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

Optimal. Leaf size=144 \[ \frac {2 (7 B+6 C) \tan (c+d x) (a \sec (c+d x)+a)^{3/2}}{35 a d}-\frac {4 (7 B+6 C) \tan (c+d x) \sqrt {a \sec (c+d x)+a}}{105 d}+\frac {2 a (7 B+6 C) \tan (c+d x)}{15 d \sqrt {a \sec (c+d x)+a}}+\frac {2 a C \tan (c+d x) \sec ^3(c+d x)}{7 d \sqrt {a \sec (c+d x)+a}} \]

[Out]

2/35*(7*B+6*C)*(a+a*sec(d*x+c))^(3/2)*tan(d*x+c)/a/d+2/15*a*(7*B+6*C)*tan(d*x+c)/d/(a+a*sec(d*x+c))^(1/2)+2/7*
a*C*sec(d*x+c)^3*tan(d*x+c)/d/(a+a*sec(d*x+c))^(1/2)-4/105*(7*B+6*C)*(a+a*sec(d*x+c))^(1/2)*tan(d*x+c)/d

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Rubi [A]  time = 0.36, antiderivative size = 144, normalized size of antiderivative = 1.00, number of steps used = 5, number of rules used = 5, integrand size = 42, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.119, Rules used = {4072, 4016, 3800, 4001, 3792} \[ \frac {2 (7 B+6 C) \tan (c+d x) (a \sec (c+d x)+a)^{3/2}}{35 a d}-\frac {4 (7 B+6 C) \tan (c+d x) \sqrt {a \sec (c+d x)+a}}{105 d}+\frac {2 a (7 B+6 C) \tan (c+d x)}{15 d \sqrt {a \sec (c+d x)+a}}+\frac {2 a C \tan (c+d x) \sec ^3(c+d x)}{7 d \sqrt {a \sec (c+d x)+a}} \]

Antiderivative was successfully verified.

[In]

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

[Out]

(2*a*(7*B + 6*C)*Tan[c + d*x])/(15*d*Sqrt[a + a*Sec[c + d*x]]) + (2*a*C*Sec[c + d*x]^3*Tan[c + d*x])/(7*d*Sqrt
[a + a*Sec[c + d*x]]) - (4*(7*B + 6*C)*Sqrt[a + a*Sec[c + d*x]]*Tan[c + d*x])/(105*d) + (2*(7*B + 6*C)*(a + a*
Sec[c + d*x])^(3/2)*Tan[c + d*x])/(35*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 3800

Int[csc[(e_.) + (f_.)*(x_)]^3*(csc[(e_.) + (f_.)*(x_)]*(b_.) + (a_))^(m_), x_Symbol] :> -Simp[(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*(b
*(m + 1) - a*Csc[e + f*x]), x], x] /; FreeQ[{a, b, e, f, m}, x] && EqQ[a^2 - b^2, 0] &&  !LtQ[m, -2^(-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 4016

Int[(csc[(e_.) + (f_.)*(x_)]*(d_.))^(n_)*Sqrt[csc[(e_.) + (f_.)*(x_)]*(b_.) + (a_)]*(csc[(e_.) + (f_.)*(x_)]*(
B_.) + (A_)), x_Symbol] :> Simp[(-2*b*B*Cot[e + f*x]*(d*Csc[e + f*x])^n)/(f*(2*n + 1)*Sqrt[a + b*Csc[e + f*x]]
), x] + Dist[(A*b*(2*n + 1) + 2*a*B*n)/(b*(2*n + 1)), Int[Sqrt[a + b*Csc[e + f*x]]*(d*Csc[e + f*x])^n, x], x]
/; FreeQ[{a, b, d, e, f, A, B, n}, x] && NeQ[A*b - a*B, 0] && EqQ[a^2 - b^2, 0] && NeQ[A*b*(2*n + 1) + 2*a*B*n
, 0] &&  !LtQ[n, 0]

Rule 4072

Int[((a_.) + csc[(e_.) + (f_.)*(x_)]*(b_.))^(m_.)*((A_.) + csc[(e_.) + (f_.)*(x_)]*(B_.) + csc[(e_.) + (f_.)*(
x_)]^2*(C_.))*((c_.) + csc[(e_.) + (f_.)*(x_)]*(d_.))^(n_.), x_Symbol] :> Dist[1/b^2, Int[(a + b*Csc[e + f*x])
^(m + 1)*(c + d*Csc[e + f*x])^n*(b*B - a*C + b*C*Csc[e + f*x]), x], x] /; FreeQ[{a, b, c, d, e, f, A, B, C, m,
n}, x] && EqQ[A*b^2 - a*b*B + a^2*C, 0]

Rubi steps

\begin {align*} \int \sec ^2(c+d x) \sqrt {a+a \sec (c+d x)} \left (B \sec (c+d x)+C \sec ^2(c+d x)\right ) \, dx &=\int \sec ^3(c+d x) \sqrt {a+a \sec (c+d x)} (B+C \sec (c+d x)) \, dx\\ &=\frac {2 a C \sec ^3(c+d x) \tan (c+d x)}{7 d \sqrt {a+a \sec (c+d x)}}+\frac {1}{7} (7 B+6 C) \int \sec ^3(c+d x) \sqrt {a+a \sec (c+d x)} \, dx\\ &=\frac {2 a C \sec ^3(c+d x) \tan (c+d x)}{7 d \sqrt {a+a \sec (c+d x)}}+\frac {2 (7 B+6 C) (a+a \sec (c+d x))^{3/2} \tan (c+d x)}{35 a d}+\frac {(2 (7 B+6 C)) \int \sec (c+d x) \left (\frac {3 a}{2}-a \sec (c+d x)\right ) \sqrt {a+a \sec (c+d x)} \, dx}{35 a}\\ &=\frac {2 a C \sec ^3(c+d x) \tan (c+d x)}{7 d \sqrt {a+a \sec (c+d x)}}-\frac {4 (7 B+6 C) \sqrt {a+a \sec (c+d x)} \tan (c+d x)}{105 d}+\frac {2 (7 B+6 C) (a+a \sec (c+d x))^{3/2} \tan (c+d x)}{35 a d}+\frac {1}{15} (7 B+6 C) \int \sec (c+d x) \sqrt {a+a \sec (c+d x)} \, dx\\ &=\frac {2 a (7 B+6 C) \tan (c+d x)}{15 d \sqrt {a+a \sec (c+d x)}}+\frac {2 a C \sec ^3(c+d x) \tan (c+d x)}{7 d \sqrt {a+a \sec (c+d x)}}-\frac {4 (7 B+6 C) \sqrt {a+a \sec (c+d x)} \tan (c+d x)}{105 d}+\frac {2 (7 B+6 C) (a+a \sec (c+d x))^{3/2} \tan (c+d x)}{35 a d}\\ \end {align*}

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Mathematica [A]  time = 0.28, size = 81, normalized size = 0.56 \[ \frac {2 a \tan (c+d x) \left (3 (7 B+6 C) \sec ^2(c+d x)+4 (7 B+6 C) \sec (c+d x)+8 (7 B+6 C)+15 C \sec ^3(c+d x)\right )}{105 d \sqrt {a (\sec (c+d x)+1)}} \]

Antiderivative was successfully verified.

[In]

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

[Out]

(2*a*(8*(7*B + 6*C) + 4*(7*B + 6*C)*Sec[c + d*x] + 3*(7*B + 6*C)*Sec[c + d*x]^2 + 15*C*Sec[c + d*x]^3)*Tan[c +
d*x])/(105*d*Sqrt[a*(1 + Sec[c + d*x])])

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fricas [A]  time = 0.46, size = 105, normalized size = 0.73 \[ \frac {2 \, {\left (8 \, {\left (7 \, B + 6 \, C\right )} \cos \left (d x + c\right )^{3} + 4 \, {\left (7 \, B + 6 \, C\right )} \cos \left (d x + c\right )^{2} + 3 \, {\left (7 \, B + 6 \, C\right )} \cos \left (d x + c\right ) + 15 \, C\right )} \sqrt {\frac {a \cos \left (d x + c\right ) + a}{\cos \left (d x + c\right )}} \sin \left (d x + c\right )}{105 \, {\left (d \cos \left (d x + c\right )^{4} + d \cos \left (d x + c\right )^{3}\right )}} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

2/105*(8*(7*B + 6*C)*cos(d*x + c)^3 + 4*(7*B + 6*C)*cos(d*x + c)^2 + 3*(7*B + 6*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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giac [A]  time = 1.68, size = 222, normalized size = 1.54 \[ -\frac {2 \, {\left (105 \, \sqrt {2} B a^{4} \mathrm {sgn}\left (\cos \left (d x + c\right )\right ) + 105 \, \sqrt {2} C a^{4} \mathrm {sgn}\left (\cos \left (d x + c\right )\right ) - {\left (175 \, \sqrt {2} B a^{4} \mathrm {sgn}\left (\cos \left (d x + c\right )\right ) + 105 \, \sqrt {2} C a^{4} \mathrm {sgn}\left (\cos \left (d x + c\right )\right ) - {\left (119 \, \sqrt {2} B a^{4} \mathrm {sgn}\left (\cos \left (d x + c\right )\right ) + 147 \, \sqrt {2} C a^{4} \mathrm {sgn}\left (\cos \left (d x + c\right )\right ) - {\left (49 \, \sqrt {2} B a^{4} \mathrm {sgn}\left (\cos \left (d x + c\right )\right ) + 27 \, \sqrt {2} C a^{4} \mathrm {sgn}\left (\cos \left (d x + c\right )\right )\right )} \tan \left (\frac {1}{2} \, d x + \frac {1}{2} \, c\right )^{2}\right )} \tan \left (\frac {1}{2} \, d x + \frac {1}{2} \, c\right )^{2}\right )} \tan \left (\frac {1}{2} \, d x + \frac {1}{2} \, c\right )^{2}\right )} \tan \left (\frac {1}{2} \, d x + \frac {1}{2} \, c\right )}{105 \, {\left (a \tan \left (\frac {1}{2} \, d x + \frac {1}{2} \, c\right )^{2} - a\right )}^{3} \sqrt {-a \tan \left (\frac {1}{2} \, d x + \frac {1}{2} \, c\right )^{2} + a} d} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

-2/105*(105*sqrt(2)*B*a^4*sgn(cos(d*x + c)) + 105*sqrt(2)*C*a^4*sgn(cos(d*x + c)) - (175*sqrt(2)*B*a^4*sgn(cos
(d*x + c)) + 105*sqrt(2)*C*a^4*sgn(cos(d*x + c)) - (119*sqrt(2)*B*a^4*sgn(cos(d*x + c)) + 147*sqrt(2)*C*a^4*sg
n(cos(d*x + c)) - (49*sqrt(2)*B*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)

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maple [A]  time = 2.12, size = 116, normalized size = 0.81 \[ -\frac {2 \left (-1+\cos \left (d x +c \right )\right ) \left (56 B \left (\cos ^{3}\left (d x +c \right )\right )+48 C \left (\cos ^{3}\left (d x +c \right )\right )+28 B \left (\cos ^{2}\left (d x +c \right )\right )+24 C \left (\cos ^{2}\left (d x +c \right )\right )+21 B \cos \left (d x +c \right )+18 C \cos \left (d x +c \right )+15 C \right ) \sqrt {\frac {a \left (1+\cos \left (d x +c \right )\right )}{\cos \left (d x +c \right )}}}{105 d \cos \left (d x +c \right )^{3} \sin \left (d x +c \right )} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

-2/105/d*(-1+cos(d*x+c))*(56*B*cos(d*x+c)^3+48*C*cos(d*x+c)^3+28*B*cos(d*x+c)^2+24*C*cos(d*x+c)^2+21*B*cos(d*x
+c)+18*C*cos(d*x+c)+15*C)*(a*(1+cos(d*x+c))/cos(d*x+c))^(1/2)/cos(d*x+c)^3/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)^2*(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

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mupad [B]  time = 6.77, size = 407, normalized size = 2.83 \[ \frac {\sqrt {a+\frac {a}{\frac {{\mathrm {e}}^{-c\,1{}\mathrm {i}-d\,x\,1{}\mathrm {i}}}{2}+\frac {{\mathrm {e}}^{c\,1{}\mathrm {i}+d\,x\,1{}\mathrm {i}}}{2}}}\,\left (-\frac {B\,8{}\mathrm {i}}{5\,d}+\frac {C\,16{}\mathrm {i}}{5\,d}+{\mathrm {e}}^{c\,1{}\mathrm {i}+d\,x\,1{}\mathrm {i}}\,\left (\frac {C\,16{}\mathrm {i}}{35\,d}+\frac {\left (56\,B+112\,C\right )\,1{}\mathrm {i}}{35\,d}\right )\right )}{\left ({\mathrm {e}}^{c\,1{}\mathrm {i}+d\,x\,1{}\mathrm {i}}+1\right )\,{\left ({\mathrm {e}}^{c\,2{}\mathrm {i}+d\,x\,2{}\mathrm {i}}+1\right )}^2}+\frac {\sqrt {a+\frac {a}{\frac {{\mathrm {e}}^{-c\,1{}\mathrm {i}-d\,x\,1{}\mathrm {i}}}{2}+\frac {{\mathrm {e}}^{c\,1{}\mathrm {i}+d\,x\,1{}\mathrm {i}}}{2}}}\,\left (\frac {B\,8{}\mathrm {i}}{7\,d}+{\mathrm {e}}^{c\,1{}\mathrm {i}+d\,x\,1{}\mathrm {i}}\,\left (\frac {B\,8{}\mathrm {i}}{7\,d}-\frac {\left (8\,B+16\,C\right )\,1{}\mathrm {i}}{7\,d}\right )-\frac {\left (8\,B+16\,C\right )\,1{}\mathrm {i}}{7\,d}\right )}{\left ({\mathrm {e}}^{c\,1{}\mathrm {i}+d\,x\,1{}\mathrm {i}}+1\right )\,{\left ({\mathrm {e}}^{c\,2{}\mathrm {i}+d\,x\,2{}\mathrm {i}}+1\right )}^3}+\frac {\left (\frac {B\,8{}\mathrm {i}}{3\,d}-\frac {{\mathrm {e}}^{c\,1{}\mathrm {i}+d\,x\,1{}\mathrm {i}}\,\left (56\,B+48\,C\right )\,1{}\mathrm {i}}{105\,d}\right )\,\sqrt {a+\frac {a}{\frac {{\mathrm {e}}^{-c\,1{}\mathrm {i}-d\,x\,1{}\mathrm {i}}}{2}+\frac {{\mathrm {e}}^{c\,1{}\mathrm {i}+d\,x\,1{}\mathrm {i}}}{2}}}}{\left ({\mathrm {e}}^{c\,1{}\mathrm {i}+d\,x\,1{}\mathrm {i}}+1\right )\,\left ({\mathrm {e}}^{c\,2{}\mathrm {i}+d\,x\,2{}\mathrm {i}}+1\right )}-\frac {{\mathrm {e}}^{c\,1{}\mathrm {i}+d\,x\,1{}\mathrm {i}}\,\sqrt {a+\frac {a}{\frac {{\mathrm {e}}^{-c\,1{}\mathrm {i}-d\,x\,1{}\mathrm {i}}}{2}+\frac {{\mathrm {e}}^{c\,1{}\mathrm {i}+d\,x\,1{}\mathrm {i}}}{2}}}\,\left (112\,B+96\,C\right )\,1{}\mathrm {i}}{105\,d\,\left ({\mathrm {e}}^{c\,1{}\mathrm {i}+d\,x\,1{}\mathrm {i}}+1\right )} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

((a + a/(exp(- c*1i - d*x*1i)/2 + exp(c*1i + d*x*1i)/2))^(1/2)*((C*16i)/(5*d) - (B*8i)/(5*d) + exp(c*1i + d*x*
1i)*((C*16i)/(35*d) + ((56*B + 112*C)*1i)/(35*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)*((B*8i)/(7*d) + exp(c*1i + d*x*1i)*((B*8i)/(7*d)
- ((8*B + 16*C)*1i)/(7*d)) - ((8*B + 16*C)*1i)/(7*d)))/((exp(c*1i + d*x*1i) + 1)*(exp(c*2i + d*x*2i) + 1)^3) +
(((B*8i)/(3*d) - (exp(c*1i + d*x*1i)*(56*B + 48*C)*1i)/(105*d))*(a + a/(exp(- c*1i - d*x*1i)/2 + exp(c*1i + d
*x*1i)/2))^(1/2))/((exp(c*1i + d*x*1i) + 1)*(exp(c*2i + d*x*2i) + 1)) - (exp(c*1i + d*x*1i)*(a + a/(exp(- c*1i
- d*x*1i)/2 + exp(c*1i + d*x*1i)/2))^(1/2)*(112*B + 96*C)*1i)/(105*d*(exp(c*1i + d*x*1i) + 1))

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sympy [F]  time = 0.00, size = 0, normalized size = 0.00 \[ \int \sqrt {a \left (\sec {\left (c + d x \right )} + 1\right )} \left (B + C \sec {\left (c + d x \right )}\right ) \sec ^{3}{\left (c + d x \right )}\, dx \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate(sec(d*x+c)**2*(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))*(B + C*sec(c + d*x))*sec(c + d*x)**3, x)

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