3.5.63 \(\int \cos ^3(c+d x) (a+b \sec (c+d x))^2 \, dx\) [463]

Optimal. Leaf size=58 \[ a b x+\frac {\left (a^2+b^2\right ) \sin (c+d x)}{d}+\frac {a b \cos (c+d x) \sin (c+d x)}{d}-\frac {a^2 \sin ^3(c+d x)}{3 d} \]

[Out]

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

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Rubi [A]
time = 0.07, antiderivative size = 58, normalized size of antiderivative = 1.00, number of steps used = 6, number of rules used = 5, integrand size = 21, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.238, Rules used = {3873, 2715, 8, 4129, 3092} \begin {gather*} \frac {\left (a^2+b^2\right ) \sin (c+d x)}{d}-\frac {a^2 \sin ^3(c+d x)}{3 d}+\frac {a b \sin (c+d x) \cos (c+d x)}{d}+a b x \end {gather*}

Antiderivative was successfully verified.

[In]

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

[Out]

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

Rule 8

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

Rule 2715

Int[((b_.)*sin[(c_.) + (d_.)*(x_)])^(n_), x_Symbol] :> Simp[(-b)*Cos[c + d*x]*((b*Sin[c + d*x])^(n - 1)/(d*n))
, x] + Dist[b^2*((n - 1)/n), Int[(b*Sin[c + d*x])^(n - 2), x], x] /; FreeQ[{b, c, d}, x] && GtQ[n, 1] && Integ
erQ[2*n]

Rule 3092

Int[sin[(e_.) + (f_.)*(x_)]^(m_.)*((A_) + (C_.)*sin[(e_.) + (f_.)*(x_)]^2), x_Symbol] :> Dist[-f^(-1), Subst[I
nt[(1 - x^2)^((m - 1)/2)*(A + C - C*x^2), x], x, Cos[e + f*x]], x] /; FreeQ[{e, f, A, C}, x] && IGtQ[(m + 1)/2
, 0]

Rule 3873

Int[(csc[(e_.) + (f_.)*(x_)]*(d_.))^(n_.)*(csc[(e_.) + (f_.)*(x_)]*(b_.) + (a_))^2, x_Symbol] :> Dist[2*a*(b/d
), Int[(d*Csc[e + f*x])^(n + 1), x], x] + Int[(d*Csc[e + f*x])^n*(a^2 + b^2*Csc[e + f*x]^2), x] /; FreeQ[{a, b
, d, e, f, n}, x]

Rule 4129

Int[csc[(e_.) + (f_.)*(x_)]^(m_.)*(csc[(e_.) + (f_.)*(x_)]^2*(C_.) + (A_)), x_Symbol] :> Int[(C + A*Sin[e + f*
x]^2)/Sin[e + f*x]^(m + 2), x] /; FreeQ[{e, f, A, C}, x] && NeQ[C*m + A*(m + 1), 0] && ILtQ[(m + 1)/2, 0]

Rubi steps

\begin {align*} \int \cos ^3(c+d x) (a+b \sec (c+d x))^2 \, dx &=(2 a b) \int \cos ^2(c+d x) \, dx+\int \cos ^3(c+d x) \left (a^2+b^2 \sec ^2(c+d x)\right ) \, dx\\ &=\frac {a b \cos (c+d x) \sin (c+d x)}{d}+(a b) \int 1 \, dx+\int \cos (c+d x) \left (b^2+a^2 \cos ^2(c+d x)\right ) \, dx\\ &=a b x+\frac {a b \cos (c+d x) \sin (c+d x)}{d}-\frac {\text {Subst}\left (\int \left (a^2+b^2-a^2 x^2\right ) \, dx,x,-\sin (c+d x)\right )}{d}\\ &=a b x+\frac {\left (a^2+b^2\right ) \sin (c+d x)}{d}+\frac {a b \cos (c+d x) \sin (c+d x)}{d}-\frac {a^2 \sin ^3(c+d x)}{3 d}\\ \end {align*}

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Mathematica [A]
time = 0.16, size = 59, normalized size = 1.02 \begin {gather*} \frac {3 \left (3 a^2+4 b^2\right ) \sin (c+d x)+a (12 b (c+d x)+6 b \sin (2 (c+d x))+a \sin (3 (c+d x)))}{12 d} \end {gather*}

Antiderivative was successfully verified.

[In]

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

[Out]

(3*(3*a^2 + 4*b^2)*Sin[c + d*x] + a*(12*b*(c + d*x) + 6*b*Sin[2*(c + d*x)] + a*Sin[3*(c + d*x)]))/(12*d)

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Maple [A]
time = 0.10, size = 63, normalized size = 1.09

method	result	size
derivativedivides	\(\frac {\frac {a^{2} \left (2+\cos ^{2}\left (d x +c \right )\right ) \sin \left (d x +c \right )}{3}+2 b a \left (\frac {\cos \left (d x +c \right ) \sin \left (d x +c \right )}{2}+\frac {d x}{2}+\frac {c}{2}\right )+b^{2} \sin \left (d x +c \right )}{d}\)	\(63\)
default	\(\frac {\frac {a^{2} \left (2+\cos ^{2}\left (d x +c \right )\right ) \sin \left (d x +c \right )}{3}+2 b a \left (\frac {\cos \left (d x +c \right ) \sin \left (d x +c \right )}{2}+\frac {d x}{2}+\frac {c}{2}\right )+b^{2} \sin \left (d x +c \right )}{d}\)	\(63\)
risch	\(a b x +\frac {3 a^{2} \sin \left (d x +c \right )}{4 d}+\frac {\sin \left (d x +c \right ) b^{2}}{d}+\frac {a^{2} \sin \left (3 d x +3 c \right )}{12 d}+\frac {b a \sin \left (2 d x +2 c \right )}{2 d}\)	\(66\)
norman	\(\frac {a b x \left (\tan ^{8}\left (\frac {d x}{2}+\frac {c}{2}\right )\right )-a b x -\frac {2 \left (a^{2}-3 b a -3 b^{2}\right ) \left (\tan ^{5}\left (\frac {d x}{2}+\frac {c}{2}\right )\right )}{3 d}+\frac {2 \left (a^{2}-b a +b^{2}\right ) \left (\tan ^{7}\left (\frac {d x}{2}+\frac {c}{2}\right )\right )}{d}-\frac {2 \left (a^{2}+b a +b^{2}\right ) \tan \left (\frac {d x}{2}+\frac {c}{2}\right )}{d}+\frac {2 \left (a^{2}+3 b a -3 b^{2}\right ) \left (\tan ^{3}\left (\frac {d x}{2}+\frac {c}{2}\right )\right )}{3 d}-2 a b x \left (\tan ^{2}\left (\frac {d x}{2}+\frac {c}{2}\right )\right )+2 a b x \left (\tan ^{6}\left (\frac {d x}{2}+\frac {c}{2}\right )\right )}{\left (1+\tan ^{2}\left (\frac {d x}{2}+\frac {c}{2}\right )\right )^{3} \left (\tan ^{2}\left (\frac {d x}{2}+\frac {c}{2}\right )-1\right )}\)	\(194\)

Verification of antiderivative is not currently implemented for this CAS.

[In]

int(cos(d*x+c)^3*(a+b*sec(d*x+c))^2,x,method=_RETURNVERBOSE)

[Out]

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

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Maxima [A]
time = 0.26, size = 60, normalized size = 1.03 \begin {gather*} -\frac {2 \, {\left (\sin \left (d x + c\right )^{3} - 3 \, \sin \left (d x + c\right )\right )} a^{2} - 3 \, {\left (2 \, d x + 2 \, c + \sin \left (2 \, d x + 2 \, c\right )\right )} a b - 6 \, b^{2} \sin \left (d x + c\right )}{6 \, d} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

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

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Fricas [A]
time = 3.35, size = 52, normalized size = 0.90 \begin {gather*} \frac {3 \, a b d x + {\left (a^{2} \cos \left (d x + c\right )^{2} + 3 \, a b \cos \left (d x + c\right ) + 2 \, a^{2} + 3 \, b^{2}\right )} \sin \left (d x + c\right )}{3 \, d} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate(cos(d*x+c)^3*(a+b*sec(d*x+c))^2,x, algorithm="fricas")

[Out]

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

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Sympy [F]
time = 0.00, size = 0, normalized size = 0.00 \begin {gather*} \int \left (a + b \sec {\left (c + d x \right )}\right )^{2} \cos ^{3}{\left (c + d x \right )}\, dx \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate(cos(d*x+c)**3*(a+b*sec(d*x+c))**2,x)

[Out]

Integral((a + b*sec(c + d*x))**2*cos(c + d*x)**3, x)

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Giac [B] Leaf count of result is larger than twice the leaf count of optimal. 153 vs. \(2 (56) = 112\).
time = 0.49, size = 153, normalized size = 2.64 \begin {gather*} \frac {3 \, {\left (d x + c\right )} a b + \frac {2 \, {\left (3 \, a^{2} \tan \left (\frac {1}{2} \, d x + \frac {1}{2} \, c\right )^{5} - 3 \, a b \tan \left (\frac {1}{2} \, d x + \frac {1}{2} \, c\right )^{5} + 3 \, b^{2} \tan \left (\frac {1}{2} \, d x + \frac {1}{2} \, c\right )^{5} + 2 \, a^{2} \tan \left (\frac {1}{2} \, d x + \frac {1}{2} \, c\right )^{3} + 6 \, b^{2} \tan \left (\frac {1}{2} \, d x + \frac {1}{2} \, c\right )^{3} + 3 \, a^{2} \tan \left (\frac {1}{2} \, d x + \frac {1}{2} \, c\right ) + 3 \, a b \tan \left (\frac {1}{2} \, d x + \frac {1}{2} \, c\right ) + 3 \, b^{2} \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}}}{3 \, d} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

1/3*(3*(d*x + c)*a*b + 2*(3*a^2*tan(1/2*d*x + 1/2*c)^5 - 3*a*b*tan(1/2*d*x + 1/2*c)^5 + 3*b^2*tan(1/2*d*x + 1/
2*c)^5 + 2*a^2*tan(1/2*d*x + 1/2*c)^3 + 6*b^2*tan(1/2*d*x + 1/2*c)^3 + 3*a^2*tan(1/2*d*x + 1/2*c) + 3*a*b*tan(
1/2*d*x + 1/2*c) + 3*b^2*tan(1/2*d*x + 1/2*c))/(tan(1/2*d*x + 1/2*c)^2 + 1)^3)/d

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Mupad [B]
time = 0.83, size = 72, normalized size = 1.24 \begin {gather*} \frac {2\,a^2\,\sin \left (c+d\,x\right )}{3\,d}+\frac {b^2\,\sin \left (c+d\,x\right )}{d}+a\,b\,x+\frac {a^2\,{\cos \left (c+d\,x\right )}^2\,\sin \left (c+d\,x\right )}{3\,d}+\frac {a\,b\,\cos \left (c+d\,x\right )\,\sin \left (c+d\,x\right )}{d} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

int(cos(c + d*x)^3*(a + b/cos(c + d*x))^2,x)

[Out]

(2*a^2*sin(c + d*x))/(3*d) + (b^2*sin(c + d*x))/d + a*b*x + (a^2*cos(c + d*x)^2*sin(c + d*x))/(3*d) + (a*b*cos
(c + d*x)*sin(c + d*x))/d

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