3.2.46 \(\int \frac {\sec ^2(c+d x) (A+B \sec (c+d x))}{\sqrt {a+a \sec (c+d x)}} \, dx\) [146]

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

[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 = 0.17, 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 = {4095, 4086, 3880, 209} \begin {gather*} -\frac {\sqrt {2} (A-B) \text {ArcTan}\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} \end {gather*}

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 209

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

Rule 3880

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 4086

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 4095

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)),
Int[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] /; Fr
eeQ[{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)) \text {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 = 0.31, size = 106, normalized size = 0.90 \begin {gather*} \frac {\left (-3 \sqrt {2} (A-B) \tanh ^{-1}\left (\frac {\sqrt {1-\sec (c+d x)}}{\sqrt {2}}\right )+2 \sqrt {1-\sec (c+d x)} (3 A-B+B \sec (c+d x))\right ) \tan (c+d x)}{3 d \sqrt {1-\sec (c+d x)} \sqrt {a (1+\sec (c+d x))}} \end {gather*}

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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Maple [B] Leaf count of result is larger than twice the leaf count of optimal. \(404\) vs. \(2(101)=202\).
time = 4.16, size = 405, normalized size = 3.43

method	result	size
default	\(\frac {\left (3 A \sin \left (d x +c \right ) \left (-\frac {2 \cos \left (d x +c \right )}{1+\cos \left (d x +c \right )}\right )^{\frac {3}{2}} \cos \left (d x +c \right ) \ln \left (\frac {\sqrt {-\frac {2 \cos \left (d x +c \right )}{1+\cos \left (d x +c \right )}}\, \sin \left (d x +c \right )-\cos \left (d x +c \right )+1}{\sin \left (d x +c \right )}\right )-3 B \sin \left (d x +c \right ) \left (-\frac {2 \cos \left (d x +c \right )}{1+\cos \left (d x +c \right )}\right )^{\frac {3}{2}} \cos \left (d x +c \right ) \ln \left (\frac {\sqrt {-\frac {2 \cos \left (d x +c \right )}{1+\cos \left (d x +c \right )}}\, \sin \left (d x +c \right )-\cos \left (d x +c \right )+1}{\sin \left (d x +c \right )}\right )+3 A \left (-\frac {2 \cos \left (d x +c \right )}{1+\cos \left (d x +c \right )}\right )^{\frac {3}{2}} \ln \left (\frac {\sqrt {-\frac {2 \cos \left (d x +c \right )}{1+\cos \left (d x +c \right )}}\, \sin \left (d x +c \right )-\cos \left (d x +c \right )+1}{\sin \left (d x +c \right )}\right ) \sin \left (d x +c \right )-3 B \left (-\frac {2 \cos \left (d x +c \right )}{1+\cos \left (d x +c \right )}\right )^{\frac {3}{2}} \ln \left (\frac {\sqrt {-\frac {2 \cos \left (d x +c \right )}{1+\cos \left (d x +c \right )}}\, \sin \left (d x +c \right )-\cos \left (d x +c \right )+1}{\sin \left (d x +c \right )}\right ) \sin \left (d x +c \right )-12 A \left (\cos ^{2}\left (d x +c \right )\right )+4 B \left (\cos ^{2}\left (d x +c \right )\right )+12 A \cos \left (d x +c \right )-8 B \cos \left (d x +c \right )+4 B \right ) \sqrt {\frac {a \left (1+\cos \left (d x +c \right )\right )}{\cos \left (d x +c \right )}}}{6 d \sin \left (d x +c \right ) \cos \left (d x +c \right ) a}\)	\(405\)

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,method=_RETURNVERBOSE)

[Out]

1/6/d*(3*A*sin(d*x+c)*(-2*cos(d*x+c)/(1+cos(d*x+c)))^(3/2)*cos(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))-3*B*sin(d*x+c)*(-2*cos(d*x+c)/(1+cos(d*x+c)))^(3/2)*cos(d*x+c)*ln(((-2*c
os(d*x+c)/(1+cos(d*x+c)))^(1/2)*sin(d*x+c)-cos(d*x+c)+1)/sin(d*x+c))+3*A*(-2*cos(d*x+c)/(1+cos(d*x+c)))^(3/2)*
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))*sin(d*x+c)-3*B*(-2*cos(d*x+c)/(1
+cos(d*x+c)))^(3/2)*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))*sin(d*x+c)-1
2*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 \begin {gather*} \text {Failed to integrate} \end {gather*}

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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Fricas [A]
time = 2.40, size = 352, normalized size = 2.98 \begin {gather*} \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 ] \end {gather*}

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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Sympy [F]
time = 0.00, size = 0, normalized size = 0.00 \begin {gather*} \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 \end {gather*}

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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Giac [A]
time = 1.35, size = 165, normalized size = 1.40 \begin {gather*} \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 (\cos \left (d x + c\right )\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 (\cos \left (d x + c\right )\right )} - \frac {3 \, \sqrt {2} A a}{\mathrm {sgn}\left (\cos \left (d x + c\right )\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 {gather*}

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(cos(d*x + c))) + 2*(sqrt(2)*(3*A*a - 2*B*a)*tan(1/2*d*x + 1/2*c)^2/sgn(cos(d*x + c)) - 3*sqrt(2)*A*a/sgn
(cos(d*x + c)))*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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Mupad [F]
time = 0.00, size = -1, normalized size = -0.01 \begin {gather*} \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 \end {gather*}

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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