3.4.44 \(\int \frac {\cos (c+d x) (B \sec (c+d x)+C \sec ^2(c+d x))}{(a+a \sec (c+d x))^2} \, dx\) [344]

Optimal. Leaf size=70 \[ \frac {B x}{a^2}-\frac {(4 B-C) \tan (c+d x)}{3 a^2 d (1+\sec (c+d x))}-\frac {(B-C) \tan (c+d x)}{3 d (a+a \sec (c+d x))^2} \]

[Out]

B*x/a^2-1/3*(4*B-C)*tan(d*x+c)/a^2/d/(1+sec(d*x+c))-1/3*(B-C)*tan(d*x+c)/d/(a+a*sec(d*x+c))^2

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Rubi [A]
time = 0.12, antiderivative size = 70, normalized size of antiderivative = 1.00, number of steps used = 4, number of rules used = 4, integrand size = 38, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.105, Rules used = {4157, 4007, 4004, 3879} \begin {gather*} -\frac {(4 B-C) \tan (c+d x)}{3 a^2 d (\sec (c+d x)+1)}+\frac {B x}{a^2}-\frac {(B-C) \tan (c+d x)}{3 d (a \sec (c+d x)+a)^2} \end {gather*}

Antiderivative was successfully verified.

[In]

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

[Out]

(B*x)/a^2 - ((4*B - C)*Tan[c + d*x])/(3*a^2*d*(1 + Sec[c + d*x])) - ((B - C)*Tan[c + d*x])/(3*d*(a + a*Sec[c +
d*x])^2)

Rule 3879

Int[csc[(e_.) + (f_.)*(x_)]/(csc[(e_.) + (f_.)*(x_)]*(b_.) + (a_)), x_Symbol] :> Simp[-Cot[e + f*x]/(f*(b + a*
Csc[e + f*x])), x] /; FreeQ[{a, b, e, f}, x] && EqQ[a^2 - b^2, 0]

Rule 4004

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

Rule 4007

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

Rule 4157

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

Rubi steps

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

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Mathematica [B] Leaf count is larger than twice the leaf count of optimal. \(153\) vs. \(2(70)=140\).
time = 0.40, size = 153, normalized size = 2.19 \begin {gather*} \frac {\sec \left (\frac {c}{2}\right ) \sec ^3\left (\frac {1}{2} (c+d x)\right ) \left (9 B d x \cos \left (\frac {d x}{2}\right )+9 B d x \cos \left (c+\frac {d x}{2}\right )+3 B d x \cos \left (c+\frac {3 d x}{2}\right )+3 B d x \cos \left (2 c+\frac {3 d x}{2}\right )-18 B \sin \left (\frac {d x}{2}\right )+6 C \sin \left (\frac {d x}{2}\right )+12 B \sin \left (c+\frac {d x}{2}\right )-6 C \sin \left (c+\frac {d x}{2}\right )-10 B \sin \left (c+\frac {3 d x}{2}\right )+4 C \sin \left (c+\frac {3 d x}{2}\right )\right )}{24 a^2 d} \end {gather*}

Antiderivative was successfully verified.

[In]

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

[Out]

(Sec[c/2]*Sec[(c + d*x)/2]^3*(9*B*d*x*Cos[(d*x)/2] + 9*B*d*x*Cos[c + (d*x)/2] + 3*B*d*x*Cos[c + (3*d*x)/2] + 3
*B*d*x*Cos[2*c + (3*d*x)/2] - 18*B*Sin[(d*x)/2] + 6*C*Sin[(d*x)/2] + 12*B*Sin[c + (d*x)/2] - 6*C*Sin[c + (d*x)
/2] - 10*B*Sin[c + (3*d*x)/2] + 4*C*Sin[c + (3*d*x)/2]))/(24*a^2*d)

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Maple [A]
time = 0.68, size = 74, normalized size = 1.06

method	result	size
derivativedivides	\(\frac {\frac {B \left (\tan ^{3}\left (\frac {d x}{2}+\frac {c}{2}\right )\right )}{3}-\frac {C \left (\tan ^{3}\left (\frac {d x}{2}+\frac {c}{2}\right )\right )}{3}-3 B \tan \left (\frac {d x}{2}+\frac {c}{2}\right )+C \tan \left (\frac {d x}{2}+\frac {c}{2}\right )+4 B \arctan \left (\tan \left (\frac {d x}{2}+\frac {c}{2}\right )\right )}{2 d \,a^{2}}\)	\(74\)
default	\(\frac {\frac {B \left (\tan ^{3}\left (\frac {d x}{2}+\frac {c}{2}\right )\right )}{3}-\frac {C \left (\tan ^{3}\left (\frac {d x}{2}+\frac {c}{2}\right )\right )}{3}-3 B \tan \left (\frac {d x}{2}+\frac {c}{2}\right )+C \tan \left (\frac {d x}{2}+\frac {c}{2}\right )+4 B \arctan \left (\tan \left (\frac {d x}{2}+\frac {c}{2}\right )\right )}{2 d \,a^{2}}\)	\(74\)
risch	\(\frac {B x}{a^{2}}-\frac {2 i \left (6 B \,{\mathrm e}^{2 i \left (d x +c \right )}-3 C \,{\mathrm e}^{2 i \left (d x +c \right )}+9 B \,{\mathrm e}^{i \left (d x +c \right )}-3 C \,{\mathrm e}^{i \left (d x +c \right )}+5 B -2 C \right )}{3 d \,a^{2} \left ({\mathrm e}^{i \left (d x +c \right )}+1\right )^{3}}\)	\(85\)
norman	\(\frac {\frac {B x \left (\tan ^{4}\left (\frac {d x}{2}+\frac {c}{2}\right )\right )}{a}-\frac {B x}{a}-\frac {\left (B -C \right ) \left (\tan ^{3}\left (\frac {d x}{2}+\frac {c}{2}\right )\right )}{6 a d}+\frac {\left (B -C \right ) \left (\tan ^{7}\left (\frac {d x}{2}+\frac {c}{2}\right )\right )}{6 a d}+\frac {\left (3 B -C \right ) \tan \left (\frac {d x}{2}+\frac {c}{2}\right )}{2 a d}-\frac {\left (3 B -C \right ) \left (\tan ^{5}\left (\frac {d x}{2}+\frac {c}{2}\right )\right )}{2 a d}}{\left (1+\tan ^{2}\left (\frac {d x}{2}+\frac {c}{2}\right )\right ) \left (\tan ^{2}\left (\frac {d x}{2}+\frac {c}{2}\right )-1\right ) a}\)	\(158\)

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

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

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Maxima [A]
time = 0.49, size = 120, normalized size = 1.71 \begin {gather*} -\frac {B {\left (\frac {\frac {9 \, \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 {12 \, \arctan \left (\frac {\sin \left (d x + c\right )}{\cos \left (d x + c\right ) + 1}\right )}{a^{2}}\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)*(B*sec(d*x+c)+C*sec(d*x+c)^2)/(a+a*sec(d*x+c))^2,x, algorithm="maxima")

[Out]

-1/6*(B*((9*sin(d*x + c)/(cos(d*x + c) + 1) - sin(d*x + c)^3/(cos(d*x + c) + 1)^3)/a^2 - 12*arctan(sin(d*x + c
)/(cos(d*x + c) + 1))/a^2) - 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 = 3.44, size = 94, normalized size = 1.34 \begin {gather*} \frac {3 \, B d x \cos \left (d x + c\right )^{2} + 6 \, B d x \cos \left (d x + c\right ) + 3 \, B d x - {\left ({\left (5 \, B - 2 \, C\right )} \cos \left (d x + c\right ) + 4 \, B - 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)*(B*sec(d*x+c)+C*sec(d*x+c)^2)/(a+a*sec(d*x+c))^2,x, algorithm="fricas")

[Out]

1/3*(3*B*d*x*cos(d*x + c)^2 + 6*B*d*x*cos(d*x + c) + 3*B*d*x - ((5*B - 2*C)*cos(d*x + c) + 4*B - 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 {B \cos {\left (c + d x \right )} \sec {\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)*(B*sec(d*x+c)+C*sec(d*x+c)**2)/(a+a*sec(d*x+c))**2,x)

[Out]

(Integral(B*cos(c + d*x)*sec(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 = 0.46, size = 85, normalized size = 1.21 \begin {gather*} \frac {\frac {6 \, {\left (d x + c\right )} B}{a^{2}} + \frac {B 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} - 9 \, B 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)*(B*sec(d*x+c)+C*sec(d*x+c)^2)/(a+a*sec(d*x+c))^2,x, algorithm="giac")

[Out]

1/6*(6*(d*x + c)*B/a^2 + (B*a^4*tan(1/2*d*x + 1/2*c)^3 - C*a^4*tan(1/2*d*x + 1/2*c)^3 - 9*B*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 = 2.73, size = 65, normalized size = 0.93 \begin {gather*} \frac {3\,C\,\mathrm {tan}\left (\frac {c}{2}+\frac {d\,x}{2}\right )-9\,B\,\mathrm {tan}\left (\frac {c}{2}+\frac {d\,x}{2}\right )+B\,{\mathrm {tan}\left (\frac {c}{2}+\frac {d\,x}{2}\right )}^3-C\,{\mathrm {tan}\left (\frac {c}{2}+\frac {d\,x}{2}\right )}^3+6\,B\,d\,x}{6\,a^2\,d} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

(3*C*tan(c/2 + (d*x)/2) - 9*B*tan(c/2 + (d*x)/2) + B*tan(c/2 + (d*x)/2)^3 - C*tan(c/2 + (d*x)/2)^3 + 6*B*d*x)/
(6*a^2*d)

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