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

Optimal. Leaf size=111 \[ \frac{5 a^3 (A+B) \tan (c+d x)}{2 d}+\frac{a^3 (7 A+5 B) \tanh ^{-1}(\sin (c+d x))}{2 d}+\frac{(3 A+5 B) \tan (c+d x) \left (a^3 \sec (c+d x)+a^3\right )}{6 d}+a^3 A x+\frac{a B \tan (c+d x) (a \sec (c+d x)+a)^2}{3 d} \]

[Out]

a^3*A*x + (a^3*(7*A + 5*B)*ArcTanh[Sin[c + d*x]])/(2*d) + (5*a^3*(A + B)*Tan[c + d*x])/(2*d) + (a*B*(a + a*Sec
[c + d*x])^2*Tan[c + d*x])/(3*d) + ((3*A + 5*B)*(a^3 + a^3*Sec[c + d*x])*Tan[c + d*x])/(6*d)

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Rubi [A]  time = 0.144216, antiderivative size = 111, normalized size of antiderivative = 1., number of steps used = 6, number of rules used = 5, integrand size = 23, \(\frac{\text{number of rules}}{\text{integrand size}}\) = 0.217, Rules used = {3917, 3914, 3767, 8, 3770} \[ \frac{5 a^3 (A+B) \tan (c+d x)}{2 d}+\frac{a^3 (7 A+5 B) \tanh ^{-1}(\sin (c+d x))}{2 d}+\frac{(3 A+5 B) \tan (c+d x) \left (a^3 \sec (c+d x)+a^3\right )}{6 d}+a^3 A x+\frac{a B \tan (c+d x) (a \sec (c+d x)+a)^2}{3 d} \]

Antiderivative was successfully verified.

[In]

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

[Out]

a^3*A*x + (a^3*(7*A + 5*B)*ArcTanh[Sin[c + d*x]])/(2*d) + (5*a^3*(A + B)*Tan[c + d*x])/(2*d) + (a*B*(a + a*Sec
[c + d*x])^2*Tan[c + d*x])/(3*d) + ((3*A + 5*B)*(a^3 + a^3*Sec[c + d*x])*Tan[c + d*x])/(6*d)

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 (A+B \sec (c+d x)) \, dx &=\frac{a B (a+a \sec (c+d x))^2 \tan (c+d x)}{3 d}+\frac{1}{3} \int (a+a \sec (c+d x))^2 (3 a A+a (3 A+5 B) \sec (c+d x)) \, dx\\ &=\frac{a B (a+a \sec (c+d x))^2 \tan (c+d x)}{3 d}+\frac{(3 A+5 B) \left (a^3+a^3 \sec (c+d x)\right ) \tan (c+d x)}{6 d}+\frac{1}{6} \int (a+a \sec (c+d x)) \left (6 a^2 A+15 a^2 (A+B) \sec (c+d x)\right ) \, dx\\ &=a^3 A x+\frac{a B (a+a \sec (c+d x))^2 \tan (c+d x)}{3 d}+\frac{(3 A+5 B) \left (a^3+a^3 \sec (c+d x)\right ) \tan (c+d x)}{6 d}+\frac{1}{2} \left (5 a^3 (A+B)\right ) \int \sec ^2(c+d x) \, dx+\frac{1}{2} \left (a^3 (7 A+5 B)\right ) \int \sec (c+d x) \, dx\\ &=a^3 A x+\frac{a^3 (7 A+5 B) \tanh ^{-1}(\sin (c+d x))}{2 d}+\frac{a B (a+a \sec (c+d x))^2 \tan (c+d x)}{3 d}+\frac{(3 A+5 B) \left (a^3+a^3 \sec (c+d x)\right ) \tan (c+d x)}{6 d}-\frac{\left (5 a^3 (A+B)\right ) \operatorname{Subst}(\int 1 \, dx,x,-\tan (c+d x))}{2 d}\\ &=a^3 A x+\frac{a^3 (7 A+5 B) \tanh ^{-1}(\sin (c+d x))}{2 d}+\frac{5 a^3 (A+B) \tan (c+d x)}{2 d}+\frac{a B (a+a \sec (c+d x))^2 \tan (c+d x)}{3 d}+\frac{(3 A+5 B) \left (a^3+a^3 \sec (c+d x)\right ) \tan (c+d x)}{6 d}\\ \end{align*}

Mathematica [B]  time = 6.38664, size = 1056, normalized size = 9.51 \[ \text{result too large to display} \]

Antiderivative was successfully verified.

[In]

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

[Out]

(A*x*Cos[c + d*x]^4*Sec[c/2 + (d*x)/2]^6*(a + a*Sec[c + d*x])^3*(A + B*Sec[c + d*x]))/(8*(B + A*Cos[c + d*x]))
+ ((-7*A - 5*B)*Cos[c + d*x]^4*Log[Cos[c/2 + (d*x)/2] - Sin[c/2 + (d*x)/2]]*Sec[c/2 + (d*x)/2]^6*(a + a*Sec[c
+ d*x])^3*(A + B*Sec[c + d*x]))/(16*d*(B + A*Cos[c + d*x])) + ((7*A + 5*B)*Cos[c + d*x]^4*Log[Cos[c/2 + (d*x)
/2] + Sin[c/2 + (d*x)/2]]*Sec[c/2 + (d*x)/2]^6*(a + a*Sec[c + d*x])^3*(A + B*Sec[c + d*x]))/(16*d*(B + A*Cos[c
+ d*x])) + (B*Cos[c + d*x]^4*Sec[c/2 + (d*x)/2]^6*(a + a*Sec[c + d*x])^3*(A + B*Sec[c + d*x])*Sin[(d*x)/2])/(
48*d*(B + A*Cos[c + d*x])*(Cos[c/2] - Sin[c/2])*(Cos[c/2 + (d*x)/2] - Sin[c/2 + (d*x)/2])^3) + (Cos[c + d*x]^4
*Sec[c/2 + (d*x)/2]^6*(a + a*Sec[c + d*x])^3*(A + B*Sec[c + d*x])*(3*A*Cos[c/2] + 10*B*Cos[c/2] - 3*A*Sin[c/2]
- 8*B*Sin[c/2]))/(96*d*(B + A*Cos[c + d*x])*(Cos[c/2] - Sin[c/2])*(Cos[c/2 + (d*x)/2] - Sin[c/2 + (d*x)/2])^2
) + (Cos[c + d*x]^4*Sec[c/2 + (d*x)/2]^6*(a + a*Sec[c + d*x])^3*(A + B*Sec[c + d*x])*(9*A*Sin[(d*x)/2] + 11*B*
Sin[(d*x)/2]))/(24*d*(B + A*Cos[c + d*x])*(Cos[c/2] - Sin[c/2])*(Cos[c/2 + (d*x)/2] - Sin[c/2 + (d*x)/2])) + (
B*Cos[c + d*x]^4*Sec[c/2 + (d*x)/2]^6*(a + a*Sec[c + d*x])^3*(A + B*Sec[c + d*x])*Sin[(d*x)/2])/(48*d*(B + A*C
os[c + d*x])*(Cos[c/2] + Sin[c/2])*(Cos[c/2 + (d*x)/2] + Sin[c/2 + (d*x)/2])^3) + (Cos[c + d*x]^4*Sec[c/2 + (d
*x)/2]^6*(a + a*Sec[c + d*x])^3*(A + B*Sec[c + d*x])*(-3*A*Cos[c/2] - 10*B*Cos[c/2] - 3*A*Sin[c/2] - 8*B*Sin[c
/2]))/(96*d*(B + A*Cos[c + d*x])*(Cos[c/2] + Sin[c/2])*(Cos[c/2 + (d*x)/2] + Sin[c/2 + (d*x)/2])^2) + (Cos[c +
d*x]^4*Sec[c/2 + (d*x)/2]^6*(a + a*Sec[c + d*x])^3*(A + B*Sec[c + d*x])*(9*A*Sin[(d*x)/2] + 11*B*Sin[(d*x)/2]
))/(24*d*(B + A*Cos[c + d*x])*(Cos[c/2] + Sin[c/2])*(Cos[c/2 + (d*x)/2] + Sin[c/2 + (d*x)/2]))

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Maple [A]  time = 0.046, size = 158, normalized size = 1.4 \begin{align*}{a}^{3}Ax+{\frac{A{a}^{3}c}{d}}+{\frac{5\,B{a}^{3}\ln \left ( \sec \left ( dx+c \right ) +\tan \left ( dx+c \right ) \right ) }{2\,d}}+{\frac{7\,A{a}^{3}\ln \left ( \sec \left ( dx+c \right ) +\tan \left ( dx+c \right ) \right ) }{2\,d}}+{\frac{11\,B{a}^{3}\tan \left ( dx+c \right ) }{3\,d}}+3\,{\frac{A{a}^{3}\tan \left ( dx+c \right ) }{d}}+{\frac{3\,B{a}^{3}\sec \left ( dx+c \right ) \tan \left ( dx+c \right ) }{2\,d}}+{\frac{A{a}^{3}\sec \left ( dx+c \right ) \tan \left ( dx+c \right ) }{2\,d}}+{\frac{B{a}^{3}\tan \left ( dx+c \right ) \left ( \sec \left ( dx+c \right ) \right ) ^{2}}{3\,d}} \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

a^3*A*x+1/d*A*a^3*c+5/2/d*B*a^3*ln(sec(d*x+c)+tan(d*x+c))+7/2/d*A*a^3*ln(sec(d*x+c)+tan(d*x+c))+11/3/d*B*a^3*t
an(d*x+c)+3/d*A*a^3*tan(d*x+c)+3/2/d*B*a^3*sec(d*x+c)*tan(d*x+c)+1/2/d*A*a^3*sec(d*x+c)*tan(d*x+c)+1/3/d*B*a^3
*tan(d*x+c)*sec(d*x+c)^2

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Maxima [A]  time = 0.995285, size = 267, normalized size = 2.41 \begin{align*} \frac{12 \,{\left (d x + c\right )} A a^{3} + 4 \,{\left (\tan \left (d x + c\right )^{3} + 3 \, \tan \left (d x + c\right )\right )} B a^{3} - 3 \, 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 )} - 9 \, 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 \, A a^{3} \log \left (\sec \left (d x + c\right ) + \tan \left (d x + c\right )\right ) + 12 \, B a^{3} \log \left (\sec \left (d x + c\right ) + \tan \left (d x + c\right )\right ) + 36 \, A a^{3} \tan \left (d x + c\right ) + 36 \, B a^{3} \tan \left (d x + c\right )}{12 \, d} \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

1/12*(12*(d*x + c)*A*a^3 + 4*(tan(d*x + c)^3 + 3*tan(d*x + c))*B*a^3 - 3*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)) - 9*B*a^3*(2*sin(d*x + c)/(sin(d*x + c)^2 - 1) - log(si
n(d*x + c) + 1) + log(sin(d*x + c) - 1)) + 36*A*a^3*log(sec(d*x + c) + tan(d*x + c)) + 12*B*a^3*log(sec(d*x +
c) + tan(d*x + c)) + 36*A*a^3*tan(d*x + c) + 36*B*a^3*tan(d*x + c))/d

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Fricas [A]  time = 0.501615, size = 356, normalized size = 3.21 \begin{align*} \frac{12 \, A a^{3} d x \cos \left (d x + c\right )^{3} + 3 \,{\left (7 \, A + 5 \, B\right )} a^{3} \cos \left (d x + c\right )^{3} \log \left (\sin \left (d x + c\right ) + 1\right ) - 3 \,{\left (7 \, A + 5 \, B\right )} a^{3} \cos \left (d x + c\right )^{3} \log \left (-\sin \left (d x + c\right ) + 1\right ) + 2 \,{\left (2 \,{\left (9 \, A + 11 \, B\right )} a^{3} \cos \left (d x + c\right )^{2} + 3 \,{\left (A + 3 \, B\right )} a^{3} \cos \left (d x + c\right ) + 2 \, B a^{3}\right )} \sin \left (d x + c\right )}{12 \, d \cos \left (d x + c\right )^{3}} \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

1/12*(12*A*a^3*d*x*cos(d*x + c)^3 + 3*(7*A + 5*B)*a^3*cos(d*x + c)^3*log(sin(d*x + c) + 1) - 3*(7*A + 5*B)*a^3
*cos(d*x + c)^3*log(-sin(d*x + c) + 1) + 2*(2*(9*A + 11*B)*a^3*cos(d*x + c)^2 + 3*(A + 3*B)*a^3*cos(d*x + c) +
2*B*a^3)*sin(d*x + c))/(d*cos(d*x + c)^3)

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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 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\right ) \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate((a+a*sec(d*x+c))**3*(A+B*sec(d*x+c)),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(B*sec(c + d*x), x) + Integral(3*B*sec(c + d*x)**2, x) + Integral(3*B*sec(c + d*x)**3, x)
+ Integral(B*sec(c + d*x)**4, x))

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Giac [A]  time = 1.3272, size = 255, normalized size = 2.3 \begin{align*} \frac{6 \,{\left (d x + c\right )} A a^{3} + 3 \,{\left (7 \, A a^{3} + 5 \, B a^{3}\right )} \log \left ({\left | \tan \left (\frac{1}{2} \, d x + \frac{1}{2} \, c\right ) + 1 \right |}\right ) - 3 \,{\left (7 \, A a^{3} + 5 \, B a^{3}\right )} \log \left ({\left | \tan \left (\frac{1}{2} \, d x + \frac{1}{2} \, c\right ) - 1 \right |}\right ) - \frac{2 \,{\left (15 \, A a^{3} \tan \left (\frac{1}{2} \, d x + \frac{1}{2} \, c\right )^{5} + 15 \, B a^{3} \tan \left (\frac{1}{2} \, d x + \frac{1}{2} \, c\right )^{5} - 36 \, A a^{3} \tan \left (\frac{1}{2} \, d x + \frac{1}{2} \, c\right )^{3} - 40 \, B a^{3} \tan \left (\frac{1}{2} \, d x + \frac{1}{2} \, c\right )^{3} + 21 \, A a^{3} \tan \left (\frac{1}{2} \, d x + \frac{1}{2} \, c\right ) + 33 \, B 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 )}^{3}}}{6 \, d} \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

1/6*(6*(d*x + c)*A*a^3 + 3*(7*A*a^3 + 5*B*a^3)*log(abs(tan(1/2*d*x + 1/2*c) + 1)) - 3*(7*A*a^3 + 5*B*a^3)*log(
abs(tan(1/2*d*x + 1/2*c) - 1)) - 2*(15*A*a^3*tan(1/2*d*x + 1/2*c)^5 + 15*B*a^3*tan(1/2*d*x + 1/2*c)^5 - 36*A*a
^3*tan(1/2*d*x + 1/2*c)^3 - 40*B*a^3*tan(1/2*d*x + 1/2*c)^3 + 21*A*a^3*tan(1/2*d*x + 1/2*c) + 33*B*a^3*tan(1/2
*d*x + 1/2*c))/(tan(1/2*d*x + 1/2*c)^2 - 1)^3)/d