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

Optimal. Leaf size=118 \[ \frac{\sqrt{2} (A-B+C) \tan ^{-1}\left (\frac{\sqrt{a} \tan (c+d x)}{\sqrt{2} \sqrt{a \sec (c+d x)+a}}\right )}{\sqrt{a} d}+\frac{2 (3 B-2 C) \tan (c+d x)}{3 d \sqrt{a \sec (c+d x)+a}}+\frac{2 C \tan (c+d x) \sqrt{a \sec (c+d x)+a}}{3 a d} \]

[Out]

(Sqrt[2]*(A - B + C)*ArcTan[(Sqrt[a]*Tan[c + d*x])/(Sqrt[2]*Sqrt[a + a*Sec[c + d*x]])])/(Sqrt[a]*d) + (2*(3*B
- 2*C)*Tan[c + d*x])/(3*d*Sqrt[a + a*Sec[c + d*x]]) + (2*C*Sqrt[a + a*Sec[c + d*x]]*Tan[c + d*x])/(3*a*d)

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Rubi [A]  time = 0.225303, antiderivative size = 118, normalized size of antiderivative = 1., number of steps used = 4, number of rules used = 4, integrand size = 41, \(\frac{\text{number of rules}}{\text{integrand size}}\) = 0.098, Rules used = {4082, 4001, 3795, 203} \[ \frac{\sqrt{2} (A-B+C) \tan ^{-1}\left (\frac{\sqrt{a} \tan (c+d x)}{\sqrt{2} \sqrt{a \sec (c+d x)+a}}\right )}{\sqrt{a} d}+\frac{2 (3 B-2 C) \tan (c+d x)}{3 d \sqrt{a \sec (c+d x)+a}}+\frac{2 C \tan (c+d x) \sqrt{a \sec (c+d x)+a}}{3 a d} \]

Antiderivative was successfully verified.

[In]

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

[Out]

(Sqrt[2]*(A - B + C)*ArcTan[(Sqrt[a]*Tan[c + d*x])/(Sqrt[2]*Sqrt[a + a*Sec[c + d*x]])])/(Sqrt[a]*d) + (2*(3*B
- 2*C)*Tan[c + d*x])/(3*d*Sqrt[a + a*Sec[c + d*x]]) + (2*C*Sqrt[a + a*Sec[c + d*x]]*Tan[c + d*x])/(3*a*d)

Rule 4082

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

Rule 4001

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

Rule 3795

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

Rule 203

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

Rubi steps

\begin{align*} \int \frac{\sec (c+d x) \left (A+B \sec (c+d x)+C \sec ^2(c+d x)\right )}{\sqrt{a+a \sec (c+d x)}} \, dx &=\frac{2 C \sqrt{a+a \sec (c+d x)} \tan (c+d x)}{3 a d}+\frac{2 \int \frac{\sec (c+d x) \left (\frac{1}{2} a (3 A+C)+\frac{1}{2} a (3 B-2 C) \sec (c+d x)\right )}{\sqrt{a+a \sec (c+d x)}} \, dx}{3 a}\\ &=\frac{2 (3 B-2 C) \tan (c+d x)}{3 d \sqrt{a+a \sec (c+d x)}}+\frac{2 C \sqrt{a+a \sec (c+d x)} \tan (c+d x)}{3 a d}+(A-B+C) \int \frac{\sec (c+d x)}{\sqrt{a+a \sec (c+d x)}} \, dx\\ &=\frac{2 (3 B-2 C) \tan (c+d x)}{3 d \sqrt{a+a \sec (c+d x)}}+\frac{2 C \sqrt{a+a \sec (c+d x)} \tan (c+d x)}{3 a d}-\frac{(2 (A-B+C)) \operatorname{Subst}\left (\int \frac{1}{2 a+x^2} \, dx,x,-\frac{a \tan (c+d x)}{\sqrt{a+a \sec (c+d x)}}\right )}{d}\\ &=\frac{\sqrt{2} (A-B+C) \tan ^{-1}\left (\frac{\sqrt{a} \tan (c+d x)}{\sqrt{2} \sqrt{a+a \sec (c+d x)}}\right )}{\sqrt{a} d}+\frac{2 (3 B-2 C) \tan (c+d x)}{3 d \sqrt{a+a \sec (c+d x)}}+\frac{2 C \sqrt{a+a \sec (c+d x)} \tan (c+d x)}{3 a d}\\ \end{align*}

Mathematica [C]  time = 7.13237, size = 628, normalized size = 5.32 \[ \frac{4 \sqrt{\frac{1}{1-2 \sin ^2\left (\frac{1}{2} (c+d x)\right )}} \sqrt{1-2 \sin ^2\left (\frac{1}{2} (c+d x)\right )} \cos \left (\frac{1}{2} (c+d x)\right ) \left (A+B \sec (c+d x)+C \sec ^2(c+d x)\right ) \left (\frac{(A-B+C) \csc ^5\left (\frac{1}{2} (c+d x)\right ) \left (-12 \sin ^8\left (\frac{1}{2} (c+d x)\right ) \cos ^4\left (\frac{1}{2} (c+d x)\right ) \text{HypergeometricPFQ}\left (\left \{2,2,\frac{7}{2}\right \},\left \{1,\frac{9}{2}\right \},-\frac{\sin ^2\left (\frac{1}{2} (c+d x)\right )}{1-2 \sin ^2\left (\frac{1}{2} (c+d x)\right )}\right )-12 \left (3 \sin ^4\left (\frac{1}{2} (c+d x)\right )-7 \sin ^2\left (\frac{1}{2} (c+d x)\right )+4\right ) \sin ^8\left (\frac{1}{2} (c+d x)\right ) \text{Hypergeometric2F1}\left (2,\frac{7}{2},\frac{9}{2},-\frac{\sin ^2\left (\frac{1}{2} (c+d x)\right )}{1-2 \sin ^2\left (\frac{1}{2} (c+d x)\right )}\right )+7 \sqrt{-\frac{\sin ^2\left (\frac{1}{2} (c+d x)\right )}{1-2 \sin ^2\left (\frac{1}{2} (c+d x)\right )}} \left (1-2 \sin ^2\left (\frac{1}{2} (c+d x)\right )\right )^3 \left (8 \sin ^4\left (\frac{1}{2} (c+d x)\right )-20 \sin ^2\left (\frac{1}{2} (c+d x)\right )+15\right ) \left (\left (3-7 \sin ^2\left (\frac{1}{2} (c+d x)\right )\right ) \sqrt{-\frac{\sin ^2\left (\frac{1}{2} (c+d x)\right )}{1-2 \sin ^2\left (\frac{1}{2} (c+d x)\right )}}-3 \left (1-2 \sin ^2\left (\frac{1}{2} (c+d x)\right )\right ) \tanh ^{-1}\left (\sqrt{-\frac{\sin ^2\left (\frac{1}{2} (c+d x)\right )}{1-2 \sin ^2\left (\frac{1}{2} (c+d x)\right )}}\right )\right )\right )}{63 \left (1-2 \sin ^2\left (\frac{1}{2} (c+d x)\right )\right )^{7/2}}-\frac{4 A \sin ^3\left (\frac{1}{2} (c+d x)\right )}{3 \left (1-2 \sin ^2\left (\frac{1}{2} (c+d x)\right )\right )^{3/2}}+\frac{4 B \sin \left (\frac{1}{2} (c+d x)\right )}{3 \sqrt{1-2 \sin ^2\left (\frac{1}{2} (c+d x)\right )}}+\frac{2 B \sin \left (\frac{1}{2} (c+d x)\right )}{3 \left (1-2 \sin ^2\left (\frac{1}{2} (c+d x)\right )\right )^{3/2}}\right )}{d \sec ^{\frac{3}{2}}(c+d x) \sqrt{a (\sec (c+d x)+1)} (A \cos (2 c+2 d x)+A+2 B \cos (c+d x)+2 C)} \]

Warning: Unable to verify antiderivative.

[In]

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

[Out]

(4*Cos[(c + d*x)/2]*(A + B*Sec[c + d*x] + C*Sec[c + d*x]^2)*Sqrt[(1 - 2*Sin[(c + d*x)/2]^2)^(-1)]*Sqrt[1 - 2*S
in[(c + d*x)/2]^2]*((2*B*Sin[(c + d*x)/2])/(3*(1 - 2*Sin[(c + d*x)/2]^2)^(3/2)) - (4*A*Sin[(c + d*x)/2]^3)/(3*
(1 - 2*Sin[(c + d*x)/2]^2)^(3/2)) + (4*B*Sin[(c + d*x)/2])/(3*Sqrt[1 - 2*Sin[(c + d*x)/2]^2]) + ((A - B + C)*C
sc[(c + d*x)/2]^5*(-12*Cos[(c + d*x)/2]^4*HypergeometricPFQ[{2, 2, 7/2}, {1, 9/2}, -(Sin[(c + d*x)/2]^2/(1 - 2
*Sin[(c + d*x)/2]^2))]*Sin[(c + d*x)/2]^8 - 12*Hypergeometric2F1[2, 7/2, 9/2, -(Sin[(c + d*x)/2]^2/(1 - 2*Sin[
(c + d*x)/2]^2))]*Sin[(c + d*x)/2]^8*(4 - 7*Sin[(c + d*x)/2]^2 + 3*Sin[(c + d*x)/2]^4) + 7*Sqrt[-(Sin[(c + d*x
)/2]^2/(1 - 2*Sin[(c + d*x)/2]^2))]*(1 - 2*Sin[(c + d*x)/2]^2)^3*(15 - 20*Sin[(c + d*x)/2]^2 + 8*Sin[(c + d*x)
/2]^4)*((3 - 7*Sin[(c + d*x)/2]^2)*Sqrt[-(Sin[(c + d*x)/2]^2/(1 - 2*Sin[(c + d*x)/2]^2))] - 3*ArcTanh[Sqrt[-(S
in[(c + d*x)/2]^2/(1 - 2*Sin[(c + d*x)/2]^2))]]*(1 - 2*Sin[(c + d*x)/2]^2))))/(63*(1 - 2*Sin[(c + d*x)/2]^2)^(
7/2))))/(d*(A + 2*C + 2*B*Cos[c + d*x] + A*Cos[2*c + 2*d*x])*Sec[c + d*x]^(3/2)*Sqrt[a*(1 + Sec[c + d*x])])

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Maple [B]  time = 0.339, size = 563, normalized size = 4.8 \begin{align*} \text{result too large to display} \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

-1/6/d/a*(3*A*cos(d*x+c)*sin(d*x+c)*(-2*cos(d*x+c)/(cos(d*x+c)+1))^(3/2)*ln(-(-(-2*cos(d*x+c)/(cos(d*x+c)+1))^
(1/2)*sin(d*x+c)+cos(d*x+c)-1)/sin(d*x+c))-3*B*cos(d*x+c)*sin(d*x+c)*(-2*cos(d*x+c)/(cos(d*x+c)+1))^(3/2)*ln(-
(-(-2*cos(d*x+c)/(cos(d*x+c)+1))^(1/2)*sin(d*x+c)+cos(d*x+c)-1)/sin(d*x+c))+3*C*cos(d*x+c)*sin(d*x+c)*(-2*cos(
d*x+c)/(cos(d*x+c)+1))^(3/2)*ln(-(-(-2*cos(d*x+c)/(cos(d*x+c)+1))^(1/2)*sin(d*x+c)+cos(d*x+c)-1)/sin(d*x+c))+3
*A*(-2*cos(d*x+c)/(cos(d*x+c)+1))^(3/2)*ln(-(-(-2*cos(d*x+c)/(cos(d*x+c)+1))^(1/2)*sin(d*x+c)+cos(d*x+c)-1)/si
n(d*x+c))*sin(d*x+c)-3*B*(-2*cos(d*x+c)/(cos(d*x+c)+1))^(3/2)*ln(-(-(-2*cos(d*x+c)/(cos(d*x+c)+1))^(1/2)*sin(d
*x+c)+cos(d*x+c)-1)/sin(d*x+c))*sin(d*x+c)+3*C*(-2*cos(d*x+c)/(cos(d*x+c)+1))^(3/2)*ln(-(-(-2*cos(d*x+c)/(cos(
d*x+c)+1))^(1/2)*sin(d*x+c)+cos(d*x+c)-1)/sin(d*x+c))*sin(d*x+c)+12*B*cos(d*x+c)^2-4*C*cos(d*x+c)^2-12*B*cos(d
*x+c)+8*C*cos(d*x+c)-4*C)*(a*(cos(d*x+c)+1)/cos(d*x+c))^(1/2)/sin(d*x+c)/cos(d*x+c)

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

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

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

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Fricas [A]  time = 0.615628, size = 937, normalized size = 7.94 \begin{align*} \left [\frac{3 \, \sqrt{2}{\left ({\left (A - B + C\right )} a \cos \left (d x + c\right )^{2} +{\left (A - B + C\right )} a \cos \left (d x + c\right )\right )} \sqrt{-\frac{1}{a}} \log \left (-\frac{2 \, \sqrt{2} \sqrt{\frac{a \cos \left (d x + c\right ) + a}{\cos \left (d x + c\right )}} \sqrt{-\frac{1}{a}} \cos \left (d x + c\right ) \sin \left (d x + c\right ) - 3 \, \cos \left (d x + c\right )^{2} - 2 \, \cos \left (d x + c\right ) + 1}{\cos \left (d x + c\right )^{2} + 2 \, \cos \left (d x + c\right ) + 1}\right ) + 4 \,{\left ({\left (3 \, B - C\right )} \cos \left (d x + c\right ) + C\right )} \sqrt{\frac{a \cos \left (d x + c\right ) + a}{\cos \left (d x + c\right )}} \sin \left (d x + c\right )}{6 \,{\left (a d \cos \left (d x + c\right )^{2} + a d \cos \left (d x + c\right )\right )}}, \frac{2 \,{\left ({\left (3 \, B - C\right )} \cos \left (d x + c\right ) + C\right )} \sqrt{\frac{a \cos \left (d x + c\right ) + a}{\cos \left (d x + c\right )}} \sin \left (d x + c\right ) - \frac{3 \, \sqrt{2}{\left ({\left (A - B + C\right )} a \cos \left (d x + c\right )^{2} +{\left (A - B + C\right )} a \cos \left (d x + c\right )\right )} \arctan \left (\frac{\sqrt{2} \sqrt{\frac{a \cos \left (d x + c\right ) + a}{\cos \left (d x + c\right )}} \cos \left (d x + c\right )}{\sqrt{a} \sin \left (d x + c\right )}\right )}{\sqrt{a}}}{3 \,{\left (a d \cos \left (d x + c\right )^{2} + a d \cos \left (d x + c\right )\right )}}\right ] \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

[1/6*(3*sqrt(2)*((A - B + C)*a*cos(d*x + c)^2 + (A - B + C)*a*cos(d*x + c))*sqrt(-1/a)*log(-(2*sqrt(2)*sqrt((a
*cos(d*x + c) + a)/cos(d*x + c))*sqrt(-1/a)*cos(d*x + c)*sin(d*x + c) - 3*cos(d*x + c)^2 - 2*cos(d*x + c) + 1)
/(cos(d*x + c)^2 + 2*cos(d*x + c) + 1)) + 4*((3*B - C)*cos(d*x + c) + C)*sqrt((a*cos(d*x + c) + a)/cos(d*x + c
))*sin(d*x + c))/(a*d*cos(d*x + c)^2 + a*d*cos(d*x + c)), 1/3*(2*((3*B - C)*cos(d*x + c) + C)*sqrt((a*cos(d*x
+ c) + a)/cos(d*x + c))*sin(d*x + c) - 3*sqrt(2)*((A - B + C)*a*cos(d*x + c)^2 + (A - B + C)*a*cos(d*x + c))*a
rctan(sqrt(2)*sqrt((a*cos(d*x + c) + a)/cos(d*x + c))*cos(d*x + c)/(sqrt(a)*sin(d*x + c)))/sqrt(a))/(a*d*cos(d
*x + c)^2 + a*d*cos(d*x + c))]

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Sympy [F]  time = 0., size = 0, normalized size = 0. \begin{align*} \int \frac{\left (A + B \sec{\left (c + d x \right )} + C \sec ^{2}{\left (c + d x \right )}\right ) \sec{\left (c + d x \right )}}{\sqrt{a \left (\sec{\left (c + d x \right )} + 1\right )}}\, dx \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

Integral((A + B*sec(c + d*x) + C*sec(c + d*x)**2)*sec(c + d*x)/sqrt(a*(sec(c + d*x) + 1)), x)

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Giac [A]  time = 8.95486, size = 252, normalized size = 2.14 \begin{align*} \frac{\frac{3 \, \sqrt{2}{\left (A - B + C\right )} \log \left ({\left | -\sqrt{-a} \tan \left (\frac{1}{2} \, d x + \frac{1}{2} \, c\right ) + \sqrt{-a \tan \left (\frac{1}{2} \, d x + \frac{1}{2} \, c\right )^{2} + a} \right |}\right )}{\sqrt{-a} \mathrm{sgn}\left (\tan \left (\frac{1}{2} \, d x + \frac{1}{2} \, c\right )^{2} - 1\right )} - \frac{2 \,{\left (\frac{\sqrt{2}{\left (3 \, B a - 2 \, C a\right )} \tan \left (\frac{1}{2} \, d x + \frac{1}{2} \, c\right )^{2}}{\mathrm{sgn}\left (\tan \left (\frac{1}{2} \, d x + \frac{1}{2} \, c\right )^{2} - 1\right )} - \frac{3 \, \sqrt{2} B a}{\mathrm{sgn}\left (\tan \left (\frac{1}{2} \, d x + \frac{1}{2} \, c\right )^{2} - 1\right )}\right )} \tan \left (\frac{1}{2} \, d x + \frac{1}{2} \, c\right )}{{\left (a \tan \left (\frac{1}{2} \, d x + \frac{1}{2} \, c\right )^{2} - a\right )} \sqrt{-a \tan \left (\frac{1}{2} \, d x + \frac{1}{2} \, c\right )^{2} + a}}}{3 \, d} \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

1/3*(3*sqrt(2)*(A - B + C)*log(abs(-sqrt(-a)*tan(1/2*d*x + 1/2*c) + sqrt(-a*tan(1/2*d*x + 1/2*c)^2 + a)))/(sqr
t(-a)*sgn(tan(1/2*d*x + 1/2*c)^2 - 1)) - 2*(sqrt(2)*(3*B*a - 2*C*a)*tan(1/2*d*x + 1/2*c)^2/sgn(tan(1/2*d*x + 1
/2*c)^2 - 1) - 3*sqrt(2)*B*a/sgn(tan(1/2*d*x + 1/2*c)^2 - 1))*tan(1/2*d*x + 1/2*c)/((a*tan(1/2*d*x + 1/2*c)^2
- a)*sqrt(-a*tan(1/2*d*x + 1/2*c)^2 + a)))/d