3.2.35 \(\int \frac {\cos (c+d x) (A+C \sec ^2(c+d x))}{(a+a \sec (c+d x))^2} \, dx\) [135]

Optimal. Leaf size=82 \[ -\frac {2 A x}{a^2}+\frac {(10 A+C) \sin (c+d x)}{3 a^2 d}-\frac {2 A \sin (c+d x)}{a^2 d (1+\sec (c+d x))}-\frac {(A+C) \sin (c+d x)}{3 d (a+a \sec (c+d x))^2} \]

[Out]

-2*A*x/a^2+1/3*(10*A+C)*sin(d*x+c)/a^2/d-2*A*sin(d*x+c)/a^2/d/(1+sec(d*x+c))-1/3*(A+C)*sin(d*x+c)/d/(a+a*sec(d
*x+c))^2

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Rubi [A]
time = 0.16, antiderivative size = 82, normalized size of antiderivative = 1.00, number of steps used = 5, number of rules used = 5, integrand size = 31, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.161, Rules used = {4170, 4105, 3872, 2717, 8} \begin {gather*} \frac {(10 A+C) \sin (c+d x)}{3 a^2 d}-\frac {2 A \sin (c+d x)}{a^2 d (\sec (c+d x)+1)}-\frac {2 A x}{a^2}-\frac {(A+C) \sin (c+d x)}{3 d (a \sec (c+d x)+a)^2} \end {gather*}

Antiderivative was successfully verified.

[In]

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

[Out]

(-2*A*x)/a^2 + ((10*A + C)*Sin[c + d*x])/(3*a^2*d) - (2*A*Sin[c + d*x])/(a^2*d*(1 + Sec[c + d*x])) - ((A + C)*
Sin[c + d*x])/(3*d*(a + a*Sec[c + d*x])^2)

Rule 8

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

Rule 2717

Int[sin[Pi/2 + (c_.) + (d_.)*(x_)], x_Symbol] :> Simp[Sin[c + d*x]/d, x] /; FreeQ[{c, d}, x]

Rule 3872

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

Rule 4105

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

Rule 4170

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

Rubi steps

\begin {align*} \int \frac {\cos (c+d x) \left (A+C \sec ^2(c+d x)\right )}{(a+a \sec (c+d x))^2} \, dx &=-\frac {(A+C) \sin (c+d x)}{3 d (a+a \sec (c+d x))^2}-\frac {\int \frac {\cos (c+d x) (-a (4 A+C)+a (2 A-C) \sec (c+d x))}{a+a \sec (c+d x)} \, dx}{3 a^2}\\ &=-\frac {2 A \sin (c+d x)}{a^2 d (1+\sec (c+d x))}-\frac {(A+C) \sin (c+d x)}{3 d (a+a \sec (c+d x))^2}-\frac {\int \cos (c+d x) \left (-a^2 (10 A+C)+6 a^2 A \sec (c+d x)\right ) \, dx}{3 a^4}\\ &=-\frac {2 A \sin (c+d x)}{a^2 d (1+\sec (c+d x))}-\frac {(A+C) \sin (c+d x)}{3 d (a+a \sec (c+d x))^2}-\frac {(2 A) \int 1 \, dx}{a^2}+\frac {(10 A+C) \int \cos (c+d x) \, dx}{3 a^2}\\ &=-\frac {2 A x}{a^2}+\frac {(10 A+C) \sin (c+d x)}{3 a^2 d}-\frac {2 A \sin (c+d x)}{a^2 d (1+\sec (c+d x))}-\frac {(A+C) \sin (c+d x)}{3 d (a+a \sec (c+d x))^2}\\ \end {align*}

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Mathematica [B] Leaf count is larger than twice the leaf count of optimal. \(195\) vs. \(2(82)=164\).
time = 0.82, size = 195, normalized size = 2.38 \begin {gather*} \frac {\sec \left (\frac {c}{2}\right ) \sec ^3\left (\frac {1}{2} (c+d x)\right ) \left (-36 A d x \cos \left (\frac {d x}{2}\right )-36 A d x \cos \left (c+\frac {d x}{2}\right )-12 A d x \cos \left (c+\frac {3 d x}{2}\right )-12 A d x \cos \left (2 c+\frac {3 d x}{2}\right )+66 A \sin \left (\frac {d x}{2}\right )+12 C \sin \left (\frac {d x}{2}\right )-30 A \sin \left (c+\frac {d x}{2}\right )-12 C \sin \left (c+\frac {d x}{2}\right )+41 A \sin \left (c+\frac {3 d x}{2}\right )+8 C \sin \left (c+\frac {3 d x}{2}\right )+9 A \sin \left (2 c+\frac {3 d x}{2}\right )+3 A \sin \left (2 c+\frac {5 d x}{2}\right )+3 A \sin \left (3 c+\frac {5 d x}{2}\right )\right )}{48 a^2 d} \end {gather*}

Antiderivative was successfully verified.

[In]

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

[Out]

(Sec[c/2]*Sec[(c + d*x)/2]^3*(-36*A*d*x*Cos[(d*x)/2] - 36*A*d*x*Cos[c + (d*x)/2] - 12*A*d*x*Cos[c + (3*d*x)/2]
- 12*A*d*x*Cos[2*c + (3*d*x)/2] + 66*A*Sin[(d*x)/2] + 12*C*Sin[(d*x)/2] - 30*A*Sin[c + (d*x)/2] - 12*C*Sin[c
+ (d*x)/2] + 41*A*Sin[c + (3*d*x)/2] + 8*C*Sin[c + (3*d*x)/2] + 9*A*Sin[2*c + (3*d*x)/2] + 3*A*Sin[2*c + (5*d*
x)/2] + 3*A*Sin[3*c + (5*d*x)/2]))/(48*a^2*d)

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Maple [A]
time = 0.52, size = 103, normalized size = 1.26

method	result	size
derivativedivides	\(\frac {-\frac {A \left (\tan ^{3}\left (\frac {d x}{2}+\frac {c}{2}\right )\right )}{3}-\frac {C \left (\tan ^{3}\left (\frac {d x}{2}+\frac {c}{2}\right )\right )}{3}+5 A \tan \left (\frac {d x}{2}+\frac {c}{2}\right )+C \tan \left (\frac {d x}{2}+\frac {c}{2}\right )-4 A \left (-\frac {\tan \left (\frac {d x}{2}+\frac {c}{2}\right )}{1+\tan ^{2}\left (\frac {d x}{2}+\frac {c}{2}\right )}+2 \arctan \left (\tan \left (\frac {d x}{2}+\frac {c}{2}\right )\right )\right )}{2 d \,a^{2}}\)	\(103\)
default	\(\frac {-\frac {A \left (\tan ^{3}\left (\frac {d x}{2}+\frac {c}{2}\right )\right )}{3}-\frac {C \left (\tan ^{3}\left (\frac {d x}{2}+\frac {c}{2}\right )\right )}{3}+5 A \tan \left (\frac {d x}{2}+\frac {c}{2}\right )+C \tan \left (\frac {d x}{2}+\frac {c}{2}\right )-4 A \left (-\frac {\tan \left (\frac {d x}{2}+\frac {c}{2}\right )}{1+\tan ^{2}\left (\frac {d x}{2}+\frac {c}{2}\right )}+2 \arctan \left (\tan \left (\frac {d x}{2}+\frac {c}{2}\right )\right )\right )}{2 d \,a^{2}}\)	\(103\)
risch	\(-\frac {2 A x}{a^{2}}-\frac {i A \,{\mathrm e}^{i \left (d x +c \right )}}{2 a^{2} d}+\frac {i A \,{\mathrm e}^{-i \left (d x +c \right )}}{2 a^{2} d}+\frac {2 i \left (9 A \,{\mathrm e}^{2 i \left (d x +c \right )}+3 C \,{\mathrm e}^{2 i \left (d x +c \right )}+15 \,{\mathrm e}^{i \left (d x +c \right )} A +3 C \,{\mathrm e}^{i \left (d x +c \right )}+8 A +2 C \right )}{3 d \,a^{2} \left ({\mathrm e}^{i \left (d x +c \right )}+1\right )^{3}}\)	\(124\)
norman	\(\frac {\frac {2 A x}{a}-\frac {2 A x \left (\tan ^{4}\left (\frac {d x}{2}+\frac {c}{2}\right )\right )}{a}-\frac {\left (A +C \right ) \left (\tan ^{7}\left (\frac {d x}{2}+\frac {c}{2}\right )\right )}{6 a d}+\frac {\left (5 A +C \right ) \left (\tan ^{5}\left (\frac {d x}{2}+\frac {c}{2}\right )\right )}{2 a d}-\frac {\left (9 A +C \right ) \tan \left (\frac {d x}{2}+\frac {c}{2}\right )}{2 a d}+\frac {\left (13 A +C \right ) \left (\tan ^{3}\left (\frac {d x}{2}+\frac {c}{2}\right )\right )}{6 a d}}{\left (1+\tan ^{2}\left (\frac {d x}{2}+\frac {c}{2}\right )\right ) \left (\tan ^{2}\left (\frac {d x}{2}+\frac {c}{2}\right )-1\right ) a}\)	\(153\)

Verification of antiderivative is not currently implemented for this CAS.

[In]

int(cos(d*x+c)*(A+C*sec(d*x+c)^2)/(a+a*sec(d*x+c))^2,x,method=_RETURNVERBOSE)

[Out]

1/2/d/a^2*(-1/3*A*tan(1/2*d*x+1/2*c)^3-1/3*C*tan(1/2*d*x+1/2*c)^3+5*A*tan(1/2*d*x+1/2*c)+C*tan(1/2*d*x+1/2*c)-
4*A*(-tan(1/2*d*x+1/2*c)/(1+tan(1/2*d*x+1/2*c)^2)+2*arctan(tan(1/2*d*x+1/2*c))))

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Maxima [B] Leaf count of result is larger than twice the leaf count of optimal. 165 vs. \(2 (78) = 156\).
time = 0.49, size = 165, normalized size = 2.01 \begin {gather*} \frac {A {\left (\frac {\frac {15 \, \sin \left (d x + c\right )}{\cos \left (d x + c\right ) + 1} - \frac {\sin \left (d x + c\right )^{3}}{{\left (\cos \left (d x + c\right ) + 1\right )}^{3}}}{a^{2}} - \frac {24 \, \arctan \left (\frac {\sin \left (d x + c\right )}{\cos \left (d x + c\right ) + 1}\right )}{a^{2}} + \frac {12 \, \sin \left (d x + c\right )}{{\left (a^{2} + \frac {a^{2} \sin \left (d x + c\right )^{2}}{{\left (\cos \left (d x + c\right ) + 1\right )}^{2}}\right )} {\left (\cos \left (d x + c\right ) + 1\right )}}\right )} + \frac {C {\left (\frac {3 \, \sin \left (d x + c\right )}{\cos \left (d x + c\right ) + 1} - \frac {\sin \left (d x + c\right )^{3}}{{\left (\cos \left (d x + c\right ) + 1\right )}^{3}}\right )}}{a^{2}}}{6 \, d} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

1/6*(A*((15*sin(d*x + c)/(cos(d*x + c) + 1) - sin(d*x + c)^3/(cos(d*x + c) + 1)^3)/a^2 - 24*arctan(sin(d*x + c
)/(cos(d*x + c) + 1))/a^2 + 12*sin(d*x + c)/((a^2 + a^2*sin(d*x + c)^2/(cos(d*x + c) + 1)^2)*(cos(d*x + c) + 1
))) + C*(3*sin(d*x + c)/(cos(d*x + c) + 1) - sin(d*x + c)^3/(cos(d*x + c) + 1)^3)/a^2)/d

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Fricas [A]
time = 2.65, size = 102, normalized size = 1.24 \begin {gather*} -\frac {6 \, A d x \cos \left (d x + c\right )^{2} + 12 \, A d x \cos \left (d x + c\right ) + 6 \, A d x - {\left (3 \, A \cos \left (d x + c\right )^{2} + 2 \, {\left (7 \, A + C\right )} \cos \left (d x + c\right ) + 10 \, A + C\right )} \sin \left (d x + c\right )}{3 \, {\left (a^{2} d \cos \left (d x + c\right )^{2} + 2 \, a^{2} d \cos \left (d x + c\right ) + a^{2} d\right )}} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate(cos(d*x+c)*(A+C*sec(d*x+c)^2)/(a+a*sec(d*x+c))^2,x, algorithm="fricas")

[Out]

-1/3*(6*A*d*x*cos(d*x + c)^2 + 12*A*d*x*cos(d*x + c) + 6*A*d*x - (3*A*cos(d*x + c)^2 + 2*(7*A + C)*cos(d*x + c
) + 10*A + C)*sin(d*x + c))/(a^2*d*cos(d*x + c)^2 + 2*a^2*d*cos(d*x + c) + a^2*d)

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Sympy [F]
time = 0.00, size = 0, normalized size = 0.00 \begin {gather*} \frac {\int \frac {A \cos {\left (c + d x \right )}}{\sec ^{2}{\left (c + d x \right )} + 2 \sec {\left (c + d x \right )} + 1}\, dx + \int \frac {C \cos {\left (c + d x \right )} \sec ^{2}{\left (c + d x \right )}}{\sec ^{2}{\left (c + d x \right )} + 2 \sec {\left (c + d x \right )} + 1}\, dx}{a^{2}} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate(cos(d*x+c)*(A+C*sec(d*x+c)**2)/(a+a*sec(d*x+c))**2,x)

[Out]

(Integral(A*cos(c + d*x)/(sec(c + d*x)**2 + 2*sec(c + d*x) + 1), x) + Integral(C*cos(c + d*x)*sec(c + d*x)**2/
(sec(c + d*x)**2 + 2*sec(c + d*x) + 1), x))/a**2

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Giac [A]
time = 0.44, size = 114, normalized size = 1.39 \begin {gather*} -\frac {\frac {12 \, {\left (d x + c\right )} A}{a^{2}} - \frac {12 \, A \tan \left (\frac {1}{2} \, d x + \frac {1}{2} \, c\right )}{{\left (\tan \left (\frac {1}{2} \, d x + \frac {1}{2} \, c\right )^{2} + 1\right )} a^{2}} + \frac {A a^{4} \tan \left (\frac {1}{2} \, d x + \frac {1}{2} \, c\right )^{3} + C a^{4} \tan \left (\frac {1}{2} \, d x + \frac {1}{2} \, c\right )^{3} - 15 \, A a^{4} \tan \left (\frac {1}{2} \, d x + \frac {1}{2} \, c\right ) - 3 \, C a^{4} \tan \left (\frac {1}{2} \, d x + \frac {1}{2} \, c\right )}{a^{6}}}{6 \, d} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

-1/6*(12*(d*x + c)*A/a^2 - 12*A*tan(1/2*d*x + 1/2*c)/((tan(1/2*d*x + 1/2*c)^2 + 1)*a^2) + (A*a^4*tan(1/2*d*x +
1/2*c)^3 + C*a^4*tan(1/2*d*x + 1/2*c)^3 - 15*A*a^4*tan(1/2*d*x + 1/2*c) - 3*C*a^4*tan(1/2*d*x + 1/2*c))/a^6)/
d

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Mupad [B]
time = 2.65, size = 99, normalized size = 1.21 \begin {gather*} \frac {\mathrm {tan}\left (\frac {c}{2}+\frac {d\,x}{2}\right )\,\left (\frac {A+C}{a^2}+\frac {3\,A-C}{2\,a^2}\right )}{d}-\frac {2\,A\,x}{a^2}+\frac {2\,A\,\mathrm {tan}\left (\frac {c}{2}+\frac {d\,x}{2}\right )}{d\,\left (a^2\,{\mathrm {tan}\left (\frac {c}{2}+\frac {d\,x}{2}\right )}^2+a^2\right )}-\frac {{\mathrm {tan}\left (\frac {c}{2}+\frac {d\,x}{2}\right )}^3\,\left (A+C\right )}{6\,a^2\,d} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

int((cos(c + d*x)*(A + C/cos(c + d*x)^2))/(a + a/cos(c + d*x))^2,x)

[Out]

(tan(c/2 + (d*x)/2)*((A + C)/a^2 + (3*A - C)/(2*a^2)))/d - (2*A*x)/a^2 + (2*A*tan(c/2 + (d*x)/2))/(d*(a^2*tan(
c/2 + (d*x)/2)^2 + a^2)) - (tan(c/2 + (d*x)/2)^3*(A + C))/(6*a^2*d)

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