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

Optimal. Leaf size=118 \[ -\frac{\sqrt{2} (B-C) \tan ^{-1}\left (\frac{\sqrt{a} \tan (c+d x)}{\sqrt{2} \sqrt{a \sec (c+d x)+a}}\right )}{\sqrt{a} d}+\frac{2 (3 B-2 C) \tan (c+d x)}{3 d \sqrt{a \sec (c+d x)+a}}+\frac{2 C \tan (c+d x) \sqrt{a \sec (c+d x)+a}}{3 a d} \]

[Out]

-((Sqrt[2]*(B - C)*ArcTan[(Sqrt[a]*Tan[c + d*x])/(Sqrt[2]*Sqrt[a + a*Sec[c + d*x]])])/(Sqrt[a]*d)) + (2*(3*B -
2*C)*Tan[c + d*x])/(3*d*Sqrt[a + a*Sec[c + d*x]]) + (2*C*Sqrt[a + a*Sec[c + d*x]]*Tan[c + d*x])/(3*a*d)

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Rubi [A]  time = 0.303758, antiderivative size = 118, normalized size of antiderivative = 1., number of steps used = 5, number of rules used = 5, integrand size = 40, \(\frac{\text{number of rules}}{\text{integrand size}}\) = 0.125, Rules used = {4072, 4010, 4001, 3795, 203} \[ -\frac{\sqrt{2} (B-C) \tan ^{-1}\left (\frac{\sqrt{a} \tan (c+d x)}{\sqrt{2} \sqrt{a \sec (c+d x)+a}}\right )}{\sqrt{a} d}+\frac{2 (3 B-2 C) \tan (c+d x)}{3 d \sqrt{a \sec (c+d x)+a}}+\frac{2 C \tan (c+d x) \sqrt{a \sec (c+d x)+a}}{3 a d} \]

Antiderivative was successfully verified.

[In]

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

[Out]

-((Sqrt[2]*(B - C)*ArcTan[(Sqrt[a]*Tan[c + d*x])/(Sqrt[2]*Sqrt[a + a*Sec[c + d*x]])])/(Sqrt[a]*d)) + (2*(3*B -
2*C)*Tan[c + d*x])/(3*d*Sqrt[a + a*Sec[c + d*x]]) + (2*C*Sqrt[a + a*Sec[c + d*x]]*Tan[c + d*x])/(3*a*d)

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]

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 3795

Int[csc[(e_.) + (f_.)*(x_)]/Sqrt[csc[(e_.) + (f_.)*(x_)]*(b_.) + (a_)], x_Symbol] :> Dist[-2/f, Subst[Int[1/(2
*a + x^2), x], x, (b*Cot[e + f*x])/Sqrt[a + b*Csc[e + f*x]]], x] /; FreeQ[{a, b, e, f}, x] && EqQ[a^2 - b^2, 0
]

Rule 203

Int[((a_) + (b_.)*(x_)^2)^(-1), x_Symbol] :> Simp[(1*ArcTan[(Rt[b, 2]*x)/Rt[a, 2]])/(Rt[a, 2]*Rt[b, 2]), x] /;
FreeQ[{a, b}, x] && PosQ[a/b] && (GtQ[a, 0] || GtQ[b, 0])

Rubi steps

\begin{align*} \int \frac{\sec (c+d x) \left (B \sec (c+d x)+C \sec ^2(c+d x)\right )}{\sqrt{a+a \sec (c+d x)}} \, dx &=\int \frac{\sec ^2(c+d x) (B+C \sec (c+d x))}{\sqrt{a+a \sec (c+d x)}} \, dx\\ &=\frac{2 C \sqrt{a+a \sec (c+d x)} \tan (c+d x)}{3 a d}+\frac{2 \int \frac{\sec (c+d x) \left (\frac{a C}{2}+\frac{1}{2} a (3 B-2 C) \sec (c+d x)\right )}{\sqrt{a+a \sec (c+d x)}} \, dx}{3 a}\\ &=\frac{2 (3 B-2 C) \tan (c+d x)}{3 d \sqrt{a+a \sec (c+d x)}}+\frac{2 C \sqrt{a+a \sec (c+d x)} \tan (c+d x)}{3 a d}+(-B+C) \int \frac{\sec (c+d x)}{\sqrt{a+a \sec (c+d x)}} \, dx\\ &=\frac{2 (3 B-2 C) \tan (c+d x)}{3 d \sqrt{a+a \sec (c+d x)}}+\frac{2 C \sqrt{a+a \sec (c+d x)} \tan (c+d x)}{3 a d}+\frac{(2 (B-C)) \operatorname{Subst}\left (\int \frac{1}{2 a+x^2} \, dx,x,-\frac{a \tan (c+d x)}{\sqrt{a+a \sec (c+d x)}}\right )}{d}\\ &=-\frac{\sqrt{2} (B-C) \tan ^{-1}\left (\frac{\sqrt{a} \tan (c+d x)}{\sqrt{2} \sqrt{a+a \sec (c+d x)}}\right )}{\sqrt{a} d}+\frac{2 (3 B-2 C) \tan (c+d x)}{3 d \sqrt{a+a \sec (c+d x)}}+\frac{2 C \sqrt{a+a \sec (c+d x)} \tan (c+d x)}{3 a d}\\ \end{align*}

Mathematica [A]  time = 0.296168, size = 106, normalized size = 0.9 \[ \frac{\tan (c+d x) \left (2 \sqrt{1-\sec (c+d x)} (3 B+C \sec (c+d x)-C)-3 \sqrt{2} (B-C) \tanh ^{-1}\left (\frac{\sqrt{1-\sec (c+d x)}}{\sqrt{2}}\right )\right )}{3 d \sqrt{1-\sec (c+d x)} \sqrt{a (\sec (c+d x)+1)}} \]

Antiderivative was successfully verified.

[In]

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

[Out]

((-3*Sqrt[2]*(B - C)*ArcTanh[Sqrt[1 - Sec[c + d*x]]/Sqrt[2]] + 2*Sqrt[1 - Sec[c + d*x]]*(3*B - C + C*Sec[c + d
*x]))*Tan[c + d*x])/(3*d*Sqrt[1 - Sec[c + d*x]]*Sqrt[a*(1 + Sec[c + d*x])])

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Maple [B]  time = 0.303, size = 405, normalized size = 3.4 \begin{align*} -{\frac{1}{6\,ad\sin \left ( dx+c \right ) \cos \left ( dx+c \right ) } \left ( -3\,B\cos \left ( dx+c \right ) \sin \left ( dx+c \right ) \left ( -2\,{\frac{\cos \left ( dx+c \right ) }{\cos \left ( dx+c \right ) +1}} \right ) ^{3/2}\ln \left ({\frac{1}{\sin \left ( dx+c \right ) } \left ( \sqrt{-2\,{\frac{\cos \left ( dx+c \right ) }{\cos \left ( dx+c \right ) +1}}}\sin \left ( dx+c \right ) -\cos \left ( dx+c \right ) +1 \right ) } \right ) +3\,C\cos \left ( dx+c \right ) \sin \left ( dx+c \right ) \left ( -2\,{\frac{\cos \left ( dx+c \right ) }{\cos \left ( dx+c \right ) +1}} \right ) ^{3/2}\ln \left ({\frac{1}{\sin \left ( dx+c \right ) } \left ( \sqrt{-2\,{\frac{\cos \left ( dx+c \right ) }{\cos \left ( dx+c \right ) +1}}}\sin \left ( dx+c \right ) -\cos \left ( dx+c \right ) +1 \right ) } \right ) -3\,B \left ( -2\,{\frac{\cos \left ( dx+c \right ) }{\cos \left ( dx+c \right ) +1}} \right ) ^{3/2}\ln \left ({\frac{1}{\sin \left ( dx+c \right ) } \left ( \sqrt{-2\,{\frac{\cos \left ( dx+c \right ) }{\cos \left ( dx+c \right ) +1}}}\sin \left ( dx+c \right ) -\cos \left ( dx+c \right ) +1 \right ) } \right ) \sin \left ( dx+c \right ) +3\,C \left ( -2\,{\frac{\cos \left ( dx+c \right ) }{\cos \left ( dx+c \right ) +1}} \right ) ^{3/2}\ln \left ({\frac{1}{\sin \left ( dx+c \right ) } \left ( \sqrt{-2\,{\frac{\cos \left ( dx+c \right ) }{\cos \left ( dx+c \right ) +1}}}\sin \left ( dx+c \right ) -\cos \left ( dx+c \right ) +1 \right ) } \right ) \sin \left ( dx+c \right ) +12\,B \left ( \cos \left ( dx+c \right ) \right ) ^{2}-4\,C \left ( \cos \left ( dx+c \right ) \right ) ^{2}-12\,B\cos \left ( dx+c \right ) +8\,C\cos \left ( dx+c \right ) -4\,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)*(B*sec(d*x+c)+C*sec(d*x+c)^2)/(a+a*sec(d*x+c))^(1/2),x)

[Out]

-1/6/d/a*(-3*B*cos(d*x+c)*sin(d*x+c)*(-2*cos(d*x+c)/(cos(d*x+c)+1))^(3/2)*ln(((-2*cos(d*x+c)/(cos(d*x+c)+1))^(
1/2)*sin(d*x+c)-cos(d*x+c)+1)/sin(d*x+c))+3*C*cos(d*x+c)*sin(d*x+c)*(-2*cos(d*x+c)/(cos(d*x+c)+1))^(3/2)*ln(((
-2*cos(d*x+c)/(cos(d*x+c)+1))^(1/2)*sin(d*x+c)-cos(d*x+c)+1)/sin(d*x+c))-3*B*(-2*cos(d*x+c)/(cos(d*x+c)+1))^(3
/2)*ln(((-2*cos(d*x+c)/(cos(d*x+c)+1))^(1/2)*sin(d*x+c)-cos(d*x+c)+1)/sin(d*x+c))*sin(d*x+c)+3*C*(-2*cos(d*x+c
)/(cos(d*x+c)+1))^(3/2)*ln(((-2*cos(d*x+c)/(cos(d*x+c)+1))^(1/2)*sin(d*x+c)-cos(d*x+c)+1)/sin(d*x+c))*sin(d*x+
c)+12*B*cos(d*x+c)^2-4*C*cos(d*x+c)^2-12*B*cos(d*x+c)+8*C*cos(d*x+c)-4*C)*(a*(cos(d*x+c)+1)/cos(d*x+c))^(1/2)/
sin(d*x+c)/cos(d*x+c)

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

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]

integrate((C*sec(d*x + c)^2 + B*sec(d*x + c))*sec(d*x + c)/sqrt(a*sec(d*x + c) + a), x)

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Fricas [A]  time = 0.642421, size = 917, normalized size = 7.77 \begin{align*} \left [-\frac{3 \, \sqrt{2}{\left ({\left (B - C\right )} a \cos \left (d x + c\right )^{2} +{\left (B - C\right )} a \cos \left (d x + c\right )\right )} \sqrt{-\frac{1}{a}} \log \left (-\frac{2 \, \sqrt{2} \sqrt{\frac{a \cos \left (d x + c\right ) + a}{\cos \left (d x + c\right )}} \sqrt{-\frac{1}{a}} \cos \left (d x + c\right ) \sin \left (d x + c\right ) - 3 \, \cos \left (d x + c\right )^{2} - 2 \, \cos \left (d x + c\right ) + 1}{\cos \left (d x + c\right )^{2} + 2 \, \cos \left (d x + c\right ) + 1}\right ) - 4 \,{\left ({\left (3 \, B - C\right )} \cos \left (d x + c\right ) + C\right )} \sqrt{\frac{a \cos \left (d x + c\right ) + a}{\cos \left (d x + c\right )}} \sin \left (d x + c\right )}{6 \,{\left (a d \cos \left (d x + c\right )^{2} + a d \cos \left (d x + c\right )\right )}}, \frac{2 \,{\left ({\left (3 \, B - C\right )} \cos \left (d x + c\right ) + C\right )} \sqrt{\frac{a \cos \left (d x + c\right ) + a}{\cos \left (d x + c\right )}} \sin \left (d x + c\right ) + \frac{3 \, \sqrt{2}{\left ({\left (B - C\right )} a \cos \left (d x + c\right )^{2} +{\left (B - C\right )} a \cos \left (d x + c\right )\right )} \arctan \left (\frac{\sqrt{2} \sqrt{\frac{a \cos \left (d x + c\right ) + a}{\cos \left (d x + c\right )}} \cos \left (d x + c\right )}{\sqrt{a} \sin \left (d x + c\right )}\right )}{\sqrt{a}}}{3 \,{\left (a d \cos \left (d x + c\right )^{2} + a d \cos \left (d x + c\right )\right )}}\right ] \end{align*}

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]

[-1/6*(3*sqrt(2)*((B - C)*a*cos(d*x + c)^2 + (B - C)*a*cos(d*x + c))*sqrt(-1/a)*log(-(2*sqrt(2)*sqrt((a*cos(d*
x + c) + a)/cos(d*x + c))*sqrt(-1/a)*cos(d*x + c)*sin(d*x + c) - 3*cos(d*x + c)^2 - 2*cos(d*x + c) + 1)/(cos(d
*x + c)^2 + 2*cos(d*x + c) + 1)) - 4*((3*B - C)*cos(d*x + c) + C)*sqrt((a*cos(d*x + c) + a)/cos(d*x + c))*sin(
d*x + c))/(a*d*cos(d*x + c)^2 + a*d*cos(d*x + c)), 1/3*(2*((3*B - C)*cos(d*x + c) + C)*sqrt((a*cos(d*x + c) +
a)/cos(d*x + c))*sin(d*x + c) + 3*sqrt(2)*((B - C)*a*cos(d*x + c)^2 + (B - C)*a*cos(d*x + c))*arctan(sqrt(2)*s
qrt((a*cos(d*x + c) + a)/cos(d*x + c))*cos(d*x + c)/(sqrt(a)*sin(d*x + c)))/sqrt(a))/(a*d*cos(d*x + c)^2 + a*d
*cos(d*x + c))]

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

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

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Giac [A]  time = 8.88951, size = 251, normalized size = 2.13 \begin{align*} -\frac{\frac{3 \, \sqrt{2}{\left (B - C\right )} \log \left ({\left | -\sqrt{-a} \tan \left (\frac{1}{2} \, d x + \frac{1}{2} \, c\right ) + \sqrt{-a \tan \left (\frac{1}{2} \, d x + \frac{1}{2} \, c\right )^{2} + a} \right |}\right )}{\sqrt{-a} \mathrm{sgn}\left (\tan \left (\frac{1}{2} \, d x + \frac{1}{2} \, c\right )^{2} - 1\right )} + \frac{2 \,{\left (\frac{\sqrt{2}{\left (3 \, B a - 2 \, C a\right )} \tan \left (\frac{1}{2} \, d x + \frac{1}{2} \, c\right )^{2}}{\mathrm{sgn}\left (\tan \left (\frac{1}{2} \, d x + \frac{1}{2} \, c\right )^{2} - 1\right )} - \frac{3 \, \sqrt{2} B a}{\mathrm{sgn}\left (\tan \left (\frac{1}{2} \, d x + \frac{1}{2} \, c\right )^{2} - 1\right )}\right )} \tan \left (\frac{1}{2} \, d x + \frac{1}{2} \, c\right )}{{\left (a \tan \left (\frac{1}{2} \, d x + \frac{1}{2} \, c\right )^{2} - a\right )} \sqrt{-a \tan \left (\frac{1}{2} \, d x + \frac{1}{2} \, c\right )^{2} + a}}}{3 \, d} \end{align*}

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]

-1/3*(3*sqrt(2)*(B - C)*log(abs(-sqrt(-a)*tan(1/2*d*x + 1/2*c) + sqrt(-a*tan(1/2*d*x + 1/2*c)^2 + a)))/(sqrt(-
a)*sgn(tan(1/2*d*x + 1/2*c)^2 - 1)) + 2*(sqrt(2)*(3*B*a - 2*C*a)*tan(1/2*d*x + 1/2*c)^2/sgn(tan(1/2*d*x + 1/2*
c)^2 - 1) - 3*sqrt(2)*B*a/sgn(tan(1/2*d*x + 1/2*c)^2 - 1))*tan(1/2*d*x + 1/2*c)/((a*tan(1/2*d*x + 1/2*c)^2 - a
)*sqrt(-a*tan(1/2*d*x + 1/2*c)^2 + a)))/d