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

Optimal. Leaf size=211 \[ \frac{16 a^2 (33 A+25 C) \tan (c+d x) \sqrt{a \sec (c+d x)+a}}{693 d}+\frac{64 a^3 (33 A+25 C) \tan (c+d x)}{693 d \sqrt{a \sec (c+d x)+a}}+\frac{2 (99 A+26 C) \tan (c+d x) (a \sec (c+d x)+a)^{5/2}}{693 d}+\frac{2 a (33 A+25 C) \tan (c+d x) (a \sec (c+d x)+a)^{3/2}}{231 d}+\frac{2 C \tan (c+d x) \sec ^2(c+d x) (a \sec (c+d x)+a)^{5/2}}{11 d}+\frac{10 C \tan (c+d x) (a \sec (c+d x)+a)^{7/2}}{99 a d} \]

[Out]

(64*a^3*(33*A + 25*C)*Tan[c + d*x])/(693*d*Sqrt[a + a*Sec[c + d*x]]) + (16*a^2*(33*A + 25*C)*Sqrt[a + a*Sec[c
+ d*x]]*Tan[c + d*x])/(693*d) + (2*a*(33*A + 25*C)*(a + a*Sec[c + d*x])^(3/2)*Tan[c + d*x])/(231*d) + (2*(99*A
+ 26*C)*(a + a*Sec[c + d*x])^(5/2)*Tan[c + d*x])/(693*d) + (2*C*Sec[c + d*x]^2*(a + a*Sec[c + d*x])^(5/2)*Tan
[c + d*x])/(11*d) + (10*C*(a + a*Sec[c + d*x])^(7/2)*Tan[c + d*x])/(99*a*d)

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Rubi [A]  time = 0.532672, antiderivative size = 211, normalized size of antiderivative = 1., number of steps used = 6, number of rules used = 5, integrand size = 35, \(\frac{\text{number of rules}}{\text{integrand size}}\) = 0.143, Rules used = {4089, 4010, 4001, 3793, 3792} \[ \frac{16 a^2 (33 A+25 C) \tan (c+d x) \sqrt{a \sec (c+d x)+a}}{693 d}+\frac{64 a^3 (33 A+25 C) \tan (c+d x)}{693 d \sqrt{a \sec (c+d x)+a}}+\frac{2 (99 A+26 C) \tan (c+d x) (a \sec (c+d x)+a)^{5/2}}{693 d}+\frac{2 a (33 A+25 C) \tan (c+d x) (a \sec (c+d x)+a)^{3/2}}{231 d}+\frac{2 C \tan (c+d x) \sec ^2(c+d x) (a \sec (c+d x)+a)^{5/2}}{11 d}+\frac{10 C \tan (c+d x) (a \sec (c+d x)+a)^{7/2}}{99 a d} \]

Antiderivative was successfully verified.

[In]

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

[Out]

(64*a^3*(33*A + 25*C)*Tan[c + d*x])/(693*d*Sqrt[a + a*Sec[c + d*x]]) + (16*a^2*(33*A + 25*C)*Sqrt[a + a*Sec[c
+ d*x]]*Tan[c + d*x])/(693*d) + (2*a*(33*A + 25*C)*(a + a*Sec[c + d*x])^(3/2)*Tan[c + d*x])/(231*d) + (2*(99*A
+ 26*C)*(a + a*Sec[c + d*x])^(5/2)*Tan[c + d*x])/(693*d) + (2*C*Sec[c + d*x]^2*(a + a*Sec[c + d*x])^(5/2)*Tan
[c + d*x])/(11*d) + (10*C*(a + a*Sec[c + d*x])^(7/2)*Tan[c + d*x])/(99*a*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 4010

Int[csc[(e_.) + (f_.)*(x_)]^2*(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 + 1))/(b*f*(m + 2)), x] + Dist[1/(b*(m + 2)), I
nt[Csc[e + f*x]*(a + b*Csc[e + f*x])^m*Simp[b*B*(m + 1) + (A*b*(m + 2) - a*B)*Csc[e + f*x], x], x], x] /; Free
Q[{a, b, e, f, A, B, m}, x] && NeQ[A*b - a*B, 0] &&  !LtQ[m, -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 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 ^2(c+d x) (a+a \sec (c+d x))^{5/2} \left (A+C \sec ^2(c+d x)\right ) \, dx &=\frac{2 C \sec ^2(c+d x) (a+a \sec (c+d x))^{5/2} \tan (c+d x)}{11 d}+\frac{2 \int \sec ^2(c+d x) (a+a \sec (c+d x))^{5/2} \left (\frac{1}{2} a (11 A+4 C)+\frac{5}{2} a C \sec (c+d x)\right ) \, dx}{11 a}\\ &=\frac{2 C \sec ^2(c+d x) (a+a \sec (c+d x))^{5/2} \tan (c+d x)}{11 d}+\frac{10 C (a+a \sec (c+d x))^{7/2} \tan (c+d x)}{99 a d}+\frac{4 \int \sec (c+d x) (a+a \sec (c+d x))^{5/2} \left (\frac{35 a^2 C}{4}+\frac{1}{4} a^2 (99 A+26 C) \sec (c+d x)\right ) \, dx}{99 a^2}\\ &=\frac{2 (99 A+26 C) (a+a \sec (c+d x))^{5/2} \tan (c+d x)}{693 d}+\frac{2 C \sec ^2(c+d x) (a+a \sec (c+d x))^{5/2} \tan (c+d x)}{11 d}+\frac{10 C (a+a \sec (c+d x))^{7/2} \tan (c+d x)}{99 a d}+\frac{1}{231} (5 (33 A+25 C)) \int \sec (c+d x) (a+a \sec (c+d x))^{5/2} \, dx\\ &=\frac{2 a (33 A+25 C) (a+a \sec (c+d x))^{3/2} \tan (c+d x)}{231 d}+\frac{2 (99 A+26 C) (a+a \sec (c+d x))^{5/2} \tan (c+d x)}{693 d}+\frac{2 C \sec ^2(c+d x) (a+a \sec (c+d x))^{5/2} \tan (c+d x)}{11 d}+\frac{10 C (a+a \sec (c+d x))^{7/2} \tan (c+d x)}{99 a d}+\frac{1}{231} (8 a (33 A+25 C)) \int \sec (c+d x) (a+a \sec (c+d x))^{3/2} \, dx\\ &=\frac{16 a^2 (33 A+25 C) \sqrt{a+a \sec (c+d x)} \tan (c+d x)}{693 d}+\frac{2 a (33 A+25 C) (a+a \sec (c+d x))^{3/2} \tan (c+d x)}{231 d}+\frac{2 (99 A+26 C) (a+a \sec (c+d x))^{5/2} \tan (c+d x)}{693 d}+\frac{2 C \sec ^2(c+d x) (a+a \sec (c+d x))^{5/2} \tan (c+d x)}{11 d}+\frac{10 C (a+a \sec (c+d x))^{7/2} \tan (c+d x)}{99 a d}+\frac{1}{693} \left (32 a^2 (33 A+25 C)\right ) \int \sec (c+d x) \sqrt{a+a \sec (c+d x)} \, dx\\ &=\frac{64 a^3 (33 A+25 C) \tan (c+d x)}{693 d \sqrt{a+a \sec (c+d x)}}+\frac{16 a^2 (33 A+25 C) \sqrt{a+a \sec (c+d x)} \tan (c+d x)}{693 d}+\frac{2 a (33 A+25 C) (a+a \sec (c+d x))^{3/2} \tan (c+d x)}{231 d}+\frac{2 (99 A+26 C) (a+a \sec (c+d x))^{5/2} \tan (c+d x)}{693 d}+\frac{2 C \sec ^2(c+d x) (a+a \sec (c+d x))^{5/2} \tan (c+d x)}{11 d}+\frac{10 C (a+a \sec (c+d x))^{7/2} \tan (c+d x)}{99 a d}\\ \end{align*}

Mathematica [A]  time = 1.49389, size = 147, normalized size = 0.7 \[ \frac{a^2 \tan \left (\frac{1}{2} (c+d x)\right ) \sec ^5(c+d x) \sqrt{a (\sec (c+d x)+1)} (2 (4983 A+5014 C) \cos (c+d x)+52 (66 A+71 C) \cos (2 (c+d x))+4587 A \cos (3 (c+d x))+759 A \cos (4 (c+d x))+759 A \cos (5 (c+d x))+2673 A+3692 C \cos (3 (c+d x))+568 C \cos (4 (c+d x))+568 C \cos (5 (c+d x))+3628 C)}{2772 d} \]

Antiderivative was successfully verified.

[In]

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

[Out]

(a^2*(2673*A + 3628*C + 2*(4983*A + 5014*C)*Cos[c + d*x] + 52*(66*A + 71*C)*Cos[2*(c + d*x)] + 4587*A*Cos[3*(c
+ d*x)] + 3692*C*Cos[3*(c + d*x)] + 759*A*Cos[4*(c + d*x)] + 568*C*Cos[4*(c + d*x)] + 759*A*Cos[5*(c + d*x)]
+ 568*C*Cos[5*(c + d*x)])*Sec[c + d*x]^5*Sqrt[a*(1 + Sec[c + d*x])]*Tan[(c + d*x)/2])/(2772*d)

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Maple [A]  time = 0.301, size = 154, normalized size = 0.7 \begin{align*} -{\frac{2\,{a}^{2} \left ( -1+\cos \left ( dx+c \right ) \right ) \left ( 1518\,A \left ( \cos \left ( dx+c \right ) \right ) ^{5}+1136\,C \left ( \cos \left ( dx+c \right ) \right ) ^{5}+759\,A \left ( \cos \left ( dx+c \right ) \right ) ^{4}+568\,C \left ( \cos \left ( dx+c \right ) \right ) ^{4}+396\,A \left ( \cos \left ( dx+c \right ) \right ) ^{3}+426\,C \left ( \cos \left ( dx+c \right ) \right ) ^{3}+99\,A \left ( \cos \left ( dx+c \right ) \right ) ^{2}+355\,C \left ( \cos \left ( dx+c \right ) \right ) ^{2}+224\,C\cos \left ( dx+c \right ) +63\,C \right ) }{693\,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)^2*(a+a*sec(d*x+c))^(5/2)*(A+C*sec(d*x+c)^2),x)

[Out]

-2/693/d*a^2*(-1+cos(d*x+c))*(1518*A*cos(d*x+c)^5+1136*C*cos(d*x+c)^5+759*A*cos(d*x+c)^4+568*C*cos(d*x+c)^4+39
6*A*cos(d*x+c)^3+426*C*cos(d*x+c)^3+99*A*cos(d*x+c)^2+355*C*cos(d*x+c)^2+224*C*cos(d*x+c)+63*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)^2*(a+a*sec(d*x+c))^(5/2)*(A+C*sec(d*x+c)^2),x, algorithm="maxima")

[Out]

Timed out

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Fricas [A]  time = 0.516659, size = 382, normalized size = 1.81 \begin{align*} \frac{2 \,{\left (2 \,{\left (759 \, A + 568 \, C\right )} a^{2} \cos \left (d x + c\right )^{5} +{\left (759 \, A + 568 \, C\right )} a^{2} \cos \left (d x + c\right )^{4} + 6 \,{\left (66 \, A + 71 \, C\right )} a^{2} \cos \left (d x + c\right )^{3} +{\left (99 \, A + 355 \, C\right )} a^{2} \cos \left (d x + c\right )^{2} + 224 \, C a^{2} \cos \left (d x + c\right ) + 63 \, C 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 )}{693 \,{\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)^2*(a+a*sec(d*x+c))^(5/2)*(A+C*sec(d*x+c)^2),x, algorithm="fricas")

[Out]

2/693*(2*(759*A + 568*C)*a^2*cos(d*x + c)^5 + (759*A + 568*C)*a^2*cos(d*x + c)^4 + 6*(66*A + 71*C)*a^2*cos(d*x
+ c)^3 + (99*A + 355*C)*a^2*cos(d*x + c)^2 + 224*C*a^2*cos(d*x + c) + 63*C*a^2)*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)**2*(a+a*sec(d*x+c))**(5/2)*(A+C*sec(d*x+c)**2),x)

[Out]

Timed out

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Giac [A]  time = 5.13425, size = 424, normalized size = 2.01 \begin{align*} -\frac{8 \,{\left (693 \, \sqrt{2} A a^{8} \mathrm{sgn}\left (\cos \left (d x + c\right )\right ) + 693 \, \sqrt{2} C a^{8} \mathrm{sgn}\left (\cos \left (d x + c\right )\right ) -{\left (2541 \, \sqrt{2} A a^{8} \mathrm{sgn}\left (\cos \left (d x + c\right )\right ) + 1617 \, \sqrt{2} C a^{8} \mathrm{sgn}\left (\cos \left (d x + c\right )\right ) -{\left (3927 \, \sqrt{2} A a^{8} \mathrm{sgn}\left (\cos \left (d x + c\right )\right ) + 3003 \, \sqrt{2} C a^{8} \mathrm{sgn}\left (\cos \left (d x + c\right )\right ) -{\left (3267 \, \sqrt{2} A a^{8} \mathrm{sgn}\left (\cos \left (d x + c\right )\right ) + 2475 \, \sqrt{2} C a^{8} \mathrm{sgn}\left (\cos \left (d x + c\right )\right ) - 4 \,{\left (363 \, \sqrt{2} A a^{8} \mathrm{sgn}\left (\cos \left (d x + c\right )\right ) + 275 \, \sqrt{2} C a^{8} \mathrm{sgn}\left (\cos \left (d x + c\right )\right ) - 2 \,{\left (33 \, \sqrt{2} A a^{8} \mathrm{sgn}\left (\cos \left (d x + c\right )\right ) + 25 \, \sqrt{2} C a^{8} \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 )}{693 \,{\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)^2*(a+a*sec(d*x+c))^(5/2)*(A+C*sec(d*x+c)^2),x, algorithm="giac")

[Out]

-8/693*(693*sqrt(2)*A*a^8*sgn(cos(d*x + c)) + 693*sqrt(2)*C*a^8*sgn(cos(d*x + c)) - (2541*sqrt(2)*A*a^8*sgn(co
s(d*x + c)) + 1617*sqrt(2)*C*a^8*sgn(cos(d*x + c)) - (3927*sqrt(2)*A*a^8*sgn(cos(d*x + c)) + 3003*sqrt(2)*C*a^
8*sgn(cos(d*x + c)) - (3267*sqrt(2)*A*a^8*sgn(cos(d*x + c)) + 2475*sqrt(2)*C*a^8*sgn(cos(d*x + c)) - 4*(363*sq
rt(2)*A*a^8*sgn(cos(d*x + c)) + 275*sqrt(2)*C*a^8*sgn(cos(d*x + c)) - 2*(33*sqrt(2)*A*a^8*sgn(cos(d*x + c)) +
25*sqrt(2)*C*a^8*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)*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)^5*sqrt(-a*t
an(1/2*d*x + 1/2*c)^2 + a)*d)