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

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

[Out]

-(A-B)*arctan(1/2*a^(1/2)*tan(d*x+c)*2^(1/2)/(a+a*sec(d*x+c))^(1/2))*2^(1/2)/d/a^(1/2)+2/3*(3*A-2*B)*tan(d*x+c
)/d/(a+a*sec(d*x+c))^(1/2)+2/3*B*(a+a*sec(d*x+c))^(1/2)*tan(d*x+c)/a/d

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Rubi [A]  time = 0.26, antiderivative size = 118, normalized size of antiderivative = 1.00, number of steps used = 4, number of rules used = 4, integrand size = 33, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.121, Rules used = {4010, 4001, 3795, 203} \[ -\frac {\sqrt {2} (A-B) \tan ^{-1}\left (\frac {\sqrt {a} \tan (c+d x)}{\sqrt {2} \sqrt {a \sec (c+d x)+a}}\right )}{\sqrt {a} d}+\frac {2 (3 A-2 B) \tan (c+d x)}{3 d \sqrt {a \sec (c+d x)+a}}+\frac {2 B \tan (c+d x) \sqrt {a \sec (c+d x)+a}}{3 a d} \]

Antiderivative was successfully verified.

[In]

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

[Out]

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

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])

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 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]

Rubi steps

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

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Mathematica [A]  time = 0.30, size = 106, normalized size = 0.90 \[ \frac {\tan (c+d x) \left (2 \sqrt {1-\sec (c+d x)} (3 A+B \sec (c+d x)-B)-3 \sqrt {2} (A-B) \tanh ^{-1}\left (\frac {\sqrt {1-\sec (c+d x)}}{\sqrt {2}}\right )\right )}{3 d \sqrt {1-\sec (c+d x)} \sqrt {a (\sec (c+d x)+1)}} \]

Antiderivative was successfully verified.

[In]

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

[Out]

((-3*Sqrt[2]*(A - B)*ArcTanh[Sqrt[1 - Sec[c + d*x]]/Sqrt[2]] + 2*Sqrt[1 - Sec[c + d*x]]*(3*A - B + B*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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fricas [A]  time = 0.51, size = 352, normalized size = 2.98 \[ \left [-\frac {3 \, \sqrt {2} {\left ({\left (A - B\right )} a \cos \left (d x + c\right )^{2} + {\left (A - B\right )} a \cos \left (d x + c\right )\right )} \sqrt {-\frac {1}{a}} \log \left (-\frac {2 \, \sqrt {2} \sqrt {\frac {a \cos \left (d x + c\right ) + a}{\cos \left (d x + c\right )}} \sqrt {-\frac {1}{a}} \cos \left (d x + c\right ) \sin \left (d x + c\right ) - 3 \, \cos \left (d x + c\right )^{2} - 2 \, \cos \left (d x + c\right ) + 1}{\cos \left (d x + c\right )^{2} + 2 \, \cos \left (d x + c\right ) + 1}\right ) - 4 \, {\left ({\left (3 \, A - B\right )} \cos \left (d x + c\right ) + B\right )} \sqrt {\frac {a \cos \left (d x + c\right ) + a}{\cos \left (d x + c\right )}} \sin \left (d x + c\right )}{6 \, {\left (a d \cos \left (d x + c\right )^{2} + a d \cos \left (d x + c\right )\right )}}, \frac {2 \, {\left ({\left (3 \, A - B\right )} \cos \left (d x + c\right ) + B\right )} \sqrt {\frac {a \cos \left (d x + c\right ) + a}{\cos \left (d x + c\right )}} \sin \left (d x + c\right ) + \frac {3 \, \sqrt {2} {\left ({\left (A - B\right )} a \cos \left (d x + c\right )^{2} + {\left (A - B\right )} a \cos \left (d x + c\right )\right )} \arctan \left (\frac {\sqrt {2} \sqrt {\frac {a \cos \left (d x + c\right ) + a}{\cos \left (d x + c\right )}} \cos \left (d x + c\right )}{\sqrt {a} \sin \left (d x + c\right )}\right )}{\sqrt {a}}}{3 \, {\left (a d \cos \left (d x + c\right )^{2} + a d \cos \left (d x + c\right )\right )}}\right ] \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

[-1/6*(3*sqrt(2)*((A - B)*a*cos(d*x + c)^2 + (A - B)*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*A - B)*cos(d*x + c) + B)*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*A - B)*cos(d*x + c) + B)*sqrt((a*cos(d*x + c) +
a)/cos(d*x + c))*sin(d*x + c) + 3*sqrt(2)*((A - B)*a*cos(d*x + c)^2 + (A - B)*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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giac [A]  time = 2.14, size = 186, normalized size = 1.58 \[ -\frac {\frac {3 \, \sqrt {2} {\left (A - B\right )} \log \left ({\left | -\sqrt {-a} \tan \left (\frac {1}{2} \, d x + \frac {1}{2} \, c\right ) + \sqrt {-a \tan \left (\frac {1}{2} \, d x + \frac {1}{2} \, c\right )^{2} + a} \right |}\right )}{\sqrt {-a} \mathrm {sgn}\left (\tan \left (\frac {1}{2} \, d x + \frac {1}{2} \, c\right )^{2} - 1\right )} + \frac {2 \, {\left (\frac {\sqrt {2} {\left (3 \, A a - 2 \, B a\right )} \tan \left (\frac {1}{2} \, d x + \frac {1}{2} \, c\right )^{2}}{\mathrm {sgn}\left (\tan \left (\frac {1}{2} \, d x + \frac {1}{2} \, c\right )^{2} - 1\right )} - \frac {3 \, \sqrt {2} A a}{\mathrm {sgn}\left (\tan \left (\frac {1}{2} \, d x + \frac {1}{2} \, c\right )^{2} - 1\right )}\right )} \tan \left (\frac {1}{2} \, d x + \frac {1}{2} \, c\right )}{{\left (a \tan \left (\frac {1}{2} \, d x + \frac {1}{2} \, c\right )^{2} - a\right )} \sqrt {-a \tan \left (\frac {1}{2} \, d x + \frac {1}{2} \, c\right )^{2} + a}}}{3 \, d} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

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

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maple [B]  time = 1.66, size = 405, normalized size = 3.43 \[ -\frac {\left (-3 A \cos \left (d x +c \right ) \sin \left (d x +c \right ) \ln \left (-\frac {-\sqrt {-\frac {2 \cos \left (d x +c \right )}{1+\cos \left (d x +c \right )}}\, \sin \left (d x +c \right )+\cos \left (d x +c \right )-1}{\sin \left (d x +c \right )}\right ) \left (-\frac {2 \cos \left (d x +c \right )}{1+\cos \left (d x +c \right )}\right )^{\frac {3}{2}}+3 B \cos \left (d x +c \right ) \sin \left (d x +c \right ) \ln \left (-\frac {-\sqrt {-\frac {2 \cos \left (d x +c \right )}{1+\cos \left (d x +c \right )}}\, \sin \left (d x +c \right )+\cos \left (d x +c \right )-1}{\sin \left (d x +c \right )}\right ) \left (-\frac {2 \cos \left (d x +c \right )}{1+\cos \left (d x +c \right )}\right )^{\frac {3}{2}}-3 A \ln \left (-\frac {-\sqrt {-\frac {2 \cos \left (d x +c \right )}{1+\cos \left (d x +c \right )}}\, \sin \left (d x +c \right )+\cos \left (d x +c \right )-1}{\sin \left (d x +c \right )}\right ) \left (-\frac {2 \cos \left (d x +c \right )}{1+\cos \left (d x +c \right )}\right )^{\frac {3}{2}} \sin \left (d x +c \right )+3 B \ln \left (-\frac {-\sqrt {-\frac {2 \cos \left (d x +c \right )}{1+\cos \left (d x +c \right )}}\, \sin \left (d x +c \right )+\cos \left (d x +c \right )-1}{\sin \left (d x +c \right )}\right ) \left (-\frac {2 \cos \left (d x +c \right )}{1+\cos \left (d x +c \right )}\right )^{\frac {3}{2}} \sin \left (d x +c \right )+12 A \left (\cos ^{2}\left (d x +c \right )\right )-4 B \left (\cos ^{2}\left (d x +c \right )\right )-12 A \cos \left (d x +c \right )+8 B \cos \left (d x +c \right )-4 B \right ) \sqrt {\frac {a \left (1+\cos \left (d x +c \right )\right )}{\cos \left (d x +c \right )}}}{6 d \sin \left (d x +c \right ) \cos \left (d x +c \right ) a} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

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

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maxima [F]  time = 0.00, size = 0, normalized size = 0.00 \[ \int \frac {{\left (B \sec \left (d x + c\right ) + A\right )} \sec \left (d x + c\right )^{2}}{\sqrt {a \sec \left (d x + c\right ) + a}}\,{d x} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

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

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mupad [F]  time = 0.00, size = -1, normalized size = -0.01 \[ \int \frac {A+\frac {B}{\cos \left (c+d\,x\right )}}{{\cos \left (c+d\,x\right )}^2\,\sqrt {a+\frac {a}{\cos \left (c+d\,x\right )}}} \,d x \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

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

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sympy [F]  time = 0.00, size = 0, normalized size = 0.00 \[ \int \frac {\left (A + B \sec {\left (c + d x \right )}\right ) \sec ^{2}{\left (c + d x \right )}}{\sqrt {a \left (\sec {\left (c + d x \right )} + 1\right )}}\, dx \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate(sec(d*x+c)**2*(A+B*sec(d*x+c))/(a+a*sec(d*x+c))**(1/2),x)

[Out]

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

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