1/48*(16*(tan(d*x + c)^3 + 3*tan(d*x + c))*A*a + 16*(tan(d*x + c)^3 + 3*tan(d*x + c))*B*a - 3*B*a*(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*a*(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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Fricas [A]  time = 0.491569, size = 339, normalized size = 3.2 \begin{align*} \frac{3 \,{\left (4 \, A + 3 \, B\right )} a \cos \left (d x + c\right )^{4} \log \left (\sin \left (d x + c\right ) + 1\right ) - 3 \,{\left (4 \, A + 3 \, B\right )} a \cos \left (d x + c\right )^{4} \log \left (-\sin \left (d x + c\right ) + 1\right ) + 2 \,{\left (16 \,{\left (A + B\right )} a \cos \left (d x + c\right )^{3} + 3 \,{\left (4 \, A + 3 \, B\right )} a \cos \left (d x + c\right )^{2} + 8 \,{\left (A + B\right )} a \cos \left (d x + c\right ) + 6 \, B a\right )} \sin \left (d x + c\right )}{48 \, d \cos \left (d x + c\right )^{4}} \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

1/48*(3*(4*A + 3*B)*a*cos(d*x + c)^4*log(sin(d*x + c) + 1) - 3*(4*A + 3*B)*a*cos(d*x + c)^4*log(-sin(d*x + c)
+ 1) + 2*(16*(A + B)*a*cos(d*x + c)^3 + 3*(4*A + 3*B)*a*cos(d*x + c)^2 + 8*(A + B)*a*cos(d*x + c) + 6*B*a)*sin
(d*x + c))/(d*cos(d*x + c)^4)

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Sympy [F]  time = 0., size = 0, normalized size = 0. \begin{align*} a \left (\int A \sec ^{3}{\left (c + d x \right )}\, dx + \int A \sec ^{4}{\left (c + d x \right )}\, dx + \int B \sec ^{4}{\left (c + d x \right )}\, dx + \int B \sec ^{5}{\left (c + d x \right )}\, dx\right ) \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

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

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Giac [A]  time = 1.28472, size = 254, normalized size = 2.4 \begin{align*} \frac{3 \,{\left (4 \, A a + 3 \, B a\right )} \log \left ({\left | \tan \left (\frac{1}{2} \, d x + \frac{1}{2} \, c\right ) + 1 \right |}\right ) - 3 \,{\left (4 \, A a + 3 \, B a\right )} \log \left ({\left | \tan \left (\frac{1}{2} \, d x + \frac{1}{2} \, c\right ) - 1 \right |}\right ) - \frac{2 \,{\left (12 \, A a \tan \left (\frac{1}{2} \, d x + \frac{1}{2} \, c\right )^{7} + 9 \, B a \tan \left (\frac{1}{2} \, d x + \frac{1}{2} \, c\right )^{7} - 28 \, A a \tan \left (\frac{1}{2} \, d x + \frac{1}{2} \, c\right )^{5} - 49 \, B a \tan \left (\frac{1}{2} \, d x + \frac{1}{2} \, c\right )^{5} + 52 \, A a \tan \left (\frac{1}{2} \, d x + \frac{1}{2} \, c\right )^{3} + 31 \, B a \tan \left (\frac{1}{2} \, d x + \frac{1}{2} \, c\right )^{3} - 36 \, A a \tan \left (\frac{1}{2} \, d x + \frac{1}{2} \, c\right ) - 39 \, B a \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} \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

1/24*(3*(4*A*a + 3*B*a)*log(abs(tan(1/2*d*x + 1/2*c) + 1)) - 3*(4*A*a + 3*B*a)*log(abs(tan(1/2*d*x + 1/2*c) -
1)) - 2*(12*A*a*tan(1/2*d*x + 1/2*c)^7 + 9*B*a*tan(1/2*d*x + 1/2*c)^7 - 28*A*a*tan(1/2*d*x + 1/2*c)^5 - 49*B*a
*tan(1/2*d*x + 1/2*c)^5 + 52*A*a*tan(1/2*d*x + 1/2*c)^3 + 31*B*a*tan(1/2*d*x + 1/2*c)^3 - 36*A*a*tan(1/2*d*x +
1/2*c) - 39*B*a*tan(1/2*d*x + 1/2*c))/(tan(1/2*d*x + 1/2*c)^2 - 1)^4)/d