3.6.9 \(\int \frac {\sec (c+d x)}{\sqrt {a+b \sin (c+d x)}} \, dx\) [509]

Optimal. Leaf size=74 \[ -\frac {\tanh ^{-1}\left (\frac {\sqrt {a+b \sin (c+d x)}}{\sqrt {a-b}}\right )}{\sqrt {a-b} d}+\frac {\tanh ^{-1}\left (\frac {\sqrt {a+b \sin (c+d x)}}{\sqrt {a+b}}\right )}{\sqrt {a+b} d} \]

[Out]

-arctanh((a+b*sin(d*x+c))^(1/2)/(a-b)^(1/2))/d/(a-b)^(1/2)+arctanh((a+b*sin(d*x+c))^(1/2)/(a+b)^(1/2))/d/(a+b)
^(1/2)

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Rubi [A]
time = 0.06, antiderivative size = 74, normalized size of antiderivative = 1.00, number of steps used = 5, number of rules used = 4, integrand size = 21, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.190, Rules used = {2747, 722, 1107, 212} \begin {gather*} \frac {\tanh ^{-1}\left (\frac {\sqrt {a+b \sin (c+d x)}}{\sqrt {a+b}}\right )}{d \sqrt {a+b}}-\frac {\tanh ^{-1}\left (\frac {\sqrt {a+b \sin (c+d x)}}{\sqrt {a-b}}\right )}{d \sqrt {a-b}} \end {gather*}

Antiderivative was successfully verified.

[In]

Int[Sec[c + d*x]/Sqrt[a + b*Sin[c + d*x]],x]

[Out]

-(ArcTanh[Sqrt[a + b*Sin[c + d*x]]/Sqrt[a - b]]/(Sqrt[a - b]*d)) + ArcTanh[Sqrt[a + b*Sin[c + d*x]]/Sqrt[a + b
]]/(Sqrt[a + b]*d)

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 722

Int[1/(Sqrt[(d_) + (e_.)*(x_)]*((a_) + (c_.)*(x_)^2)), x_Symbol] :> Dist[2*e, Subst[Int[1/(c*d^2 + a*e^2 - 2*c
*d*x^2 + c*x^4), x], x, Sqrt[d + e*x]], x] /; FreeQ[{a, c, d, e}, x] && NeQ[c*d^2 + a*e^2, 0]

Rule 1107

Int[((a_) + (b_.)*(x_)^2 + (c_.)*(x_)^4)^(-1), x_Symbol] :> With[{q = Rt[b^2 - 4*a*c, 2]}, Dist[c/q, Int[1/(b/
2 - q/2 + c*x^2), x], x] - Dist[c/q, Int[1/(b/2 + q/2 + c*x^2), x], x]] /; FreeQ[{a, b, c}, x] && NeQ[b^2 - 4*
a*c, 0] && PosQ[b^2 - 4*a*c]

Rule 2747

Int[cos[(e_.) + (f_.)*(x_)]^(p_.)*((a_) + (b_.)*sin[(e_.) + (f_.)*(x_)])^(m_.), x_Symbol] :> Dist[1/(b^p*f), S
ubst[Int[(a + x)^m*(b^2 - x^2)^((p - 1)/2), x], x, b*Sin[e + f*x]], x] /; FreeQ[{a, b, e, f, m}, x] && Integer
Q[(p - 1)/2] && NeQ[a^2 - b^2, 0]

Rubi steps

\begin {align*} \int \frac {\sec (c+d x)}{\sqrt {a+b \sin (c+d x)}} \, dx &=\frac {b \text {Subst}\left (\int \frac {1}{\sqrt {a+x} \left (b^2-x^2\right )} \, dx,x,b \sin (c+d x)\right )}{d}\\ &=\frac {(2 b) \text {Subst}\left (\int \frac {1}{-a^2+b^2+2 a x^2-x^4} \, dx,x,\sqrt {a+b \sin (c+d x)}\right )}{d}\\ &=-\frac {\text {Subst}\left (\int \frac {1}{a-b-x^2} \, dx,x,\sqrt {a+b \sin (c+d x)}\right )}{d}+\frac {\text {Subst}\left (\int \frac {1}{a+b-x^2} \, dx,x,\sqrt {a+b \sin (c+d x)}\right )}{d}\\ &=-\frac {\tanh ^{-1}\left (\frac {\sqrt {a+b \sin (c+d x)}}{\sqrt {a-b}}\right )}{\sqrt {a-b} d}+\frac {\tanh ^{-1}\left (\frac {\sqrt {a+b \sin (c+d x)}}{\sqrt {a+b}}\right )}{\sqrt {a+b} d}\\ \end {align*}

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Mathematica [A]
time = 0.06, size = 74, normalized size = 1.00 \begin {gather*} -\frac {\tanh ^{-1}\left (\frac {\sqrt {a+b \sin (c+d x)}}{\sqrt {a-b}}\right )}{\sqrt {a-b} d}+\frac {\tanh ^{-1}\left (\frac {\sqrt {a+b \sin (c+d x)}}{\sqrt {a+b}}\right )}{\sqrt {a+b} d} \end {gather*}

Antiderivative was successfully verified.

[In]

Integrate[Sec[c + d*x]/Sqrt[a + b*Sin[c + d*x]],x]

[Out]

-(ArcTanh[Sqrt[a + b*Sin[c + d*x]]/Sqrt[a - b]]/(Sqrt[a - b]*d)) + ArcTanh[Sqrt[a + b*Sin[c + d*x]]/Sqrt[a + b
]]/(Sqrt[a + b]*d)

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Maple [A]
time = 1.27, size = 60, normalized size = 0.81

method	result	size
default	\(\frac {\frac {\arctanh \left (\frac {\sqrt {a +b \sin \left (d x +c \right )}}{\sqrt {a +b}}\right )}{\sqrt {a +b}}+\frac {\arctan \left (\frac {\sqrt {a +b \sin \left (d x +c \right )}}{\sqrt {-a +b}}\right )}{\sqrt {-a +b}}}{d}\)	\(60\)

Verification of antiderivative is not currently implemented for this CAS.

[In]

int(sec(d*x+c)/(a+b*sin(d*x+c))^(1/2),x,method=_RETURNVERBOSE)

[Out]

(1/(a+b)^(1/2)*arctanh((a+b*sin(d*x+c))^(1/2)/(a+b)^(1/2))+1/(-a+b)^(1/2)*arctan((a+b*sin(d*x+c))^(1/2)/(-a+b)
^(1/2)))/d

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Maxima [F(-2)]
time = 0.00, size = 0, normalized size = 0.00 \begin {gather*} \text {Exception raised: ValueError} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

Exception raised: ValueError >> Computation failed since Maxima requested additional constraints; using the 'a
ssume' command before evaluation *may* help (example of legal syntax is 'assume(4*a-4*b>0)', see `assume?` for
more detail

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Fricas [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(sec(d*x+c)/(a+b*sin(d*x+c))^(1/2),x, algorithm="fricas")

[Out]

integral(sec(d*x + c)/sqrt(b*sin(d*x + c) + a), x)

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Sympy [F]
time = 0.00, size = 0, normalized size = 0.00 \begin {gather*} \int \frac {\sec {\left (c + d x \right )}}{\sqrt {a + b \sin {\left (c + d x \right )}}}\, dx \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate(sec(d*x+c)/(a+b*sin(d*x+c))**(1/2),x)

[Out]

Integral(sec(c + d*x)/sqrt(a + b*sin(c + d*x)), x)

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Giac [A]
time = 4.49, size = 75, normalized size = 1.01 \begin {gather*} \frac {b {\left (\frac {\arctan \left (\frac {\sqrt {b \sin \left (d x + c\right ) + a}}{\sqrt {-a + b}}\right )}{\sqrt {-a + b} b} - \frac {\arctan \left (\frac {\sqrt {b \sin \left (d x + c\right ) + a}}{\sqrt {-a - b}}\right )}{\sqrt {-a - b} b}\right )}}{d} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

b*(arctan(sqrt(b*sin(d*x + c) + a)/sqrt(-a + b))/(sqrt(-a + b)*b) - arctan(sqrt(b*sin(d*x + c) + a)/sqrt(-a -
b))/(sqrt(-a - b)*b))/d

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

Verification of antiderivative is not currently implemented for this CAS.

[In]

int(1/(cos(c + d*x)*(a + b*sin(c + d*x))^(1/2)),x)

[Out]

int(1/(cos(c + d*x)*(a + b*sin(c + d*x))^(1/2)), x)

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