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

Optimal. Leaf size=100 \[ \frac{2 \sqrt{a} A \tan ^{-1}\left (\frac{\sqrt{a} \tan (c+d x)}{\sqrt{a \sec (c+d x)+a}}\right )}{d}+\frac{2 a (3 B+C) \tan (c+d x)}{3 d \sqrt{a \sec (c+d x)+a}}+\frac{2 C \tan (c+d x) \sqrt{a \sec (c+d x)+a}}{3 d} \]

[Out]

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

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Rubi [A]  time = 0.152305, antiderivative size = 100, normalized size of antiderivative = 1., number of steps used = 5, number of rules used = 5, integrand size = 35, \(\frac{\text{number of rules}}{\text{integrand size}}\) = 0.143, Rules used = {4054, 3915, 3774, 203, 3792} \[ \frac{2 \sqrt{a} A \tan ^{-1}\left (\frac{\sqrt{a} \tan (c+d x)}{\sqrt{a \sec (c+d x)+a}}\right )}{d}+\frac{2 a (3 B+C) \tan (c+d x)}{3 d \sqrt{a \sec (c+d x)+a}}+\frac{2 C \tan (c+d x) \sqrt{a \sec (c+d x)+a}}{3 d} \]

Antiderivative was successfully verified.

[In]

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

[Out]

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

Rule 4054

Int[((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)/(f*(m + 1)), x] + Dist[1/(b*(m + 1)),
Int[(a + b*Csc[e + f*x])^m*Simp[A*b*(m + 1) + (a*C*m + b*B*(m + 1))*Csc[e + f*x], x], x], x] /; FreeQ[{a, b,
e, f, A, B, C, m}, x] && EqQ[a^2 - b^2, 0] &&  !LtQ[m, -2^(-1)]

Rule 3915

Int[Sqrt[csc[(e_.) + (f_.)*(x_)]*(b_.) + (a_)]*(csc[(e_.) + (f_.)*(x_)]*(d_.) + (c_)), x_Symbol] :> Dist[c, In
t[Sqrt[a + b*Csc[e + f*x]], x], x] + Dist[d, Int[Sqrt[a + b*Csc[e + f*x]]*Csc[e + f*x], x], x] /; FreeQ[{a, b,
c, d, e, f}, x] && NeQ[b*c - a*d, 0] && EqQ[a^2 - b^2, 0]

Rule 3774

Int[Sqrt[csc[(c_.) + (d_.)*(x_)]*(b_.) + (a_)], x_Symbol] :> Dist[(-2*b)/d, Subst[Int[1/(a + x^2), x], x, (b*C
ot[c + d*x])/Sqrt[a + b*Csc[c + d*x]]], x] /; FreeQ[{a, b, c, d}, 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])

Rule 3792

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

Rubi steps

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

Mathematica [A]  time = 0.687608, size = 101, normalized size = 1.01 \[ \frac{\sec \left (\frac{1}{2} (c+d x)\right ) \sec (c+d x) \sqrt{a (\sec (c+d x)+1)} \left (3 \sqrt{2} A \sin ^{-1}\left (\sqrt{2} \sin \left (\frac{1}{2} (c+d x)\right )\right ) \cos ^{\frac{3}{2}}(c+d x)+2 \sin \left (\frac{1}{2} (c+d x)\right ) ((3 B+2 C) \cos (c+d x)+C)\right )}{3 d} \]

Antiderivative was successfully verified.

[In]

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

[Out]

(Sec[(c + d*x)/2]*Sec[c + d*x]*Sqrt[a*(1 + Sec[c + d*x])]*(3*Sqrt[2]*A*ArcSin[Sqrt[2]*Sin[(c + d*x)/2]]*Cos[c
+ d*x]^(3/2) + 2*(C + (3*B + 2*C)*Cos[c + d*x])*Sin[(c + d*x)/2]))/(3*d)

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Maple [B]  time = 0.321, size = 236, normalized size = 2.4 \begin{align*}{\frac{1}{6\,d\sin \left ( dx+c \right ) \cos \left ( dx+c \right ) }\sqrt{{\frac{a \left ( \cos \left ( dx+c \right ) +1 \right ) }{\cos \left ( dx+c \right ) }}} \left ( 3\,A\sqrt{2}\sin \left ( dx+c \right ) \cos \left ( dx+c \right ){\it Artanh} \left ( 1/2\,{\frac{\sqrt{2}\sin \left ( dx+c \right ) }{\cos \left ( dx+c \right ) }\sqrt{-2\,{\frac{\cos \left ( dx+c \right ) }{\cos \left ( dx+c \right ) +1}}}} \right ) \left ( -2\,{\frac{\cos \left ( dx+c \right ) }{\cos \left ( dx+c \right ) +1}} \right ) ^{3/2}+3\,A\sqrt{2}{\it Artanh} \left ( 1/2\,{\frac{\sqrt{2}\sin \left ( dx+c \right ) }{\cos \left ( dx+c \right ) }\sqrt{-2\,{\frac{\cos \left ( dx+c \right ) }{\cos \left ( dx+c \right ) +1}}}} \right ) \left ( -2\,{\frac{\cos \left ( dx+c \right ) }{\cos \left ( dx+c \right ) +1}} \right ) ^{3/2}\sin \left ( dx+c \right ) -12\,B \left ( \cos \left ( dx+c \right ) \right ) ^{2}-8\,C \left ( \cos \left ( dx+c \right ) \right ) ^{2}+12\,B\cos \left ( dx+c \right ) +4\,C\cos \left ( dx+c \right ) +4\,C \right ) } \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)*(a+a*sec(d*x+c))^(1/2),x)

[Out]

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

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Maxima [F(-1)]  time = 0., size = 0, normalized size = 0. \begin{align*} \text{Timed out} \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)*(a+a*sec(d*x+c))^(1/2),x, algorithm="maxima")

[Out]

Timed out

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Fricas [A]  time = 0.551226, size = 792, normalized size = 7.92 \begin{align*} \left [\frac{3 \,{\left (A \cos \left (d x + c\right )^{2} + A \cos \left (d x + c\right )\right )} \sqrt{-a} \log \left (\frac{2 \, a \cos \left (d x + c\right )^{2} - 2 \, \sqrt{-a} \sqrt{\frac{a \cos \left (d x + c\right ) + a}{\cos \left (d x + c\right )}} \cos \left (d x + c\right ) \sin \left (d x + c\right ) + a \cos \left (d x + c\right ) - a}{\cos \left (d x + c\right ) + 1}\right ) + 2 \,{\left ({\left (3 \, B + 2 \, 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 )}{3 \,{\left (d \cos \left (d x + c\right )^{2} + d \cos \left (d x + c\right )\right )}}, -\frac{2 \,{\left (3 \,{\left (A \cos \left (d x + c\right )^{2} + A \cos \left (d x + c\right )\right )} \sqrt{a} \arctan \left (\frac{\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 ) -{\left ({\left (3 \, B + 2 \, 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 )\right )}}{3 \,{\left (d \cos \left (d x + c\right )^{2} + d \cos \left (d x + c\right )\right )}}\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)*(a+a*sec(d*x+c))^(1/2),x, algorithm="fricas")

[Out]

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

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

[Out]

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

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Giac [F(-2)]  time = 0., size = 0, normalized size = 0. \begin{align*} \text{Exception raised: NotImplementedError} \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)*(a+a*sec(d*x+c))^(1/2),x, algorithm="giac")

[Out]

Exception raised: NotImplementedError