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

Optimal. Leaf size=108 \[ \frac{a^3 (6 A+7 B) \tanh ^{-1}(\sin (c+d x))}{2 d}+\frac{(A+2 B) \sin (c+d x) \left (a^3 \sec (c+d x)+a^3\right )}{d}+a^3 x (3 A+B)-\frac{5 a^3 B \sin (c+d x)}{2 d}+\frac{a B \sin (c+d x) (a \sec (c+d x)+a)^2}{2 d} \]

[Out]

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

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Rubi [A]  time = 0.238215, antiderivative size = 108, normalized size of antiderivative = 1., number of steps used = 5, number of rules used = 3, integrand size = 29, \(\frac{\text{number of rules}}{\text{integrand size}}\) = 0.103, Rules used = {4018, 3996, 3770} \[ \frac{a^3 (6 A+7 B) \tanh ^{-1}(\sin (c+d x))}{2 d}+\frac{(A+2 B) \sin (c+d x) \left (a^3 \sec (c+d x)+a^3\right )}{d}+a^3 x (3 A+B)-\frac{5 a^3 B \sin (c+d x)}{2 d}+\frac{a B \sin (c+d x) (a \sec (c+d x)+a)^2}{2 d} \]

Antiderivative was successfully verified.

[In]

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

[Out]

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

Rule 4018

Int[(csc[(e_.) + (f_.)*(x_)]*(d_.))^(n_)*(csc[(e_.) + (f_.)*(x_)]*(b_.) + (a_))^(m_)*(csc[(e_.) + (f_.)*(x_)]*
(B_.) + (A_)), x_Symbol] :> -Simp[(b*B*Cot[e + f*x]*(a + b*Csc[e + f*x])^(m - 1)*(d*Csc[e + f*x])^n)/(f*(m + n
)), x] + Dist[1/(d*(m + n)), Int[(a + b*Csc[e + f*x])^(m - 1)*(d*Csc[e + f*x])^n*Simp[a*A*d*(m + n) + B*(b*d*n
) + (A*b*d*(m + n) + a*B*d*(2*m + n - 1))*Csc[e + f*x], x], x], x] /; FreeQ[{a, b, d, e, f, A, B, n}, x] && Ne
Q[A*b - a*B, 0] && EqQ[a^2 - b^2, 0] && GtQ[m, 1/2] &&  !LtQ[n, -1]

Rule 3996

Int[(csc[(e_.) + (f_.)*(x_)]*(d_.))^(n_)*(csc[(e_.) + (f_.)*(x_)]*(b_.) + (a_))*(csc[(e_.) + (f_.)*(x_)]*(B_.)
+ (A_)), x_Symbol] :> Simp[(A*a*Cot[e + f*x]*(d*Csc[e + f*x])^n)/(f*n), x] + Dist[1/(d*n), Int[(d*Csc[e + f*x
])^(n + 1)*Simp[n*(B*a + A*b) + (B*b*n + A*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 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 \cos (c+d x) (a+a \sec (c+d x))^3 (A+B \sec (c+d x)) \, dx &=\frac{a B (a+a \sec (c+d x))^2 \sin (c+d x)}{2 d}+\frac{1}{2} \int \cos (c+d x) (a+a \sec (c+d x))^2 (a (2 A-B)+2 a (A+2 B) \sec (c+d x)) \, dx\\ &=\frac{a B (a+a \sec (c+d x))^2 \sin (c+d x)}{2 d}+\frac{(A+2 B) \left (a^3+a^3 \sec (c+d x)\right ) \sin (c+d x)}{d}+\frac{1}{2} \int \cos (c+d x) (a+a \sec (c+d x)) \left (-5 a^2 B+a^2 (6 A+7 B) \sec (c+d x)\right ) \, dx\\ &=-\frac{5 a^3 B \sin (c+d x)}{2 d}+\frac{a B (a+a \sec (c+d x))^2 \sin (c+d x)}{2 d}+\frac{(A+2 B) \left (a^3+a^3 \sec (c+d x)\right ) \sin (c+d x)}{d}-\frac{1}{2} \int \left (-2 a^3 (3 A+B)-a^3 (6 A+7 B) \sec (c+d x)\right ) \, dx\\ &=a^3 (3 A+B) x-\frac{5 a^3 B \sin (c+d x)}{2 d}+\frac{a B (a+a \sec (c+d x))^2 \sin (c+d x)}{2 d}+\frac{(A+2 B) \left (a^3+a^3 \sec (c+d x)\right ) \sin (c+d x)}{d}+\frac{1}{2} \left (a^3 (6 A+7 B)\right ) \int \sec (c+d x) \, dx\\ &=a^3 (3 A+B) x+\frac{a^3 (6 A+7 B) \tanh ^{-1}(\sin (c+d x))}{2 d}-\frac{5 a^3 B \sin (c+d x)}{2 d}+\frac{a B (a+a \sec (c+d x))^2 \sin (c+d x)}{2 d}+\frac{(A+2 B) \left (a^3+a^3 \sec (c+d x)\right ) \sin (c+d x)}{d}\\ \end{align*}

Mathematica [B]  time = 2.52613, size = 335, normalized size = 3.1 \[ \frac{a^3 \cos ^4(c+d x) \sec ^6\left (\frac{1}{2} (c+d x)\right ) (\sec (c+d x)+1)^3 (A+B \sec (c+d x)) \left (\frac{4 (A+3 B) \sin \left (\frac{d x}{2}\right )}{d \left (\cos \left (\frac{c}{2}\right )-\sin \left (\frac{c}{2}\right )\right ) \left (\cos \left (\frac{1}{2} (c+d x)\right )-\sin \left (\frac{1}{2} (c+d x)\right )\right )}+\frac{4 (A+3 B) \sin \left (\frac{d x}{2}\right )}{d \left (\sin \left (\frac{c}{2}\right )+\cos \left (\frac{c}{2}\right )\right ) \left (\sin \left (\frac{1}{2} (c+d x)\right )+\cos \left (\frac{1}{2} (c+d x)\right )\right )}-\frac{2 (6 A+7 B) \log \left (\cos \left (\frac{1}{2} (c+d x)\right )-\sin \left (\frac{1}{2} (c+d x)\right )\right )}{d}+\frac{2 (6 A+7 B) \log \left (\sin \left (\frac{1}{2} (c+d x)\right )+\cos \left (\frac{1}{2} (c+d x)\right )\right )}{d}+4 x (3 A+B)+\frac{4 A \sin (c) \cos (d x)}{d}+\frac{4 A \cos (c) \sin (d x)}{d}+\frac{B}{d \left (\cos \left (\frac{1}{2} (c+d x)\right )-\sin \left (\frac{1}{2} (c+d x)\right )\right )^2}-\frac{B}{d \left (\sin \left (\frac{1}{2} (c+d x)\right )+\cos \left (\frac{1}{2} (c+d x)\right )\right )^2}\right )}{32 (A \cos (c+d x)+B)} \]

Antiderivative was successfully verified.

[In]

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

[Out]

(a^3*Cos[c + d*x]^4*Sec[(c + d*x)/2]^6*(1 + Sec[c + d*x])^3*(A + B*Sec[c + d*x])*(4*(3*A + B)*x - (2*(6*A + 7*
B)*Log[Cos[(c + d*x)/2] - Sin[(c + d*x)/2]])/d + (2*(6*A + 7*B)*Log[Cos[(c + d*x)/2] + Sin[(c + d*x)/2]])/d +
(4*A*Cos[d*x]*Sin[c])/d + (4*A*Cos[c]*Sin[d*x])/d + B/(d*(Cos[(c + d*x)/2] - Sin[(c + d*x)/2])^2) + (4*(A + 3*
B)*Sin[(d*x)/2])/(d*(Cos[c/2] - Sin[c/2])*(Cos[(c + d*x)/2] - Sin[(c + d*x)/2])) - B/(d*(Cos[(c + d*x)/2] + Si
n[(c + d*x)/2])^2) + (4*(A + 3*B)*Sin[(d*x)/2])/(d*(Cos[c/2] + Sin[c/2])*(Cos[(c + d*x)/2] + Sin[(c + d*x)/2])
)))/(32*(B + A*Cos[c + d*x]))

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Maple [A]  time = 0.073, size = 144, normalized size = 1.3 \begin{align*}{\frac{A{a}^{3}\sin \left ( dx+c \right ) }{d}}+B{a}^{3}x+{\frac{B{a}^{3}c}{d}}+3\,{a}^{3}Ax+3\,{\frac{A{a}^{3}c}{d}}+{\frac{7\,B{a}^{3}\ln \left ( \sec \left ( dx+c \right ) +\tan \left ( dx+c \right ) \right ) }{2\,d}}+3\,{\frac{A{a}^{3}\ln \left ( \sec \left ( dx+c \right ) +\tan \left ( dx+c \right ) \right ) }{d}}+3\,{\frac{B{a}^{3}\tan \left ( dx+c \right ) }{d}}+{\frac{A{a}^{3}\tan \left ( dx+c \right ) }{d}}+{\frac{B{a}^{3}\sec \left ( dx+c \right ) \tan \left ( dx+c \right ) }{2\,d}} \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

a^3*A*sin(d*x+c)/d+B*a^3*x+1/d*B*a^3*c+3*a^3*A*x+3/d*A*a^3*c+7/2/d*B*a^3*ln(sec(d*x+c)+tan(d*x+c))+3/d*A*a^3*l
n(sec(d*x+c)+tan(d*x+c))+3/d*B*a^3*tan(d*x+c)+1/d*A*a^3*tan(d*x+c)+1/2/d*B*a^3*sec(d*x+c)*tan(d*x+c)

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Maxima [A]  time = 0.988914, size = 223, normalized size = 2.06 \begin{align*} \frac{12 \,{\left (d x + c\right )} A a^{3} + 4 \,{\left (d x + c\right )} B a^{3} - 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 )} + 6 \, A a^{3}{\left (\log \left (\sin \left (d x + c\right ) + 1\right ) - \log \left (\sin \left (d x + c\right ) - 1\right )\right )} + 6 \, B a^{3}{\left (\log \left (\sin \left (d x + c\right ) + 1\right ) - \log \left (\sin \left (d x + c\right ) - 1\right )\right )} + 4 \, A a^{3} \sin \left (d x + c\right ) + 4 \, A a^{3} \tan \left (d x + c\right ) + 12 \, B a^{3} \tan \left (d x + c\right )}{4 \, d} \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

1/4*(12*(d*x + c)*A*a^3 + 4*(d*x + c)*B*a^3 - 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)) + 6*A*a^3*(log(sin(d*x + c) + 1) - log(sin(d*x + c) - 1)) + 6*B*a^3*(log(sin(d*x +
c) + 1) - log(sin(d*x + c) - 1)) + 4*A*a^3*sin(d*x + c) + 4*A*a^3*tan(d*x + c) + 12*B*a^3*tan(d*x + c))/d

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Fricas [A]  time = 0.504092, size = 342, normalized size = 3.17 \begin{align*} \frac{4 \,{\left (3 \, A + B\right )} a^{3} d x \cos \left (d x + c\right )^{2} +{\left (6 \, A + 7 \, B\right )} a^{3} \cos \left (d x + c\right )^{2} \log \left (\sin \left (d x + c\right ) + 1\right ) -{\left (6 \, A + 7 \, B\right )} a^{3} \cos \left (d x + c\right )^{2} \log \left (-\sin \left (d x + c\right ) + 1\right ) + 2 \,{\left (2 \, A a^{3} \cos \left (d x + c\right )^{2} + 2 \,{\left (A + 3 \, B\right )} a^{3} \cos \left (d x + c\right ) + B a^{3}\right )} \sin \left (d x + c\right )}{4 \, d \cos \left (d x + c\right )^{2}} \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

1/4*(4*(3*A + B)*a^3*d*x*cos(d*x + c)^2 + (6*A + 7*B)*a^3*cos(d*x + c)^2*log(sin(d*x + c) + 1) - (6*A + 7*B)*a
^3*cos(d*x + c)^2*log(-sin(d*x + c) + 1) + 2*(2*A*a^3*cos(d*x + c)^2 + 2*(A + 3*B)*a^3*cos(d*x + c) + B*a^3)*s
in(d*x + c))/(d*cos(d*x + c)^2)

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Sympy [F(-1)]  time = 0., size = 0, normalized size = 0. \begin{align*} \text{Timed out} \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

Timed out

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Giac [A]  time = 1.38422, size = 259, normalized size = 2.4 \begin{align*} \frac{\frac{4 \, A a^{3} \tan \left (\frac{1}{2} \, d x + \frac{1}{2} \, c\right )}{\tan \left (\frac{1}{2} \, d x + \frac{1}{2} \, c\right )^{2} + 1} + 2 \,{\left (3 \, A a^{3} + B a^{3}\right )}{\left (d x + c\right )} +{\left (6 \, A a^{3} + 7 \, B a^{3}\right )} \log \left ({\left | \tan \left (\frac{1}{2} \, d x + \frac{1}{2} \, c\right ) + 1 \right |}\right ) -{\left (6 \, A a^{3} + 7 \, 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 (2 \, A a^{3} \tan \left (\frac{1}{2} \, d x + \frac{1}{2} \, c\right )^{3} + 5 \, B a^{3} \tan \left (\frac{1}{2} \, d x + \frac{1}{2} \, c\right )^{3} - 2 \, A a^{3} \tan \left (\frac{1}{2} \, d x + \frac{1}{2} \, c\right ) - 7 \, 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 )}^{2}}}{2 \, d} \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

1/2*(4*A*a^3*tan(1/2*d*x + 1/2*c)/(tan(1/2*d*x + 1/2*c)^2 + 1) + 2*(3*A*a^3 + B*a^3)*(d*x + c) + (6*A*a^3 + 7*
B*a^3)*log(abs(tan(1/2*d*x + 1/2*c) + 1)) - (6*A*a^3 + 7*B*a^3)*log(abs(tan(1/2*d*x + 1/2*c) - 1)) - 2*(2*A*a^
3*tan(1/2*d*x + 1/2*c)^3 + 5*B*a^3*tan(1/2*d*x + 1/2*c)^3 - 2*A*a^3*tan(1/2*d*x + 1/2*c) - 7*B*a^3*tan(1/2*d*x
+ 1/2*c))/(tan(1/2*d*x + 1/2*c)^2 - 1)^2)/d