3.43 \(\int \sec ^4(c+d x) (a+a \sec (c+d x)) (A+B \sec (c+d x)) \, dx\)

Optimal. Leaf size=134 \[ \frac{a (5 A+4 B) \tan ^3(c+d x)}{15 d}+\frac{a (5 A+4 B) \tan (c+d x)}{5 d}+\frac{3 a (A+B) \tanh ^{-1}(\sin (c+d x))}{8 d}+\frac{a (A+B) \tan (c+d x) \sec ^3(c+d x)}{4 d}+\frac{3 a (A+B) \tan (c+d x) \sec (c+d x)}{8 d}+\frac{a B \tan (c+d x) \sec ^4(c+d x)}{5 d} \]

[Out]

(3*a*(A + B)*ArcTanh[Sin[c + d*x]])/(8*d) + (a*(5*A + 4*B)*Tan[c + d*x])/(5*d) + (3*a*(A + B)*Sec[c + d*x]*Tan
[c + d*x])/(8*d) + (a*(A + B)*Sec[c + d*x]^3*Tan[c + d*x])/(4*d) + (a*B*Sec[c + d*x]^4*Tan[c + d*x])/(5*d) + (
a*(5*A + 4*B)*Tan[c + d*x]^3)/(15*d)

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Rubi [A]  time = 0.141087, antiderivative size = 134, normalized size of antiderivative = 1., number of steps used = 7, number of rules used = 5, integrand size = 29, \(\frac{\text{number of rules}}{\text{integrand size}}\) = 0.172, Rules used = {3997, 3787, 3767, 3768, 3770} \[ \frac{a (5 A+4 B) \tan ^3(c+d x)}{15 d}+\frac{a (5 A+4 B) \tan (c+d x)}{5 d}+\frac{3 a (A+B) \tanh ^{-1}(\sin (c+d x))}{8 d}+\frac{a (A+B) \tan (c+d x) \sec ^3(c+d x)}{4 d}+\frac{3 a (A+B) \tan (c+d x) \sec (c+d x)}{8 d}+\frac{a B \tan (c+d x) \sec ^4(c+d x)}{5 d} \]

Antiderivative was successfully verified.

[In]

Int[Sec[c + d*x]^4*(a + a*Sec[c + d*x])*(A + B*Sec[c + d*x]),x]

[Out]

(3*a*(A + B)*ArcTanh[Sin[c + d*x]])/(8*d) + (a*(5*A + 4*B)*Tan[c + d*x])/(5*d) + (3*a*(A + B)*Sec[c + d*x]*Tan
[c + d*x])/(8*d) + (a*(A + B)*Sec[c + d*x]^3*Tan[c + d*x])/(4*d) + (a*B*Sec[c + d*x]^4*Tan[c + d*x])/(5*d) + (
a*(5*A + 4*B)*Tan[c + d*x]^3)/(15*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 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]

Rubi steps

\begin{align*} \int \sec ^4(c+d x) (a+a \sec (c+d x)) (A+B \sec (c+d x)) \, dx &=\frac{a B \sec ^4(c+d x) \tan (c+d x)}{5 d}+\frac{1}{5} \int \sec ^4(c+d x) (a (5 A+4 B)+5 a (A+B) \sec (c+d x)) \, dx\\ &=\frac{a B \sec ^4(c+d x) \tan (c+d x)}{5 d}+(a (A+B)) \int \sec ^5(c+d x) \, dx+\frac{1}{5} (a (5 A+4 B)) \int \sec ^4(c+d x) \, dx\\ &=\frac{a (A+B) \sec ^3(c+d x) \tan (c+d x)}{4 d}+\frac{a B \sec ^4(c+d x) \tan (c+d x)}{5 d}+\frac{1}{4} (3 a (A+B)) \int \sec ^3(c+d x) \, dx-\frac{(a (5 A+4 B)) \operatorname{Subst}\left (\int \left (1+x^2\right ) \, dx,x,-\tan (c+d x)\right )}{5 d}\\ &=\frac{a (5 A+4 B) \tan (c+d x)}{5 d}+\frac{3 a (A+B) \sec (c+d x) \tan (c+d x)}{8 d}+\frac{a (A+B) \sec ^3(c+d x) \tan (c+d x)}{4 d}+\frac{a B \sec ^4(c+d x) \tan (c+d x)}{5 d}+\frac{a (5 A+4 B) \tan ^3(c+d x)}{15 d}+\frac{1}{8} (3 a (A+B)) \int \sec (c+d x) \, dx\\ &=\frac{3 a (A+B) \tanh ^{-1}(\sin (c+d x))}{8 d}+\frac{a (5 A+4 B) \tan (c+d x)}{5 d}+\frac{3 a (A+B) \sec (c+d x) \tan (c+d x)}{8 d}+\frac{a (A+B) \sec ^3(c+d x) \tan (c+d x)}{4 d}+\frac{a B \sec ^4(c+d x) \tan (c+d x)}{5 d}+\frac{a (5 A+4 B) \tan ^3(c+d x)}{15 d}\\ \end{align*}

Mathematica [A]  time = 0.744289, size = 87, normalized size = 0.65 \[ \frac{a \left (45 (A+B) \tanh ^{-1}(\sin (c+d x))+\tan (c+d x) \left (8 \left (5 (A+2 B) \tan ^2(c+d x)+15 (A+B)+3 B \tan ^4(c+d x)\right )+30 (A+B) \sec ^3(c+d x)+45 (A+B) \sec (c+d x)\right )\right )}{120 d} \]

Antiderivative was successfully verified.

[In]

Integrate[Sec[c + d*x]^4*(a + a*Sec[c + d*x])*(A + B*Sec[c + d*x]),x]

[Out]

(a*(45*(A + B)*ArcTanh[Sin[c + d*x]] + Tan[c + d*x]*(45*(A + B)*Sec[c + d*x] + 30*(A + B)*Sec[c + d*x]^3 + 8*(
15*(A + B) + 5*(A + 2*B)*Tan[c + d*x]^2 + 3*B*Tan[c + d*x]^4))))/(120*d)

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Maple [A]  time = 0.045, size = 213, normalized size = 1.6 \begin{align*}{\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}}+{\frac{Aa\tan \left ( dx+c \right ) \left ( \sec \left ( dx+c \right ) \right ) ^{3}}{4\,d}}+{\frac{3\,Aa\tan \left ( dx+c \right ) \sec \left ( dx+c \right ) }{8\,d}}+{\frac{3\,Aa\ln \left ( \sec \left ( dx+c \right ) +\tan \left ( dx+c \right ) \right ) }{8\,d}}+{\frac{8\,Ba\tan \left ( dx+c \right ) }{15\,d}}+{\frac{Ba \left ( \sec \left ( dx+c \right ) \right ) ^{4}\tan \left ( dx+c \right ) }{5\,d}}+{\frac{4\,Ba \left ( \sec \left ( dx+c \right ) \right ) ^{2}\tan \left ( dx+c \right ) }{15\,d}} \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

int(sec(d*x+c)^4*(a+a*sec(d*x+c))*(A+B*sec(d*x+c)),x)

[Out]

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)*ta
n(d*x+c)/d+3/8/d*B*a*ln(sec(d*x+c)+tan(d*x+c))+1/4/d*A*a*tan(d*x+c)*sec(d*x+c)^3+3/8/d*A*a*tan(d*x+c)*sec(d*x+
c)+3/8/d*A*a*ln(sec(d*x+c)+tan(d*x+c))+8/15*a*B*tan(d*x+c)/d+1/5*a*B*sec(d*x+c)^4*tan(d*x+c)/d+4/15*a*B*sec(d*
x+c)^2*tan(d*x+c)/d

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Maxima [A]  time = 0.984584, size = 270, normalized size = 2.01 \begin{align*} \frac{80 \,{\left (\tan \left (d x + c\right )^{3} + 3 \, \tan \left (d x + c\right )\right )} A a + 16 \,{\left (3 \, \tan \left (d x + c\right )^{5} + 10 \, \tan \left (d x + c\right )^{3} + 15 \, \tan \left (d x + c\right )\right )} B a - 15 \, A 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 )} - 15 \, 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 )}}{240 \, d} \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate(sec(d*x+c)^4*(a+a*sec(d*x+c))*(A+B*sec(d*x+c)),x, algorithm="maxima")

[Out]

1/240*(80*(tan(d*x + c)^3 + 3*tan(d*x + c))*A*a + 16*(3*tan(d*x + c)^5 + 10*tan(d*x + c)^3 + 15*tan(d*x + c))*
B*a - 15*A*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)) - 15*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)))/d

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Fricas [A]  time = 0.497098, size = 381, normalized size = 2.84 \begin{align*} \frac{45 \,{\left (A + B\right )} a \cos \left (d x + c\right )^{5} \log \left (\sin \left (d x + c\right ) + 1\right ) - 45 \,{\left (A + B\right )} a \cos \left (d x + c\right )^{5} \log \left (-\sin \left (d x + c\right ) + 1\right ) + 2 \,{\left (16 \,{\left (5 \, A + 4 \, B\right )} a \cos \left (d x + c\right )^{4} + 45 \,{\left (A + B\right )} a \cos \left (d x + c\right )^{3} + 8 \,{\left (5 \, A + 4 \, B\right )} a \cos \left (d x + c\right )^{2} + 30 \,{\left (A + B\right )} a \cos \left (d x + c\right ) + 24 \, B a\right )} \sin \left (d x + c\right )}{240 \, d \cos \left (d x + c\right )^{5}} \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate(sec(d*x+c)^4*(a+a*sec(d*x+c))*(A+B*sec(d*x+c)),x, algorithm="fricas")

[Out]

1/240*(45*(A + B)*a*cos(d*x + c)^5*log(sin(d*x + c) + 1) - 45*(A + B)*a*cos(d*x + c)^5*log(-sin(d*x + c) + 1)
+ 2*(16*(5*A + 4*B)*a*cos(d*x + c)^4 + 45*(A + B)*a*cos(d*x + c)^3 + 8*(5*A + 4*B)*a*cos(d*x + c)^2 + 30*(A +
B)*a*cos(d*x + c) + 24*B*a)*sin(d*x + c))/(d*cos(d*x + c)^5)

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Sympy [F]  time = 0., size = 0, normalized size = 0. \begin{align*} a \left (\int A \sec ^{4}{\left (c + d x \right )}\, dx + \int A \sec ^{5}{\left (c + d x \right )}\, dx + \int B \sec ^{5}{\left (c + d x \right )}\, dx + \int B \sec ^{6}{\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)**4*(a+a*sec(d*x+c))*(A+B*sec(d*x+c)),x)

[Out]

a*(Integral(A*sec(c + d*x)**4, x) + Integral(A*sec(c + d*x)**5, x) + Integral(B*sec(c + d*x)**5, x) + Integral
(B*sec(c + d*x)**6, x))

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Giac [A]  time = 1.36729, size = 289, normalized size = 2.16 \begin{align*} \frac{45 \,{\left (A a + B a\right )} \log \left ({\left | \tan \left (\frac{1}{2} \, d x + \frac{1}{2} \, c\right ) + 1 \right |}\right ) - 45 \,{\left (A a + B a\right )} \log \left ({\left | \tan \left (\frac{1}{2} \, d x + \frac{1}{2} \, c\right ) - 1 \right |}\right ) - \frac{2 \,{\left (45 \, A a \tan \left (\frac{1}{2} \, d x + \frac{1}{2} \, c\right )^{9} + 45 \, B a \tan \left (\frac{1}{2} \, d x + \frac{1}{2} \, c\right )^{9} - 290 \, A a \tan \left (\frac{1}{2} \, d x + \frac{1}{2} \, c\right )^{7} - 130 \, B a \tan \left (\frac{1}{2} \, d x + \frac{1}{2} \, c\right )^{7} + 400 \, A a \tan \left (\frac{1}{2} \, d x + \frac{1}{2} \, c\right )^{5} + 464 \, B a \tan \left (\frac{1}{2} \, d x + \frac{1}{2} \, c\right )^{5} - 350 \, A a \tan \left (\frac{1}{2} \, d x + \frac{1}{2} \, c\right )^{3} - 190 \, B a \tan \left (\frac{1}{2} \, d x + \frac{1}{2} \, c\right )^{3} + 195 \, A a \tan \left (\frac{1}{2} \, d x + \frac{1}{2} \, c\right ) + 195 \, 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 )}^{5}}}{120 \, d} \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate(sec(d*x+c)^4*(a+a*sec(d*x+c))*(A+B*sec(d*x+c)),x, algorithm="giac")

[Out]

1/120*(45*(A*a + B*a)*log(abs(tan(1/2*d*x + 1/2*c) + 1)) - 45*(A*a + B*a)*log(abs(tan(1/2*d*x + 1/2*c) - 1)) -
2*(45*A*a*tan(1/2*d*x + 1/2*c)^9 + 45*B*a*tan(1/2*d*x + 1/2*c)^9 - 290*A*a*tan(1/2*d*x + 1/2*c)^7 - 130*B*a*t
an(1/2*d*x + 1/2*c)^7 + 400*A*a*tan(1/2*d*x + 1/2*c)^5 + 464*B*a*tan(1/2*d*x + 1/2*c)^5 - 350*A*a*tan(1/2*d*x
+ 1/2*c)^3 - 190*B*a*tan(1/2*d*x + 1/2*c)^3 + 195*A*a*tan(1/2*d*x + 1/2*c) + 195*B*a*tan(1/2*d*x + 1/2*c))/(ta
n(1/2*d*x + 1/2*c)^2 - 1)^5)/d