3.352 \(\int \frac{B \sec (c+d x)+C \sec ^2(c+d x)}{(a+a \sec (c+d x))^3} \, dx\)

Optimal. Leaf size=102 \[ \frac{(2 B+3 C) \tan (c+d x)}{15 d \left (a^3 \sec (c+d x)+a^3\right )}+\frac{(2 B+3 C) \tan (c+d x)}{15 a d (a \sec (c+d x)+a)^2}+\frac{(B-C) \tan (c+d x)}{5 d (a \sec (c+d x)+a)^3} \]

[Out]

((B - C)*Tan[c + d*x])/(5*d*(a + a*Sec[c + d*x])^3) + ((2*B + 3*C)*Tan[c + d*x])/(15*a*d*(a + a*Sec[c + d*x])^
2) + ((2*B + 3*C)*Tan[c + d*x])/(15*d*(a^3 + a^3*Sec[c + d*x]))

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Rubi [A]  time = 0.109576, antiderivative size = 102, normalized size of antiderivative = 1., number of steps used = 4, number of rules used = 4, integrand size = 32, \(\frac{\text{number of rules}}{\text{integrand size}}\) = 0.125, Rules used = {4052, 12, 3796, 3794} \[ \frac{(2 B+3 C) \tan (c+d x)}{15 d \left (a^3 \sec (c+d x)+a^3\right )}+\frac{(2 B+3 C) \tan (c+d x)}{15 a d (a \sec (c+d x)+a)^2}+\frac{(B-C) \tan (c+d x)}{5 d (a \sec (c+d x)+a)^3} \]

Antiderivative was successfully verified.

[In]

Int[(B*Sec[c + d*x] + C*Sec[c + d*x]^2)/(a + a*Sec[c + d*x])^3,x]

[Out]

((B - C)*Tan[c + d*x])/(5*d*(a + a*Sec[c + d*x])^3) + ((2*B + 3*C)*Tan[c + d*x])/(15*a*d*(a + a*Sec[c + d*x])^
2) + ((2*B + 3*C)*Tan[c + d*x])/(15*d*(a^3 + a^3*Sec[c + d*x]))

Rule 4052

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

Rule 12

Int[(a_)*(u_), x_Symbol] :> Dist[a, Int[u, x], x] /; FreeQ[a, x] &&  !MatchQ[u, (b_)*(v_) /; FreeQ[b, x]]

Rule 3796

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

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{B \sec (c+d x)+C \sec ^2(c+d x)}{(a+a \sec (c+d x))^3} \, dx &=\frac{(B-C) \tan (c+d x)}{5 d (a+a \sec (c+d x))^3}+\frac{\int \frac{a (2 B+3 C) \sec (c+d x)}{(a+a \sec (c+d x))^2} \, dx}{5 a^2}\\ &=\frac{(B-C) \tan (c+d x)}{5 d (a+a \sec (c+d x))^3}+\frac{(2 B+3 C) \int \frac{\sec (c+d x)}{(a+a \sec (c+d x))^2} \, dx}{5 a}\\ &=\frac{(B-C) \tan (c+d x)}{5 d (a+a \sec (c+d x))^3}+\frac{(2 B+3 C) \tan (c+d x)}{15 a d (a+a \sec (c+d x))^2}+\frac{(2 B+3 C) \int \frac{\sec (c+d x)}{a+a \sec (c+d x)} \, dx}{15 a^2}\\ &=\frac{(B-C) \tan (c+d x)}{5 d (a+a \sec (c+d x))^3}+\frac{(2 B+3 C) \tan (c+d x)}{15 a d (a+a \sec (c+d x))^2}+\frac{(2 B+3 C) \tan (c+d x)}{15 d \left (a^3+a^3 \sec (c+d x)\right )}\\ \end{align*}

Mathematica [A]  time = 0.359352, size = 70, normalized size = 0.69 \[ \frac{\tan \left (\frac{1}{2} (c+d x)\right ) \sec ^4\left (\frac{1}{2} (c+d x)\right ) (6 (2 B+3 C) \cos (c+d x)+(7 B+3 C) \cos (2 (c+d x))+11 B+9 C)}{120 a^3 d} \]

Antiderivative was successfully verified.

[In]

Integrate[(B*Sec[c + d*x] + C*Sec[c + d*x]^2)/(a + a*Sec[c + d*x])^3,x]

[Out]

((11*B + 9*C + 6*(2*B + 3*C)*Cos[c + d*x] + (7*B + 3*C)*Cos[2*(c + d*x)])*Sec[(c + d*x)/2]^4*Tan[(c + d*x)/2])
/(120*a^3*d)

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Maple [A]  time = 0.056, size = 64, normalized size = 0.6 \begin{align*}{\frac{1}{4\,d{a}^{3}} \left ({\frac{B-C}{5} \left ( \tan \left ({\frac{dx}{2}}+{\frac{c}{2}} \right ) \right ) ^{5}}-{\frac{2\,B}{3} \left ( \tan \left ({\frac{dx}{2}}+{\frac{c}{2}} \right ) \right ) ^{3}}+C\tan \left ({\frac{dx}{2}}+{\frac{c}{2}} \right ) +B\tan \left ({\frac{dx}{2}}+{\frac{c}{2}} \right ) \right ) } \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

int((B*sec(d*x+c)+C*sec(d*x+c)^2)/(a+a*sec(d*x+c))^3,x)

[Out]

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

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Maxima [A]  time = 0.957598, size = 155, normalized size = 1.52 \begin{align*} \frac{\frac{B{\left (\frac{15 \, \sin \left (d x + c\right )}{\cos \left (d x + c\right ) + 1} - \frac{10 \, \sin \left (d x + c\right )^{3}}{{\left (\cos \left (d x + c\right ) + 1\right )}^{3}} + \frac{3 \, \sin \left (d x + c\right )^{5}}{{\left (\cos \left (d x + c\right ) + 1\right )}^{5}}\right )}}{a^{3}} + \frac{3 \, C{\left (\frac{5 \, \sin \left (d x + c\right )}{\cos \left (d x + c\right ) + 1} - \frac{\sin \left (d x + c\right )^{5}}{{\left (\cos \left (d x + c\right ) + 1\right )}^{5}}\right )}}{a^{3}}}{60 \, d} \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate((B*sec(d*x+c)+C*sec(d*x+c)^2)/(a+a*sec(d*x+c))^3,x, algorithm="maxima")

[Out]

1/60*(B*(15*sin(d*x + c)/(cos(d*x + c) + 1) - 10*sin(d*x + c)^3/(cos(d*x + c) + 1)^3 + 3*sin(d*x + c)^5/(cos(d
*x + c) + 1)^5)/a^3 + 3*C*(5*sin(d*x + c)/(cos(d*x + c) + 1) - sin(d*x + c)^5/(cos(d*x + c) + 1)^5)/a^3)/d

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Fricas [A]  time = 0.461763, size = 227, normalized size = 2.23 \begin{align*} \frac{{\left ({\left (7 \, B + 3 \, C\right )} \cos \left (d x + c\right )^{2} + 3 \,{\left (2 \, B + 3 \, C\right )} \cos \left (d x + c\right ) + 2 \, B + 3 \, C\right )} \sin \left (d x + c\right )}{15 \,{\left (a^{3} d \cos \left (d x + c\right )^{3} + 3 \, a^{3} d \cos \left (d x + c\right )^{2} + 3 \, a^{3} d \cos \left (d x + c\right ) + a^{3} d\right )}} \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate((B*sec(d*x+c)+C*sec(d*x+c)^2)/(a+a*sec(d*x+c))^3,x, algorithm="fricas")

[Out]

1/15*((7*B + 3*C)*cos(d*x + c)^2 + 3*(2*B + 3*C)*cos(d*x + c) + 2*B + 3*C)*sin(d*x + c)/(a^3*d*cos(d*x + c)^3
+ 3*a^3*d*cos(d*x + c)^2 + 3*a^3*d*cos(d*x + c) + a^3*d)

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Sympy [F]  time = 0., size = 0, normalized size = 0. \begin{align*} \frac{\int \frac{B \sec{\left (c + d x \right )}}{\sec ^{3}{\left (c + d x \right )} + 3 \sec ^{2}{\left (c + d x \right )} + 3 \sec{\left (c + d x \right )} + 1}\, dx + \int \frac{C \sec ^{2}{\left (c + d x \right )}}{\sec ^{3}{\left (c + d x \right )} + 3 \sec ^{2}{\left (c + d x \right )} + 3 \sec{\left (c + d x \right )} + 1}\, dx}{a^{3}} \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

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

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Giac [A]  time = 1.16376, size = 101, normalized size = 0.99 \begin{align*} \frac{3 \, B \tan \left (\frac{1}{2} \, d x + \frac{1}{2} \, c\right )^{5} - 3 \, C \tan \left (\frac{1}{2} \, d x + \frac{1}{2} \, c\right )^{5} - 10 \, B \tan \left (\frac{1}{2} \, d x + \frac{1}{2} \, c\right )^{3} + 15 \, B \tan \left (\frac{1}{2} \, d x + \frac{1}{2} \, c\right ) + 15 \, C \tan \left (\frac{1}{2} \, d x + \frac{1}{2} \, c\right )}{60 \, a^{3} d} \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate((B*sec(d*x+c)+C*sec(d*x+c)^2)/(a+a*sec(d*x+c))^3,x, algorithm="giac")

[Out]

1/60*(3*B*tan(1/2*d*x + 1/2*c)^5 - 3*C*tan(1/2*d*x + 1/2*c)^5 - 10*B*tan(1/2*d*x + 1/2*c)^3 + 15*B*tan(1/2*d*x
+ 1/2*c) + 15*C*tan(1/2*d*x + 1/2*c))/(a^3*d)