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

Optimal. Leaf size=147 \[ \frac{5 a^3 (4 A+3 C) \tan (c+d x)}{8 d}+\frac{a^3 (28 A+15 C) \tanh ^{-1}(\sin (c+d x))}{8 d}+\frac{(4 A+5 C) \tan (c+d x) \left (a^3 \sec (c+d x)+a^3\right )}{8 d}+a^3 A x+\frac{C \tan (c+d x) \left (a^2 \sec (c+d x)+a^2\right )^2}{4 a d}+\frac{C \tan (c+d x) (a \sec (c+d x)+a)^3}{4 d} \]

[Out]

a^3*A*x + (a^3*(28*A + 15*C)*ArcTanh[Sin[c + d*x]])/(8*d) + (5*a^3*(4*A + 3*C)*Tan[c + d*x])/(8*d) + (C*(a + a
*Sec[c + d*x])^3*Tan[c + d*x])/(4*d) + (C*(a^2 + a^2*Sec[c + d*x])^2*Tan[c + d*x])/(4*a*d) + ((4*A + 5*C)*(a^3
+ a^3*Sec[c + d*x])*Tan[c + d*x])/(8*d)

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Rubi [A]  time = 0.218977, antiderivative size = 147, normalized size of antiderivative = 1., number of steps used = 7, number of rules used = 6, integrand size = 25, \(\frac{\text{number of rules}}{\text{integrand size}}\) = 0.24, Rules used = {4055, 3917, 3914, 3767, 8, 3770} \[ \frac{5 a^3 (4 A+3 C) \tan (c+d x)}{8 d}+\frac{a^3 (28 A+15 C) \tanh ^{-1}(\sin (c+d x))}{8 d}+\frac{(4 A+5 C) \tan (c+d x) \left (a^3 \sec (c+d x)+a^3\right )}{8 d}+a^3 A x+\frac{C \tan (c+d x) \left (a^2 \sec (c+d x)+a^2\right )^2}{4 a d}+\frac{C \tan (c+d x) (a \sec (c+d x)+a)^3}{4 d} \]

Antiderivative was successfully verified.

[In]

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

[Out]

a^3*A*x + (a^3*(28*A + 15*C)*ArcTanh[Sin[c + d*x]])/(8*d) + (5*a^3*(4*A + 3*C)*Tan[c + d*x])/(8*d) + (C*(a + a
*Sec[c + d*x])^3*Tan[c + d*x])/(4*d) + (C*(a^2 + a^2*Sec[c + d*x])^2*Tan[c + d*x])/(4*a*d) + ((4*A + 5*C)*(a^3
+ a^3*Sec[c + d*x])*Tan[c + d*x])/(8*d)

Rule 4055

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

Rule 3917

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 - 1)*Simp[a*c*m
+ (b*c*m + a*d*(2*m - 1))*Csc[e + f*x], x], x], x] /; FreeQ[{a, b, c, d, e, f}, x] && NeQ[b*c - a*d, 0] && Gt
Q[m, 1] && EqQ[a^2 - b^2, 0] && IntegerQ[2*m]

Rule 3914

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

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 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 (a+a \sec (c+d x))^3 \left (A+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+a \sec (c+d x))^3 (4 a A+3 a C \sec (c+d x)) \, dx}{4 a}\\ &=\frac{C (a+a \sec (c+d x))^3 \tan (c+d x)}{4 d}+\frac{C \left (a^2+a^2 \sec (c+d x)\right )^2 \tan (c+d x)}{4 a d}+\frac{\int (a+a \sec (c+d x))^2 \left (12 a^2 A+3 a^2 (4 A+5 C) \sec (c+d x)\right ) \, dx}{12 a}\\ &=\frac{C (a+a \sec (c+d x))^3 \tan (c+d x)}{4 d}+\frac{C \left (a^2+a^2 \sec (c+d x)\right )^2 \tan (c+d x)}{4 a d}+\frac{(4 A+5 C) \left (a^3+a^3 \sec (c+d x)\right ) \tan (c+d x)}{8 d}+\frac{\int (a+a \sec (c+d x)) \left (24 a^3 A+15 a^3 (4 A+3 C) \sec (c+d x)\right ) \, dx}{24 a}\\ &=a^3 A x+\frac{C (a+a \sec (c+d x))^3 \tan (c+d x)}{4 d}+\frac{C \left (a^2+a^2 \sec (c+d x)\right )^2 \tan (c+d x)}{4 a d}+\frac{(4 A+5 C) \left (a^3+a^3 \sec (c+d x)\right ) \tan (c+d x)}{8 d}+\frac{1}{8} \left (5 a^3 (4 A+3 C)\right ) \int \sec ^2(c+d x) \, dx+\frac{1}{8} \left (a^3 (28 A+15 C)\right ) \int \sec (c+d x) \, dx\\ &=a^3 A x+\frac{a^3 (28 A+15 C) \tanh ^{-1}(\sin (c+d x))}{8 d}+\frac{C (a+a \sec (c+d x))^3 \tan (c+d x)}{4 d}+\frac{C \left (a^2+a^2 \sec (c+d x)\right )^2 \tan (c+d x)}{4 a d}+\frac{(4 A+5 C) \left (a^3+a^3 \sec (c+d x)\right ) \tan (c+d x)}{8 d}-\frac{\left (5 a^3 (4 A+3 C)\right ) \operatorname{Subst}(\int 1 \, dx,x,-\tan (c+d x))}{8 d}\\ &=a^3 A x+\frac{a^3 (28 A+15 C) \tanh ^{-1}(\sin (c+d x))}{8 d}+\frac{5 a^3 (4 A+3 C) \tan (c+d x)}{8 d}+\frac{C (a+a \sec (c+d x))^3 \tan (c+d x)}{4 d}+\frac{C \left (a^2+a^2 \sec (c+d x)\right )^2 \tan (c+d x)}{4 a d}+\frac{(4 A+5 C) \left (a^3+a^3 \sec (c+d x)\right ) \tan (c+d x)}{8 d}\\ \end{align*}

Mathematica [B]  time = 1.94526, size = 363, normalized size = 2.47 \[ \frac{a^3 (\cos (c+d x)+1)^3 \sec ^6\left (\frac{1}{2} (c+d x)\right ) \sec ^4(c+d x) \left (A \cos ^2(c+d x)+C\right ) \left (\sec (c) (4 A \sin (2 c+d x)+72 A \sin (c+2 d x)-24 A \sin (3 c+2 d x)+4 A \sin (2 c+3 d x)+4 A \sin (4 c+3 d x)+24 A \sin (3 c+4 d x)+24 A d x \cos (c)+16 A d x \cos (c+2 d x)+16 A d x \cos (3 c+2 d x)+4 A d x \cos (3 c+4 d x)+4 A d x \cos (5 c+4 d x)-72 A \sin (c)+4 A \sin (d x)+23 C \sin (2 c+d x)+88 C \sin (c+2 d x)-8 C \sin (3 c+2 d x)+15 C \sin (2 c+3 d x)+15 C \sin (4 c+3 d x)+24 C \sin (3 c+4 d x)-72 C \sin (c)+23 C \sin (d x))-8 (28 A+15 C) \cos ^4(c+d x) \left (\log \left (\cos \left (\frac{1}{2} (c+d x)\right )-\sin \left (\frac{1}{2} (c+d x)\right )\right )-\log \left (\sin \left (\frac{1}{2} (c+d x)\right )+\cos \left (\frac{1}{2} (c+d x)\right )\right )\right )\right )}{256 d (A \cos (2 (c+d x))+A+2 C)} \]

Antiderivative was successfully verified.

[In]

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

[Out]

(a^3*(1 + Cos[c + d*x])^3*(C + A*Cos[c + d*x]^2)*Sec[(c + d*x)/2]^6*Sec[c + d*x]^4*(-8*(28*A + 15*C)*Cos[c + d
*x]^4*(Log[Cos[(c + d*x)/2] - Sin[(c + d*x)/2]] - Log[Cos[(c + d*x)/2] + Sin[(c + d*x)/2]]) + Sec[c]*(24*A*d*x
*Cos[c] + 16*A*d*x*Cos[c + 2*d*x] + 16*A*d*x*Cos[3*c + 2*d*x] + 4*A*d*x*Cos[3*c + 4*d*x] + 4*A*d*x*Cos[5*c + 4
*d*x] - 72*A*Sin[c] - 72*C*Sin[c] + 4*A*Sin[d*x] + 23*C*Sin[d*x] + 4*A*Sin[2*c + d*x] + 23*C*Sin[2*c + d*x] +
72*A*Sin[c + 2*d*x] + 88*C*Sin[c + 2*d*x] - 24*A*Sin[3*c + 2*d*x] - 8*C*Sin[3*c + 2*d*x] + 4*A*Sin[2*c + 3*d*x
] + 15*C*Sin[2*c + 3*d*x] + 4*A*Sin[4*c + 3*d*x] + 15*C*Sin[4*c + 3*d*x] + 24*A*Sin[3*c + 4*d*x] + 24*C*Sin[3*
c + 4*d*x])))/(256*d*(A + 2*C + A*Cos[2*(c + d*x)]))

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Maple [A]  time = 0.054, size = 180, normalized size = 1.2 \begin{align*}{a}^{3}Ax+{\frac{A{a}^{3}c}{d}}+3\,{\frac{{a}^{3}C\tan \left ( dx+c \right ) }{d}}+{\frac{7\,A{a}^{3}\ln \left ( \sec \left ( dx+c \right ) +\tan \left ( dx+c \right ) \right ) }{2\,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}}+3\,{\frac{A{a}^{3}\tan \left ( dx+c \right ) }{d}}+{\frac{{a}^{3}C\tan \left ( dx+c \right ) \left ( \sec \left ( dx+c \right ) \right ) ^{2}}{d}}+{\frac{A{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 ) ^{3}}{4\,d}} \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

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

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Maxima [A]  time = 0.94449, size = 338, normalized size = 2.3 \begin{align*} \frac{16 \,{\left (d x + c\right )} A a^{3} + 16 \,{\left (\tan \left (d x + c\right )^{3} + 3 \, \tan \left (d x + c\right )\right )} C a^{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 )} - 4 \, A 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 )} - 12 \, 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 \, A a^{3} \log \left (\sec \left (d x + c\right ) + \tan \left (d x + c\right )\right ) + 48 \, A a^{3} \tan \left (d x + c\right ) + 16 \, C a^{3} \tan \left (d x + c\right )}{16 \, d} \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

1/16*(16*(d*x + c)*A*a^3 + 16*(tan(d*x + c)^3 + 3*tan(d*x + c))*C*a^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)) - 4*A*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)) - 12*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*A*a^3*log(sec(d*x + c) + tan(d*x
+ c)) + 48*A*a^3*tan(d*x + c) + 16*C*a^3*tan(d*x + c))/d

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Fricas [A]  time = 0.533892, size = 386, normalized size = 2.63 \begin{align*} \frac{16 \, A a^{3} d x \cos \left (d x + c\right )^{4} +{\left (28 \, A + 15 \, C\right )} a^{3} \cos \left (d x + c\right )^{4} \log \left (\sin \left (d x + c\right ) + 1\right ) -{\left (28 \, A + 15 \, C\right )} a^{3} \cos \left (d x + c\right )^{4} \log \left (-\sin \left (d x + c\right ) + 1\right ) + 2 \,{\left (24 \,{\left (A + C\right )} a^{3} \cos \left (d x + c\right )^{3} +{\left (4 \, A + 15 \, C\right )} a^{3} \cos \left (d x + c\right )^{2} + 8 \, C a^{3} \cos \left (d x + c\right ) + 2 \, C a^{3}\right )} \sin \left (d x + c\right )}{16 \, 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*(A+C*sec(d*x+c)^2),x, algorithm="fricas")

[Out]

1/16*(16*A*a^3*d*x*cos(d*x + c)^4 + (28*A + 15*C)*a^3*cos(d*x + c)^4*log(sin(d*x + c) + 1) - (28*A + 15*C)*a^3
*cos(d*x + c)^4*log(-sin(d*x + c) + 1) + 2*(24*(A + C)*a^3*cos(d*x + c)^3 + (4*A + 15*C)*a^3*cos(d*x + c)^2 +
8*C*a^3*cos(d*x + c) + 2*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 A\, dx + \int 3 A \sec{\left (c + d x \right )}\, dx + \int 3 A \sec ^{2}{\left (c + d x \right )}\, dx + \int A \sec ^{3}{\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*(A+C*sec(d*x+c)**2),x)

[Out]

a**3*(Integral(A, x) + Integral(3*A*sec(c + d*x), x) + Integral(3*A*sec(c + d*x)**2, x) + Integral(A*sec(c + d
*x)**3, 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 = 1.25785, size = 300, normalized size = 2.04 \begin{align*} \frac{8 \,{\left (d x + c\right )} A a^{3} +{\left (28 \, A a^{3} + 15 \, C a^{3}\right )} \log \left ({\left | \tan \left (\frac{1}{2} \, d x + \frac{1}{2} \, c\right ) + 1 \right |}\right ) -{\left (28 \, A a^{3} + 15 \, 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 (20 \, A a^{3} \tan \left (\frac{1}{2} \, d x + \frac{1}{2} \, c\right )^{7} + 15 \, C a^{3} \tan \left (\frac{1}{2} \, d x + \frac{1}{2} \, c\right )^{7} - 68 \, A a^{3} \tan \left (\frac{1}{2} \, d x + \frac{1}{2} \, c\right )^{5} - 55 \, C a^{3} \tan \left (\frac{1}{2} \, d x + \frac{1}{2} \, c\right )^{5} + 76 \, A a^{3} \tan \left (\frac{1}{2} \, d x + \frac{1}{2} \, c\right )^{3} + 73 \, C a^{3} \tan \left (\frac{1}{2} \, d x + \frac{1}{2} \, c\right )^{3} - 28 \, A a^{3} \tan \left (\frac{1}{2} \, d x + \frac{1}{2} \, c\right ) - 49 \, 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}}}{8 \, d} \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

1/8*(8*(d*x + c)*A*a^3 + (28*A*a^3 + 15*C*a^3)*log(abs(tan(1/2*d*x + 1/2*c) + 1)) - (28*A*a^3 + 15*C*a^3)*log(
abs(tan(1/2*d*x + 1/2*c) - 1)) - 2*(20*A*a^3*tan(1/2*d*x + 1/2*c)^7 + 15*C*a^3*tan(1/2*d*x + 1/2*c)^7 - 68*A*a
^3*tan(1/2*d*x + 1/2*c)^5 - 55*C*a^3*tan(1/2*d*x + 1/2*c)^5 + 76*A*a^3*tan(1/2*d*x + 1/2*c)^3 + 73*C*a^3*tan(1
/2*d*x + 1/2*c)^3 - 28*A*a^3*tan(1/2*d*x + 1/2*c) - 49*C*a^3*tan(1/2*d*x + 1/2*c))/(tan(1/2*d*x + 1/2*c)^2 - 1
)^4)/d