3.324 \(\int (a+a \sec (c+d x))^3 (B \sec (c+d x)+C \sec ^2(c+d x)) \, dx\)

Optimal. Leaf size=125 \[ \frac{a^3 (4 B+3 C) \tan ^3(c+d x)}{12 d}+\frac{a^3 (4 B+3 C) \tan (c+d x)}{d}+\frac{5 a^3 (4 B+3 C) \tanh ^{-1}(\sin (c+d x))}{8 d}+\frac{3 a^3 (4 B+3 C) \tan (c+d x) \sec (c+d x)}{8 d}+\frac{C \tan (c+d x) (a \sec (c+d x)+a)^3}{4 d} \]

[Out]

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

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Rubi [A]  time = 0.139477, antiderivative size = 125, normalized size of antiderivative = 1., number of steps used = 11, number of rules used = 7, integrand size = 32, \(\frac{\text{number of rules}}{\text{integrand size}}\) = 0.219, Rules used = {4054, 12, 3791, 3770, 3767, 8, 3768} \[ \frac{a^3 (4 B+3 C) \tan ^3(c+d x)}{12 d}+\frac{a^3 (4 B+3 C) \tan (c+d x)}{d}+\frac{5 a^3 (4 B+3 C) \tanh ^{-1}(\sin (c+d x))}{8 d}+\frac{3 a^3 (4 B+3 C) \tan (c+d x) \sec (c+d x)}{8 d}+\frac{C \tan (c+d x) (a \sec (c+d x)+a)^3}{4 d} \]

Antiderivative was successfully verified.

[In]

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

[Out]

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

Rule 4054

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/(b*(m + 1)),
Int[(a + b*Csc[e + f*x])^m*Simp[A*b*(m + 1) + (a*C*m + b*B*(m + 1))*Csc[e + f*x], x], x], x] /; FreeQ[{a, b,
e, f, A, B, C, m}, x] && EqQ[a^2 - b^2, 0] &&  !LtQ[m, -2^(-1)]

Rule 12

Int[(a_)*(u_), x_Symbol] :> Dist[a, Int[u, x], x] /; FreeQ[a, x] &&  !MatchQ[u, (b_)*(v_) /; FreeQ[b, x]]

Rule 3791

Int[(csc[(e_.) + (f_.)*(x_)]*(d_.))^(n_.)*(csc[(e_.) + (f_.)*(x_)]*(b_.) + (a_))^(m_), x_Symbol] :> Int[Expand
Trig[(a + b*csc[e + f*x])^m*(d*csc[e + f*x])^n, x], x] /; FreeQ[{a, b, d, e, f, m, n}, x] && EqQ[a^2 - b^2, 0]
&& IGtQ[m, 0] && RationalQ[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]

Rule 8

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

Rubi steps

\begin{align*} \int (a+a \sec (c+d x))^3 \left (B \sec (c+d x)+C \sec ^2(c+d x)\right ) \, dx &=\frac{C (a+a \sec (c+d x))^3 \tan (c+d x)}{4 d}+\frac{\int a (4 B+3 C) \sec (c+d x) (a+a \sec (c+d x))^3 \, dx}{4 a}\\ &=\frac{C (a+a \sec (c+d x))^3 \tan (c+d x)}{4 d}+\frac{1}{4} (4 B+3 C) \int \sec (c+d x) (a+a \sec (c+d x))^3 \, dx\\ &=\frac{C (a+a \sec (c+d x))^3 \tan (c+d x)}{4 d}+\frac{1}{4} (4 B+3 C) \int \left (a^3 \sec (c+d x)+3 a^3 \sec ^2(c+d x)+3 a^3 \sec ^3(c+d x)+a^3 \sec ^4(c+d x)\right ) \, dx\\ &=\frac{C (a+a \sec (c+d x))^3 \tan (c+d x)}{4 d}+\frac{1}{4} \left (a^3 (4 B+3 C)\right ) \int \sec (c+d x) \, dx+\frac{1}{4} \left (a^3 (4 B+3 C)\right ) \int \sec ^4(c+d x) \, dx+\frac{1}{4} \left (3 a^3 (4 B+3 C)\right ) \int \sec ^2(c+d x) \, dx+\frac{1}{4} \left (3 a^3 (4 B+3 C)\right ) \int \sec ^3(c+d x) \, dx\\ &=\frac{a^3 (4 B+3 C) \tanh ^{-1}(\sin (c+d x))}{4 d}+\frac{3 a^3 (4 B+3 C) \sec (c+d x) \tan (c+d x)}{8 d}+\frac{C (a+a \sec (c+d x))^3 \tan (c+d x)}{4 d}+\frac{1}{8} \left (3 a^3 (4 B+3 C)\right ) \int \sec (c+d x) \, dx-\frac{\left (a^3 (4 B+3 C)\right ) \operatorname{Subst}\left (\int \left (1+x^2\right ) \, dx,x,-\tan (c+d x)\right )}{4 d}-\frac{\left (3 a^3 (4 B+3 C)\right ) \operatorname{Subst}(\int 1 \, dx,x,-\tan (c+d x))}{4 d}\\ &=\frac{5 a^3 (4 B+3 C) \tanh ^{-1}(\sin (c+d x))}{8 d}+\frac{a^3 (4 B+3 C) \tan (c+d x)}{d}+\frac{3 a^3 (4 B+3 C) \sec (c+d x) \tan (c+d x)}{8 d}+\frac{C (a+a \sec (c+d x))^3 \tan (c+d x)}{4 d}+\frac{a^3 (4 B+3 C) \tan ^3(c+d x)}{12 d}\\ \end{align*}

Mathematica [A]  time = 0.450373, size = 81, normalized size = 0.65 \[ \frac{a^3 \left (15 (4 B+3 C) \tanh ^{-1}(\sin (c+d x))+\tan (c+d x) \left (8 (B+3 C) \tan ^2(c+d x)+9 (4 B+5 C) \sec (c+d x)+96 (B+C)+6 C \sec ^3(c+d x)\right )\right )}{24 d} \]

Antiderivative was successfully verified.

[In]

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

[Out]

(a^3*(15*(4*B + 3*C)*ArcTanh[Sin[c + d*x]] + Tan[c + d*x]*(96*(B + C) + 9*(4*B + 5*C)*Sec[c + d*x] + 6*C*Sec[c
+ d*x]^3 + 8*(B + 3*C)*Tan[c + d*x]^2)))/(24*d)

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Maple [A]  time = 0.051, size = 188, normalized size = 1.5 \begin{align*}{\frac{5\,B{a}^{3}\ln \left ( \sec \left ( dx+c \right ) +\tan \left ( dx+c \right ) \right ) }{2\,d}}+3\,{\frac{{a}^{3}C\tan \left ( dx+c \right ) }{d}}+{\frac{11\,B{a}^{3}\tan \left ( dx+c \right ) }{3\,d}}+{\frac{15\,{a}^{3}C\sec \left ( dx+c \right ) \tan \left ( dx+c \right ) }{8\,d}}+{\frac{15\,{a}^{3}C\ln \left ( \sec \left ( dx+c \right ) +\tan \left ( dx+c \right ) \right ) }{8\,d}}+{\frac{3\,B{a}^{3}\sec \left ( dx+c \right ) \tan \left ( dx+c \right ) }{2\,d}}+{\frac{{a}^{3}C\tan \left ( dx+c \right ) \left ( \sec \left ( dx+c \right ) \right ) ^{2}}{d}}+{\frac{B{a}^{3}\tan \left ( dx+c \right ) \left ( \sec \left ( dx+c \right ) \right ) ^{2}}{3\,d}}+{\frac{{a}^{3}C\tan \left ( dx+c \right ) \left ( \sec \left ( dx+c \right ) \right ) ^{3}}{4\,d}} \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

5/2/d*B*a^3*ln(sec(d*x+c)+tan(d*x+c))+3*a^3*C*tan(d*x+c)/d+11/3/d*B*a^3*tan(d*x+c)+15/8/d*a^3*C*sec(d*x+c)*tan
(d*x+c)+15/8/d*a^3*C*ln(sec(d*x+c)+tan(d*x+c))+3/2/d*B*a^3*sec(d*x+c)*tan(d*x+c)+1/d*a^3*C*tan(d*x+c)*sec(d*x+
c)^2+1/3/d*B*a^3*tan(d*x+c)*sec(d*x+c)^2+1/4/d*a^3*C*tan(d*x+c)*sec(d*x+c)^3

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Maxima [B]  time = 0.954426, size = 354, normalized size = 2.83 \begin{align*} \frac{16 \,{\left (\tan \left (d x + c\right )^{3} + 3 \, \tan \left (d x + c\right )\right )} B a^{3} + 48 \,{\left (\tan \left (d x + c\right )^{3} + 3 \, \tan \left (d x + c\right )\right )} C a^{3} - 3 \, C a^{3}{\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 )} - 36 \, B a^{3}{\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 )} - 36 \, C a^{3}{\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 \, B a^{3} \log \left (\sec \left (d x + c\right ) + \tan \left (d x + c\right )\right ) + 144 \, B a^{3} \tan \left (d x + c\right ) + 48 \, C a^{3} \tan \left (d x + c\right )}{48 \, d} \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

1/48*(16*(tan(d*x + c)^3 + 3*tan(d*x + c))*B*a^3 + 48*(tan(d*x + c)^3 + 3*tan(d*x + c))*C*a^3 - 3*C*a^3*(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)) - 36*B*a^3*(2*sin(d*x + c)/(sin(d*x + c)^2 - 1) - log(sin(d*x + c) + 1) + log(sin(d*x + c) - 1
)) - 36*C*a^3*(2*sin(d*x + c)/(sin(d*x + c)^2 - 1) - log(sin(d*x + c) + 1) + log(sin(d*x + c) - 1)) + 48*B*a^3
*log(sec(d*x + c) + tan(d*x + c)) + 144*B*a^3*tan(d*x + c) + 48*C*a^3*tan(d*x + c))/d

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Fricas [A]  time = 0.524279, size = 366, normalized size = 2.93 \begin{align*} \frac{15 \,{\left (4 \, B + 3 \, C\right )} a^{3} \cos \left (d x + c\right )^{4} \log \left (\sin \left (d x + c\right ) + 1\right ) - 15 \,{\left (4 \, B + 3 \, C\right )} a^{3} \cos \left (d x + c\right )^{4} \log \left (-\sin \left (d x + c\right ) + 1\right ) + 2 \,{\left (8 \,{\left (11 \, B + 9 \, C\right )} a^{3} \cos \left (d x + c\right )^{3} + 9 \,{\left (4 \, B + 5 \, C\right )} a^{3} \cos \left (d x + c\right )^{2} + 8 \,{\left (B + 3 \, C\right )} a^{3} \cos \left (d x + c\right ) + 6 \, C a^{3}\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((a+a*sec(d*x+c))^3*(B*sec(d*x+c)+C*sec(d*x+c)^2),x, algorithm="fricas")

[Out]

1/48*(15*(4*B + 3*C)*a^3*cos(d*x + c)^4*log(sin(d*x + c) + 1) - 15*(4*B + 3*C)*a^3*cos(d*x + c)^4*log(-sin(d*x
+ c) + 1) + 2*(8*(11*B + 9*C)*a^3*cos(d*x + c)^3 + 9*(4*B + 5*C)*a^3*cos(d*x + c)^2 + 8*(B + 3*C)*a^3*cos(d*x
+ c) + 6*C*a^3)*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^{3} \left (\int B \sec{\left (c + d x \right )}\, dx + \int 3 B \sec ^{2}{\left (c + d x \right )}\, dx + \int 3 B \sec ^{3}{\left (c + d x \right )}\, dx + \int B \sec ^{4}{\left (c + d x \right )}\, dx + \int C \sec ^{2}{\left (c + d x \right )}\, dx + \int 3 C \sec ^{3}{\left (c + d x \right )}\, dx + \int 3 C \sec ^{4}{\left (c + d x \right )}\, dx + \int C \sec ^{5}{\left (c + d x \right )}\, dx\right ) \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

a**3*(Integral(B*sec(c + d*x), x) + Integral(3*B*sec(c + d*x)**2, x) + Integral(3*B*sec(c + d*x)**3, x) + Inte
gral(B*sec(c + d*x)**4, x) + Integral(C*sec(c + d*x)**2, x) + Integral(3*C*sec(c + d*x)**3, x) + Integral(3*C*
sec(c + d*x)**4, x) + Integral(C*sec(c + d*x)**5, x))

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Giac [A]  time = 1.19281, size = 286, normalized size = 2.29 \begin{align*} \frac{15 \,{\left (4 \, B a^{3} + 3 \, C a^{3}\right )} \log \left ({\left | \tan \left (\frac{1}{2} \, d x + \frac{1}{2} \, c\right ) + 1 \right |}\right ) - 15 \,{\left (4 \, B a^{3} + 3 \, C a^{3}\right )} \log \left ({\left | \tan \left (\frac{1}{2} \, d x + \frac{1}{2} \, c\right ) - 1 \right |}\right ) - \frac{2 \,{\left (60 \, B a^{3} \tan \left (\frac{1}{2} \, d x + \frac{1}{2} \, c\right )^{7} + 45 \, C a^{3} \tan \left (\frac{1}{2} \, d x + \frac{1}{2} \, c\right )^{7} - 220 \, B a^{3} \tan \left (\frac{1}{2} \, d x + \frac{1}{2} \, c\right )^{5} - 165 \, C a^{3} \tan \left (\frac{1}{2} \, d x + \frac{1}{2} \, c\right )^{5} + 292 \, B a^{3} \tan \left (\frac{1}{2} \, d x + \frac{1}{2} \, c\right )^{3} + 219 \, C a^{3} \tan \left (\frac{1}{2} \, d x + \frac{1}{2} \, c\right )^{3} - 132 \, B a^{3} \tan \left (\frac{1}{2} \, d x + \frac{1}{2} \, c\right ) - 147 \, C a^{3} \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((a+a*sec(d*x+c))^3*(B*sec(d*x+c)+C*sec(d*x+c)^2),x, algorithm="giac")

[Out]

1/24*(15*(4*B*a^3 + 3*C*a^3)*log(abs(tan(1/2*d*x + 1/2*c) + 1)) - 15*(4*B*a^3 + 3*C*a^3)*log(abs(tan(1/2*d*x +
1/2*c) - 1)) - 2*(60*B*a^3*tan(1/2*d*x + 1/2*c)^7 + 45*C*a^3*tan(1/2*d*x + 1/2*c)^7 - 220*B*a^3*tan(1/2*d*x +
1/2*c)^5 - 165*C*a^3*tan(1/2*d*x + 1/2*c)^5 + 292*B*a^3*tan(1/2*d*x + 1/2*c)^3 + 219*C*a^3*tan(1/2*d*x + 1/2*
c)^3 - 132*B*a^3*tan(1/2*d*x + 1/2*c) - 147*C*a^3*tan(1/2*d*x + 1/2*c))/(tan(1/2*d*x + 1/2*c)^2 - 1)^4)/d