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

Optimal. Leaf size=225 \[ \frac{2 a^2 (33 A+28 C) \tan (c+d x) \sec ^3(c+d x)}{231 d \sqrt{a \sec (c+d x)+a}}+\frac{2 a^2 (143 A+112 C) \tan (c+d x)}{165 d \sqrt{a \sec (c+d x)+a}}+\frac{2 (143 A+112 C) \tan (c+d x) (a \sec (c+d x)+a)^{3/2}}{385 d}-\frac{4 a (143 A+112 C) \tan (c+d x) \sqrt{a \sec (c+d x)+a}}{1155 d}+\frac{2 C \tan (c+d x) \sec ^3(c+d x) (a \sec (c+d x)+a)^{3/2}}{11 d}+\frac{2 a C \tan (c+d x) \sec ^3(c+d x) \sqrt{a \sec (c+d x)+a}}{33 d} \]

[Out]

(2*a^2*(143*A + 112*C)*Tan[c + d*x])/(165*d*Sqrt[a + a*Sec[c + d*x]]) + (2*a^2*(33*A + 28*C)*Sec[c + d*x]^3*Ta
n[c + d*x])/(231*d*Sqrt[a + a*Sec[c + d*x]]) - (4*a*(143*A + 112*C)*Sqrt[a + a*Sec[c + d*x]]*Tan[c + d*x])/(11
55*d) + (2*a*C*Sec[c + d*x]^3*Sqrt[a + a*Sec[c + d*x]]*Tan[c + d*x])/(33*d) + (2*(143*A + 112*C)*(a + a*Sec[c
+ d*x])^(3/2)*Tan[c + d*x])/(385*d) + (2*C*Sec[c + d*x]^3*(a + a*Sec[c + d*x])^(3/2)*Tan[c + d*x])/(11*d)

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Rubi [A]  time = 0.654957, antiderivative size = 225, normalized size of antiderivative = 1., number of steps used = 6, number of rules used = 6, integrand size = 35, \(\frac{\text{number of rules}}{\text{integrand size}}\) = 0.171, Rules used = {4089, 4018, 4016, 3800, 4001, 3792} \[ \frac{2 a^2 (33 A+28 C) \tan (c+d x) \sec ^3(c+d x)}{231 d \sqrt{a \sec (c+d x)+a}}+\frac{2 a^2 (143 A+112 C) \tan (c+d x)}{165 d \sqrt{a \sec (c+d x)+a}}+\frac{2 (143 A+112 C) \tan (c+d x) (a \sec (c+d x)+a)^{3/2}}{385 d}-\frac{4 a (143 A+112 C) \tan (c+d x) \sqrt{a \sec (c+d x)+a}}{1155 d}+\frac{2 C \tan (c+d x) \sec ^3(c+d x) (a \sec (c+d x)+a)^{3/2}}{11 d}+\frac{2 a C \tan (c+d x) \sec ^3(c+d x) \sqrt{a \sec (c+d x)+a}}{33 d} \]

Antiderivative was successfully verified.

[In]

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

[Out]

(2*a^2*(143*A + 112*C)*Tan[c + d*x])/(165*d*Sqrt[a + a*Sec[c + d*x]]) + (2*a^2*(33*A + 28*C)*Sec[c + d*x]^3*Ta
n[c + d*x])/(231*d*Sqrt[a + a*Sec[c + d*x]]) - (4*a*(143*A + 112*C)*Sqrt[a + a*Sec[c + d*x]]*Tan[c + d*x])/(11
55*d) + (2*a*C*Sec[c + d*x]^3*Sqrt[a + a*Sec[c + d*x]]*Tan[c + d*x])/(33*d) + (2*(143*A + 112*C)*(a + a*Sec[c
+ d*x])^(3/2)*Tan[c + d*x])/(385*d) + (2*C*Sec[c + d*x]^3*(a + a*Sec[c + d*x])^(3/2)*Tan[c + d*x])/(11*d)

Rule 4089

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

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} \left (A+C \sec ^2(c+d x)\right ) \, dx &=\frac{2 C \sec ^3(c+d x) (a+a \sec (c+d x))^{3/2} \tan (c+d x)}{11 d}+\frac{2 \int \sec ^3(c+d x) (a+a \sec (c+d x))^{3/2} \left (\frac{1}{2} a (11 A+6 C)+\frac{3}{2} a C \sec (c+d x)\right ) \, dx}{11 a}\\ &=\frac{2 a C \sec ^3(c+d x) \sqrt{a+a \sec (c+d x)} \tan (c+d x)}{33 d}+\frac{2 C \sec ^3(c+d x) (a+a \sec (c+d x))^{3/2} \tan (c+d x)}{11 d}+\frac{4 \int \sec ^3(c+d x) \sqrt{a+a \sec (c+d x)} \left (\frac{9}{4} a^2 (11 A+8 C)+\frac{3}{4} a^2 (33 A+28 C) \sec (c+d x)\right ) \, dx}{99 a}\\ &=\frac{2 a^2 (33 A+28 C) \sec ^3(c+d x) \tan (c+d x)}{231 d \sqrt{a+a \sec (c+d x)}}+\frac{2 a C \sec ^3(c+d x) \sqrt{a+a \sec (c+d x)} \tan (c+d x)}{33 d}+\frac{2 C \sec ^3(c+d x) (a+a \sec (c+d x))^{3/2} \tan (c+d x)}{11 d}+\frac{1}{77} (a (143 A+112 C)) \int \sec ^3(c+d x) \sqrt{a+a \sec (c+d x)} \, dx\\ &=\frac{2 a^2 (33 A+28 C) \sec ^3(c+d x) \tan (c+d x)}{231 d \sqrt{a+a \sec (c+d x)}}+\frac{2 a C \sec ^3(c+d x) \sqrt{a+a \sec (c+d x)} \tan (c+d x)}{33 d}+\frac{2 (143 A+112 C) (a+a \sec (c+d x))^{3/2} \tan (c+d x)}{385 d}+\frac{2 C \sec ^3(c+d x) (a+a \sec (c+d x))^{3/2} \tan (c+d x)}{11 d}+\frac{1}{385} (2 (143 A+112 C)) \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 (33 A+28 C) \sec ^3(c+d x) \tan (c+d x)}{231 d \sqrt{a+a \sec (c+d x)}}-\frac{4 a (143 A+112 C) \sqrt{a+a \sec (c+d x)} \tan (c+d x)}{1155 d}+\frac{2 a C \sec ^3(c+d x) \sqrt{a+a \sec (c+d x)} \tan (c+d x)}{33 d}+\frac{2 (143 A+112 C) (a+a \sec (c+d x))^{3/2} \tan (c+d x)}{385 d}+\frac{2 C \sec ^3(c+d x) (a+a \sec (c+d x))^{3/2} \tan (c+d x)}{11 d}+\frac{1}{165} (a (143 A+112 C)) \int \sec (c+d x) \sqrt{a+a \sec (c+d x)} \, dx\\ &=\frac{2 a^2 (143 A+112 C) \tan (c+d x)}{165 d \sqrt{a+a \sec (c+d x)}}+\frac{2 a^2 (33 A+28 C) \sec ^3(c+d x) \tan (c+d x)}{231 d \sqrt{a+a \sec (c+d x)}}-\frac{4 a (143 A+112 C) \sqrt{a+a \sec (c+d x)} \tan (c+d x)}{1155 d}+\frac{2 a C \sec ^3(c+d x) \sqrt{a+a \sec (c+d x)} \tan (c+d x)}{33 d}+\frac{2 (143 A+112 C) (a+a \sec (c+d x))^{3/2} \tan (c+d x)}{385 d}+\frac{2 C \sec ^3(c+d x) (a+a \sec (c+d x))^{3/2} \tan (c+d x)}{11 d}\\ \end{align*}

Mathematica [A]  time = 1.365, size = 144, normalized size = 0.64 \[ \frac{a \tan \left (\frac{1}{2} (c+d x)\right ) \sec ^5(c+d x) \sqrt{a (\sec (c+d x)+1)} ((4147 A+4228 C) \cos (c+d x)+2 (737 A+728 C) \cos (2 (c+d x))+1859 A \cos (3 (c+d x))+286 A \cos (4 (c+d x))+286 A \cos (5 (c+d x))+1188 A+1456 C \cos (3 (c+d x))+224 C \cos (4 (c+d x))+224 C \cos (5 (c+d x))+1652 C)}{2310 d} \]

Antiderivative was successfully verified.

[In]

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

[Out]

(a*(1188*A + 1652*C + (4147*A + 4228*C)*Cos[c + d*x] + 2*(737*A + 728*C)*Cos[2*(c + d*x)] + 1859*A*Cos[3*(c +
d*x)] + 1456*C*Cos[3*(c + d*x)] + 286*A*Cos[4*(c + d*x)] + 224*C*Cos[4*(c + d*x)] + 286*A*Cos[5*(c + d*x)] + 2
24*C*Cos[5*(c + d*x)])*Sec[c + d*x]^5*Sqrt[a*(1 + Sec[c + d*x])]*Tan[(c + d*x)/2])/(2310*d)

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Maple [A]  time = 0.327, size = 152, normalized size = 0.7 \begin{align*} -{\frac{2\,a \left ( -1+\cos \left ( dx+c \right ) \right ) \left ( 1144\,A \left ( \cos \left ( dx+c \right ) \right ) ^{5}+896\,C \left ( \cos \left ( dx+c \right ) \right ) ^{5}+572\,A \left ( \cos \left ( dx+c \right ) \right ) ^{4}+448\,C \left ( \cos \left ( dx+c \right ) \right ) ^{4}+429\,A \left ( \cos \left ( dx+c \right ) \right ) ^{3}+336\,C \left ( \cos \left ( dx+c \right ) \right ) ^{3}+165\,A \left ( \cos \left ( dx+c \right ) \right ) ^{2}+280\,C \left ( \cos \left ( dx+c \right ) \right ) ^{2}+245\,C\cos \left ( dx+c \right ) +105\,C \right ) }{1155\,d \left ( \cos \left ( dx+c \right ) \right ) ^{5}\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+C*sec(d*x+c)^2),x)

[Out]

-2/1155/d*a*(-1+cos(d*x+c))*(1144*A*cos(d*x+c)^5+896*C*cos(d*x+c)^5+572*A*cos(d*x+c)^4+448*C*cos(d*x+c)^4+429*
A*cos(d*x+c)^3+336*C*cos(d*x+c)^3+165*A*cos(d*x+c)^2+280*C*cos(d*x+c)^2+245*C*cos(d*x+c)+105*C)*(a*(cos(d*x+c)
+1)/cos(d*x+c))^(1/2)/cos(d*x+c)^5/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+C*sec(d*x+c)^2),x, algorithm="maxima")

[Out]

Timed out

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Fricas [A]  time = 0.515857, size = 375, normalized size = 1.67 \begin{align*} \frac{2 \,{\left (8 \,{\left (143 \, A + 112 \, C\right )} a \cos \left (d x + c\right )^{5} + 4 \,{\left (143 \, A + 112 \, C\right )} a \cos \left (d x + c\right )^{4} + 3 \,{\left (143 \, A + 112 \, C\right )} a \cos \left (d x + c\right )^{3} + 5 \,{\left (33 \, A + 56 \, C\right )} a \cos \left (d x + c\right )^{2} + 245 \, C a \cos \left (d x + c\right ) + 105 \, C a\right )} \sqrt{\frac{a \cos \left (d x + c\right ) + a}{\cos \left (d x + c\right )}} \sin \left (d x + c\right )}{1155 \,{\left (d \cos \left (d x + c\right )^{6} + d \cos \left (d x + c\right )^{5}\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+C*sec(d*x+c)^2),x, algorithm="fricas")

[Out]

2/1155*(8*(143*A + 112*C)*a*cos(d*x + c)^5 + 4*(143*A + 112*C)*a*cos(d*x + c)^4 + 3*(143*A + 112*C)*a*cos(d*x
+ c)^3 + 5*(33*A + 56*C)*a*cos(d*x + c)^2 + 245*C*a*cos(d*x + c) + 105*C*a)*sqrt((a*cos(d*x + c) + a)/cos(d*x
+ c))*sin(d*x + c)/(d*cos(d*x + c)^6 + d*cos(d*x + c)^5)

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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+C*sec(d*x+c)**2),x)

[Out]

Timed out

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Giac [A]  time = 4.82788, size = 424, normalized size = 1.88 \begin{align*} -\frac{4 \,{\left (1155 \, \sqrt{2} A a^{7} \mathrm{sgn}\left (\cos \left (d x + c\right )\right ) + 1155 \, \sqrt{2} C a^{7} \mathrm{sgn}\left (\cos \left (d x + c\right )\right ) -{\left (3850 \, \sqrt{2} A a^{7} \mathrm{sgn}\left (\cos \left (d x + c\right )\right ) + 2310 \, \sqrt{2} C a^{7} \mathrm{sgn}\left (\cos \left (d x + c\right )\right ) -{\left (5698 \, \sqrt{2} A a^{7} \mathrm{sgn}\left (\cos \left (d x + c\right )\right ) + 5082 \, \sqrt{2} C a^{7} \mathrm{sgn}\left (\cos \left (d x + c\right )\right ) -{\left (4884 \, \sqrt{2} A a^{7} \mathrm{sgn}\left (\cos \left (d x + c\right )\right ) + 3696 \, \sqrt{2} C a^{7} \mathrm{sgn}\left (\cos \left (d x + c\right )\right ) -{\left (2299 \, \sqrt{2} A a^{7} \mathrm{sgn}\left (\cos \left (d x + c\right )\right ) + 1771 \, \sqrt{2} C a^{7} \mathrm{sgn}\left (\cos \left (d x + c\right )\right ) - 2 \,{\left (209 \, \sqrt{2} A a^{7} \mathrm{sgn}\left (\cos \left (d x + c\right )\right ) + 161 \, \sqrt{2} C a^{7} \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 )^{2}\right )} \tan \left (\frac{1}{2} \, d x + \frac{1}{2} \, c\right )}{1155 \,{\left (a \tan \left (\frac{1}{2} \, d x + \frac{1}{2} \, c\right )^{2} - a\right )}^{5} \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+C*sec(d*x+c)^2),x, algorithm="giac")

[Out]

-4/1155*(1155*sqrt(2)*A*a^7*sgn(cos(d*x + c)) + 1155*sqrt(2)*C*a^7*sgn(cos(d*x + c)) - (3850*sqrt(2)*A*a^7*sgn
(cos(d*x + c)) + 2310*sqrt(2)*C*a^7*sgn(cos(d*x + c)) - (5698*sqrt(2)*A*a^7*sgn(cos(d*x + c)) + 5082*sqrt(2)*C
*a^7*sgn(cos(d*x + c)) - (4884*sqrt(2)*A*a^7*sgn(cos(d*x + c)) + 3696*sqrt(2)*C*a^7*sgn(cos(d*x + c)) - (2299*
sqrt(2)*A*a^7*sgn(cos(d*x + c)) + 1771*sqrt(2)*C*a^7*sgn(cos(d*x + c)) - 2*(209*sqrt(2)*A*a^7*sgn(cos(d*x + c)
) + 161*sqrt(2)*C*a^7*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)^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)^5*sqrt
(-a*tan(1/2*d*x + 1/2*c)^2 + a)*d)