3.127 \(\int \sec ^3(c+d x) (a+a \sec (c+d x))^{3/2} (A+B \sec (c+d x)) \, dx\)

Optimal. Leaf size=189 \[ \frac{2 a^2 (9 A+10 B) \tan (c+d x) \sec ^3(c+d x)}{63 d \sqrt{a \sec (c+d x)+a}}+\frac{2 a^2 (39 A+34 B) \tan (c+d x)}{45 d \sqrt{a \sec (c+d x)+a}}+\frac{2 (39 A+34 B) \tan (c+d x) (a \sec (c+d x)+a)^{3/2}}{105 d}-\frac{4 a (39 A+34 B) \tan (c+d x) \sqrt{a \sec (c+d x)+a}}{315 d}+\frac{2 a B \tan (c+d x) \sec ^3(c+d x) \sqrt{a \sec (c+d x)+a}}{9 d} \]

[Out]

(2*a^2*(39*A + 34*B)*Tan[c + d*x])/(45*d*Sqrt[a + a*Sec[c + d*x]]) + (2*a^2*(9*A + 10*B)*Sec[c + d*x]^3*Tan[c
+ d*x])/(63*d*Sqrt[a + a*Sec[c + d*x]]) - (4*a*(39*A + 34*B)*Sqrt[a + a*Sec[c + d*x]]*Tan[c + d*x])/(315*d) +
(2*a*B*Sec[c + d*x]^3*Sqrt[a + a*Sec[c + d*x]]*Tan[c + d*x])/(9*d) + (2*(39*A + 34*B)*(a + a*Sec[c + d*x])^(3/
2)*Tan[c + d*x])/(105*d)

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Rubi [A]  time = 0.461041, antiderivative size = 189, normalized size of antiderivative = 1., number of steps used = 5, number of rules used = 5, integrand size = 33, \(\frac{\text{number of rules}}{\text{integrand size}}\) = 0.152, Rules used = {4018, 4016, 3800, 4001, 3792} \[ \frac{2 a^2 (9 A+10 B) \tan (c+d x) \sec ^3(c+d x)}{63 d \sqrt{a \sec (c+d x)+a}}+\frac{2 a^2 (39 A+34 B) \tan (c+d x)}{45 d \sqrt{a \sec (c+d x)+a}}+\frac{2 (39 A+34 B) \tan (c+d x) (a \sec (c+d x)+a)^{3/2}}{105 d}-\frac{4 a (39 A+34 B) \tan (c+d x) \sqrt{a \sec (c+d x)+a}}{315 d}+\frac{2 a B \tan (c+d x) \sec ^3(c+d x) \sqrt{a \sec (c+d x)+a}}{9 d} \]

Antiderivative was successfully verified.

[In]

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

[Out]

(2*a^2*(39*A + 34*B)*Tan[c + d*x])/(45*d*Sqrt[a + a*Sec[c + d*x]]) + (2*a^2*(9*A + 10*B)*Sec[c + d*x]^3*Tan[c
+ d*x])/(63*d*Sqrt[a + a*Sec[c + d*x]]) - (4*a*(39*A + 34*B)*Sqrt[a + a*Sec[c + d*x]]*Tan[c + d*x])/(315*d) +
(2*a*B*Sec[c + d*x]^3*Sqrt[a + a*Sec[c + d*x]]*Tan[c + d*x])/(9*d) + (2*(39*A + 34*B)*(a + a*Sec[c + d*x])^(3/
2)*Tan[c + d*x])/(105*d)

Rule 4018

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

Rule 4016

Int[(csc[(e_.) + (f_.)*(x_)]*(d_.))^(n_)*Sqrt[csc[(e_.) + (f_.)*(x_)]*(b_.) + (a_)]*(csc[(e_.) + (f_.)*(x_)]*(
B_.) + (A_)), x_Symbol] :> Simp[(-2*b*B*Cot[e + f*x]*(d*Csc[e + f*x])^n)/(f*(2*n + 1)*Sqrt[a + b*Csc[e + f*x]]
), x] + Dist[(A*b*(2*n + 1) + 2*a*B*n)/(b*(2*n + 1)), Int[Sqrt[a + b*Csc[e + f*x]]*(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] && NeQ[A*b*(2*n + 1) + 2*a*B*n
, 0] &&  !LtQ[n, 0]

Rule 3800

Int[csc[(e_.) + (f_.)*(x_)]^3*(csc[(e_.) + (f_.)*(x_)]*(b_.) + (a_))^(m_), x_Symbol] :> -Simp[(Cot[e + f*x]*(a
+ b*Csc[e + f*x])^(m + 1))/(b*f*(m + 2)), x] + Dist[1/(b*(m + 2)), Int[Csc[e + f*x]*(a + b*Csc[e + f*x])^m*(b
*(m + 1) - a*Csc[e + f*x]), x], x] /; FreeQ[{a, b, e, f, m}, x] && EqQ[a^2 - b^2, 0] &&  !LtQ[m, -2^(-1)]

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 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 ^3(c+d x) (a+a \sec (c+d x))^{3/2} (A+B \sec (c+d x)) \, dx &=\frac{2 a B \sec ^3(c+d x) \sqrt{a+a \sec (c+d x)} \tan (c+d x)}{9 d}+\frac{2}{9} \int \sec ^3(c+d x) \sqrt{a+a \sec (c+d x)} \left (\frac{3}{2} a (3 A+2 B)+\frac{1}{2} a (9 A+10 B) \sec (c+d x)\right ) \, dx\\ &=\frac{2 a^2 (9 A+10 B) \sec ^3(c+d x) \tan (c+d x)}{63 d \sqrt{a+a \sec (c+d x)}}+\frac{2 a B \sec ^3(c+d x) \sqrt{a+a \sec (c+d x)} \tan (c+d x)}{9 d}+\frac{1}{21} (a (39 A+34 B)) \int \sec ^3(c+d x) \sqrt{a+a \sec (c+d x)} \, dx\\ &=\frac{2 a^2 (9 A+10 B) \sec ^3(c+d x) \tan (c+d x)}{63 d \sqrt{a+a \sec (c+d x)}}+\frac{2 a B \sec ^3(c+d x) \sqrt{a+a \sec (c+d x)} \tan (c+d x)}{9 d}+\frac{2 (39 A+34 B) (a+a \sec (c+d x))^{3/2} \tan (c+d x)}{105 d}+\frac{1}{105} (2 (39 A+34 B)) \int \sec (c+d x) \left (\frac{3 a}{2}-a \sec (c+d x)\right ) \sqrt{a+a \sec (c+d x)} \, dx\\ &=\frac{2 a^2 (9 A+10 B) \sec ^3(c+d x) \tan (c+d x)}{63 d \sqrt{a+a \sec (c+d x)}}-\frac{4 a (39 A+34 B) \sqrt{a+a \sec (c+d x)} \tan (c+d x)}{315 d}+\frac{2 a B \sec ^3(c+d x) \sqrt{a+a \sec (c+d x)} \tan (c+d x)}{9 d}+\frac{2 (39 A+34 B) (a+a \sec (c+d x))^{3/2} \tan (c+d x)}{105 d}+\frac{1}{45} (a (39 A+34 B)) \int \sec (c+d x) \sqrt{a+a \sec (c+d x)} \, dx\\ &=\frac{2 a^2 (39 A+34 B) \tan (c+d x)}{45 d \sqrt{a+a \sec (c+d x)}}+\frac{2 a^2 (9 A+10 B) \sec ^3(c+d x) \tan (c+d x)}{63 d \sqrt{a+a \sec (c+d x)}}-\frac{4 a (39 A+34 B) \sqrt{a+a \sec (c+d x)} \tan (c+d x)}{315 d}+\frac{2 a B \sec ^3(c+d x) \sqrt{a+a \sec (c+d x)} \tan (c+d x)}{9 d}+\frac{2 (39 A+34 B) (a+a \sec (c+d x))^{3/2} \tan (c+d x)}{105 d}\\ \end{align*}

Mathematica [A]  time = 0.734687, size = 100, normalized size = 0.53 \[ \frac{2 a^2 \tan (c+d x) \left (5 (9 A+17 B) \sec ^3(c+d x)+3 (39 A+34 B) \sec ^2(c+d x)+4 (39 A+34 B) \sec (c+d x)+8 (39 A+34 B)+35 B \sec ^4(c+d x)\right )}{315 d \sqrt{a (\sec (c+d x)+1)}} \]

Antiderivative was successfully verified.

[In]

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

[Out]

(2*a^2*(8*(39*A + 34*B) + 4*(39*A + 34*B)*Sec[c + d*x] + 3*(39*A + 34*B)*Sec[c + d*x]^2 + 5*(9*A + 17*B)*Sec[c
+ d*x]^3 + 35*B*Sec[c + d*x]^4)*Tan[c + d*x])/(315*d*Sqrt[a*(1 + Sec[c + d*x])])

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Maple [A]  time = 0.282, size = 139, normalized size = 0.7 \begin{align*} -{\frac{2\,a \left ( -1+\cos \left ( dx+c \right ) \right ) \left ( 312\,A \left ( \cos \left ( dx+c \right ) \right ) ^{4}+272\,B \left ( \cos \left ( dx+c \right ) \right ) ^{4}+156\,A \left ( \cos \left ( dx+c \right ) \right ) ^{3}+136\,B \left ( \cos \left ( dx+c \right ) \right ) ^{3}+117\,A \left ( \cos \left ( dx+c \right ) \right ) ^{2}+102\,B \left ( \cos \left ( dx+c \right ) \right ) ^{2}+45\,A\cos \left ( dx+c \right ) +85\,B\cos \left ( dx+c \right ) +35\,B \right ) }{315\,d \left ( \cos \left ( dx+c \right ) \right ) ^{4}\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)^3*(a+a*sec(d*x+c))^(3/2)*(A+B*sec(d*x+c)),x)

[Out]

-2/315/d*a*(-1+cos(d*x+c))*(312*A*cos(d*x+c)^4+272*B*cos(d*x+c)^4+156*A*cos(d*x+c)^3+136*B*cos(d*x+c)^3+117*A*
cos(d*x+c)^2+102*B*cos(d*x+c)^2+45*A*cos(d*x+c)+85*B*cos(d*x+c)+35*B)*(a*(cos(d*x+c)+1)/cos(d*x+c))^(1/2)/cos(
d*x+c)^4/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)^3*(a+a*sec(d*x+c))^(3/2)*(A+B*sec(d*x+c)),x, algorithm="maxima")

[Out]

Timed out

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Fricas [A]  time = 0.486077, size = 329, normalized size = 1.74 \begin{align*} \frac{2 \,{\left (8 \,{\left (39 \, A + 34 \, B\right )} a \cos \left (d x + c\right )^{4} + 4 \,{\left (39 \, A + 34 \, B\right )} a \cos \left (d x + c\right )^{3} + 3 \,{\left (39 \, A + 34 \, B\right )} a \cos \left (d x + c\right )^{2} + 5 \,{\left (9 \, A + 17 \, B\right )} a \cos \left (d x + c\right ) + 35 \, B a\right )} \sqrt{\frac{a \cos \left (d x + c\right ) + a}{\cos \left (d x + c\right )}} \sin \left (d x + c\right )}{315 \,{\left (d \cos \left (d x + c\right )^{5} + d \cos \left (d x + c\right )^{4}\right )}} \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

2/315*(8*(39*A + 34*B)*a*cos(d*x + c)^4 + 4*(39*A + 34*B)*a*cos(d*x + c)^3 + 3*(39*A + 34*B)*a*cos(d*x + c)^2
+ 5*(9*A + 17*B)*a*cos(d*x + c) + 35*B*a)*sqrt((a*cos(d*x + c) + a)/cos(d*x + c))*sin(d*x + c)/(d*cos(d*x + c)
^5 + d*cos(d*x + c)^4)

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

[Out]

Timed out

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Giac [A]  time = 5.12644, size = 362, normalized size = 1.92 \begin{align*} \frac{4 \,{\left (315 \, \sqrt{2} A a^{6} \mathrm{sgn}\left (\cos \left (d x + c\right )\right ) + 315 \, \sqrt{2} B a^{6} \mathrm{sgn}\left (\cos \left (d x + c\right )\right ) -{\left (735 \, \sqrt{2} A a^{6} \mathrm{sgn}\left (\cos \left (d x + c\right )\right ) + 525 \, \sqrt{2} B a^{6} \mathrm{sgn}\left (\cos \left (d x + c\right )\right ) -{\left (819 \, \sqrt{2} A a^{6} \mathrm{sgn}\left (\cos \left (d x + c\right )\right ) + 819 \, \sqrt{2} B a^{6} \mathrm{sgn}\left (\cos \left (d x + c\right )\right ) -{\left (513 \, \sqrt{2} A a^{6} \mathrm{sgn}\left (\cos \left (d x + c\right )\right ) + 423 \, \sqrt{2} B a^{6} \mathrm{sgn}\left (\cos \left (d x + c\right )\right ) - 2 \,{\left (57 \, \sqrt{2} A a^{6} \mathrm{sgn}\left (\cos \left (d x + c\right )\right ) + 47 \, \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 )^{2}\right )} \tan \left (\frac{1}{2} \, d x + \frac{1}{2} \, c\right )}{315 \,{\left (a \tan \left (\frac{1}{2} \, d x + \frac{1}{2} \, c\right )^{2} - a\right )}^{4} \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)^3*(a+a*sec(d*x+c))^(3/2)*(A+B*sec(d*x+c)),x, algorithm="giac")

[Out]

4/315*(315*sqrt(2)*A*a^6*sgn(cos(d*x + c)) + 315*sqrt(2)*B*a^6*sgn(cos(d*x + c)) - (735*sqrt(2)*A*a^6*sgn(cos(
d*x + c)) + 525*sqrt(2)*B*a^6*sgn(cos(d*x + c)) - (819*sqrt(2)*A*a^6*sgn(cos(d*x + c)) + 819*sqrt(2)*B*a^6*sgn
(cos(d*x + c)) - (513*sqrt(2)*A*a^6*sgn(cos(d*x + c)) + 423*sqrt(2)*B*a^6*sgn(cos(d*x + c)) - 2*(57*sqrt(2)*A*
a^6*sgn(cos(d*x + c)) + 47*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)*ta
n(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)^4*sqrt(-a*t
an(1/2*d*x + 1/2*c)^2 + a)*d)