3.6.44 \(\int \frac {\sqrt {\cos (c+d x)} (A+B \sec (c+d x))}{\sqrt {a+a \sec (c+d x)}} \, dx\) [544]

Optimal. Leaf size=119 \[ -\frac {\sqrt {2} (A-B) \tanh ^{-1}\left (\frac {\sqrt {a} \sqrt {\sec (c+d x)} \sin (c+d x)}{\sqrt {2} \sqrt {a+a \sec (c+d x)}}\right ) \sqrt {\cos (c+d x)} \sqrt {\sec (c+d x)}}{\sqrt {a} d}+\frac {2 A \sin (c+d x)}{d \sqrt {\cos (c+d x)} \sqrt {a+a \sec (c+d x)}} \]

[Out]

-(A-B)*arctanh(1/2*sin(d*x+c)*a^(1/2)*sec(d*x+c)^(1/2)*2^(1/2)/(a+a*sec(d*x+c))^(1/2))*2^(1/2)*cos(d*x+c)^(1/2
)*sec(d*x+c)^(1/2)/d/a^(1/2)+2*A*sin(d*x+c)/d/cos(d*x+c)^(1/2)/(a+a*sec(d*x+c))^(1/2)

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Rubi [A]
time = 0.19, antiderivative size = 119, normalized size of antiderivative = 1.00, number of steps used = 4, number of rules used = 4, integrand size = 35, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.114, Rules used = {3034, 4098, 3893, 212} \begin {gather*} \frac {2 A \sin (c+d x)}{d \sqrt {\cos (c+d x)} \sqrt {a \sec (c+d x)+a}}-\frac {\sqrt {2} (A-B) \sqrt {\cos (c+d x)} \sqrt {\sec (c+d x)} \tanh ^{-1}\left (\frac {\sqrt {a} \sin (c+d x) \sqrt {\sec (c+d x)}}{\sqrt {2} \sqrt {a \sec (c+d x)+a}}\right )}{\sqrt {a} d} \end {gather*}

Antiderivative was successfully verified.

[In]

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

[Out]

-((Sqrt[2]*(A - B)*ArcTanh[(Sqrt[a]*Sqrt[Sec[c + d*x]]*Sin[c + d*x])/(Sqrt[2]*Sqrt[a + a*Sec[c + d*x]])]*Sqrt[
Cos[c + d*x]]*Sqrt[Sec[c + d*x]])/(Sqrt[a]*d)) + (2*A*Sin[c + d*x])/(d*Sqrt[Cos[c + d*x]]*Sqrt[a + a*Sec[c + d
*x]])

Rule 212

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

Rule 3034

Int[((a_.) + csc[(e_.) + (f_.)*(x_)]*(b_.))^(m_.)*(csc[(e_.) + (f_.)*(x_)]*(d_.) + (c_))^(n_.)*((g_.)*sin[(e_.
) + (f_.)*(x_)])^(p_.), x_Symbol] :> Dist[(g*Csc[e + f*x])^p*(g*Sin[e + f*x])^p, Int[(a + b*Csc[e + f*x])^m*((
c + d*Csc[e + f*x])^n/(g*Csc[e + f*x])^p), x], x] /; FreeQ[{a, b, c, d, e, f, g, m, n, p}, x] && NeQ[b*c - a*d
, 0] &&  !IntegerQ[p] &&  !(IntegerQ[m] && IntegerQ[n])

Rule 3893

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

Rule 4098

Int[(csc[(e_.) + (f_.)*(x_)]*(d_.))^(n_)*(csc[(e_.) + (f_.)*(x_)]*(b_.) + (a_))^(m_)*(csc[(e_.) + (f_.)*(x_)]*
(B_.) + (A_)), x_Symbol] :> Simp[A*Cot[e + f*x]*(a + b*Csc[e + f*x])^m*((d*Csc[e + f*x])^n/(f*n)), x] - Dist[(
a*A*m - b*B*n)/(b*d*n), Int[(a + b*Csc[e + f*x])^m*(d*Csc[e + f*x])^(n + 1), x], x] /; FreeQ[{a, b, d, e, f, A
, B, m, n}, x] && NeQ[A*b - a*B, 0] && EqQ[a^2 - b^2, 0] && EqQ[m + n + 1, 0] &&  !LeQ[m, -1]

Rubi steps

\begin {align*} \int \frac {\sqrt {\cos (c+d x)} (A+B \sec (c+d x))}{\sqrt {a+a \sec (c+d x)}} \, dx &=\left (\sqrt {\cos (c+d x)} \sqrt {\sec (c+d x)}\right ) \int \frac {A+B \sec (c+d x)}{\sqrt {\sec (c+d x)} \sqrt {a+a \sec (c+d x)}} \, dx\\ &=\frac {2 A \sin (c+d x)}{d \sqrt {\cos (c+d x)} \sqrt {a+a \sec (c+d x)}}+\left ((-A+B) \sqrt {\cos (c+d x)} \sqrt {\sec (c+d x)}\right ) \int \frac {\sqrt {\sec (c+d x)}}{\sqrt {a+a \sec (c+d x)}} \, dx\\ &=\frac {2 A \sin (c+d x)}{d \sqrt {\cos (c+d x)} \sqrt {a+a \sec (c+d x)}}-\frac {\left (2 (-A+B) \sqrt {\cos (c+d x)} \sqrt {\sec (c+d x)}\right ) \text {Subst}\left (\int \frac {1}{2 a-x^2} \, dx,x,-\frac {a \sqrt {\sec (c+d x)} \sin (c+d x)}{\sqrt {a+a \sec (c+d x)}}\right )}{d}\\ &=-\frac {\sqrt {2} (A-B) \tanh ^{-1}\left (\frac {\sqrt {a} \sqrt {\sec (c+d x)} \sin (c+d x)}{\sqrt {2} \sqrt {a+a \sec (c+d x)}}\right ) \sqrt {\cos (c+d x)} \sqrt {\sec (c+d x)}}{\sqrt {a} d}+\frac {2 A \sin (c+d x)}{d \sqrt {\cos (c+d x)} \sqrt {a+a \sec (c+d x)}}\\ \end {align*}

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Mathematica [A]
time = 0.31, size = 140, normalized size = 1.18 \begin {gather*} \frac {\sqrt {\cos (c+d x)} \left (2 A \sqrt {1-\sec (c+d x)}+\sqrt {2} (A-B) \text {ArcTan}\left (\frac {\sqrt {2} \sqrt {\sec (c+d x)}}{\sqrt {1-\sec (c+d x)}}\right ) \sqrt {\sec (c+d x)}\right ) (A+B \sec (c+d x)) \sin (c+d x)}{d (B+A \cos (c+d x)) \sqrt {1-\sec (c+d x)} \sqrt {a (1+\sec (c+d x))}} \end {gather*}

Antiderivative was successfully verified.

[In]

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

[Out]

(Sqrt[Cos[c + d*x]]*(2*A*Sqrt[1 - Sec[c + d*x]] + Sqrt[2]*(A - B)*ArcTan[(Sqrt[2]*Sqrt[Sec[c + d*x]])/Sqrt[1 -
Sec[c + d*x]]]*Sqrt[Sec[c + d*x]])*(A + B*Sec[c + d*x])*Sin[c + d*x])/(d*(B + A*Cos[c + d*x])*Sqrt[1 - Sec[c
+ d*x]]*Sqrt[a*(1 + Sec[c + d*x])])

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Maple [A]
time = 13.44, size = 142, normalized size = 1.19

method	result	size
default	\(\frac {\left (A \sin \left (d x +c \right ) \sqrt {-\frac {2}{1+\cos \left (d x +c \right )}}-A \arctan \left (\frac {\sin \left (d x +c \right ) \sqrt {-\frac {2}{1+\cos \left (d x +c \right )}}}{2}\right )+B \arctan \left (\frac {\sin \left (d x +c \right ) \sqrt {-\frac {2}{1+\cos \left (d x +c \right )}}}{2}\right )\right ) \left (\sqrt {\cos }\left (d x +c \right )\right ) \sqrt {\frac {a \left (1+\cos \left (d x +c \right )\right )}{\cos \left (d x +c \right )}}\, \sqrt {-\frac {2}{1+\cos \left (d x +c \right )}}\, \left (\cos ^{2}\left (d x +c \right )-1\right )}{d a \sin \left (d x +c \right )^{2}}\)	\(142\)

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

1/d*(A*sin(d*x+c)*(-2/(1+cos(d*x+c)))^(1/2)-A*arctan(1/2*sin(d*x+c)*(-2/(1+cos(d*x+c)))^(1/2))+B*arctan(1/2*si
n(d*x+c)*(-2/(1+cos(d*x+c)))^(1/2)))*cos(d*x+c)^(1/2)*(a*(1+cos(d*x+c))/cos(d*x+c))^(1/2)*(-2/(1+cos(d*x+c)))^
(1/2)/a/sin(d*x+c)^2*(cos(d*x+c)^2-1)

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Maxima [A]
time = 0.86, size = 195, normalized size = 1.64 \begin {gather*} -\frac {\frac {{\left (\sqrt {2} \log \left (\cos \left (\frac {1}{2} \, d x + \frac {1}{2} \, c\right )^{2} + \sin \left (\frac {1}{2} \, d x + \frac {1}{2} \, c\right )^{2} + 2 \, \sin \left (\frac {1}{2} \, d x + \frac {1}{2} \, c\right ) + 1\right ) - \sqrt {2} \log \left (\cos \left (\frac {1}{2} \, d x + \frac {1}{2} \, c\right )^{2} + \sin \left (\frac {1}{2} \, d x + \frac {1}{2} \, c\right )^{2} - 2 \, \sin \left (\frac {1}{2} \, d x + \frac {1}{2} \, c\right ) + 1\right ) - 4 \, \sqrt {2} \sin \left (\frac {1}{2} \, d x + \frac {1}{2} \, c\right )\right )} A}{\sqrt {a}} - \frac {{\left (\sqrt {2} \log \left (\cos \left (\frac {1}{2} \, d x + \frac {1}{2} \, c\right )^{2} + \sin \left (\frac {1}{2} \, d x + \frac {1}{2} \, c\right )^{2} + 2 \, \sin \left (\frac {1}{2} \, d x + \frac {1}{2} \, c\right ) + 1\right ) - \sqrt {2} \log \left (\cos \left (\frac {1}{2} \, d x + \frac {1}{2} \, c\right )^{2} + \sin \left (\frac {1}{2} \, d x + \frac {1}{2} \, c\right )^{2} - 2 \, \sin \left (\frac {1}{2} \, d x + \frac {1}{2} \, c\right ) + 1\right )\right )} B}{\sqrt {a}}}{2 \, d} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

-1/2*((sqrt(2)*log(cos(1/2*d*x + 1/2*c)^2 + sin(1/2*d*x + 1/2*c)^2 + 2*sin(1/2*d*x + 1/2*c) + 1) - sqrt(2)*log
(cos(1/2*d*x + 1/2*c)^2 + sin(1/2*d*x + 1/2*c)^2 - 2*sin(1/2*d*x + 1/2*c) + 1) - 4*sqrt(2)*sin(1/2*d*x + 1/2*c
))*A/sqrt(a) - (sqrt(2)*log(cos(1/2*d*x + 1/2*c)^2 + sin(1/2*d*x + 1/2*c)^2 + 2*sin(1/2*d*x + 1/2*c) + 1) - sq
rt(2)*log(cos(1/2*d*x + 1/2*c)^2 + sin(1/2*d*x + 1/2*c)^2 - 2*sin(1/2*d*x + 1/2*c) + 1))*B/sqrt(a))/d

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Fricas [A]
time = 1.88, size = 306, normalized size = 2.57 \begin {gather*} \left [\frac {4 \, A \sqrt {\frac {a \cos \left (d x + c\right ) + a}{\cos \left (d x + c\right )}} \sqrt {\cos \left (d x + c\right )} \sin \left (d x + c\right ) - \frac {\sqrt {2} {\left ({\left (A - B\right )} a \cos \left (d x + c\right ) + {\left (A - B\right )} a\right )} \log \left (-\frac {\cos \left (d x + c\right )^{2} - \frac {2 \, \sqrt {2} \sqrt {\frac {a \cos \left (d x + c\right ) + a}{\cos \left (d x + c\right )}} \sqrt {\cos \left (d x + c\right )} \sin \left (d x + c\right )}{\sqrt {a}} - 2 \, \cos \left (d x + c\right ) - 3}{\cos \left (d x + c\right )^{2} + 2 \, \cos \left (d x + c\right ) + 1}\right )}{\sqrt {a}}}{2 \, {\left (a d \cos \left (d x + c\right ) + a d\right )}}, \frac {\sqrt {2} {\left ({\left (A - B\right )} a \cos \left (d x + c\right ) + {\left (A - B\right )} a\right )} \sqrt {-\frac {1}{a}} \arctan \left (\frac {\sqrt {2} \sqrt {\frac {a \cos \left (d x + c\right ) + a}{\cos \left (d x + c\right )}} \sqrt {-\frac {1}{a}} \sqrt {\cos \left (d x + c\right )}}{\sin \left (d x + c\right )}\right ) + 2 \, A \sqrt {\frac {a \cos \left (d x + c\right ) + a}{\cos \left (d x + c\right )}} \sqrt {\cos \left (d x + c\right )} \sin \left (d x + c\right )}{a d \cos \left (d x + c\right ) + a d}\right ] \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

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

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

[Out]

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

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Giac [F]
time = 0.00, size = 0, normalized size = 0.00 \begin {gather*} \text {could not integrate} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

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

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

[Out]

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

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