3.67 \(\int \frac{A+B \sec (c+d x)+C \sec ^2(c+d x)}{\sqrt{b \sec (c+d x)}} \, dx\)

Optimal. Leaf size=110 \[ \frac{2 B \sqrt{\cos (c+d x)} \text{EllipticF}\left (\frac{1}{2} (c+d x),2\right ) \sqrt{b \sec (c+d x)}}{b d}+\frac{2 (A-C) E\left (\left .\frac{1}{2} (c+d x)\right |2\right )}{d \sqrt{\cos (c+d x)} \sqrt{b \sec (c+d x)}}+\frac{2 C \tan (c+d x)}{d \sqrt{b \sec (c+d x)}} \]

[Out]

(2*(A - C)*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]]*El
lipticF[(c + d*x)/2, 2]*Sqrt[b*Sec[c + d*x]])/(b*d) + (2*C*Tan[c + d*x])/(d*Sqrt[b*Sec[c + d*x]])

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Rubi [A]  time = 0.111177, antiderivative size = 110, normalized size of antiderivative = 1., number of steps used = 6, number of rules used = 5, integrand size = 33, \(\frac{\text{number of rules}}{\text{integrand size}}\) = 0.152, Rules used = {4047, 3771, 2641, 4046, 2639} \[ \frac{2 (A-C) E\left (\left .\frac{1}{2} (c+d x)\right |2\right )}{d \sqrt{\cos (c+d x)} \sqrt{b \sec (c+d x)}}+\frac{2 B \sqrt{\cos (c+d x)} F\left (\left .\frac{1}{2} (c+d x)\right |2\right ) \sqrt{b \sec (c+d x)}}{b d}+\frac{2 C \tan (c+d x)}{d \sqrt{b \sec (c+d x)}} \]

Antiderivative was successfully verified.

[In]

Int[(A + B*Sec[c + d*x] + C*Sec[c + d*x]^2)/Sqrt[b*Sec[c + d*x]],x]

[Out]

(2*(A - C)*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]]*El
lipticF[(c + d*x)/2, 2]*Sqrt[b*Sec[c + d*x]])/(b*d) + (2*C*Tan[c + d*x])/(d*Sqrt[b*Sec[c + d*x]])

Rule 4047

Int[(csc[(e_.) + (f_.)*(x_)]*(b_.))^(m_.)*((A_.) + csc[(e_.) + (f_.)*(x_)]*(B_.) + csc[(e_.) + (f_.)*(x_)]^2*(
C_.)), x_Symbol] :> Dist[B/b, Int[(b*Csc[e + f*x])^(m + 1), x], x] + Int[(b*Csc[e + f*x])^m*(A + C*Csc[e + f*x
]^2), x] /; FreeQ[{b, e, f, A, B, C, m}, 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 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 4046

Int[(csc[(e_.) + (f_.)*(x_)]*(b_.))^(m_.)*(csc[(e_.) + (f_.)*(x_)]^2*(C_.) + (A_)), x_Symbol] :> -Simp[(C*Cot[
e + f*x]*(b*Csc[e + f*x])^m)/(f*(m + 1)), x] + Dist[(C*m + A*(m + 1))/(m + 1), Int[(b*Csc[e + f*x])^m, x], x]
/; FreeQ[{b, e, f, A, C, m}, x] && NeQ[C*m + A*(m + 1), 0] &&  !LeQ[m, -1]

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]

Rubi steps

\begin{align*} \int \frac{A+B \sec (c+d x)+C \sec ^2(c+d x)}{\sqrt{b \sec (c+d x)}} \, dx &=\frac{B \int \sqrt{b \sec (c+d x)} \, dx}{b}+\int \frac{A+C \sec ^2(c+d x)}{\sqrt{b \sec (c+d x)}} \, dx\\ &=\frac{2 C \tan (c+d x)}{d \sqrt{b \sec (c+d x)}}+(A-C) \int \frac{1}{\sqrt{b \sec (c+d x)}} \, dx+\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 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}+\frac{2 C \tan (c+d x)}{d \sqrt{b \sec (c+d x)}}+\frac{(A-C) \int \sqrt{\cos (c+d x)} \, dx}{\sqrt{\cos (c+d x)} \sqrt{b \sec (c+d x)}}\\ &=\frac{2 (A-C) 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}+\frac{2 C \tan (c+d x)}{d \sqrt{b \sec (c+d x)}}\\ \end{align*}

Mathematica [C]  time = 1.38564, size = 156, normalized size = 1.42 \[ \frac{2 e^{-i d x} (\sin (d x)-i \cos (d x)) \sqrt{b \sec (c+d x)} \left ((A-C) e^{i (c+d x)} \sqrt{1+e^{2 i (c+d x)}} \text{Hypergeometric2F1}\left (\frac{1}{2},\frac{3}{4},\frac{7}{4},-e^{2 i (c+d x)}\right )+3 i B \sqrt{\cos (c+d x)} \text{EllipticF}\left (\frac{1}{2} (c+d x),2\right )-3 A \cos (c+d x)+3 i C \sin (c+d x)+3 C \cos (c+d x)\right )}{3 b d} \]

Antiderivative was successfully verified.

[In]

Integrate[(A + B*Sec[c + d*x] + C*Sec[c + d*x]^2)/Sqrt[b*Sec[c + d*x]],x]

[Out]

(2*Sqrt[b*Sec[c + d*x]]*((-I)*Cos[d*x] + Sin[d*x])*(-3*A*Cos[c + d*x] + 3*C*Cos[c + d*x] + (3*I)*B*Sqrt[Cos[c
+ d*x]]*EllipticF[(c + d*x)/2, 2] + (A - C)*E^(I*(c + d*x))*Sqrt[1 + E^((2*I)*(c + d*x))]*Hypergeometric2F1[1/
2, 3/4, 7/4, -E^((2*I)*(c + d*x))] + (3*I)*C*Sin[c + d*x]))/(3*b*d*E^(I*d*x))

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Maple [C]  time = 0.292, size = 719, normalized size = 6.5 \begin{align*} \text{result too large to display} \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

2/d*(I*A*EllipticF(I*(-1+cos(d*x+c))/sin(d*x+c),I)*cos(d*x+c)*(1/(cos(d*x+c)+1))^(1/2)*(cos(d*x+c)/(cos(d*x+c)
+1))^(1/2)*sin(d*x+c)-I*A*sin(d*x+c)*cos(d*x+c)*(1/(cos(d*x+c)+1))^(1/2)*(cos(d*x+c)/(cos(d*x+c)+1))^(1/2)*Ell
ipticE(I*(-1+cos(d*x+c))/sin(d*x+c),I)+I*B*EllipticF(I*(-1+cos(d*x+c))/sin(d*x+c),I)*cos(d*x+c)*(1/(cos(d*x+c)
+1))^(1/2)*(cos(d*x+c)/(cos(d*x+c)+1))^(1/2)*sin(d*x+c)-I*C*cos(d*x+c)*(1/(cos(d*x+c)+1))^(1/2)*(cos(d*x+c)/(c
os(d*x+c)+1))^(1/2)*EllipticF(I*(-1+cos(d*x+c))/sin(d*x+c),I)*sin(d*x+c)+I*C*EllipticE(I*(-1+cos(d*x+c))/sin(d
*x+c),I)*cos(d*x+c)*(1/(cos(d*x+c)+1))^(1/2)*(cos(d*x+c)/(cos(d*x+c)+1))^(1/2)*sin(d*x+c)+I*A*(1/(cos(d*x+c)+1
))^(1/2)*(cos(d*x+c)/(cos(d*x+c)+1))^(1/2)*EllipticF(I*(-1+cos(d*x+c))/sin(d*x+c),I)*sin(d*x+c)-I*A*sin(d*x+c)
*EllipticE(I*(-1+cos(d*x+c))/sin(d*x+c),I)*(1/(cos(d*x+c)+1))^(1/2)*(cos(d*x+c)/(cos(d*x+c)+1))^(1/2)+I*B*Elli
pticF(I*(-1+cos(d*x+c))/sin(d*x+c),I)*sin(d*x+c)*(1/(cos(d*x+c)+1))^(1/2)*(cos(d*x+c)/(cos(d*x+c)+1))^(1/2)-I*
C*EllipticF(I*(-1+cos(d*x+c))/sin(d*x+c),I)*(1/(cos(d*x+c)+1))^(1/2)*(cos(d*x+c)/(cos(d*x+c)+1))^(1/2)*sin(d*x
+c)+I*C*EllipticE(I*(-1+cos(d*x+c))/sin(d*x+c),I)*(1/(cos(d*x+c)+1))^(1/2)*(cos(d*x+c)/(cos(d*x+c)+1))^(1/2)*s
in(d*x+c)-A*cos(d*x+c)^2+A*cos(d*x+c)-C*cos(d*x+c)+C)*(b/cos(d*x+c))^(1/2)/sin(d*x+c)/b

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Maxima [F]  time = 0., size = 0, normalized size = 0. \begin{align*} \int \frac{C \sec \left (d x + c\right )^{2} + B \sec \left (d x + c\right ) + A}{\sqrt{b \sec \left (d x + c\right )}}\,{d x} \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

integrate((C*sec(d*x + c)^2 + B*sec(d*x + c) + A)/sqrt(b*sec(d*x + c)), x)

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Fricas [F]  time = 0., size = 0, normalized size = 0. \begin{align*}{\rm integral}\left (\frac{{\left (C \sec \left (d x + c\right )^{2} + 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 ) \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

integral((C*sec(d*x + c)^2 + B*sec(d*x + c) + A)*sqrt(b*sec(d*x + c))/(b*sec(d*x + c)), x)

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Sympy [F]  time = 0., size = 0, normalized size = 0. \begin{align*} \int \frac{A + B \sec{\left (c + d x \right )} + C \sec ^{2}{\left (c + d x \right )}}{\sqrt{b \sec{\left (c + d x \right )}}}\, dx \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

Integral((A + B*sec(c + d*x) + C*sec(c + d*x)**2)/sqrt(b*sec(c + d*x)), x)

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Giac [F]  time = 0., size = 0, normalized size = 0. \begin{align*} \int \frac{C \sec \left (d x + c\right )^{2} + B \sec \left (d x + c\right ) + A}{\sqrt{b \sec \left (d x + c\right )}}\,{d x} \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

integrate((C*sec(d*x + c)^2 + B*sec(d*x + c) + A)/sqrt(b*sec(d*x + c)), x)