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

Optimal. Leaf size=137 \[ \frac{2 (35 A+18 C) \tan (c+d x) \sqrt{a \sec (c+d x)+a}}{105 d}+\frac{2 a (35 A+27 C) \tan (c+d x)}{105 d \sqrt{a \sec (c+d x)+a}}+\frac{2 C \tan (c+d x) \sec ^2(c+d x) \sqrt{a \sec (c+d x)+a}}{7 d}+\frac{2 C \tan (c+d x) (a \sec (c+d x)+a)^{3/2}}{35 a d} \]

[Out]

(2*a*(35*A + 27*C)*Tan[c + d*x])/(105*d*Sqrt[a + a*Sec[c + d*x]]) + (2*(35*A + 18*C)*Sqrt[a + a*Sec[c + d*x]]*
Tan[c + d*x])/(105*d) + (2*C*Sec[c + d*x]^2*Sqrt[a + a*Sec[c + d*x]]*Tan[c + d*x])/(7*d) + (2*C*(a + a*Sec[c +
d*x])^(3/2)*Tan[c + d*x])/(35*a*d)

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Rubi [A]  time = 0.3905, antiderivative size = 137, normalized size of antiderivative = 1., number of steps used = 4, number of rules used = 4, integrand size = 35, \(\frac{\text{number of rules}}{\text{integrand size}}\) = 0.114, Rules used = {4089, 4010, 4001, 3792} \[ \frac{2 (35 A+18 C) \tan (c+d x) \sqrt{a \sec (c+d x)+a}}{105 d}+\frac{2 a (35 A+27 C) \tan (c+d x)}{105 d \sqrt{a \sec (c+d x)+a}}+\frac{2 C \tan (c+d x) \sec ^2(c+d x) \sqrt{a \sec (c+d x)+a}}{7 d}+\frac{2 C \tan (c+d x) (a \sec (c+d x)+a)^{3/2}}{35 a d} \]

Antiderivative was successfully verified.

[In]

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

[Out]

(2*a*(35*A + 27*C)*Tan[c + d*x])/(105*d*Sqrt[a + a*Sec[c + d*x]]) + (2*(35*A + 18*C)*Sqrt[a + a*Sec[c + d*x]]*
Tan[c + d*x])/(105*d) + (2*C*Sec[c + d*x]^2*Sqrt[a + a*Sec[c + d*x]]*Tan[c + d*x])/(7*d) + (2*C*(a + a*Sec[c +
d*x])^(3/2)*Tan[c + d*x])/(35*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 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) \sqrt{a+a \sec (c+d x)} \left (A+C \sec ^2(c+d x)\right ) \, dx &=\frac{2 C \sec ^2(c+d x) \sqrt{a+a \sec (c+d x)} \tan (c+d x)}{7 d}+\frac{2 \int \sec ^2(c+d x) \sqrt{a+a \sec (c+d x)} \left (\frac{1}{2} a (7 A+4 C)+\frac{1}{2} a C \sec (c+d x)\right ) \, dx}{7 a}\\ &=\frac{2 C \sec ^2(c+d x) \sqrt{a+a \sec (c+d x)} \tan (c+d x)}{7 d}+\frac{2 C (a+a \sec (c+d x))^{3/2} \tan (c+d x)}{35 a d}+\frac{4 \int \sec (c+d x) \sqrt{a+a \sec (c+d x)} \left (\frac{3 a^2 C}{4}+\frac{1}{4} a^2 (35 A+18 C) \sec (c+d x)\right ) \, dx}{35 a^2}\\ &=\frac{2 (35 A+18 C) \sqrt{a+a \sec (c+d x)} \tan (c+d x)}{105 d}+\frac{2 C \sec ^2(c+d x) \sqrt{a+a \sec (c+d x)} \tan (c+d x)}{7 d}+\frac{2 C (a+a \sec (c+d x))^{3/2} \tan (c+d x)}{35 a d}+\frac{1}{105} (35 A+27 C) \int \sec (c+d x) \sqrt{a+a \sec (c+d x)} \, dx\\ &=\frac{2 a (35 A+27 C) \tan (c+d x)}{105 d \sqrt{a+a \sec (c+d x)}}+\frac{2 (35 A+18 C) \sqrt{a+a \sec (c+d x)} \tan (c+d x)}{105 d}+\frac{2 C \sec ^2(c+d x) \sqrt{a+a \sec (c+d x)} \tan (c+d x)}{7 d}+\frac{2 C (a+a \sec (c+d x))^{3/2} \tan (c+d x)}{35 a d}\\ \end{align*}

Mathematica [A]  time = 0.783647, size = 99, normalized size = 0.72 \[ \frac{\tan \left (\frac{1}{2} (c+d x)\right ) \sec ^3(c+d x) \sqrt{a (\sec (c+d x)+1)} (3 (35 A+36 C) \cos (c+d x)+(35 A+24 C) \cos (2 (c+d x))+35 A \cos (3 (c+d x))+35 A+24 C \cos (3 (c+d x))+54 C)}{105 d} \]

Antiderivative was successfully verified.

[In]

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

[Out]

((35*A + 54*C + 3*(35*A + 36*C)*Cos[c + d*x] + (35*A + 24*C)*Cos[2*(c + d*x)] + 35*A*Cos[3*(c + d*x)] + 24*C*C
os[3*(c + d*x)])*Sec[c + d*x]^3*Sqrt[a*(1 + Sec[c + d*x])]*Tan[(c + d*x)/2])/(105*d)

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Maple [A]  time = 0.331, size = 107, normalized size = 0.8 \begin{align*} -{\frac{ \left ( -2+2\,\cos \left ( dx+c \right ) \right ) \left ( 70\,A \left ( \cos \left ( dx+c \right ) \right ) ^{3}+48\,C \left ( \cos \left ( dx+c \right ) \right ) ^{3}+35\,A \left ( \cos \left ( dx+c \right ) \right ) ^{2}+24\,C \left ( \cos \left ( dx+c \right ) \right ) ^{2}+18\,C\cos \left ( dx+c \right ) +15\,C \right ) }{105\,d \left ( \cos \left ( dx+c \right ) \right ) ^{3}\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+C*sec(d*x+c)^2)*(a+a*sec(d*x+c))^(1/2),x)

[Out]

-2/105/d*(-1+cos(d*x+c))*(70*A*cos(d*x+c)^3+48*C*cos(d*x+c)^3+35*A*cos(d*x+c)^2+24*C*cos(d*x+c)^2+18*C*cos(d*x
+c)+15*C)*(a*(cos(d*x+c)+1)/cos(d*x+c))^(1/2)/cos(d*x+c)^3/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+C*sec(d*x+c)^2)*(a+a*sec(d*x+c))^(1/2),x, algorithm="maxima")

[Out]

Timed out

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Fricas [A]  time = 0.486188, size = 255, normalized size = 1.86 \begin{align*} \frac{2 \,{\left (2 \,{\left (35 \, A + 24 \, C\right )} \cos \left (d x + c\right )^{3} +{\left (35 \, A + 24 \, C\right )} \cos \left (d x + c\right )^{2} + 18 \, C \cos \left (d x + c\right ) + 15 \, C\right )} \sqrt{\frac{a \cos \left (d x + c\right ) + a}{\cos \left (d x + c\right )}} \sin \left (d x + c\right )}{105 \,{\left (d \cos \left (d x + c\right )^{4} + d \cos \left (d x + c\right )^{3}\right )}} \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

2/105*(2*(35*A + 24*C)*cos(d*x + c)^3 + (35*A + 24*C)*cos(d*x + c)^2 + 18*C*cos(d*x + c) + 15*C)*sqrt((a*cos(d
*x + c) + a)/cos(d*x + c))*sin(d*x + c)/(d*cos(d*x + c)^4 + d*cos(d*x + c)^3)

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Sympy [F]  time = 0., size = 0, normalized size = 0. \begin{align*} \int \sqrt{a \left (\sec{\left (c + d x \right )} + 1\right )} \left (A + C \sec ^{2}{\left (c + d x \right )}\right ) \sec ^{2}{\left (c + d x \right )}\, dx \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate(sec(d*x+c)**2*(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)**2, x)

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Giac [A]  time = 4.56029, size = 300, normalized size = 2.19 \begin{align*} -\frac{2 \,{\left (105 \, \sqrt{2} A a^{4} \mathrm{sgn}\left (\cos \left (d x + c\right )\right ) + 105 \, \sqrt{2} C a^{4} \mathrm{sgn}\left (\cos \left (d x + c\right )\right ) -{\left (245 \, \sqrt{2} A a^{4} \mathrm{sgn}\left (\cos \left (d x + c\right )\right ) + 105 \, \sqrt{2} C a^{4} \mathrm{sgn}\left (\cos \left (d x + c\right )\right ) -{\left (175 \, \sqrt{2} A a^{4} \mathrm{sgn}\left (\cos \left (d x + c\right )\right ) + 147 \, \sqrt{2} C a^{4} \mathrm{sgn}\left (\cos \left (d x + c\right )\right ) -{\left (35 \, \sqrt{2} A a^{4} \mathrm{sgn}\left (\cos \left (d x + c\right )\right ) + 27 \, \sqrt{2} C a^{4} \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 )}{105 \,{\left (a \tan \left (\frac{1}{2} \, d x + \frac{1}{2} \, c\right )^{2} - a\right )}^{3} \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+C*sec(d*x+c)^2)*(a+a*sec(d*x+c))^(1/2),x, algorithm="giac")

[Out]

-2/105*(105*sqrt(2)*A*a^4*sgn(cos(d*x + c)) + 105*sqrt(2)*C*a^4*sgn(cos(d*x + c)) - (245*sqrt(2)*A*a^4*sgn(cos
(d*x + c)) + 105*sqrt(2)*C*a^4*sgn(cos(d*x + c)) - (175*sqrt(2)*A*a^4*sgn(cos(d*x + c)) + 147*sqrt(2)*C*a^4*sg
n(cos(d*x + c)) - (35*sqrt(2)*A*a^4*sgn(cos(d*x + c)) + 27*sqrt(2)*C*a^4*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)/((a*tan(1/2*d*x + 1/2*c)^2 - a)^3*s
qrt(-a*tan(1/2*d*x + 1/2*c)^2 + a)*d)