3.111 \(\int \frac{\sec ^3(c+d x) (A+B \sec (c+d x))}{(a+a \sec (c+d x))^4} \, dx\)

Optimal. Leaf size=146 \[ \frac{(4 A+3 B) \tan (c+d x)}{15 d \left (a^4 \sec (c+d x)+a^4\right )}-\frac{8 (4 A+3 B) \tan (c+d x)}{105 d \left (a^2 \sec (c+d x)+a^2\right )^2}-\frac{(A-B) \tan (c+d x) \sec ^3(c+d x)}{7 d (a \sec (c+d x)+a)^4}+\frac{(4 A+3 B) \tan (c+d x)}{35 a d (a \sec (c+d x)+a)^3} \]

[Out]

-((A - B)*Sec[c + d*x]^3*Tan[c + d*x])/(7*d*(a + a*Sec[c + d*x])^4) + ((4*A + 3*B)*Tan[c + d*x])/(35*a*d*(a +
a*Sec[c + d*x])^3) - (8*(4*A + 3*B)*Tan[c + d*x])/(105*d*(a^2 + a^2*Sec[c + d*x])^2) + ((4*A + 3*B)*Tan[c + d*
x])/(15*d*(a^4 + a^4*Sec[c + d*x]))

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Rubi [A]  time = 0.228662, antiderivative size = 146, normalized size of antiderivative = 1., number of steps used = 4, number of rules used = 4, integrand size = 31, \(\frac{\text{number of rules}}{\text{integrand size}}\) = 0.129, Rules used = {4012, 3799, 4000, 3794} \[ \frac{(4 A+3 B) \tan (c+d x)}{15 d \left (a^4 \sec (c+d x)+a^4\right )}-\frac{8 (4 A+3 B) \tan (c+d x)}{105 d \left (a^2 \sec (c+d x)+a^2\right )^2}-\frac{(A-B) \tan (c+d x) \sec ^3(c+d x)}{7 d (a \sec (c+d x)+a)^4}+\frac{(4 A+3 B) \tan (c+d x)}{35 a d (a \sec (c+d x)+a)^3} \]

Antiderivative was successfully verified.

[In]

Int[(Sec[c + d*x]^3*(A + B*Sec[c + d*x]))/(a + a*Sec[c + d*x])^4,x]

[Out]

-((A - B)*Sec[c + d*x]^3*Tan[c + d*x])/(7*d*(a + a*Sec[c + d*x])^4) + ((4*A + 3*B)*Tan[c + d*x])/(35*a*d*(a +
a*Sec[c + d*x])^3) - (8*(4*A + 3*B)*Tan[c + d*x])/(105*d*(a^2 + a^2*Sec[c + d*x])^2) + ((4*A + 3*B)*Tan[c + d*
x])/(15*d*(a^4 + a^4*Sec[c + d*x]))

Rule 4012

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[(a*A*m + b*B*(m + 1))/(a^2*(2*m + 1)), Int[(a + b*Csc[e + f*x])^(m + 1)*(d*Csc[e + f*x])^n
, x], x] /; FreeQ[{a, b, d, e, f, A, B, n}, x] && NeQ[A*b - a*B, 0] && EqQ[a^2 - b^2, 0] && EqQ[m + n + 1, 0]
&& LeQ[m, -1]

Rule 3799

Int[csc[(e_.) + (f_.)*(x_)]^3*(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[1/(a^2*(2*m + 1)), Int[Csc[e + f*x]*(a + b*Csc[e + f*x])^(m
+ 1)*(a*m - b*(2*m + 1)*Csc[e + f*x]), x], x] /; FreeQ[{a, b, e, f}, x] && EqQ[a^2 - b^2, 0] && LtQ[m, -2^(-1)
]

Rule 4000

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

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 ^3(c+d x) (A+B \sec (c+d x))}{(a+a \sec (c+d x))^4} \, dx &=-\frac{(A-B) \sec ^3(c+d x) \tan (c+d x)}{7 d (a+a \sec (c+d x))^4}+\frac{(4 A+3 B) \int \frac{\sec ^3(c+d x)}{(a+a \sec (c+d x))^3} \, dx}{7 a}\\ &=-\frac{(A-B) \sec ^3(c+d x) \tan (c+d x)}{7 d (a+a \sec (c+d x))^4}+\frac{(4 A+3 B) \tan (c+d x)}{35 a d (a+a \sec (c+d x))^3}+\frac{(4 A+3 B) \int \frac{\sec (c+d x) (-3 a+5 a \sec (c+d x))}{(a+a \sec (c+d x))^2} \, dx}{35 a^3}\\ &=-\frac{(A-B) \sec ^3(c+d x) \tan (c+d x)}{7 d (a+a \sec (c+d x))^4}+\frac{(4 A+3 B) \tan (c+d x)}{35 a d (a+a \sec (c+d x))^3}-\frac{8 (4 A+3 B) \tan (c+d x)}{105 d \left (a^2+a^2 \sec (c+d x)\right )^2}+\frac{(4 A+3 B) \int \frac{\sec (c+d x)}{a+a \sec (c+d x)} \, dx}{15 a^3}\\ &=-\frac{(A-B) \sec ^3(c+d x) \tan (c+d x)}{7 d (a+a \sec (c+d x))^4}+\frac{(4 A+3 B) \tan (c+d x)}{35 a d (a+a \sec (c+d x))^3}-\frac{8 (4 A+3 B) \tan (c+d x)}{105 d \left (a^2+a^2 \sec (c+d x)\right )^2}+\frac{(4 A+3 B) \tan (c+d x)}{15 d \left (a^4+a^4 \sec (c+d x)\right )}\\ \end{align*}

Mathematica [A]  time = 0.340026, size = 109, normalized size = 0.75 \[ \frac{\sec \left (\frac{c}{2}\right ) \cos \left (\frac{1}{2} (c+d x)\right ) \left ((4 A+3 B) \left (21 \sin \left (c+\frac{3 d x}{2}\right )+7 \sin \left (2 c+\frac{5 d x}{2}\right )+\sin \left (3 c+\frac{7 d x}{2}\right )\right )+35 (2 A+3 B) \sin \left (\frac{d x}{2}\right )-70 A \sin \left (c+\frac{d x}{2}\right )\right )}{210 a^4 d (\cos (c+d x)+1)^4} \]

Antiderivative was successfully verified.

[In]

Integrate[(Sec[c + d*x]^3*(A + B*Sec[c + d*x]))/(a + a*Sec[c + d*x])^4,x]

[Out]

(Cos[(c + d*x)/2]*Sec[c/2]*(35*(2*A + 3*B)*Sin[(d*x)/2] - 70*A*Sin[c + (d*x)/2] + (4*A + 3*B)*(21*Sin[c + (3*d
*x)/2] + 7*Sin[2*c + (5*d*x)/2] + Sin[3*c + (7*d*x)/2])))/(210*a^4*d*(1 + Cos[c + d*x])^4)

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Maple [A]  time = 0.058, size = 88, normalized size = 0.6 \begin{align*}{\frac{1}{8\,d{a}^{4}} \left ({\frac{-A+B}{7} \left ( \tan \left ({\frac{dx}{2}}+{\frac{c}{2}} \right ) \right ) ^{7}}+{\frac{-A+3\,B}{5} \left ( \tan \left ({\frac{dx}{2}}+{\frac{c}{2}} \right ) \right ) ^{5}}+{\frac{A+3\,B}{3} \left ( \tan \left ({\frac{dx}{2}}+{\frac{c}{2}} \right ) \right ) ^{3}}+A\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(sec(d*x+c)^3*(A+B*sec(d*x+c))/(a+a*sec(d*x+c))^4,x)

[Out]

1/8/d/a^4*(1/7*(-A+B)*tan(1/2*d*x+1/2*c)^7+1/5*(-A+3*B)*tan(1/2*d*x+1/2*c)^5+1/3*(A+3*B)*tan(1/2*d*x+1/2*c)^3+
A*tan(1/2*d*x+1/2*c)+B*tan(1/2*d*x+1/2*c))

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Maxima [A]  time = 1.00143, size = 236, normalized size = 1.62 \begin{align*} \frac{\frac{A{\left (\frac{105 \, \sin \left (d x + c\right )}{\cos \left (d x + c\right ) + 1} + \frac{35 \, \sin \left (d x + c\right )^{3}}{{\left (\cos \left (d x + c\right ) + 1\right )}^{3}} - \frac{21 \, \sin \left (d x + c\right )^{5}}{{\left (\cos \left (d x + c\right ) + 1\right )}^{5}} - \frac{15 \, \sin \left (d x + c\right )^{7}}{{\left (\cos \left (d x + c\right ) + 1\right )}^{7}}\right )}}{a^{4}} + \frac{3 \, B{\left (\frac{35 \, \sin \left (d x + c\right )}{\cos \left (d x + c\right ) + 1} + \frac{35 \, \sin \left (d x + c\right )^{3}}{{\left (\cos \left (d x + c\right ) + 1\right )}^{3}} + \frac{21 \, \sin \left (d x + c\right )^{5}}{{\left (\cos \left (d x + c\right ) + 1\right )}^{5}} + \frac{5 \, \sin \left (d x + c\right )^{7}}{{\left (\cos \left (d x + c\right ) + 1\right )}^{7}}\right )}}{a^{4}}}{840 \, d} \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate(sec(d*x+c)^3*(A+B*sec(d*x+c))/(a+a*sec(d*x+c))^4,x, algorithm="maxima")

[Out]

1/840*(A*(105*sin(d*x + c)/(cos(d*x + c) + 1) + 35*sin(d*x + c)^3/(cos(d*x + c) + 1)^3 - 21*sin(d*x + c)^5/(co
s(d*x + c) + 1)^5 - 15*sin(d*x + c)^7/(cos(d*x + c) + 1)^7)/a^4 + 3*B*(35*sin(d*x + c)/(cos(d*x + c) + 1) + 35
*sin(d*x + c)^3/(cos(d*x + c) + 1)^3 + 21*sin(d*x + c)^5/(cos(d*x + c) + 1)^5 + 5*sin(d*x + c)^7/(cos(d*x + c)
+ 1)^7)/a^4)/d

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Fricas [A]  time = 0.444594, size = 311, normalized size = 2.13 \begin{align*} \frac{{\left (2 \,{\left (4 \, A + 3 \, B\right )} \cos \left (d x + c\right )^{3} + 8 \,{\left (4 \, A + 3 \, B\right )} \cos \left (d x + c\right )^{2} + 13 \,{\left (4 \, A + 3 \, B\right )} \cos \left (d x + c\right ) + 13 \, A + 36 \, B\right )} \sin \left (d x + c\right )}{105 \,{\left (a^{4} d \cos \left (d x + c\right )^{4} + 4 \, a^{4} d \cos \left (d x + c\right )^{3} + 6 \, a^{4} d \cos \left (d x + c\right )^{2} + 4 \, a^{4} d \cos \left (d x + c\right ) + a^{4} d\right )}} \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate(sec(d*x+c)^3*(A+B*sec(d*x+c))/(a+a*sec(d*x+c))^4,x, algorithm="fricas")

[Out]

1/105*(2*(4*A + 3*B)*cos(d*x + c)^3 + 8*(4*A + 3*B)*cos(d*x + c)^2 + 13*(4*A + 3*B)*cos(d*x + c) + 13*A + 36*B
)*sin(d*x + c)/(a^4*d*cos(d*x + c)^4 + 4*a^4*d*cos(d*x + c)^3 + 6*a^4*d*cos(d*x + c)^2 + 4*a^4*d*cos(d*x + c)
+ a^4*d)

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Sympy [F]  time = 0., size = 0, normalized size = 0. \begin{align*} \frac{\int \frac{A \sec ^{3}{\left (c + d x \right )}}{\sec ^{4}{\left (c + d x \right )} + 4 \sec ^{3}{\left (c + d x \right )} + 6 \sec ^{2}{\left (c + d x \right )} + 4 \sec{\left (c + d x \right )} + 1}\, dx + \int \frac{B \sec ^{4}{\left (c + d x \right )}}{\sec ^{4}{\left (c + d x \right )} + 4 \sec ^{3}{\left (c + d x \right )} + 6 \sec ^{2}{\left (c + d x \right )} + 4 \sec{\left (c + d x \right )} + 1}\, dx}{a^{4}} \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate(sec(d*x+c)**3*(A+B*sec(d*x+c))/(a+a*sec(d*x+c))**4,x)

[Out]

(Integral(A*sec(c + d*x)**3/(sec(c + d*x)**4 + 4*sec(c + d*x)**3 + 6*sec(c + d*x)**2 + 4*sec(c + d*x) + 1), x)
+ Integral(B*sec(c + d*x)**4/(sec(c + d*x)**4 + 4*sec(c + d*x)**3 + 6*sec(c + d*x)**2 + 4*sec(c + d*x) + 1),
x))/a**4

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Giac [A]  time = 1.38166, size = 158, normalized size = 1.08 \begin{align*} -\frac{15 \, A \tan \left (\frac{1}{2} \, d x + \frac{1}{2} \, c\right )^{7} - 15 \, B \tan \left (\frac{1}{2} \, d x + \frac{1}{2} \, c\right )^{7} + 21 \, A \tan \left (\frac{1}{2} \, d x + \frac{1}{2} \, c\right )^{5} - 63 \, B \tan \left (\frac{1}{2} \, d x + \frac{1}{2} \, c\right )^{5} - 35 \, A \tan \left (\frac{1}{2} \, d x + \frac{1}{2} \, c\right )^{3} - 105 \, B \tan \left (\frac{1}{2} \, d x + \frac{1}{2} \, c\right )^{3} - 105 \, A \tan \left (\frac{1}{2} \, d x + \frac{1}{2} \, c\right ) - 105 \, B \tan \left (\frac{1}{2} \, d x + \frac{1}{2} \, c\right )}{840 \, a^{4} d} \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate(sec(d*x+c)^3*(A+B*sec(d*x+c))/(a+a*sec(d*x+c))^4,x, algorithm="giac")

[Out]

-1/840*(15*A*tan(1/2*d*x + 1/2*c)^7 - 15*B*tan(1/2*d*x + 1/2*c)^7 + 21*A*tan(1/2*d*x + 1/2*c)^5 - 63*B*tan(1/2
*d*x + 1/2*c)^5 - 35*A*tan(1/2*d*x + 1/2*c)^3 - 105*B*tan(1/2*d*x + 1/2*c)^3 - 105*A*tan(1/2*d*x + 1/2*c) - 10
5*B*tan(1/2*d*x + 1/2*c))/(a^4*d)