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

Optimal. Leaf size=101 \[ \frac {2 (5 B-2 C) \tan (c+d x) \sqrt {a \sec (c+d x)+a}}{15 d}+\frac {2 a (5 B+7 C) \tan (c+d x)}{15 d \sqrt {a \sec (c+d x)+a}}+\frac {2 C \tan (c+d x) (a \sec (c+d x)+a)^{3/2}}{5 a d} \]

[Out]

2/5*C*(a+a*sec(d*x+c))^(3/2)*tan(d*x+c)/a/d+2/15*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

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Rubi [A]  time = 0.28, antiderivative size = 101, normalized size of antiderivative = 1.00, number of steps used = 4, number of rules used = 4, integrand size = 40, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.100, Rules used = {4072, 4010, 4001, 3792} \[ \frac {2 (5 B-2 C) \tan (c+d x) \sqrt {a \sec (c+d x)+a}}{15 d}+\frac {2 a (5 B+7 C) \tan (c+d x)}{15 d \sqrt {a \sec (c+d x)+a}}+\frac {2 C \tan (c+d x) (a \sec (c+d x)+a)^{3/2}}{5 a d} \]

Antiderivative was successfully verified.

[In]

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

[Out]

(2*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 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 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 (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 ^2(c+d x) \sqrt {a+a \sec (c+d x)} (B+C \sec (c+d x)) \, 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 {3 a C}{2}+\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} (5 B+7 C) \int \sec (c+d x) \sqrt {a+a \sec (c+d x)} \, dx\\ &=\frac {2 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*}

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Mathematica [A]  time = 0.33, size = 80, normalized size = 0.79 \[ \frac {2 \tan (c+d x) \sec (c+d x) \sqrt {a (\sec (c+d x)+1)} ((5 B+4 C) \cos (c+d x)+(5 B+4 C) \cos (2 (c+d x))+5 B+7 C)}{15 d (\cos (c+d x)+1)} \]

Antiderivative was successfully verified.

[In]

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

[Out]

(2*(5*B + 7*C + (5*B + 4*C)*Cos[c + d*x] + (5*B + 4*C)*Cos[2*(c + d*x)])*Sec[c + d*x]*Sqrt[a*(1 + Sec[c + d*x]
)]*Tan[c + d*x])/(15*d*(1 + Cos[c + d*x]))

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fricas [A]  time = 0.44, size = 87, normalized size = 0.86 \[ \frac {2 \, {\left (2 \, {\left (5 \, B + 4 \, C\right )} \cos \left (d x + c\right )^{2} + {\left (5 \, B + 4 \, C\right )} \cos \left (d x + c\right ) + 3 \, C\right )} \sqrt {\frac {a \cos \left (d x + c\right ) + a}{\cos \left (d x + c\right )}} \sin \left (d x + c\right )}{15 \, {\left (d \cos \left (d x + c\right )^{3} + d \cos \left (d x + c\right )^{2}\right )}} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate(sec(d*x+c)*(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*(2*(5*B + 4*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))*s
in(d*x + c)/(d*cos(d*x + c)^3 + d*cos(d*x + c)^2)

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giac [A]  time = 4.98, size = 176, normalized size = 1.74 \[ \frac {2 \, {\left (15 \, \sqrt {2} B a^{3} \mathrm {sgn}\left (\cos \left (d x + c\right )\right ) + 15 \, \sqrt {2} C a^{3} \mathrm {sgn}\left (\cos \left (d x + c\right )\right ) - {\left (20 \, \sqrt {2} B a^{3} \mathrm {sgn}\left (\cos \left (d x + c\right )\right ) + 10 \, \sqrt {2} C a^{3} \mathrm {sgn}\left (\cos \left (d x + c\right )\right ) - {\left (5 \, \sqrt {2} B a^{3} \mathrm {sgn}\left (\cos \left (d x + c\right )\right ) + 7 \, \sqrt {2} C a^{3} \mathrm {sgn}\left (\cos \left (d x + c\right )\right )\right )} \tan \left (\frac {1}{2} \, d x + \frac {1}{2} \, c\right )^{2}\right )} \tan \left (\frac {1}{2} \, d x + \frac {1}{2} \, c\right )^{2}\right )} \tan \left (\frac {1}{2} \, d x + \frac {1}{2} \, c\right )}{15 \, {\left (a \tan \left (\frac {1}{2} \, d x + \frac {1}{2} \, c\right )^{2} - a\right )}^{2} \sqrt {-a \tan \left (\frac {1}{2} \, d x + \frac {1}{2} \, c\right )^{2} + a} d} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate(sec(d*x+c)*(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*(15*sqrt(2)*B*a^3*sgn(cos(d*x + c)) + 15*sqrt(2)*C*a^3*sgn(cos(d*x + c)) - (20*sqrt(2)*B*a^3*sgn(cos(d*x
+ c)) + 10*sqrt(2)*C*a^3*sgn(cos(d*x + c)) - (5*sqrt(2)*B*a^3*sgn(cos(d*x + c)) + 7*sqrt(2)*C*a^3*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)/((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 = 1.75, size = 94, normalized size = 0.93 \[ -\frac {2 \left (-1+\cos \left (d x +c \right )\right ) \left (10 B \left (\cos ^{2}\left (d x +c \right )\right )+8 C \left (\cos ^{2}\left (d x +c \right )\right )+5 B \cos \left (d x +c \right )+4 C \cos \left (d x +c \right )+3 C \right ) \sqrt {\frac {a \left (1+\cos \left (d x +c \right )\right )}{\cos \left (d x +c \right )}}}{15 d \cos \left (d x +c \right )^{2} \sin \left (d x +c \right )} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

int(sec(d*x+c)*(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))*(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)

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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)*(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 = 7.25, size = 212, normalized size = 2.10 \[ -\frac {4\,\left ({\mathrm {e}}^{c\,1{}\mathrm {i}+d\,x\,1{}\mathrm {i}}-1\right )\,\sqrt {a+\frac {a}{\frac {{\mathrm {e}}^{-c\,1{}\mathrm {i}-d\,x\,1{}\mathrm {i}}}{2}+\frac {{\mathrm {e}}^{c\,1{}\mathrm {i}+d\,x\,1{}\mathrm {i}}}{2}}}\,\left (B\,5{}\mathrm {i}+C\,4{}\mathrm {i}+B\,{\mathrm {e}}^{c\,1{}\mathrm {i}+d\,x\,1{}\mathrm {i}}\,5{}\mathrm {i}+B\,{\mathrm {e}}^{c\,2{}\mathrm {i}+d\,x\,2{}\mathrm {i}}\,10{}\mathrm {i}+B\,{\mathrm {e}}^{c\,3{}\mathrm {i}+d\,x\,3{}\mathrm {i}}\,5{}\mathrm {i}+B\,{\mathrm {e}}^{c\,4{}\mathrm {i}+d\,x\,4{}\mathrm {i}}\,5{}\mathrm {i}+C\,{\mathrm {e}}^{c\,1{}\mathrm {i}+d\,x\,1{}\mathrm {i}}\,4{}\mathrm {i}+C\,{\mathrm {e}}^{c\,2{}\mathrm {i}+d\,x\,2{}\mathrm {i}}\,14{}\mathrm {i}+C\,{\mathrm {e}}^{c\,3{}\mathrm {i}+d\,x\,3{}\mathrm {i}}\,4{}\mathrm {i}+C\,{\mathrm {e}}^{c\,4{}\mathrm {i}+d\,x\,4{}\mathrm {i}}\,4{}\mathrm {i}\right )}{15\,d\,\left ({\mathrm {e}}^{c\,1{}\mathrm {i}+d\,x\,1{}\mathrm {i}}+1\right )\,{\left ({\mathrm {e}}^{c\,2{}\mathrm {i}+d\,x\,2{}\mathrm {i}}+1\right )}^2} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

-(4*(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)*(B*5i + C*4i + B*ex
p(c*1i + d*x*1i)*5i + B*exp(c*2i + d*x*2i)*10i + B*exp(c*3i + d*x*3i)*5i + B*exp(c*4i + d*x*4i)*5i + C*exp(c*1
i + d*x*1i)*4i + C*exp(c*2i + d*x*2i)*14i + C*exp(c*3i + d*x*3i)*4i + C*exp(c*4i + d*x*4i)*4i))/(15*d*(exp(c*1
i + 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 \[ \int \sqrt {a \left (\sec {\left (c + d x \right )} + 1\right )} \left (B + C \sec {\left (c + d x \right )}\right ) \sec ^{2}{\left (c + d x \right )}\, dx \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate(sec(d*x+c)*(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)**2, x)

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