3.509 \(\int \frac {\sec (c+d x)}{\sqrt {a+b \sin (c+d x)}} \, dx\)

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

[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 = 0.09, 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 = {2668, 708, 1093, 206} \[ \frac {\tanh ^{-1}\left (\frac {\sqrt {a+b \sin (c+d x)}}{\sqrt {a+b}}\right )}{d \sqrt {a+b}}-\frac {\tanh ^{-1}\left (\frac {\sqrt {a+b \sin (c+d x)}}{\sqrt {a-b}}\right )}{d \sqrt {a-b}} \]

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 206

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

Rule 708

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 1093

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 2668

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 \operatorname {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) \operatorname {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 {\operatorname {Subst}\left (\int \frac {1}{a-b-x^2} \, dx,x,\sqrt {a+b \sin (c+d x)}\right )}{d}+\frac {\operatorname {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 = 0.05, size = 74, normalized size = 1.00 \[ \frac {\tanh ^{-1}\left (\frac {\sqrt {a+b \sin (c+d x)}}{\sqrt {a+b}}\right )}{d \sqrt {a+b}}-\frac {\tanh ^{-1}\left (\frac {\sqrt {a+b \sin (c+d x)}}{\sqrt {a-b}}\right )}{d \sqrt {a-b}} \]

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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fricas [F]  time = 0.68, size = 0, normalized size = 0.00 \[ {\rm integral}\left (\frac {\sec \left (d x + c\right )}{\sqrt {b \sin \left (d x + c\right ) + a}}, x\right ) \]

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

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]

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

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maple [A]  time = 0.47, size = 62, normalized size = 0.84 \[ \frac {\arctan \left (\frac {\sqrt {a +b \sin \left (d x +c \right )}}{\sqrt {-a +b}}\right )}{d \sqrt {-a +b}}+\frac {\arctanh \left (\frac {\sqrt {a +b \sin \left (d x +c \right )}}{\sqrt {a +b}}\right )}{d \sqrt {a +b}} \]

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)

[Out]

1/d/(-a+b)^(1/2)*arctan((a+b*sin(d*x+c))^(1/2)/(-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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maxima [F(-2)]  time = 0.00, size = 0, normalized size = 0.00 \[ \text {Exception raised: ValueError} \]

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 details)Is 4*a-4*b positive or negative?

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

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

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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