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

Optimal. Leaf size=147 \[ \frac {2 (35 A-14 B+18 C) \tan (c+d x) \sqrt {a \sec (c+d x)+a}}{105 d}+\frac {2 a (35 A+49 B+27 C) \tan (c+d x)}{105 d \sqrt {a \sec (c+d x)+a}}+\frac {2 (7 B+C) \tan (c+d x) (a \sec (c+d x)+a)^{3/2}}{35 a d}+\frac {2 C \tan (c+d x) \sec ^2(c+d x) \sqrt {a \sec (c+d x)+a}}{7 d} \]

[Out]

2/35*(7*B+C)*(a+a*sec(d*x+c))^(3/2)*tan(d*x+c)/a/d+2/105*a*(35*A+49*B+27*C)*tan(d*x+c)/d/(a+a*sec(d*x+c))^(1/2
)+2/105*(35*A-14*B+18*C)*(a+a*sec(d*x+c))^(1/2)*tan(d*x+c)/d+2/7*C*sec(d*x+c)^2*(a+a*sec(d*x+c))^(1/2)*tan(d*x
+c)/d

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Rubi [A]  time = 0.43, antiderivative size = 147, normalized size of antiderivative = 1.00, number of steps used = 4, number of rules used = 4, integrand size = 43, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.093, Rules used = {4088, 4010, 4001, 3792} \[ \frac {2 (35 A-14 B+18 C) \tan (c+d x) \sqrt {a \sec (c+d x)+a}}{105 d}+\frac {2 a (35 A+49 B+27 C) \tan (c+d x)}{105 d \sqrt {a \sec (c+d x)+a}}+\frac {2 (7 B+C) \tan (c+d x) (a \sec (c+d x)+a)^{3/2}}{35 a d}+\frac {2 C \tan (c+d x) \sec ^2(c+d x) \sqrt {a \sec (c+d x)+a}}{7 d} \]

Antiderivative was successfully verified.

[In]

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

[Out]

(2*a*(35*A + 49*B + 27*C)*Tan[c + d*x])/(105*d*Sqrt[a + a*Sec[c + d*x]]) + (2*(35*A - 14*B + 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*(7
*B + 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 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 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 4088

Int[((A_.) + csc[(e_.) + (f_.)*(x_)]*(B_.) + 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 + b*B*(m + n + 1))*Csc[e + f*x], x], x], x] /; FreeQ[{a, b, d, e, f, A,
B, C, m, n}, x] && EqQ[a^2 - b^2, 0] &&  !LtQ[m, -2^(-1)] &&  !LtQ[n, -2^(-1)] && NeQ[m + n + 1, 0]

Rubi steps

\begin {align*} \int \sec ^2(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 \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 (7 B+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 (7 B+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}{4} a^2 (7 B+C)+\frac {1}{4} a^2 (35 A-14 B+18 C) \sec (c+d x)\right ) \, dx}{35 a^2}\\ &=\frac {2 (35 A-14 B+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 (7 B+C) (a+a \sec (c+d x))^{3/2} \tan (c+d x)}{35 a d}+\frac {1}{105} (35 A+49 B+27 C) \int \sec (c+d x) \sqrt {a+a \sec (c+d x)} \, dx\\ &=\frac {2 a (35 A+49 B+27 C) \tan (c+d x)}{105 d \sqrt {a+a \sec (c+d x)}}+\frac {2 (35 A-14 B+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 (7 B+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 = 1.34, size = 119, normalized size = 0.81 \[ \frac {\tan \left (\frac {1}{2} (c+d x)\right ) \sec ^3(c+d x) \sqrt {a (\sec (c+d x)+1)} (3 (35 A+42 B+36 C) \cos (c+d x)+(35 A+28 B+24 C) \cos (2 (c+d x))+35 A \cos (3 (c+d x))+35 A+28 B \cos (3 (c+d x))+28 B+24 C \cos (3 (c+d x))+54 C)}{105 d} \]

Antiderivative was successfully verified.

[In]

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

[Out]

((35*A + 28*B + 54*C + 3*(35*A + 42*B + 36*C)*Cos[c + d*x] + (35*A + 28*B + 24*C)*Cos[2*(c + d*x)] + 35*A*Cos[
3*(c + d*x)] + 28*B*Cos[3*(c + d*x)] + 24*C*Cos[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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fricas [A]  time = 0.43, size = 110, normalized size = 0.75 \[ \frac {2 \, {\left (2 \, {\left (35 \, A + 28 \, B + 24 \, C\right )} \cos \left (d x + c\right )^{3} + {\left (35 \, A + 28 \, B + 24 \, C\right )} \cos \left (d x + c\right )^{2} + 3 \, {\left (7 \, B + 6 \, C\right )} \cos \left (d x + c\right ) + 15 \, C\right )} \sqrt {\frac {a \cos \left (d x + c\right ) + a}{\cos \left (d x + c\right )}} \sin \left (d x + c\right )}{105 \, {\left (d \cos \left (d x + c\right )^{4} + d \cos \left (d x + c\right )^{3}\right )}} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

2/105*(2*(35*A + 28*B + 24*C)*cos(d*x + c)^3 + (35*A + 28*B + 24*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 [B]  time = 1.39, size = 286, normalized size = 1.95 \[ -\frac {2 \, {\left (105 \, \sqrt {2} A a^{4} \mathrm {sgn}\left (\cos \left (d x + c\right )\right ) + 105 \, \sqrt {2} B a^{4} \mathrm {sgn}\left (\cos \left (d x + c\right )\right ) + 105 \, \sqrt {2} C a^{4} \mathrm {sgn}\left (\cos \left (d x + c\right )\right ) - {\left (245 \, \sqrt {2} A a^{4} \mathrm {sgn}\left (\cos \left (d x + c\right )\right ) + 175 \, \sqrt {2} B a^{4} \mathrm {sgn}\left (\cos \left (d x + c\right )\right ) + 105 \, \sqrt {2} C a^{4} \mathrm {sgn}\left (\cos \left (d x + c\right )\right ) - {\left (175 \, \sqrt {2} A a^{4} \mathrm {sgn}\left (\cos \left (d x + c\right )\right ) + 119 \, \sqrt {2} B a^{4} \mathrm {sgn}\left (\cos \left (d x + c\right )\right ) + 147 \, \sqrt {2} C a^{4} \mathrm {sgn}\left (\cos \left (d x + c\right )\right ) - {\left (35 \, \sqrt {2} A a^{4} \mathrm {sgn}\left (\cos \left (d x + c\right )\right ) + 49 \, \sqrt {2} B a^{4} \mathrm {sgn}\left (\cos \left (d x + c\right )\right ) + 27 \, \sqrt {2} C a^{4} \mathrm {sgn}\left (\cos \left (d x + c\right )\right )\right )} \tan \left (\frac {1}{2} \, d x + \frac {1}{2} \, c\right )^{2}\right )} \tan \left (\frac {1}{2} \, d x + \frac {1}{2} \, c\right )^{2}\right )} \tan \left (\frac {1}{2} \, d x + \frac {1}{2} \, c\right )^{2}\right )} \tan \left (\frac {1}{2} \, d x + \frac {1}{2} \, c\right )}{105 \, {\left (a \tan \left (\frac {1}{2} \, d x + \frac {1}{2} \, c\right )^{2} - a\right )}^{3} \sqrt {-a \tan \left (\frac {1}{2} \, d x + \frac {1}{2} \, c\right )^{2} + a} d} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

-2/105*(105*sqrt(2)*A*a^4*sgn(cos(d*x + c)) + 105*sqrt(2)*B*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)) + 175*sqrt(2)*B*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)) + 119*sqrt(2)*B*a^4*sgn(cos(d*x + c)) + 147*sqrt(2)*C*a^
4*sgn(cos(d*x + c)) - (35*sqrt(2)*A*a^4*sgn(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*sqrt(-a*tan(1/2*d*x + 1/2*c)^2 + a)*d)

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maple [A]  time = 2.08, size = 138, normalized size = 0.94 \[ -\frac {2 \left (-1+\cos \left (d x +c \right )\right ) \left (70 A \left (\cos ^{3}\left (d x +c \right )\right )+56 B \left (\cos ^{3}\left (d x +c \right )\right )+48 C \left (\cos ^{3}\left (d x +c \right )\right )+35 A \left (\cos ^{2}\left (d x +c \right )\right )+28 B \left (\cos ^{2}\left (d x +c \right )\right )+24 C \left (\cos ^{2}\left (d x +c \right )\right )+21 B \cos \left (d x +c \right )+18 C \cos \left (d x +c \right )+15 C \right ) \sqrt {\frac {a \left (1+\cos \left (d x +c \right )\right )}{\cos \left (d x +c \right )}}}{105 d \cos \left (d x +c \right )^{3} \sin \left (d x +c \right )} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

int(sec(d*x+c)^2*(A+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))*(70*A*cos(d*x+c)^3+56*B*cos(d*x+c)^3+48*C*cos(d*x+c)^3+35*A*cos(d*x+c)^2+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*(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

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mupad [B]  time = 7.64, size = 479, normalized size = 3.26 \[ \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 ({\mathrm {e}}^{c\,1{}\mathrm {i}+d\,x\,1{}\mathrm {i}}\,\left (\frac {A\,4{}\mathrm {i}}{3\,d}-\frac {\left (56\,B+48\,C\right )\,1{}\mathrm {i}}{105\,d}\right )+\frac {\left (140\,A+280\,B\right )\,1{}\mathrm {i}}{105\,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 )}+\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 ({\mathrm {e}}^{c\,1{}\mathrm {i}+d\,x\,1{}\mathrm {i}}\,\left (\frac {A\,4{}\mathrm {i}}{7\,d}-\frac {\left (8\,A+8\,B+16\,C\right )\,1{}\mathrm {i}}{7\,d}+\frac {\left (4\,A+8\,B\right )\,1{}\mathrm {i}}{7\,d}\right )+\frac {A\,4{}\mathrm {i}}{7\,d}-\frac {\left (8\,A+8\,B+16\,C\right )\,1{}\mathrm {i}}{7\,d}+\frac {\left (4\,A+8\,B\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 {\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}}\,\left (-\frac {A\,4{}\mathrm {i}}{5\,d}+\frac {C\,16{}\mathrm {i}}{35\,d}+\frac {\left (28\,A+56\,B+112\,C\right )\,1{}\mathrm {i}}{35\,d}\right )-\frac {\left (28\,A+56\,B\right )\,1{}\mathrm {i}}{35\,d}+\frac {\left (28\,A+112\,C\right )\,1{}\mathrm {i}}{35\,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 )}^2}-\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 (140\,A+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(((a + a/cos(c + d*x))^(1/2)*(A + B/cos(c + d*x) + C/cos(c + d*x)^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)*(exp(c*1i + d*x*1i)*((A*4i)/(3*d) - ((56*B + 48
*C)*1i)/(105*d)) + ((140*A + 280*B)*1i)/(105*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*4i)/(7*d) - ((8*A + 8*B + 16*
C)*1i)/(7*d) + ((4*A + 8*B)*1i)/(7*d)) + (A*4i)/(7*d) - ((8*A + 8*B + 16*C)*1i)/(7*d) + ((4*A + 8*B)*1i)/(7*d)
))/((exp(c*1i + d*x*1i) + 1)*(exp(c*2i + d*x*2i) + 1)^3) + ((a + a/(exp(- c*1i - d*x*1i)/2 + exp(c*1i + d*x*1i
)/2))^(1/2)*(exp(c*1i + d*x*1i)*((C*16i)/(35*d) - (A*4i)/(5*d) + ((28*A + 56*B + 112*C)*1i)/(35*d)) - ((28*A +
56*B)*1i)/(35*d) + ((28*A + 112*C)*1i)/(35*d)))/((exp(c*1i + d*x*1i) + 1)*(exp(c*2i + d*x*2i) + 1)^2) - (exp(
c*1i + d*x*1i)*(a + a/(exp(- c*1i - d*x*1i)/2 + exp(c*1i + d*x*1i)/2))^(1/2)*(140*A + 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 \[ \int \sqrt {a \left (\sec {\left (c + d x \right )} + 1\right )} \left (A + B \sec {\left (c + d x \right )} + C \sec ^{2}{\left (c + d x \right )}\right ) \sec ^{2}{\left (c + d x \right )}\, dx \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate(sec(d*x+c)**2*(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)**2, x)

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