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

Optimal. Leaf size=169 \[ \frac {64 a^3 (21 A+13 C) \tan (c+d x)}{315 d \sqrt {a \sec (c+d x)+a}}+\frac {16 a^2 (21 A+13 C) \tan (c+d x) \sqrt {a \sec (c+d x)+a}}{315 d}+\frac {2 a (21 A+13 C) \tan (c+d x) (a \sec (c+d x)+a)^{3/2}}{105 d}+\frac {2 C \tan (c+d x) (a \sec (c+d x)+a)^{7/2}}{9 a d}-\frac {4 C \tan (c+d x) (a \sec (c+d x)+a)^{5/2}}{63 d} \]

[Out]

2/105*a*(21*A+13*C)*(a+a*sec(d*x+c))^(3/2)*tan(d*x+c)/d-4/63*C*(a+a*sec(d*x+c))^(5/2)*tan(d*x+c)/d+2/9*C*(a+a*
sec(d*x+c))^(7/2)*tan(d*x+c)/a/d+64/315*a^3*(21*A+13*C)*tan(d*x+c)/d/(a+a*sec(d*x+c))^(1/2)+16/315*a^2*(21*A+1
3*C)*(a+a*sec(d*x+c))^(1/2)*tan(d*x+c)/d

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Rubi [A]  time = 0.32, antiderivative size = 169, normalized size of antiderivative = 1.00, number of steps used = 5, number of rules used = 4, integrand size = 33, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.121, Rules used = {4083, 4001, 3793, 3792} \[ \frac {16 a^2 (21 A+13 C) \tan (c+d x) \sqrt {a \sec (c+d x)+a}}{315 d}+\frac {64 a^3 (21 A+13 C) \tan (c+d x)}{315 d \sqrt {a \sec (c+d x)+a}}+\frac {2 a (21 A+13 C) \tan (c+d x) (a \sec (c+d x)+a)^{3/2}}{105 d}+\frac {2 C \tan (c+d x) (a \sec (c+d x)+a)^{7/2}}{9 a d}-\frac {4 C \tan (c+d x) (a \sec (c+d x)+a)^{5/2}}{63 d} \]

Antiderivative was successfully verified.

[In]

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

[Out]

(64*a^3*(21*A + 13*C)*Tan[c + d*x])/(315*d*Sqrt[a + a*Sec[c + d*x]]) + (16*a^2*(21*A + 13*C)*Sqrt[a + a*Sec[c
+ d*x]]*Tan[c + d*x])/(315*d) + (2*a*(21*A + 13*C)*(a + a*Sec[c + d*x])^(3/2)*Tan[c + d*x])/(105*d) - (4*C*(a
+ a*Sec[c + d*x])^(5/2)*Tan[c + d*x])/(63*d) + (2*C*(a + a*Sec[c + d*x])^(7/2)*Tan[c + d*x])/(9*a*d)

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]

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 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 4083

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

Rubi steps

\begin {align*} \int \sec (c+d x) (a+a \sec (c+d x))^{5/2} \left (A+C \sec ^2(c+d x)\right ) \, dx &=\frac {2 C (a+a \sec (c+d x))^{7/2} \tan (c+d x)}{9 a d}+\frac {2 \int \sec (c+d x) (a+a \sec (c+d x))^{5/2} \left (\frac {1}{2} a (9 A+7 C)-a C \sec (c+d x)\right ) \, dx}{9 a}\\ &=-\frac {4 C (a+a \sec (c+d x))^{5/2} \tan (c+d x)}{63 d}+\frac {2 C (a+a \sec (c+d x))^{7/2} \tan (c+d x)}{9 a d}+\frac {1}{21} (21 A+13 C) \int \sec (c+d x) (a+a \sec (c+d x))^{5/2} \, dx\\ &=\frac {2 a (21 A+13 C) (a+a \sec (c+d x))^{3/2} \tan (c+d x)}{105 d}-\frac {4 C (a+a \sec (c+d x))^{5/2} \tan (c+d x)}{63 d}+\frac {2 C (a+a \sec (c+d x))^{7/2} \tan (c+d x)}{9 a d}+\frac {1}{105} (8 a (21 A+13 C)) \int \sec (c+d x) (a+a \sec (c+d x))^{3/2} \, dx\\ &=\frac {16 a^2 (21 A+13 C) \sqrt {a+a \sec (c+d x)} \tan (c+d x)}{315 d}+\frac {2 a (21 A+13 C) (a+a \sec (c+d x))^{3/2} \tan (c+d x)}{105 d}-\frac {4 C (a+a \sec (c+d x))^{5/2} \tan (c+d x)}{63 d}+\frac {2 C (a+a \sec (c+d x))^{7/2} \tan (c+d x)}{9 a d}+\frac {1}{315} \left (32 a^2 (21 A+13 C)\right ) \int \sec (c+d x) \sqrt {a+a \sec (c+d x)} \, dx\\ &=\frac {64 a^3 (21 A+13 C) \tan (c+d x)}{315 d \sqrt {a+a \sec (c+d x)}}+\frac {16 a^2 (21 A+13 C) \sqrt {a+a \sec (c+d x)} \tan (c+d x)}{315 d}+\frac {2 a (21 A+13 C) (a+a \sec (c+d x))^{3/2} \tan (c+d x)}{105 d}-\frac {4 C (a+a \sec (c+d x))^{5/2} \tan (c+d x)}{63 d}+\frac {2 C (a+a \sec (c+d x))^{7/2} \tan (c+d x)}{9 a d}\\ \end {align*}

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Mathematica [A]  time = 1.54, size = 125, normalized size = 0.74 \[ \frac {a^2 \tan \left (\frac {1}{2} (c+d x)\right ) \sec ^4(c+d x) \sqrt {a (\sec (c+d x)+1)} (4 (441 A+698 C) \cos (c+d x)+4 (966 A+803 C) \cos (2 (c+d x))+588 A \cos (3 (c+d x))+903 A \cos (4 (c+d x))+2961 A+584 C \cos (3 (c+d x))+584 C \cos (4 (c+d x))+2908 C)}{1260 d} \]

Antiderivative was successfully verified.

[In]

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

[Out]

(a^2*(2961*A + 2908*C + 4*(441*A + 698*C)*Cos[c + d*x] + 4*(966*A + 803*C)*Cos[2*(c + d*x)] + 588*A*Cos[3*(c +
d*x)] + 584*C*Cos[3*(c + d*x)] + 903*A*Cos[4*(c + d*x)] + 584*C*Cos[4*(c + d*x)])*Sec[c + d*x]^4*Sqrt[a*(1 +
Sec[c + d*x])]*Tan[(c + d*x)/2])/(1260*d)

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fricas [A]  time = 0.44, size = 130, normalized size = 0.77 \[ \frac {2 \, {\left ({\left (903 \, A + 584 \, C\right )} a^{2} \cos \left (d x + c\right )^{4} + 2 \, {\left (147 \, A + 146 \, C\right )} a^{2} \cos \left (d x + c\right )^{3} + 3 \, {\left (21 \, A + 73 \, C\right )} a^{2} \cos \left (d x + c\right )^{2} + 130 \, C a^{2} \cos \left (d x + c\right ) + 35 \, C a^{2}\right )} \sqrt {\frac {a \cos \left (d x + c\right ) + a}{\cos \left (d x + c\right )}} \sin \left (d x + c\right )}{315 \, {\left (d \cos \left (d x + c\right )^{5} + d \cos \left (d x + c\right )^{4}\right )}} \]

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+C*sec(d*x+c)^2),x, algorithm="fricas")

[Out]

2/315*((903*A + 584*C)*a^2*cos(d*x + c)^4 + 2*(147*A + 146*C)*a^2*cos(d*x + c)^3 + 3*(21*A + 73*C)*a^2*cos(d*x
+ c)^2 + 130*C*a^2*cos(d*x + c) + 35*C*a^2)*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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giac [A]  time = 2.13, size = 261, normalized size = 1.54 \[ \frac {8 \, {\left ({\left ({\left (4 \, {\left (2 \, \sqrt {2} {\left (21 \, A a^{7} \mathrm {sgn}\left (\cos \left (d x + c\right )\right ) + 13 \, C a^{7} \mathrm {sgn}\left (\cos \left (d x + c\right )\right )\right )} \tan \left (\frac {1}{2} \, d x + \frac {1}{2} \, c\right )^{2} - 9 \, \sqrt {2} {\left (21 \, A a^{7} \mathrm {sgn}\left (\cos \left (d x + c\right )\right ) + 13 \, C a^{7} \mathrm {sgn}\left (\cos \left (d x + c\right )\right )\right )}\right )} \tan \left (\frac {1}{2} \, d x + \frac {1}{2} \, c\right )^{2} + 63 \, \sqrt {2} {\left (21 \, A a^{7} \mathrm {sgn}\left (\cos \left (d x + c\right )\right ) + 13 \, C a^{7} \mathrm {sgn}\left (\cos \left (d x + c\right )\right )\right )}\right )} \tan \left (\frac {1}{2} \, d x + \frac {1}{2} \, c\right )^{2} - 210 \, \sqrt {2} {\left (5 \, A a^{7} \mathrm {sgn}\left (\cos \left (d x + c\right )\right ) + 3 \, C a^{7} \mathrm {sgn}\left (\cos \left (d x + c\right )\right )\right )}\right )} \tan \left (\frac {1}{2} \, d x + \frac {1}{2} \, c\right )^{2} + 315 \, \sqrt {2} {\left (A a^{7} \mathrm {sgn}\left (\cos \left (d x + c\right )\right ) + C a^{7} \mathrm {sgn}\left (\cos \left (d x + c\right )\right )\right )}\right )} \tan \left (\frac {1}{2} \, d x + \frac {1}{2} \, c\right )}{315 \, {\left (a \tan \left (\frac {1}{2} \, d x + \frac {1}{2} \, c\right )^{2} - a\right )}^{4} \sqrt {-a \tan \left (\frac {1}{2} \, d x + \frac {1}{2} \, c\right )^{2} + a} d} \]

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+C*sec(d*x+c)^2),x, algorithm="giac")

[Out]

8/315*(((4*(2*sqrt(2)*(21*A*a^7*sgn(cos(d*x + c)) + 13*C*a^7*sgn(cos(d*x + c)))*tan(1/2*d*x + 1/2*c)^2 - 9*sqr
t(2)*(21*A*a^7*sgn(cos(d*x + c)) + 13*C*a^7*sgn(cos(d*x + c))))*tan(1/2*d*x + 1/2*c)^2 + 63*sqrt(2)*(21*A*a^7*
sgn(cos(d*x + c)) + 13*C*a^7*sgn(cos(d*x + c))))*tan(1/2*d*x + 1/2*c)^2 - 210*sqrt(2)*(5*A*a^7*sgn(cos(d*x + c
)) + 3*C*a^7*sgn(cos(d*x + c))))*tan(1/2*d*x + 1/2*c)^2 + 315*sqrt(2)*(A*a^7*sgn(cos(d*x + c)) + C*a^7*sgn(cos
(d*x + c))))*tan(1/2*d*x + 1/2*c)/((a*tan(1/2*d*x + 1/2*c)^2 - a)^4*sqrt(-a*tan(1/2*d*x + 1/2*c)^2 + a)*d)

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maple [A]  time = 1.72, size = 132, normalized size = 0.78 \[ -\frac {2 \left (-1+\cos \left (d x +c \right )\right ) \left (903 A \left (\cos ^{4}\left (d x +c \right )\right )+584 C \left (\cos ^{4}\left (d x +c \right )\right )+294 A \left (\cos ^{3}\left (d x +c \right )\right )+292 C \left (\cos ^{3}\left (d x +c \right )\right )+63 A \left (\cos ^{2}\left (d x +c \right )\right )+219 C \left (\cos ^{2}\left (d x +c \right )\right )+130 C \cos \left (d x +c \right )+35 C \right ) \sqrt {\frac {a \left (1+\cos \left (d x +c \right )\right )}{\cos \left (d x +c \right )}}\, a^{2}}{315 d \cos \left (d x +c \right )^{4} \sin \left (d x +c \right )} \]

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

[Out]

-2/315/d*(-1+cos(d*x+c))*(903*A*cos(d*x+c)^4+584*C*cos(d*x+c)^4+294*A*cos(d*x+c)^3+292*C*cos(d*x+c)^3+63*A*cos
(d*x+c)^2+219*C*cos(d*x+c)^2+130*C*cos(d*x+c)+35*C)*(a*(1+cos(d*x+c))/cos(d*x+c))^(1/2)/cos(d*x+c)^4/sin(d*x+c
)*a^2

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maxima [F(-1)]  time = 0.00, size = 0, normalized size = 0.00 \[ \text {Timed out} \]

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+C*sec(d*x+c)^2),x, algorithm="maxima")

[Out]

Timed out

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mupad [B]  time = 11.20, size = 766, normalized size = 4.53 \[ \text {result too large to display} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

int(((A + C/cos(c + d*x)^2)*(a + a/cos(c + d*x))^(5/2))/cos(c + d*x),x)

[Out]

((a + a/(exp(- c*1i - d*x*1i)/2 + exp(c*1i + d*x*1i)/2))^(1/2)*((A*a^2*2i)/d - (a^2*exp(c*1i + d*x*1i)*(903*A
+ 584*C)*2i)/(315*d)))/(exp(c*1i + d*x*1i) + 1) - ((a + a/(exp(- c*1i - d*x*1i)/2 + exp(c*1i + d*x*1i)/2))^(1/
2)*(exp(c*1i + d*x*1i)*((A*a^2*2i)/d - (a^2*(A + 2*C)*4i)/d + (a^2*(21*A - 32*C)*2i)/(105*d)) - (A*a^2*2i)/(5*
d) + (a^2*(5*A + 2*C)*4i)/(5*d) - (a^2*(5*A + 32*C)*2i)/(5*d)))/((exp(c*1i + d*x*1i) + 1)*(exp(c*2i + d*x*2i)
+ 1)^2) - ((a + a/(exp(- c*1i - d*x*1i)/2 + exp(c*1i + d*x*1i)/2))^(1/2)*(exp(c*1i + d*x*1i)*((A*a^2*4i)/(3*d)
- (a^2*(3*A + C)*8i)/(9*d) + (a^2*(13*A + 20*C)*4i)/(9*d) - (a^2*(A + C)*40i)/(9*d)) - (A*a^2*4i)/(3*d) + (a^
2*(3*A + C)*8i)/(9*d) - (a^2*(13*A + 20*C)*4i)/(9*d) + (a^2*(A + C)*40i)/(9*d)))/((exp(c*1i + d*x*1i) + 1)*(ex
p(c*2i + d*x*2i) + 1)^4) + ((a + a/(exp(- c*1i - d*x*1i)/2 + exp(c*1i + d*x*1i)/2))^(1/2)*(exp(c*1i + d*x*1i)*
((A*a^2*10i)/(3*d) - (a^2*(189*A + 292*C)*2i)/(315*d)) - (A*a^2*2i)/(3*d) + (a^2*(9*A + 4*C)*2i)/(3*d)))/((exp
(c*1i + d*x*1i) + 1)*(exp(c*2i + d*x*2i) + 1)) + ((a + a/(exp(- c*1i - d*x*1i)/2 + exp(c*1i + d*x*1i)/2))^(1/2
)*(exp(c*1i + d*x*1i)*((A*a^2*10i)/(7*d) - (a^2*(3*A + 4*C)*10i)/(7*d) - (a^2*(9*A - 16*C)*2i)/(63*d) + (a^2*(
11*A + 20*C)*2i)/(7*d)) - (A*a^2*2i)/(7*d) + (a^2*(11*A + 4*C)*2i)/(7*d) - (a^2*(5*A + 12*C)*6i)/(7*d) + (a^2*
(5*A - 16*C)*2i)/(7*d)))/((exp(c*1i + d*x*1i) + 1)*(exp(c*2i + d*x*2i) + 1)^3)

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sympy [F(-1)]  time = 0.00, size = 0, normalized size = 0.00 \[ \text {Timed out} \]

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

[Out]

Timed out

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