3.65 \(\int \sec ^3(c+d x) (a \cos (c+d x)+b \sin (c+d x))^3 \, dx\)

Optimal. Leaf size=72 \[ -\frac {b \left (3 a^2-b^2\right ) \log (\cos (c+d x))}{d}+a x \left (a^2-3 b^2\right )+\frac {2 a b^2 \tan (c+d x)}{d}+\frac {b (a+b \tan (c+d x))^2}{2 d} \]

[Out]

a*(a^2-3*b^2)*x-b*(3*a^2-b^2)*ln(cos(d*x+c))/d+2*a*b^2*tan(d*x+c)/d+1/2*b*(a+b*tan(d*x+c))^2/d

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Rubi [A]  time = 0.09, antiderivative size = 72, normalized size of antiderivative = 1.00, number of steps used = 4, number of rules used = 4, integrand size = 28, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.143, Rules used = {3086, 3482, 3525, 3475} \[ -\frac {b \left (3 a^2-b^2\right ) \log (\cos (c+d x))}{d}+a x \left (a^2-3 b^2\right )+\frac {2 a b^2 \tan (c+d x)}{d}+\frac {b (a+b \tan (c+d x))^2}{2 d} \]

Antiderivative was successfully verified.

[In]

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

[Out]

a*(a^2 - 3*b^2)*x - (b*(3*a^2 - b^2)*Log[Cos[c + d*x]])/d + (2*a*b^2*Tan[c + d*x])/d + (b*(a + b*Tan[c + d*x])
^2)/(2*d)

Rule 3086

Int[cos[(c_.) + (d_.)*(x_)]^(m_)*(cos[(c_.) + (d_.)*(x_)]*(a_.) + (b_.)*sin[(c_.) + (d_.)*(x_)])^(n_.), x_Symb
ol] :> Int[(a + b*Tan[c + d*x])^n, x] /; FreeQ[{a, b, c, d}, x] && EqQ[m + n, 0] && IntegerQ[n] && NeQ[a^2 + b
^2, 0]

Rule 3475

Int[tan[(c_.) + (d_.)*(x_)], x_Symbol] :> -Simp[Log[RemoveContent[Cos[c + d*x], x]]/d, x] /; FreeQ[{c, d}, x]

Rule 3482

Int[((a_) + (b_.)*tan[(c_.) + (d_.)*(x_)])^(n_), x_Symbol] :> Simp[(b*(a + b*Tan[c + d*x])^(n - 1))/(d*(n - 1)
), x] + Int[(a^2 - b^2 + 2*a*b*Tan[c + d*x])*(a + b*Tan[c + d*x])^(n - 2), x] /; FreeQ[{a, b, c, d}, x] && NeQ
[a^2 + b^2, 0] && GtQ[n, 1]

Rule 3525

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

Rubi steps

\begin {align*} \int \sec ^3(c+d x) (a \cos (c+d x)+b \sin (c+d x))^3 \, dx &=\int (a+b \tan (c+d x))^3 \, dx\\ &=\frac {b (a+b \tan (c+d x))^2}{2 d}+\int (a+b \tan (c+d x)) \left (a^2-b^2+2 a b \tan (c+d x)\right ) \, dx\\ &=a \left (a^2-3 b^2\right ) x+\frac {2 a b^2 \tan (c+d x)}{d}+\frac {b (a+b \tan (c+d x))^2}{2 d}+\left (b \left (3 a^2-b^2\right )\right ) \int \tan (c+d x) \, dx\\ &=a \left (a^2-3 b^2\right ) x-\frac {b \left (3 a^2-b^2\right ) \log (\cos (c+d x))}{d}+\frac {2 a b^2 \tan (c+d x)}{d}+\frac {b (a+b \tan (c+d x))^2}{2 d}\\ \end {align*}

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Mathematica [C]  time = 0.27, size = 79, normalized size = 1.10 \[ \frac {6 a b^2 \tan (c+d x)+(-b+i a)^3 \log (-\tan (c+d x)+i)-(b+i a)^3 \log (\tan (c+d x)+i)+b^3 \tan ^2(c+d x)}{2 d} \]

Antiderivative was successfully verified.

[In]

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

[Out]

((I*a - b)^3*Log[I - Tan[c + d*x]] - (I*a + b)^3*Log[I + Tan[c + d*x]] + 6*a*b^2*Tan[c + d*x] + b^3*Tan[c + d*
x]^2)/(2*d)

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fricas [A]  time = 0.72, size = 88, normalized size = 1.22 \[ \frac {2 \, {\left (a^{3} - 3 \, a b^{2}\right )} d x \cos \left (d x + c\right )^{2} + 6 \, a b^{2} \cos \left (d x + c\right ) \sin \left (d x + c\right ) - 2 \, {\left (3 \, a^{2} b - b^{3}\right )} \cos \left (d x + c\right )^{2} \log \left (-\cos \left (d x + c\right )\right ) + b^{3}}{2 \, d \cos \left (d x + c\right )^{2}} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

1/2*(2*(a^3 - 3*a*b^2)*d*x*cos(d*x + c)^2 + 6*a*b^2*cos(d*x + c)*sin(d*x + c) - 2*(3*a^2*b - b^3)*cos(d*x + c)
^2*log(-cos(d*x + c)) + b^3)/(d*cos(d*x + c)^2)

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giac [A]  time = 0.32, size = 71, normalized size = 0.99 \[ \frac {b^{3} \tan \left (d x + c\right )^{2} + 6 \, a b^{2} \tan \left (d x + c\right ) + 2 \, {\left (a^{3} - 3 \, a b^{2}\right )} {\left (d x + c\right )} + {\left (3 \, a^{2} b - b^{3}\right )} \log \left (\tan \left (d x + c\right )^{2} + 1\right )}{2 \, d} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

1/2*(b^3*tan(d*x + c)^2 + 6*a*b^2*tan(d*x + c) + 2*(a^3 - 3*a*b^2)*(d*x + c) + (3*a^2*b - b^3)*log(tan(d*x + c
)^2 + 1))/d

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maple [A]  time = 1.80, size = 93, normalized size = 1.29 \[ a^{3} x +\frac {a^{3} c}{d}-\frac {3 a^{2} b \ln \left (\cos \left (d x +c \right )\right )}{d}-3 a \,b^{2} x +\frac {3 a \,b^{2} \tan \left (d x +c \right )}{d}-\frac {3 a \,b^{2} c}{d}+\frac {b^{3} \left (\tan ^{2}\left (d x +c \right )\right )}{2 d}+\frac {b^{3} \ln \left (\cos \left (d x +c \right )\right )}{d} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

int(sec(d*x+c)^3*(a*cos(d*x+c)+b*sin(d*x+c))^3,x)

[Out]

a^3*x+1/d*a^3*c-3*a^2*b*ln(cos(d*x+c))/d-3*a*b^2*x+3*a*b^2*tan(d*x+c)/d-3/d*a*b^2*c+1/2*b^3*tan(d*x+c)^2/d+b^3
*ln(cos(d*x+c))/d

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maxima [A]  time = 0.42, size = 85, normalized size = 1.18 \[ \frac {2 \, {\left (d x + c\right )} a^{3} - 6 \, {\left (d x + c - \tan \left (d x + c\right )\right )} a b^{2} - b^{3} {\left (\frac {1}{\sin \left (d x + c\right )^{2} - 1} - \log \left (\sin \left (d x + c\right )^{2} - 1\right )\right )} - 3 \, a^{2} b \log \left (-\sin \left (d x + c\right )^{2} + 1\right )}{2 \, d} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

1/2*(2*(d*x + c)*a^3 - 6*(d*x + c - tan(d*x + c))*a*b^2 - b^3*(1/(sin(d*x + c)^2 - 1) - log(sin(d*x + c)^2 - 1
)) - 3*a^2*b*log(-sin(d*x + c)^2 + 1))/d

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mupad [B]  time = 1.86, size = 183, normalized size = 2.54 \[ \frac {2\,\left (\frac {b^3\,\ln \left (\frac {\cos \left (c+d\,x\right )}{\cos \left (c+d\,x\right )+1}\right )}{2}-\frac {b^3\,\ln \left (\frac {1}{{\cos \left (\frac {c}{2}+\frac {d\,x}{2}\right )}^2}\right )}{2}+a^3\,\mathrm {atan}\left (\frac {\sin \left (\frac {c}{2}+\frac {d\,x}{2}\right )}{\cos \left (\frac {c}{2}+\frac {d\,x}{2}\right )}\right )+\frac {3\,a^2\,b\,\ln \left (\frac {1}{{\cos \left (\frac {c}{2}+\frac {d\,x}{2}\right )}^2}\right )}{2}-3\,a\,b^2\,\mathrm {atan}\left (\frac {\sin \left (\frac {c}{2}+\frac {d\,x}{2}\right )}{\cos \left (\frac {c}{2}+\frac {d\,x}{2}\right )}\right )-\frac {3\,a^2\,b\,\ln \left (\frac {\cos \left (c+d\,x\right )}{\cos \left (c+d\,x\right )+1}\right )}{2}\right )}{d}+\frac {\frac {b^3}{2}+\frac {3\,a\,\sin \left (2\,c+2\,d\,x\right )\,b^2}{2}}{d\,\left (\frac {\cos \left (2\,c+2\,d\,x\right )}{2}+\frac {1}{2}\right )} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

int((a*cos(c + d*x) + b*sin(c + d*x))^3/cos(c + d*x)^3,x)

[Out]

(2*((b^3*log(cos(c + d*x)/(cos(c + d*x) + 1)))/2 - (b^3*log(1/cos(c/2 + (d*x)/2)^2))/2 + a^3*atan(sin(c/2 + (d
*x)/2)/cos(c/2 + (d*x)/2)) + (3*a^2*b*log(1/cos(c/2 + (d*x)/2)^2))/2 - 3*a*b^2*atan(sin(c/2 + (d*x)/2)/cos(c/2
+ (d*x)/2)) - (3*a^2*b*log(cos(c + d*x)/(cos(c + d*x) + 1)))/2))/d + (b^3/2 + (3*a*b^2*sin(2*c + 2*d*x))/2)/(
d*(cos(2*c + 2*d*x)/2 + 1/2))

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sympy [F]  time = 0.00, size = 0, normalized size = 0.00 \[ \int \left (a \cos {\left (c + d x \right )} + b \sin {\left (c + d x \right )}\right )^{3} \sec ^{3}{\left (c + d x \right )}\, dx \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate(sec(d*x+c)**3*(a*cos(d*x+c)+b*sin(d*x+c))**3,x)

[Out]

Integral((a*cos(c + d*x) + b*sin(c + d*x))**3*sec(c + d*x)**3, x)

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