3.485 \(\int \sec (c+d x) \sqrt {a+a \sec (c+d x)} (A+B \sec (c+d x)+C \sec ^2(c+d x)) \, dx\)

Optimal. Leaf size=104 \[ \frac {2 a (15 A+5 B+7 C) \tan (c+d x)}{15 d \sqrt {a \sec (c+d x)+a}}+\frac {2 (5 B-2 C) \tan (c+d x) \sqrt {a \sec (c+d x)+a}}{15 d}+\frac {2 C \tan (c+d x) (a \sec (c+d x)+a)^{3/2}}{5 a d} \]

[Out]

2/5*C*(a+a*sec(d*x+c))^(3/2)*tan(d*x+c)/a/d+2/15*a*(15*A+5*B+7*C)*tan(d*x+c)/d/(a+a*sec(d*x+c))^(1/2)+2/15*(5*
B-2*C)*(a+a*sec(d*x+c))^(1/2)*tan(d*x+c)/d

________________________________________________________________________________________

Rubi [A]  time = 0.21, antiderivative size = 104, normalized size of antiderivative = 1.00, number of steps used = 3, number of rules used = 3, integrand size = 41, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.073, Rules used = {4082, 4001, 3792} \[ \frac {2 a (15 A+5 B+7 C) \tan (c+d x)}{15 d \sqrt {a \sec (c+d x)+a}}+\frac {2 (5 B-2 C) \tan (c+d x) \sqrt {a \sec (c+d x)+a}}{15 d}+\frac {2 C \tan (c+d x) (a \sec (c+d x)+a)^{3/2}}{5 a d} \]

Antiderivative was successfully verified.

[In]

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

[Out]

(2*a*(15*A + 5*B + 7*C)*Tan[c + d*x])/(15*d*Sqrt[a + a*Sec[c + d*x]]) + (2*(5*B - 2*C)*Sqrt[a + a*Sec[c + d*x]
]*Tan[c + d*x])/(15*d) + (2*C*(a + a*Sec[c + d*x])^(3/2)*Tan[c + d*x])/(5*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 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 4082

Int[csc[(e_.) + (f_.)*(x_)]*((A_.) + csc[(e_.) + (f_.)*(x_)]*(B_.) + 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) + (b*B*
(m + 2) - a*C)*Csc[e + f*x], x], x], x] /; FreeQ[{a, b, e, f, A, B, C, m}, x] &&  !LtQ[m, -1]

Rubi steps

\begin {align*} \int \sec (c+d x) \sqrt {a+a \sec (c+d x)} \left (A+B \sec (c+d x)+C \sec ^2(c+d x)\right ) \, dx &=\frac {2 C (a+a \sec (c+d x))^{3/2} \tan (c+d x)}{5 a d}+\frac {2 \int \sec (c+d x) \sqrt {a+a \sec (c+d x)} \left (\frac {1}{2} a (5 A+3 C)+\frac {1}{2} a (5 B-2 C) \sec (c+d x)\right ) \, dx}{5 a}\\ &=\frac {2 (5 B-2 C) \sqrt {a+a \sec (c+d x)} \tan (c+d x)}{15 d}+\frac {2 C (a+a \sec (c+d x))^{3/2} \tan (c+d x)}{5 a d}+\frac {1}{15} (15 A+5 B+7 C) \int \sec (c+d x) \sqrt {a+a \sec (c+d x)} \, dx\\ &=\frac {2 a (15 A+5 B+7 C) \tan (c+d x)}{15 d \sqrt {a+a \sec (c+d x)}}+\frac {2 (5 B-2 C) \sqrt {a+a \sec (c+d x)} \tan (c+d x)}{15 d}+\frac {2 C (a+a \sec (c+d x))^{3/2} \tan (c+d x)}{5 a d}\\ \end {align*}

________________________________________________________________________________________

Mathematica [A]  time = 0.84, size = 83, normalized size = 0.80 \[ \frac {\tan \left (\frac {1}{2} (c+d x)\right ) \sec ^2(c+d x) \sqrt {a (\sec (c+d x)+1)} ((15 A+10 B+8 C) \cos (2 (c+d x))+15 A+2 (5 B+4 C) \cos (c+d x)+10 B+14 C)}{15 d} \]

Antiderivative was successfully verified.

[In]

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

[Out]

((15*A + 10*B + 14*C + 2*(5*B + 4*C)*Cos[c + d*x] + (15*A + 10*B + 8*C)*Cos[2*(c + d*x)])*Sec[c + d*x]^2*Sqrt[
a*(1 + Sec[c + d*x])]*Tan[(c + d*x)/2])/(15*d)

________________________________________________________________________________________

fricas [A]  time = 0.43, size = 89, normalized size = 0.86 \[ \frac {2 \, {\left ({\left (15 \, A + 10 \, B + 8 \, C\right )} \cos \left (d x + c\right )^{2} + {\left (5 \, B + 4 \, C\right )} \cos \left (d x + c\right ) + 3 \, C\right )} \sqrt {\frac {a \cos \left (d x + c\right ) + a}{\cos \left (d x + c\right )}} \sin \left (d x + c\right )}{15 \, {\left (d \cos \left (d x + c\right )^{3} + d \cos \left (d x + c\right )^{2}\right )}} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate(sec(d*x+c)*(A+B*sec(d*x+c)+C*sec(d*x+c)^2)*(a+a*sec(d*x+c))^(1/2),x, algorithm="fricas")

[Out]

2/15*((15*A + 10*B + 8*C)*cos(d*x + c)^2 + (5*B + 4*C)*cos(d*x + c) + 3*C)*sqrt((a*cos(d*x + c) + a)/cos(d*x +
c))*sin(d*x + c)/(d*cos(d*x + c)^3 + d*cos(d*x + c)^2)

________________________________________________________________________________________

giac [B]  time = 1.26, size = 206, normalized size = 1.98 \[ \frac {2 \, {\left ({\left (\sqrt {2} {\left (15 \, A a^{3} \mathrm {sgn}\left (\cos \left (d x + c\right )\right ) + 5 \, B a^{3} \mathrm {sgn}\left (\cos \left (d x + c\right )\right ) + 7 \, C a^{3} \mathrm {sgn}\left (\cos \left (d x + c\right )\right )\right )} \tan \left (\frac {1}{2} \, d x + \frac {1}{2} \, c\right )^{2} - 10 \, \sqrt {2} {\left (3 \, A a^{3} \mathrm {sgn}\left (\cos \left (d x + c\right )\right ) + 2 \, B a^{3} \mathrm {sgn}\left (\cos \left (d x + c\right )\right ) + C a^{3} \mathrm {sgn}\left (\cos \left (d x + c\right )\right )\right )}\right )} \tan \left (\frac {1}{2} \, d x + \frac {1}{2} \, c\right )^{2} + 15 \, \sqrt {2} {\left (A a^{3} \mathrm {sgn}\left (\cos \left (d x + c\right )\right ) + B a^{3} \mathrm {sgn}\left (\cos \left (d x + c\right )\right ) + C a^{3} \mathrm {sgn}\left (\cos \left (d x + c\right )\right )\right )}\right )} \tan \left (\frac {1}{2} \, d x + \frac {1}{2} \, c\right )}{15 \, {\left (a \tan \left (\frac {1}{2} \, d x + \frac {1}{2} \, c\right )^{2} - a\right )}^{2} \sqrt {-a \tan \left (\frac {1}{2} \, d x + \frac {1}{2} \, c\right )^{2} + a} d} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate(sec(d*x+c)*(A+B*sec(d*x+c)+C*sec(d*x+c)^2)*(a+a*sec(d*x+c))^(1/2),x, algorithm="giac")

[Out]

2/15*((sqrt(2)*(15*A*a^3*sgn(cos(d*x + c)) + 5*B*a^3*sgn(cos(d*x + c)) + 7*C*a^3*sgn(cos(d*x + c)))*tan(1/2*d*
x + 1/2*c)^2 - 10*sqrt(2)*(3*A*a^3*sgn(cos(d*x + c)) + 2*B*a^3*sgn(cos(d*x + c)) + C*a^3*sgn(cos(d*x + c))))*t
an(1/2*d*x + 1/2*c)^2 + 15*sqrt(2)*(A*a^3*sgn(cos(d*x + c)) + B*a^3*sgn(cos(d*x + c)) + C*a^3*sgn(cos(d*x + c)
)))*tan(1/2*d*x + 1/2*c)/((a*tan(1/2*d*x + 1/2*c)^2 - a)^2*sqrt(-a*tan(1/2*d*x + 1/2*c)^2 + a)*d)

________________________________________________________________________________________

maple [A]  time = 2.43, size = 105, normalized size = 1.01 \[ -\frac {2 \left (-1+\cos \left (d x +c \right )\right ) \left (15 A \left (\cos ^{2}\left (d x +c \right )\right )+10 B \left (\cos ^{2}\left (d x +c \right )\right )+8 C \left (\cos ^{2}\left (d x +c \right )\right )+5 B \cos \left (d x +c \right )+4 C \cos \left (d x +c \right )+3 C \right ) \sqrt {\frac {a \left (1+\cos \left (d x +c \right )\right )}{\cos \left (d x +c \right )}}}{15 d \cos \left (d x +c \right )^{2} \sin \left (d x +c \right )} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

int(sec(d*x+c)*(A+B*sec(d*x+c)+C*sec(d*x+c)^2)*(a+a*sec(d*x+c))^(1/2),x)

[Out]

-2/15/d*(-1+cos(d*x+c))*(15*A*cos(d*x+c)^2+10*B*cos(d*x+c)^2+8*C*cos(d*x+c)^2+5*B*cos(d*x+c)+4*C*cos(d*x+c)+3*
C)*(a*(1+cos(d*x+c))/cos(d*x+c))^(1/2)/cos(d*x+c)^2/sin(d*x+c)

________________________________________________________________________________________

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+B*sec(d*x+c)+C*sec(d*x+c)^2)*(a+a*sec(d*x+c))^(1/2),x, algorithm="maxima")

[Out]

Timed out

________________________________________________________________________________________

mupad [B]  time = 6.64, size = 246, normalized size = 2.37 \[ -\frac {2\,\left ({\mathrm {e}}^{c\,1{}\mathrm {i}+d\,x\,1{}\mathrm {i}}-1\right )\,\sqrt {a+\frac {a}{\frac {{\mathrm {e}}^{-c\,1{}\mathrm {i}-d\,x\,1{}\mathrm {i}}}{2}+\frac {{\mathrm {e}}^{c\,1{}\mathrm {i}+d\,x\,1{}\mathrm {i}}}{2}}}\,\left (A\,15{}\mathrm {i}+B\,10{}\mathrm {i}+C\,8{}\mathrm {i}+A\,{\mathrm {e}}^{c\,2{}\mathrm {i}+d\,x\,2{}\mathrm {i}}\,30{}\mathrm {i}+A\,{\mathrm {e}}^{c\,4{}\mathrm {i}+d\,x\,4{}\mathrm {i}}\,15{}\mathrm {i}+B\,{\mathrm {e}}^{c\,1{}\mathrm {i}+d\,x\,1{}\mathrm {i}}\,10{}\mathrm {i}+B\,{\mathrm {e}}^{c\,2{}\mathrm {i}+d\,x\,2{}\mathrm {i}}\,20{}\mathrm {i}+B\,{\mathrm {e}}^{c\,3{}\mathrm {i}+d\,x\,3{}\mathrm {i}}\,10{}\mathrm {i}+B\,{\mathrm {e}}^{c\,4{}\mathrm {i}+d\,x\,4{}\mathrm {i}}\,10{}\mathrm {i}+C\,{\mathrm {e}}^{c\,1{}\mathrm {i}+d\,x\,1{}\mathrm {i}}\,8{}\mathrm {i}+C\,{\mathrm {e}}^{c\,2{}\mathrm {i}+d\,x\,2{}\mathrm {i}}\,28{}\mathrm {i}+C\,{\mathrm {e}}^{c\,3{}\mathrm {i}+d\,x\,3{}\mathrm {i}}\,8{}\mathrm {i}+C\,{\mathrm {e}}^{c\,4{}\mathrm {i}+d\,x\,4{}\mathrm {i}}\,8{}\mathrm {i}\right )}{15\,d\,\left ({\mathrm {e}}^{c\,1{}\mathrm {i}+d\,x\,1{}\mathrm {i}}+1\right )\,{\left ({\mathrm {e}}^{c\,2{}\mathrm {i}+d\,x\,2{}\mathrm {i}}+1\right )}^2} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

int(((a + a/cos(c + d*x))^(1/2)*(A + B/cos(c + d*x) + C/cos(c + d*x)^2))/cos(c + d*x),x)

[Out]

-(2*(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)*(A*15i + B*10i + C*
8i + A*exp(c*2i + d*x*2i)*30i + A*exp(c*4i + d*x*4i)*15i + B*exp(c*1i + d*x*1i)*10i + B*exp(c*2i + d*x*2i)*20i
+ B*exp(c*3i + d*x*3i)*10i + B*exp(c*4i + d*x*4i)*10i + C*exp(c*1i + d*x*1i)*8i + C*exp(c*2i + d*x*2i)*28i +
C*exp(c*3i + d*x*3i)*8i + C*exp(c*4i + d*x*4i)*8i))/(15*d*(exp(c*1i + d*x*1i) + 1)*(exp(c*2i + d*x*2i) + 1)^2)

________________________________________________________________________________________

sympy [F]  time = 0.00, size = 0, normalized size = 0.00 \[ \int \sqrt {a \left (\sec {\left (c + d x \right )} + 1\right )} \left (A + B \sec {\left (c + d x \right )} + C \sec ^{2}{\left (c + d x \right )}\right ) \sec {\left (c + d x \right )}\, dx \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate(sec(d*x+c)*(A+B*sec(d*x+c)+C*sec(d*x+c)**2)*(a+a*sec(d*x+c))**(1/2),x)

[Out]

Integral(sqrt(a*(sec(c + d*x) + 1))*(A + B*sec(c + d*x) + C*sec(c + d*x)**2)*sec(c + d*x), x)

________________________________________________________________________________________