3.4 \(\int \frac {A+B \sec (c+d x)}{\sqrt {b \sec (c+d x)}} \, dx\)

Optimal. Leaf size=82 \[ \frac {2 A E\left (\left .\frac {1}{2} (c+d x)\right |2\right )}{d \sqrt {\cos (c+d x)} \sqrt {b \sec (c+d x)}}+\frac {2 B \sqrt {\cos (c+d x)} F\left (\left .\frac {1}{2} (c+d x)\right |2\right ) \sqrt {b \sec (c+d x)}}{b d} \]

[Out]

2*A*(cos(1/2*d*x+1/2*c)^2)^(1/2)/cos(1/2*d*x+1/2*c)*EllipticE(sin(1/2*d*x+1/2*c),2^(1/2))/d/cos(d*x+c)^(1/2)/(
b*sec(d*x+c))^(1/2)+2*B*(cos(1/2*d*x+1/2*c)^2)^(1/2)/cos(1/2*d*x+1/2*c)*EllipticF(sin(1/2*d*x+1/2*c),2^(1/2))*
cos(d*x+c)^(1/2)*(b*sec(d*x+c))^(1/2)/b/d

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Rubi [A]  time = 0.07, antiderivative size = 82, normalized size of antiderivative = 1.00, number of steps used = 5, number of rules used = 4, integrand size = 23, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.174, Rules used = {3787, 3771, 2639, 2641} \[ \frac {2 A E\left (\left .\frac {1}{2} (c+d x)\right |2\right )}{d \sqrt {\cos (c+d x)} \sqrt {b \sec (c+d x)}}+\frac {2 B \sqrt {\cos (c+d x)} F\left (\left .\frac {1}{2} (c+d x)\right |2\right ) \sqrt {b \sec (c+d x)}}{b d} \]

Antiderivative was successfully verified.

[In]

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

[Out]

(2*A*EllipticE[(c + d*x)/2, 2])/(d*Sqrt[Cos[c + d*x]]*Sqrt[b*Sec[c + d*x]]) + (2*B*Sqrt[Cos[c + d*x]]*Elliptic
F[(c + d*x)/2, 2]*Sqrt[b*Sec[c + d*x]])/(b*d)

Rule 2639

Int[Sqrt[sin[(c_.) + (d_.)*(x_)]], x_Symbol] :> Simp[(2*EllipticE[(1*(c - Pi/2 + d*x))/2, 2])/d, x] /; FreeQ[{
c, d}, x]

Rule 2641

Int[1/Sqrt[sin[(c_.) + (d_.)*(x_)]], x_Symbol] :> Simp[(2*EllipticF[(1*(c - Pi/2 + d*x))/2, 2])/d, x] /; FreeQ
[{c, d}, x]

Rule 3771

Int[(csc[(c_.) + (d_.)*(x_)]*(b_.))^(n_), x_Symbol] :> Dist[(b*Csc[c + d*x])^n*Sin[c + d*x]^n, Int[1/Sin[c + d
*x]^n, x], x] /; FreeQ[{b, c, d}, x] && EqQ[n^2, 1/4]

Rule 3787

Int[(csc[(e_.) + (f_.)*(x_)]*(d_.))^(n_.)*(csc[(e_.) + (f_.)*(x_)]*(b_.) + (a_)), x_Symbol] :> Dist[a, Int[(d*
Csc[e + f*x])^n, x], x] + Dist[b/d, Int[(d*Csc[e + f*x])^(n + 1), x], x] /; FreeQ[{a, b, d, e, f, n}, x]

Rubi steps

\begin {align*} \int \frac {A+B \sec (c+d x)}{\sqrt {b \sec (c+d x)}} \, dx &=A \int \frac {1}{\sqrt {b \sec (c+d x)}} \, dx+\frac {B \int \sqrt {b \sec (c+d x)} \, dx}{b}\\ &=\frac {A \int \sqrt {\cos (c+d x)} \, dx}{\sqrt {\cos (c+d x)} \sqrt {b \sec (c+d x)}}+\frac {\left (B \sqrt {\cos (c+d x)} \sqrt {b \sec (c+d x)}\right ) \int \frac {1}{\sqrt {\cos (c+d x)}} \, dx}{b}\\ &=\frac {2 A E\left (\left .\frac {1}{2} (c+d x)\right |2\right )}{d \sqrt {\cos (c+d x)} \sqrt {b \sec (c+d x)}}+\frac {2 B \sqrt {\cos (c+d x)} F\left (\left .\frac {1}{2} (c+d x)\right |2\right ) \sqrt {b \sec (c+d x)}}{b d}\\ \end {align*}

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Mathematica [A]  time = 0.09, size = 54, normalized size = 0.66 \[ \frac {2 \left (A E\left (\left .\frac {1}{2} (c+d x)\right |2\right )+B F\left (\left .\frac {1}{2} (c+d x)\right |2\right )\right )}{d \sqrt {\cos (c+d x)} \sqrt {b \sec (c+d x)}} \]

Antiderivative was successfully verified.

[In]

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

[Out]

(2*(A*EllipticE[(c + d*x)/2, 2] + B*EllipticF[(c + d*x)/2, 2]))/(d*Sqrt[Cos[c + d*x]]*Sqrt[b*Sec[c + d*x]])

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fricas [F]  time = 0.46, size = 0, normalized size = 0.00 \[ {\rm integral}\left (\frac {{\left (B \sec \left (d x + c\right ) + A\right )} \sqrt {b \sec \left (d x + c\right )}}{b \sec \left (d x + c\right )}, x\right ) \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

integral((B*sec(d*x + c) + A)*sqrt(b*sec(d*x + c))/(b*sec(d*x + c)), x)

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

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

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

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maple [C]  time = 1.52, size = 445, normalized size = 5.43 \[ \frac {2 \left (i A \cos \left (d x +c \right ) \sqrt {\frac {1}{1+\cos \left (d x +c \right )}}\, \sqrt {\frac {\cos \left (d x +c \right )}{1+\cos \left (d x +c \right )}}\, \EllipticF \left (\frac {i \left (-1+\cos \left (d x +c \right )\right )}{\sin \left (d x +c \right )}, i\right ) \sin \left (d x +c \right )-i A \cos \left (d x +c \right ) \sqrt {\frac {1}{1+\cos \left (d x +c \right )}}\, \sqrt {\frac {\cos \left (d x +c \right )}{1+\cos \left (d x +c \right )}}\, \EllipticE \left (\frac {i \left (-1+\cos \left (d x +c \right )\right )}{\sin \left (d x +c \right )}, i\right ) \sin \left (d x +c \right )+i B \cos \left (d x +c \right ) \sqrt {\frac {1}{1+\cos \left (d x +c \right )}}\, \sqrt {\frac {\cos \left (d x +c \right )}{1+\cos \left (d x +c \right )}}\, \EllipticF \left (\frac {i \left (-1+\cos \left (d x +c \right )\right )}{\sin \left (d x +c \right )}, i\right ) \sin \left (d x +c \right )+i A \EllipticF \left (\frac {i \left (-1+\cos \left (d x +c \right )\right )}{\sin \left (d x +c \right )}, i\right ) \sin \left (d x +c \right ) \sqrt {\frac {1}{1+\cos \left (d x +c \right )}}\, \sqrt {\frac {\cos \left (d x +c \right )}{1+\cos \left (d x +c \right )}}-i A \sin \left (d x +c \right ) \EllipticE \left (\frac {i \left (-1+\cos \left (d x +c \right )\right )}{\sin \left (d x +c \right )}, i\right ) \sqrt {\frac {1}{1+\cos \left (d x +c \right )}}\, \sqrt {\frac {\cos \left (d x +c \right )}{1+\cos \left (d x +c \right )}}+i B \sqrt {\frac {1}{1+\cos \left (d x +c \right )}}\, \sqrt {\frac {\cos \left (d x +c \right )}{1+\cos \left (d x +c \right )}}\, \sin \left (d x +c \right ) \EllipticF \left (\frac {i \left (-1+\cos \left (d x +c \right )\right )}{\sin \left (d x +c \right )}, i\right )-A \left (\cos ^{2}\left (d x +c \right )\right )+A \cos \left (d x +c \right )\right ) \sqrt {\frac {b}{\cos \left (d x +c \right )}}}{d \sin \left (d x +c \right ) b} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

2/d*(I*A*sin(d*x+c)*cos(d*x+c)*(1/(1+cos(d*x+c)))^(1/2)*(cos(d*x+c)/(1+cos(d*x+c)))^(1/2)*EllipticF(I*(-1+cos(
d*x+c))/sin(d*x+c),I)-I*A*EllipticE(I*(-1+cos(d*x+c))/sin(d*x+c),I)*cos(d*x+c)*sin(d*x+c)*(1/(1+cos(d*x+c)))^(
1/2)*(cos(d*x+c)/(1+cos(d*x+c)))^(1/2)+I*B*cos(d*x+c)*(1/(1+cos(d*x+c)))^(1/2)*(cos(d*x+c)/(1+cos(d*x+c)))^(1/
2)*EllipticF(I*(-1+cos(d*x+c))/sin(d*x+c),I)*sin(d*x+c)+I*A*EllipticF(I*(-1+cos(d*x+c))/sin(d*x+c),I)*sin(d*x+
c)*(1/(1+cos(d*x+c)))^(1/2)*(cos(d*x+c)/(1+cos(d*x+c)))^(1/2)-I*A*sin(d*x+c)*EllipticE(I*(-1+cos(d*x+c))/sin(d
*x+c),I)*(1/(1+cos(d*x+c)))^(1/2)*(cos(d*x+c)/(1+cos(d*x+c)))^(1/2)+I*B*sin(d*x+c)*EllipticF(I*(-1+cos(d*x+c))
/sin(d*x+c),I)*(1/(1+cos(d*x+c)))^(1/2)*(cos(d*x+c)/(1+cos(d*x+c)))^(1/2)-A*cos(d*x+c)^2+A*cos(d*x+c))*(b/cos(
d*x+c))^(1/2)/sin(d*x+c)/b

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

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

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

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

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

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

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

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

Integral((A + B*sec(c + d*x))/sqrt(b*sec(c + d*x)), x)

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