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

Optimal. Leaf size=95 \[ \frac {2 a (15 A+7 C) \tan (c+d x)}{15 d \sqrt {a \sec (c+d x)+a}}+\frac {2 C \tan (c+d x) (a \sec (c+d x)+a)^{3/2}}{5 a d}-\frac {4 C \tan (c+d x) \sqrt {a \sec (c+d x)+a}}{15 d} \]

[Out]

2/5*C*(a+a*sec(d*x+c))^(3/2)*tan(d*x+c)/a/d+2/15*a*(15*A+7*C)*tan(d*x+c)/d/(a+a*sec(d*x+c))^(1/2)-4/15*C*(a+a*
sec(d*x+c))^(1/2)*tan(d*x+c)/d

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Rubi [A]  time = 0.20, antiderivative size = 95, normalized size of antiderivative = 1.00, number of steps used = 3, number of rules used = 3, integrand size = 33, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.091, Rules used = {4083, 4001, 3792} \[ \frac {2 a (15 A+7 C) \tan (c+d x)}{15 d \sqrt {a \sec (c+d x)+a}}+\frac {2 C \tan (c+d x) (a \sec (c+d x)+a)^{3/2}}{5 a d}-\frac {4 C \tan (c+d x) \sqrt {a \sec (c+d x)+a}}{15 d} \]

Antiderivative was successfully verified.

[In]

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

[Out]

(2*a*(15*A + 7*C)*Tan[c + d*x])/(15*d*Sqrt[a + a*Sec[c + d*x]]) - (4*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 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) \sqrt {a+a \sec (c+d x)} \left (A+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)-a C \sec (c+d x)\right ) \, dx}{5 a}\\ &=-\frac {4 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+7 C) \int \sec (c+d x) \sqrt {a+a \sec (c+d x)} \, dx\\ &=\frac {2 a (15 A+7 C) \tan (c+d x)}{15 d \sqrt {a+a \sec (c+d x)}}-\frac {4 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*}

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Mathematica [A]  time = 1.07, size = 71, normalized size = 0.75 \[ \frac {\tan \left (\frac {1}{2} (c+d x)\right ) \sec ^2(c+d x) \sqrt {a (\sec (c+d x)+1)} ((15 A+8 C) \cos (2 (c+d x))+15 A+8 C \cos (c+d x)+14 C)}{15 d} \]

Antiderivative was successfully verified.

[In]

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

[Out]

((15*A + 14*C + 8*C*Cos[c + d*x] + (15*A + 8*C)*Cos[2*(c + d*x)])*Sec[c + d*x]^2*Sqrt[a*(1 + Sec[c + d*x])]*Ta
n[(c + d*x)/2])/(15*d)

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fricas [A]  time = 0.44, size = 81, normalized size = 0.85 \[ \frac {2 \, {\left ({\left (15 \, A + 8 \, C\right )} \cos \left (d x + c\right )^{2} + 4 \, C \cos \left (d x + c\right ) + 3 \, C\right )} \sqrt {\frac {a \cos \left (d x + c\right ) + a}{\cos \left (d x + c\right )}} \sin \left (d x + c\right )}{15 \, {\left (d \cos \left (d x + c\right )^{3} + d \cos \left (d x + c\right )^{2}\right )}} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

2/15*((15*A + 8*C)*cos(d*x + c)^2 + 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)

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giac [B]  time = 1.19, size = 168, normalized size = 1.77 \[ \frac {2 \, {\left ({\left (\sqrt {2} {\left (15 \, A a^{3} \mathrm {sgn}\left (\cos \left (d x + c\right )\right ) + 7 \, C a^{3} \mathrm {sgn}\left (\cos \left (d x + c\right )\right )\right )} \tan \left (\frac {1}{2} \, d x + \frac {1}{2} \, c\right )^{2} - 10 \, \sqrt {2} {\left (3 \, A a^{3} \mathrm {sgn}\left (\cos \left (d x + c\right )\right ) + C a^{3} \mathrm {sgn}\left (\cos \left (d x + c\right )\right )\right )}\right )} \tan \left (\frac {1}{2} \, d x + \frac {1}{2} \, c\right )^{2} + 15 \, \sqrt {2} {\left (A a^{3} \mathrm {sgn}\left (\cos \left (d x + c\right )\right ) + C a^{3} \mathrm {sgn}\left (\cos \left (d x + c\right )\right )\right )}\right )} \tan \left (\frac {1}{2} \, d x + \frac {1}{2} \, c\right )}{15 \, {\left (a \tan \left (\frac {1}{2} \, d x + \frac {1}{2} \, c\right )^{2} - a\right )}^{2} \sqrt {-a \tan \left (\frac {1}{2} \, d x + \frac {1}{2} \, c\right )^{2} + a} d} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate(sec(d*x+c)*(A+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)) + 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)) + C*a^3*sgn(cos(d*x + c))))*tan(1/2*d*x + 1/2*c)^2 + 15*sqrt(2)*(A*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)

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maple [A]  time = 1.63, size = 85, normalized size = 0.89 \[ -\frac {2 \left (-1+\cos \left (d x +c \right )\right ) \left (15 A \left (\cos ^{2}\left (d x +c \right )\right )+8 C \left (\cos ^{2}\left (d x +c \right )\right )+4 C \cos \left (d x +c \right )+3 C \right ) \sqrt {\frac {a \left (1+\cos \left (d x +c \right )\right )}{\cos \left (d x +c \right )}}}{15 d \cos \left (d x +c \right )^{2} \sin \left (d x +c \right )} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

int(sec(d*x+c)*(A+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+8*C*cos(d*x+c)^2+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)

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

[Out]

Timed out

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mupad [B]  time = 6.96, size = 182, normalized size = 1.92 \[ -\frac {2\,\left ({\mathrm {e}}^{c\,1{}\mathrm {i}+d\,x\,1{}\mathrm {i}}-1\right )\,\sqrt {a+\frac {a}{\frac {{\mathrm {e}}^{-c\,1{}\mathrm {i}-d\,x\,1{}\mathrm {i}}}{2}+\frac {{\mathrm {e}}^{c\,1{}\mathrm {i}+d\,x\,1{}\mathrm {i}}}{2}}}\,\left (A\,15{}\mathrm {i}+C\,8{}\mathrm {i}+A\,{\mathrm {e}}^{c\,2{}\mathrm {i}+d\,x\,2{}\mathrm {i}}\,30{}\mathrm {i}+A\,{\mathrm {e}}^{c\,4{}\mathrm {i}+d\,x\,4{}\mathrm {i}}\,15{}\mathrm {i}+C\,{\mathrm {e}}^{c\,1{}\mathrm {i}+d\,x\,1{}\mathrm {i}}\,8{}\mathrm {i}+C\,{\mathrm {e}}^{c\,2{}\mathrm {i}+d\,x\,2{}\mathrm {i}}\,28{}\mathrm {i}+C\,{\mathrm {e}}^{c\,3{}\mathrm {i}+d\,x\,3{}\mathrm {i}}\,8{}\mathrm {i}+C\,{\mathrm {e}}^{c\,4{}\mathrm {i}+d\,x\,4{}\mathrm {i}}\,8{}\mathrm {i}\right )}{15\,d\,\left ({\mathrm {e}}^{c\,1{}\mathrm {i}+d\,x\,1{}\mathrm {i}}+1\right )\,{\left ({\mathrm {e}}^{c\,2{}\mathrm {i}+d\,x\,2{}\mathrm {i}}+1\right )}^2} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

int(((A + C/cos(c + d*x)^2)*(a + a/cos(c + d*x))^(1/2))/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 + C*8i + A*e
xp(c*2i + d*x*2i)*30i + A*exp(c*4i + d*x*4i)*15i + 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)

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

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

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