3.137 \(\int \sec (c+d x) (a+a \sec (c+d x))^{5/2} (A+B \sec (c+d x)) \, dx\)

Optimal. Leaf size=138 \[ \frac{64 a^3 (7 A+5 B) \tan (c+d x)}{105 d \sqrt{a \sec (c+d x)+a}}+\frac{16 a^2 (7 A+5 B) \tan (c+d x) \sqrt{a \sec (c+d x)+a}}{105 d}+\frac{2 a (7 A+5 B) \tan (c+d x) (a \sec (c+d x)+a)^{3/2}}{35 d}+\frac{2 B \tan (c+d x) (a \sec (c+d x)+a)^{5/2}}{7 d} \]

[Out]

(64*a^3*(7*A + 5*B)*Tan[c + d*x])/(105*d*Sqrt[a + a*Sec[c + d*x]]) + (16*a^2*(7*A + 5*B)*Sqrt[a + a*Sec[c + d*
x]]*Tan[c + d*x])/(105*d) + (2*a*(7*A + 5*B)*(a + a*Sec[c + d*x])^(3/2)*Tan[c + d*x])/(35*d) + (2*B*(a + a*Sec
[c + d*x])^(5/2)*Tan[c + d*x])/(7*d)

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Rubi [A]  time = 0.183963, antiderivative size = 138, normalized size of antiderivative = 1., number of steps used = 4, number of rules used = 3, integrand size = 31, \(\frac{\text{number of rules}}{\text{integrand size}}\) = 0.097, Rules used = {4001, 3793, 3792} \[ \frac{64 a^3 (7 A+5 B) \tan (c+d x)}{105 d \sqrt{a \sec (c+d x)+a}}+\frac{16 a^2 (7 A+5 B) \tan (c+d x) \sqrt{a \sec (c+d x)+a}}{105 d}+\frac{2 a (7 A+5 B) \tan (c+d x) (a \sec (c+d x)+a)^{3/2}}{35 d}+\frac{2 B \tan (c+d x) (a \sec (c+d x)+a)^{5/2}}{7 d} \]

Antiderivative was successfully verified.

[In]

Int[Sec[c + d*x]*(a + a*Sec[c + d*x])^(5/2)*(A + B*Sec[c + d*x]),x]

[Out]

(64*a^3*(7*A + 5*B)*Tan[c + d*x])/(105*d*Sqrt[a + a*Sec[c + d*x]]) + (16*a^2*(7*A + 5*B)*Sqrt[a + a*Sec[c + d*
x]]*Tan[c + d*x])/(105*d) + (2*a*(7*A + 5*B)*(a + a*Sec[c + d*x])^(3/2)*Tan[c + d*x])/(35*d) + (2*B*(a + a*Sec
[c + d*x])^(5/2)*Tan[c + d*x])/(7*d)

Rule 4001

Int[csc[(e_.) + (f_.)*(x_)]*(csc[(e_.) + (f_.)*(x_)]*(b_.) + (a_))^(m_)*(csc[(e_.) + (f_.)*(x_)]*(B_.) + (A_))
, x_Symbol] :> -Simp[(B*Cot[e + f*x]*(a + b*Csc[e + f*x])^m)/(f*(m + 1)), x] + Dist[(a*B*m + A*b*(m + 1))/(b*(
m + 1)), Int[Csc[e + f*x]*(a + b*Csc[e + f*x])^m, x], x] /; FreeQ[{a, b, A, B, e, f, m}, 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 3793

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 - 1))/(f*m), x] + Dist[(a*(2*m - 1))/m, 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] && GtQ[m, 1/2] && IntegerQ[2*m]

Rule 3792

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

Rubi steps

\begin{align*} \int \sec (c+d x) (a+a \sec (c+d x))^{5/2} (A+B \sec (c+d x)) \, dx &=\frac{2 B (a+a \sec (c+d x))^{5/2} \tan (c+d x)}{7 d}+\frac{1}{7} (7 A+5 B) \int \sec (c+d x) (a+a \sec (c+d x))^{5/2} \, dx\\ &=\frac{2 a (7 A+5 B) (a+a \sec (c+d x))^{3/2} \tan (c+d x)}{35 d}+\frac{2 B (a+a \sec (c+d x))^{5/2} \tan (c+d x)}{7 d}+\frac{1}{35} (8 a (7 A+5 B)) \int \sec (c+d x) (a+a \sec (c+d x))^{3/2} \, dx\\ &=\frac{16 a^2 (7 A+5 B) \sqrt{a+a \sec (c+d x)} \tan (c+d x)}{105 d}+\frac{2 a (7 A+5 B) (a+a \sec (c+d x))^{3/2} \tan (c+d x)}{35 d}+\frac{2 B (a+a \sec (c+d x))^{5/2} \tan (c+d x)}{7 d}+\frac{1}{105} \left (32 a^2 (7 A+5 B)\right ) \int \sec (c+d x) \sqrt{a+a \sec (c+d x)} \, dx\\ &=\frac{64 a^3 (7 A+5 B) \tan (c+d x)}{105 d \sqrt{a+a \sec (c+d x)}}+\frac{16 a^2 (7 A+5 B) \sqrt{a+a \sec (c+d x)} \tan (c+d x)}{105 d}+\frac{2 a (7 A+5 B) (a+a \sec (c+d x))^{3/2} \tan (c+d x)}{35 d}+\frac{2 B (a+a \sec (c+d x))^{5/2} \tan (c+d x)}{7 d}\\ \end{align*}

Mathematica [A]  time = 0.47565, size = 89, normalized size = 0.64 \[ \frac{2 a^2 \sqrt{a (\sec (c+d x)+1)} \left ((301 A+230 B) \sin (c+d x)+\tan (c+d x) \left (3 (7 A+20 B) \sec (c+d x)+98 A+15 B \sec ^2(c+d x)+115 B\right )\right )}{105 d (\cos (c+d x)+1)} \]

Antiderivative was successfully verified.

[In]

Integrate[Sec[c + d*x]*(a + a*Sec[c + d*x])^(5/2)*(A + B*Sec[c + d*x]),x]

[Out]

(2*a^2*Sqrt[a*(1 + Sec[c + d*x])]*((301*A + 230*B)*Sin[c + d*x] + (98*A + 115*B + 3*(7*A + 20*B)*Sec[c + d*x]
+ 15*B*Sec[c + d*x]^2)*Tan[c + d*x]))/(105*d*(1 + Cos[c + d*x]))

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Maple [A]  time = 0.231, size = 119, normalized size = 0.9 \begin{align*} -{\frac{2\,{a}^{2} \left ( -1+\cos \left ( dx+c \right ) \right ) \left ( 301\,A \left ( \cos \left ( dx+c \right ) \right ) ^{3}+230\,B \left ( \cos \left ( dx+c \right ) \right ) ^{3}+98\,A \left ( \cos \left ( dx+c \right ) \right ) ^{2}+115\,B \left ( \cos \left ( dx+c \right ) \right ) ^{2}+21\,A\cos \left ( dx+c \right ) +60\,B\cos \left ( dx+c \right ) +15\,B \right ) }{105\,d \left ( \cos \left ( dx+c \right ) \right ) ^{3}\sin \left ( dx+c \right ) }\sqrt{{\frac{a \left ( \cos \left ( dx+c \right ) +1 \right ) }{\cos \left ( dx+c \right ) }}}} \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

int(sec(d*x+c)*(a+a*sec(d*x+c))^(5/2)*(A+B*sec(d*x+c)),x)

[Out]

-2/105/d*a^2*(-1+cos(d*x+c))*(301*A*cos(d*x+c)^3+230*B*cos(d*x+c)^3+98*A*cos(d*x+c)^2+115*B*cos(d*x+c)^2+21*A*
cos(d*x+c)+60*B*cos(d*x+c)+15*B)*(a*(cos(d*x+c)+1)/cos(d*x+c))^(1/2)/cos(d*x+c)^3/sin(d*x+c)

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Maxima [F(-1)]  time = 0., size = 0, normalized size = 0. \begin{align*} \text{Timed out} \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate(sec(d*x+c)*(a+a*sec(d*x+c))^(5/2)*(A+B*sec(d*x+c)),x, algorithm="maxima")

[Out]

Timed out

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Fricas [A]  time = 0.479549, size = 292, normalized size = 2.12 \begin{align*} \frac{2 \,{\left ({\left (301 \, A + 230 \, B\right )} a^{2} \cos \left (d x + c\right )^{3} +{\left (98 \, A + 115 \, B\right )} a^{2} \cos \left (d x + c\right )^{2} + 3 \,{\left (7 \, A + 20 \, B\right )} a^{2} \cos \left (d x + c\right ) + 15 \, B a^{2}\right )} \sqrt{\frac{a \cos \left (d x + c\right ) + a}{\cos \left (d x + c\right )}} \sin \left (d x + c\right )}{105 \,{\left (d \cos \left (d x + c\right )^{4} + d \cos \left (d x + c\right )^{3}\right )}} \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate(sec(d*x+c)*(a+a*sec(d*x+c))^(5/2)*(A+B*sec(d*x+c)),x, algorithm="fricas")

[Out]

2/105*((301*A + 230*B)*a^2*cos(d*x + c)^3 + (98*A + 115*B)*a^2*cos(d*x + c)^2 + 3*(7*A + 20*B)*a^2*cos(d*x + c
) + 15*B*a^2)*sqrt((a*cos(d*x + c) + a)/cos(d*x + c))*sin(d*x + c)/(d*cos(d*x + c)^4 + d*cos(d*x + c)^3)

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Sympy [F(-1)]  time = 0., size = 0, normalized size = 0. \begin{align*} \text{Timed out} \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate(sec(d*x+c)*(a+a*sec(d*x+c))**(5/2)*(A+B*sec(d*x+c)),x)

[Out]

Timed out

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Giac [A]  time = 5.07358, size = 300, normalized size = 2.17 \begin{align*} -\frac{8 \,{\left (105 \, \sqrt{2} A a^{6} \mathrm{sgn}\left (\cos \left (d x + c\right )\right ) + 105 \, \sqrt{2} B a^{6} \mathrm{sgn}\left (\cos \left (d x + c\right )\right ) -{\left (245 \, \sqrt{2} A a^{6} \mathrm{sgn}\left (\cos \left (d x + c\right )\right ) + 175 \, \sqrt{2} B a^{6} \mathrm{sgn}\left (\cos \left (d x + c\right )\right ) - 4 \,{\left (49 \, \sqrt{2} A a^{6} \mathrm{sgn}\left (\cos \left (d x + c\right )\right ) + 35 \, \sqrt{2} B a^{6} \mathrm{sgn}\left (\cos \left (d x + c\right )\right ) - 2 \,{\left (7 \, \sqrt{2} A a^{6} \mathrm{sgn}\left (\cos \left (d x + c\right )\right ) + 5 \, \sqrt{2} B a^{6} \mathrm{sgn}\left (\cos \left (d x + c\right )\right )\right )} \tan \left (\frac{1}{2} \, d x + \frac{1}{2} \, c\right )^{2}\right )} \tan \left (\frac{1}{2} \, d x + \frac{1}{2} \, c\right )^{2}\right )} \tan \left (\frac{1}{2} \, d x + \frac{1}{2} \, c\right )^{2}\right )} \tan \left (\frac{1}{2} \, d x + \frac{1}{2} \, c\right )}{105 \,{\left (a \tan \left (\frac{1}{2} \, d x + \frac{1}{2} \, c\right )^{2} - a\right )}^{3} \sqrt{-a \tan \left (\frac{1}{2} \, d x + \frac{1}{2} \, c\right )^{2} + a} d} \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate(sec(d*x+c)*(a+a*sec(d*x+c))^(5/2)*(A+B*sec(d*x+c)),x, algorithm="giac")

[Out]

-8/105*(105*sqrt(2)*A*a^6*sgn(cos(d*x + c)) + 105*sqrt(2)*B*a^6*sgn(cos(d*x + c)) - (245*sqrt(2)*A*a^6*sgn(cos
(d*x + c)) + 175*sqrt(2)*B*a^6*sgn(cos(d*x + c)) - 4*(49*sqrt(2)*A*a^6*sgn(cos(d*x + c)) + 35*sqrt(2)*B*a^6*sg
n(cos(d*x + c)) - 2*(7*sqrt(2)*A*a^6*sgn(cos(d*x + c)) + 5*sqrt(2)*B*a^6*sgn(cos(d*x + c)))*tan(1/2*d*x + 1/2*
c)^2)*tan(1/2*d*x + 1/2*c)^2)*tan(1/2*d*x + 1/2*c)^2)*tan(1/2*d*x + 1/2*c)/((a*tan(1/2*d*x + 1/2*c)^2 - a)^3*s
qrt(-a*tan(1/2*d*x + 1/2*c)^2 + a)*d)