3.342 \(\int \frac{\sec (c+d x) (B \sec (c+d x)+C \sec ^2(c+d x))}{(a+a \sec (c+d x))^2} \, dx\)

Optimal. Leaf size=79 \[ \frac{(2 B-5 C) \tan (c+d x)}{3 a^2 d (\sec (c+d x)+1)}+\frac{C \tanh ^{-1}(\sin (c+d x))}{a^2 d}-\frac{(B-C) \tan (c+d x)}{3 d (a \sec (c+d x)+a)^2} \]

[Out]

(C*ArcTanh[Sin[c + d*x]])/(a^2*d) + ((2*B - 5*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)

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Rubi [A]  time = 0.233121, antiderivative size = 79, normalized size of antiderivative = 1., number of steps used = 5, number of rules used = 5, integrand size = 38, \(\frac{\text{number of rules}}{\text{integrand size}}\) = 0.132, Rules used = {4072, 4008, 3998, 3770, 3794} \[ \frac{(2 B-5 C) \tan (c+d x)}{3 a^2 d (\sec (c+d x)+1)}+\frac{C \tanh ^{-1}(\sin (c+d x))}{a^2 d}-\frac{(B-C) \tan (c+d x)}{3 d (a \sec (c+d x)+a)^2} \]

Antiderivative was successfully verified.

[In]

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

[Out]

(C*ArcTanh[Sin[c + d*x]])/(a^2*d) + ((2*B - 5*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 4072

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]

Rule 4008

Int[csc[(e_.) + (f_.)*(x_)]^2*(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)/(b*f*(2*m + 1)), x] + Dist[1/(b^2*(2*
m + 1)), Int[Csc[e + f*x]*(a + b*Csc[e + f*x])^(m + 1)*Simp[A*b*m - a*B*m + b*B*(2*m + 1)*Csc[e + f*x], x], x]
, x] /; FreeQ[{a, b, e, f, A, B}, x] && NeQ[A*b - a*B, 0] && EqQ[a^2 - b^2, 0] && LtQ[m, -2^(-1)]

Rule 3998

Int[(csc[(e_.) + (f_.)*(x_)]*(csc[(e_.) + (f_.)*(x_)]*(B_.) + (A_)))/(csc[(e_.) + (f_.)*(x_)]*(b_.) + (a_)), x
_Symbol] :> Dist[B/b, Int[Csc[e + f*x], x], x] + Dist[(A*b - a*B)/b, Int[Csc[e + f*x]/(a + b*Csc[e + f*x]), x]
, x] /; FreeQ[{a, b, e, f, A, B}, x] && NeQ[A*b - a*B, 0]

Rule 3770

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

Rule 3794

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]

Rubi steps

\begin{align*} \int \frac{\sec (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{\sec ^2(c+d x) (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{\sec (c+d x) (-2 a (B-C)-3 a C \sec (c+d x))}{a+a \sec (c+d x)} \, dx}{3 a^2}\\ &=-\frac{(B-C) \tan (c+d x)}{3 d (a+a \sec (c+d x))^2}+\frac{(2 B-5 C) \int \frac{\sec (c+d x)}{a+a \sec (c+d x)} \, dx}{3 a}+\frac{C \int \sec (c+d x) \, dx}{a^2}\\ &=\frac{C \tanh ^{-1}(\sin (c+d x))}{a^2 d}-\frac{(B-C) \tan (c+d x)}{3 d (a+a \sec (c+d x))^2}+\frac{(2 B-5 C) \tan (c+d x)}{3 d \left (a^2+a^2 \sec (c+d x)\right )}\\ \end{align*}

Mathematica [A]  time = 0.758428, size = 106, normalized size = 1.34 \[ \frac{(B-4 C) \tan \left (\frac{1}{2} (c+d x)\right )+4 (B-C) \sin ^4\left (\frac{1}{2} (c+d x)\right ) \csc ^3(c+d x)+3 C \left (\log \left (\sin \left (\frac{1}{2} (c+d x)\right )+\cos \left (\frac{1}{2} (c+d x)\right )\right )-\log \left (\cos \left (\frac{1}{2} (c+d x)\right )-\sin \left (\frac{1}{2} (c+d x)\right )\right )\right )}{3 a^2 d} \]

Antiderivative was successfully verified.

[In]

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

[Out]

(3*C*(-Log[Cos[(c + d*x)/2] - Sin[(c + d*x)/2]] + Log[Cos[(c + d*x)/2] + Sin[(c + d*x)/2]]) + 4*(B - C)*Csc[c
+ d*x]^3*Sin[(c + d*x)/2]^4 + (B - 4*C)*Tan[(c + d*x)/2])/(3*a^2*d)

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Maple [A]  time = 0.049, size = 119, normalized size = 1.5 \begin{align*}{\frac{B}{6\,d{a}^{2}} \left ( \tan \left ({\frac{dx}{2}}+{\frac{c}{2}} \right ) \right ) ^{3}}-{\frac{C}{6\,d{a}^{2}} \left ( \tan \left ({\frac{dx}{2}}+{\frac{c}{2}} \right ) \right ) ^{3}}+{\frac{B}{2\,d{a}^{2}}\tan \left ({\frac{dx}{2}}+{\frac{c}{2}} \right ) }-{\frac{3\,C}{2\,d{a}^{2}}\tan \left ({\frac{dx}{2}}+{\frac{c}{2}} \right ) }-{\frac{C}{d{a}^{2}}\ln \left ( \tan \left ({\frac{dx}{2}}+{\frac{c}{2}} \right ) -1 \right ) }+{\frac{C}{d{a}^{2}}\ln \left ( \tan \left ({\frac{dx}{2}}+{\frac{c}{2}} \right ) +1 \right ) } \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

1/6/d/a^2*tan(1/2*d*x+1/2*c)^3*B-1/6/d/a^2*C*tan(1/2*d*x+1/2*c)^3+1/2/d/a^2*B*tan(1/2*d*x+1/2*c)-3/2/d/a^2*C*t
an(1/2*d*x+1/2*c)-1/d/a^2*ln(tan(1/2*d*x+1/2*c)-1)*C+1/d/a^2*ln(tan(1/2*d*x+1/2*c)+1)*C

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Maxima [A]  time = 0.957943, size = 196, normalized size = 2.48 \begin{align*} -\frac{C{\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{6 \, \log \left (\frac{\sin \left (d x + c\right )}{\cos \left (d x + c\right ) + 1} + 1\right )}{a^{2}} + \frac{6 \, \log \left (\frac{\sin \left (d x + c\right )}{\cos \left (d x + c\right ) + 1} - 1\right )}{a^{2}}\right )} - \frac{B{\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{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate(sec(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*(C*((9*sin(d*x + c)/(cos(d*x + c) + 1) + sin(d*x + c)^3/(cos(d*x + c) + 1)^3)/a^2 - 6*log(sin(d*x + c)/(c
os(d*x + c) + 1) + 1)/a^2 + 6*log(sin(d*x + c)/(cos(d*x + c) + 1) - 1)/a^2) - B*(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 = 0.502423, size = 338, normalized size = 4.28 \begin{align*} \frac{3 \,{\left (C \cos \left (d x + c\right )^{2} + 2 \, C \cos \left (d x + c\right ) + C\right )} \log \left (\sin \left (d x + c\right ) + 1\right ) - 3 \,{\left (C \cos \left (d x + c\right )^{2} + 2 \, C \cos \left (d x + c\right ) + C\right )} \log \left (-\sin \left (d x + c\right ) + 1\right ) + 2 \,{\left ({\left (B - 4 \, C\right )} \cos \left (d x + c\right ) + 2 \, B - 5 \, C\right )} \sin \left (d x + c\right )}{6 \,{\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{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate(sec(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]

1/6*(3*(C*cos(d*x + c)^2 + 2*C*cos(d*x + c) + C)*log(sin(d*x + c) + 1) - 3*(C*cos(d*x + c)^2 + 2*C*cos(d*x + c
) + C)*log(-sin(d*x + c) + 1) + 2*((B - 4*C)*cos(d*x + c) + 2*B - 5*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., size = 0, normalized size = 0. \begin{align*} \frac{\int \frac{B \sec ^{2}{\left (c + d x \right )}}{\sec ^{2}{\left (c + d x \right )} + 2 \sec{\left (c + d x \right )} + 1}\, dx + \int \frac{C \sec ^{3}{\left (c + d x \right )}}{\sec ^{2}{\left (c + d x \right )} + 2 \sec{\left (c + d x \right )} + 1}\, dx}{a^{2}} \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate(sec(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*sec(c + d*x)**2/(sec(c + d*x)**2 + 2*sec(c + d*x) + 1), x) + Integral(C*sec(c + d*x)**3/(sec(c + d
*x)**2 + 2*sec(c + d*x) + 1), x))/a**2

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Giac [A]  time = 1.16945, size = 151, normalized size = 1.91 \begin{align*} \frac{\frac{6 \, C \log \left ({\left | \tan \left (\frac{1}{2} \, d x + \frac{1}{2} \, c\right ) + 1 \right |}\right )}{a^{2}} - \frac{6 \, C \log \left ({\left | \tan \left (\frac{1}{2} \, d x + \frac{1}{2} \, c\right ) - 1 \right |}\right )}{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} + 3 \, B a^{4} \tan \left (\frac{1}{2} \, d x + \frac{1}{2} \, c\right ) - 9 \, C a^{4} \tan \left (\frac{1}{2} \, d x + \frac{1}{2} \, c\right )}{a^{6}}}{6 \, d} \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate(sec(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*C*log(abs(tan(1/2*d*x + 1/2*c) + 1))/a^2 - 6*C*log(abs(tan(1/2*d*x + 1/2*c) - 1))/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 + 3*B*a^4*tan(1/2*d*x + 1/2*c) - 9*C*a^4*tan(1/2*d*x + 1/2*c))/a
^6)/d