3.496 \(\int \cos ^2(c+d x) (a+a \sec (c+d x))^{3/2} (A+B \sec (c+d x)+C \sec ^2(c+d x)) \, dx\)

Optimal. Leaf size=157 \[ \frac{a^2 (5 A+4 B-8 C) \sin (c+d x)}{4 d \sqrt{a \sec (c+d x)+a}}+\frac{a^{3/2} (7 A+12 B+8 C) \tan ^{-1}\left (\frac{\sqrt{a} \tan (c+d x)}{\sqrt{a \sec (c+d x)+a}}\right )}{4 d}-\frac{a (A-4 C) \sin (c+d x) \sqrt{a \sec (c+d x)+a}}{2 d}+\frac{A \sin (c+d x) \cos (c+d x) (a \sec (c+d x)+a)^{3/2}}{2 d} \]

[Out]

(a^(3/2)*(7*A + 12*B + 8*C)*ArcTan[(Sqrt[a]*Tan[c + d*x])/Sqrt[a + a*Sec[c + d*x]]])/(4*d) + (a^2*(5*A + 4*B -
8*C)*Sin[c + d*x])/(4*d*Sqrt[a + a*Sec[c + d*x]]) - (a*(A - 4*C)*Sqrt[a + a*Sec[c + d*x]]*Sin[c + d*x])/(2*d)
+ (A*Cos[c + d*x]*(a + a*Sec[c + d*x])^(3/2)*Sin[c + d*x])/(2*d)

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Rubi [A]  time = 0.450759, antiderivative size = 157, normalized size of antiderivative = 1., number of steps used = 5, number of rules used = 5, integrand size = 43, \(\frac{\text{number of rules}}{\text{integrand size}}\) = 0.116, Rules used = {4086, 4018, 4015, 3774, 203} \[ \frac{a^2 (5 A+4 B-8 C) \sin (c+d x)}{4 d \sqrt{a \sec (c+d x)+a}}+\frac{a^{3/2} (7 A+12 B+8 C) \tan ^{-1}\left (\frac{\sqrt{a} \tan (c+d x)}{\sqrt{a \sec (c+d x)+a}}\right )}{4 d}-\frac{a (A-4 C) \sin (c+d x) \sqrt{a \sec (c+d x)+a}}{2 d}+\frac{A \sin (c+d x) \cos (c+d x) (a \sec (c+d x)+a)^{3/2}}{2 d} \]

Antiderivative was successfully verified.

[In]

Int[Cos[c + d*x]^2*(a + a*Sec[c + d*x])^(3/2)*(A + B*Sec[c + d*x] + C*Sec[c + d*x]^2),x]

[Out]

(a^(3/2)*(7*A + 12*B + 8*C)*ArcTan[(Sqrt[a]*Tan[c + d*x])/Sqrt[a + a*Sec[c + d*x]]])/(4*d) + (a^2*(5*A + 4*B -
8*C)*Sin[c + d*x])/(4*d*Sqrt[a + a*Sec[c + d*x]]) - (a*(A - 4*C)*Sqrt[a + a*Sec[c + d*x]]*Sin[c + d*x])/(2*d)
+ (A*Cos[c + d*x]*(a + a*Sec[c + d*x])^(3/2)*Sin[c + d*x])/(2*d)

Rule 4086

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

Rule 4018

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

Rule 4015

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

Rubi steps

\begin{align*} \int \cos ^2(c+d x) (a+a \sec (c+d x))^{3/2} \left (A+B \sec (c+d x)+C \sec ^2(c+d x)\right ) \, dx &=\frac{A \cos (c+d x) (a+a \sec (c+d x))^{3/2} \sin (c+d x)}{2 d}+\frac{\int \cos (c+d x) (a+a \sec (c+d x))^{3/2} \left (\frac{1}{2} a (3 A+4 B)-\frac{1}{2} a (A-4 C) \sec (c+d x)\right ) \, dx}{2 a}\\ &=-\frac{a (A-4 C) \sqrt{a+a \sec (c+d x)} \sin (c+d x)}{2 d}+\frac{A \cos (c+d x) (a+a \sec (c+d x))^{3/2} \sin (c+d x)}{2 d}+\frac{\int \cos (c+d x) \sqrt{a+a \sec (c+d x)} \left (\frac{1}{4} a^2 (5 A+4 B-8 C)+\frac{1}{4} a^2 (A+4 B+8 C) \sec (c+d x)\right ) \, dx}{a}\\ &=\frac{a^2 (5 A+4 B-8 C) \sin (c+d x)}{4 d \sqrt{a+a \sec (c+d x)}}-\frac{a (A-4 C) \sqrt{a+a \sec (c+d x)} \sin (c+d x)}{2 d}+\frac{A \cos (c+d x) (a+a \sec (c+d x))^{3/2} \sin (c+d x)}{2 d}+\frac{1}{8} (a (7 A+12 B+8 C)) \int \sqrt{a+a \sec (c+d x)} \, dx\\ &=\frac{a^2 (5 A+4 B-8 C) \sin (c+d x)}{4 d \sqrt{a+a \sec (c+d x)}}-\frac{a (A-4 C) \sqrt{a+a \sec (c+d x)} \sin (c+d x)}{2 d}+\frac{A \cos (c+d x) (a+a \sec (c+d x))^{3/2} \sin (c+d x)}{2 d}-\frac{\left (a^2 (7 A+12 B+8 C)\right ) \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 )}{4 d}\\ &=\frac{a^{3/2} (7 A+12 B+8 C) \tan ^{-1}\left (\frac{\sqrt{a} \tan (c+d x)}{\sqrt{a+a \sec (c+d x)}}\right )}{4 d}+\frac{a^2 (5 A+4 B-8 C) \sin (c+d x)}{4 d \sqrt{a+a \sec (c+d x)}}-\frac{a (A-4 C) \sqrt{a+a \sec (c+d x)} \sin (c+d x)}{2 d}+\frac{A \cos (c+d x) (a+a \sec (c+d x))^{3/2} \sin (c+d x)}{2 d}\\ \end{align*}

Mathematica [A]  time = 0.760699, size = 117, normalized size = 0.75 \[ \frac{a \sec \left (\frac{1}{2} (c+d x)\right ) \sqrt{a (\sec (c+d x)+1)} \left (\sqrt{2} (7 A+12 B+8 C) \sin ^{-1}\left (\sqrt{2} \sin \left (\frac{1}{2} (c+d x)\right )\right ) \sqrt{\cos (c+d x)}+2 \sin \left (\frac{1}{2} (c+d x)\right ) ((7 A+4 B) \cos (c+d x)+A \cos (2 (c+d x))+A+8 C)\right )}{8 d} \]

Antiderivative was successfully verified.

[In]

Integrate[Cos[c + d*x]^2*(a + a*Sec[c + d*x])^(3/2)*(A + B*Sec[c + d*x] + C*Sec[c + d*x]^2),x]

[Out]

(a*Sec[(c + d*x)/2]*Sqrt[a*(1 + Sec[c + d*x])]*(Sqrt[2]*(7*A + 12*B + 8*C)*ArcSin[Sqrt[2]*Sin[(c + d*x)/2]]*Sq
rt[Cos[c + d*x]] + 2*(A + 8*C + (7*A + 4*B)*Cos[c + d*x] + A*Cos[2*(c + d*x)])*Sin[(c + d*x)/2]))/(8*d)

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Maple [B]  time = 0.395, size = 569, normalized size = 3.6 \begin{align*} \text{result too large to display} \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

int(cos(d*x+c)^2*(a+a*sec(d*x+c))^(3/2)*(A+B*sec(d*x+c)+C*sec(d*x+c)^2),x)

[Out]

1/16/d*a*(a*(cos(d*x+c)+1)/cos(d*x+c))^(1/2)*(7*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)+12*B*cos(d*x+c)*sin(d*x
+c)*2^(1/2)*(-2*cos(d*x+c)/(cos(d*x+c)+1))^(3/2)*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))+8*C*cos(d*x+c)*sin(d*x+c)*2^(1/2)*(-2*cos(d*x+c)/(cos(d*x+c)+1))^(3/2)*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))+7*A*2^(1/2)*arctanh(1/2*2^(1/2)*(-2*cos(d*x+c)/(co
s(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*(-2*cos(d*x+c)/
(cos(d*x+c)+1))^(3/2)*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^(1/2)*
sin(d*x+c)+8*C*(-2*cos(d*x+c)/(cos(d*x+c)+1))^(3/2)*arctanh(1/2*2^(1/2)*(-2*cos(d*x+c)/(cos(d*x+c)+1))^(1/2)*s
in(d*x+c)/cos(d*x+c))*2^(1/2)*sin(d*x+c)-8*A*cos(d*x+c)^4-20*A*cos(d*x+c)^3-16*B*cos(d*x+c)^3+28*A*cos(d*x+c)^
2+16*B*cos(d*x+c)^2-32*C*cos(d*x+c)^2+32*C*cos(d*x+c))/cos(d*x+c)/sin(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(cos(d*x+c)^2*(a+a*sec(d*x+c))^(3/2)*(A+B*sec(d*x+c)+C*sec(d*x+c)^2),x, algorithm="maxima")

[Out]

Timed out

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Fricas [A]  time = 0.924091, size = 887, normalized size = 5.65 \begin{align*} \left [\frac{{\left ({\left (7 \, A + 12 \, B + 8 \, C\right )} a \cos \left (d x + c\right ) +{\left (7 \, A + 12 \, B + 8 \, C\right )} a\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 (2 \, A a \cos \left (d x + c\right )^{2} +{\left (7 \, A + 4 \, B\right )} a \cos \left (d x + c\right ) + 8 \, C a\right )} \sqrt{\frac{a \cos \left (d x + c\right ) + a}{\cos \left (d x + c\right )}} \sin \left (d x + c\right )}{8 \,{\left (d \cos \left (d x + c\right ) + d\right )}}, -\frac{{\left ({\left (7 \, A + 12 \, B + 8 \, C\right )} a \cos \left (d x + c\right ) +{\left (7 \, A + 12 \, B + 8 \, C\right )} a\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 (2 \, A a \cos \left (d x + c\right )^{2} +{\left (7 \, A + 4 \, B\right )} a \cos \left (d x + c\right ) + 8 \, C a\right )} \sqrt{\frac{a \cos \left (d x + c\right ) + a}{\cos \left (d x + c\right )}} \sin \left (d x + c\right )}{4 \,{\left (d \cos \left (d x + c\right ) + d\right )}}\right ] \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate(cos(d*x+c)^2*(a+a*sec(d*x+c))^(3/2)*(A+B*sec(d*x+c)+C*sec(d*x+c)^2),x, algorithm="fricas")

[Out]

[1/8*(((7*A + 12*B + 8*C)*a*cos(d*x + c) + (7*A + 12*B + 8*C)*a)*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*(2*A*a*cos(d*x + c)^2 + (7*A + 4*B)*a*cos(d*x + c) + 8*C*a)*sqrt((a*cos(d*x + c) + a)/cos(d*x + c))*sin(d*x
+ c))/(d*cos(d*x + c) + d), -1/4*(((7*A + 12*B + 8*C)*a*cos(d*x + c) + (7*A + 12*B + 8*C)*a)*sqrt(a)*arctan(s
qrt((a*cos(d*x + c) + a)/cos(d*x + c))*cos(d*x + c)/(sqrt(a)*sin(d*x + c))) - (2*A*a*cos(d*x + c)^2 + (7*A + 4
*B)*a*cos(d*x + c) + 8*C*a)*sqrt((a*cos(d*x + c) + a)/cos(d*x + c))*sin(d*x + c))/(d*cos(d*x + c) + d)]

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Sympy [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(cos(d*x+c)**2*(a+a*sec(d*x+c))**(3/2)*(A+B*sec(d*x+c)+C*sec(d*x+c)**2),x)

[Out]

Timed out

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Giac [B]  time = 6.76439, size = 990, normalized size = 6.31 \begin{align*} \text{result too large to display} \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate(cos(d*x+c)^2*(a+a*sec(d*x+c))^(3/2)*(A+B*sec(d*x+c)+C*sec(d*x+c)^2),x, algorithm="giac")

[Out]

-1/8*(16*sqrt(2)*sqrt(-a*tan(1/2*d*x + 1/2*c)^2 + a)*C*a^2*sgn(cos(d*x + c))*tan(1/2*d*x + 1/2*c)/(a*tan(1/2*d
*x + 1/2*c)^2 - a) + (7*A*sqrt(-a)*a*sgn(cos(d*x + c)) + 12*B*sqrt(-a)*a*sgn(cos(d*x + c)) + 8*C*sqrt(-a)*a*sg
n(cos(d*x + 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))^2 - a*(2*sqrt(2
) + 3))) - (7*A*sqrt(-a)*a*sgn(cos(d*x + c)) + 12*B*sqrt(-a)*a*sgn(cos(d*x + c)) + 8*C*sqrt(-a)*a*sgn(cos(d*x
+ 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))^2 + a*(2*sqrt(2) - 3))) +
4*sqrt(2)*(7*(sqrt(-a)*tan(1/2*d*x + 1/2*c) - sqrt(-a*tan(1/2*d*x + 1/2*c)^2 + a))^6*A*sqrt(-a)*a^2*sgn(cos(d
*x + c)) + 12*(sqrt(-a)*tan(1/2*d*x + 1/2*c) - sqrt(-a*tan(1/2*d*x + 1/2*c)^2 + a))^6*B*sqrt(-a)*a^2*sgn(cos(d
*x + c)) - 95*(sqrt(-a)*tan(1/2*d*x + 1/2*c) - sqrt(-a*tan(1/2*d*x + 1/2*c)^2 + a))^4*A*sqrt(-a)*a^3*sgn(cos(d
*x + c)) - 76*(sqrt(-a)*tan(1/2*d*x + 1/2*c) - sqrt(-a*tan(1/2*d*x + 1/2*c)^2 + a))^4*B*sqrt(-a)*a^3*sgn(cos(d
*x + c)) + 53*(sqrt(-a)*tan(1/2*d*x + 1/2*c) - sqrt(-a*tan(1/2*d*x + 1/2*c)^2 + a))^2*A*sqrt(-a)*a^4*sgn(cos(d
*x + c)) + 36*(sqrt(-a)*tan(1/2*d*x + 1/2*c) - sqrt(-a*tan(1/2*d*x + 1/2*c)^2 + a))^2*B*sqrt(-a)*a^4*sgn(cos(d
*x + c)) - 5*A*sqrt(-a)*a^5*sgn(cos(d*x + c)) - 4*B*sqrt(-a)*a^5*sgn(cos(d*x + c)))/((sqrt(-a)*tan(1/2*d*x + 1
/2*c) - sqrt(-a*tan(1/2*d*x + 1/2*c)^2 + a))^4 - 6*(sqrt(-a)*tan(1/2*d*x + 1/2*c) - sqrt(-a*tan(1/2*d*x + 1/2*
c)^2 + a))^2*a + a^2)^2)/d