3.391 \(\int \frac{\cos ^3(c+d x) (B \sec (c+d x)+C \sec ^2(c+d x))}{\sqrt{a+a \sec (c+d x)}} \, dx\)

Optimal. Leaf size=165 \[ -\frac{(B-4 C) \sin (c+d x)}{4 d \sqrt{a \sec (c+d x)+a}}+\frac{(7 B-4 C) \tan ^{-1}\left (\frac{\sqrt{a} \tan (c+d x)}{\sqrt{a \sec (c+d x)+a}}\right )}{4 \sqrt{a} d}-\frac{\sqrt{2} (B-C) \tan ^{-1}\left (\frac{\sqrt{a} \tan (c+d x)}{\sqrt{2} \sqrt{a \sec (c+d x)+a}}\right )}{\sqrt{a} d}+\frac{B \sin (c+d x) \cos (c+d x)}{2 d \sqrt{a \sec (c+d x)+a}} \]

[Out]

((7*B - 4*C)*ArcTan[(Sqrt[a]*Tan[c + d*x])/Sqrt[a + a*Sec[c + d*x]]])/(4*Sqrt[a]*d) - (Sqrt[2]*(B - C)*ArcTan[
(Sqrt[a]*Tan[c + d*x])/(Sqrt[2]*Sqrt[a + a*Sec[c + d*x]])])/(Sqrt[a]*d) - ((B - 4*C)*Sin[c + d*x])/(4*d*Sqrt[a
+ a*Sec[c + d*x]]) + (B*Cos[c + d*x]*Sin[c + d*x])/(2*d*Sqrt[a + a*Sec[c + d*x]])

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Rubi [A]  time = 0.470853, antiderivative size = 165, normalized size of antiderivative = 1., number of steps used = 8, number of rules used = 6, integrand size = 42, \(\frac{\text{number of rules}}{\text{integrand size}}\) = 0.143, Rules used = {4072, 4022, 3920, 3774, 203, 3795} \[ -\frac{(B-4 C) \sin (c+d x)}{4 d \sqrt{a \sec (c+d x)+a}}+\frac{(7 B-4 C) \tan ^{-1}\left (\frac{\sqrt{a} \tan (c+d x)}{\sqrt{a \sec (c+d x)+a}}\right )}{4 \sqrt{a} d}-\frac{\sqrt{2} (B-C) \tan ^{-1}\left (\frac{\sqrt{a} \tan (c+d x)}{\sqrt{2} \sqrt{a \sec (c+d x)+a}}\right )}{\sqrt{a} d}+\frac{B \sin (c+d x) \cos (c+d x)}{2 d \sqrt{a \sec (c+d x)+a}} \]

Antiderivative was successfully verified.

[In]

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

[Out]

((7*B - 4*C)*ArcTan[(Sqrt[a]*Tan[c + d*x])/Sqrt[a + a*Sec[c + d*x]]])/(4*Sqrt[a]*d) - (Sqrt[2]*(B - C)*ArcTan[
(Sqrt[a]*Tan[c + d*x])/(Sqrt[2]*Sqrt[a + a*Sec[c + d*x]])])/(Sqrt[a]*d) - ((B - 4*C)*Sin[c + d*x])/(4*d*Sqrt[a
+ a*Sec[c + d*x]]) + (B*Cos[c + d*x]*Sin[c + d*x])/(2*d*Sqrt[a + a*Sec[c + d*x]])

Rule 4072

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

Rule 4022

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

Rule 3920

Int[(csc[(e_.) + (f_.)*(x_)]*(d_.) + (c_))/Sqrt[csc[(e_.) + (f_.)*(x_)]*(b_.) + (a_)], x_Symbol] :> Dist[c/a,
Int[Sqrt[a + b*Csc[e + f*x]], x], x] - Dist[(b*c - a*d)/a, Int[Csc[e + f*x]/Sqrt[a + b*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 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
]

Rubi steps

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

Mathematica [A]  time = 0.411016, size = 135, normalized size = 0.82 \[ \frac{\tan (c+d x) \left (\cos (c+d x) \sqrt{1-\sec (c+d x)} (2 B \cos (c+d x)-B+4 C)+(7 B-4 C) \tanh ^{-1}\left (\sqrt{1-\sec (c+d x)}\right )-4 \sqrt{2} (B-C) \tanh ^{-1}\left (\frac{\sqrt{1-\sec (c+d x)}}{\sqrt{2}}\right )\right )}{4 d \sqrt{1-\sec (c+d x)} \sqrt{a (\sec (c+d x)+1)}} \]

Antiderivative was successfully verified.

[In]

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

[Out]

(((7*B - 4*C)*ArcTanh[Sqrt[1 - Sec[c + d*x]]] - 4*Sqrt[2]*(B - C)*ArcTanh[Sqrt[1 - Sec[c + d*x]]/Sqrt[2]] + Co
s[c + d*x]*(-B + 4*C + 2*B*Cos[c + d*x])*Sqrt[1 - Sec[c + d*x]])*Tan[c + d*x])/(4*d*Sqrt[1 - Sec[c + d*x]]*Sqr
t[a*(1 + Sec[c + d*x])])

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Maple [B]  time = 0.388, size = 717, normalized size = 4.4 \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)^3*(B*sec(d*x+c)+C*sec(d*x+c)^2)/(a+a*sec(d*x+c))^(1/2),x)

[Out]

-1/16/d/a*(-7*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))+4*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))-8*B*cos(d*x+c)*s
in(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))-7*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*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))+4*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)*sin(d*x+c)/cos(d*x+c))*2^(1/2)*sin(d*x+c)-8*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)+8*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)-c
os(d*x+c)+1)/sin(d*x+c))*sin(d*x+c)+8*B*cos(d*x+c)^4-12*B*cos(d*x+c)^3+16*C*cos(d*x+c)^3+4*B*cos(d*x+c)^2-16*C
*cos(d*x+c)^2)*(a*(cos(d*x+c)+1)/cos(d*x+c))^(1/2)/cos(d*x+c)/sin(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 )\right )} \cos \left (d x + c\right )^{3}}{\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(cos(d*x+c)^3*(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))*cos(d*x + c)^3/sqrt(a*sec(d*x + c) + a), x)

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Fricas [A]  time = 5.94974, size = 1323, normalized size = 8.02 \begin{align*} \left [-\frac{4 \, \sqrt{2}{\left ({\left (B - C\right )} a \cos \left (d x + c\right ) +{\left (B - C\right )} a\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 ) -{\left ({\left (7 \, B - 4 \, C\right )} \cos \left (d x + c\right ) + 7 \, B - 4 \, C\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 \, B \cos \left (d x + c\right )^{2} -{\left (B - 4 \, C\right )} \cos \left (d x + c\right )\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 (a d \cos \left (d x + c\right ) + a d\right )}}, -\frac{{\left ({\left (7 \, B - 4 \, C\right )} \cos \left (d x + c\right ) + 7 \, B - 4 \, C\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 \, B \cos \left (d x + c\right )^{2} -{\left (B - 4 \, C\right )} \cos \left (d x + c\right )\right )} \sqrt{\frac{a \cos \left (d x + c\right ) + a}{\cos \left (d x + c\right )}} \sin \left (d x + c\right ) - \frac{4 \, \sqrt{2}{\left ({\left (B - C\right )} a \cos \left (d x + c\right ) +{\left (B - C\right )} a\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}}}{4 \,{\left (a d \cos \left (d x + c\right ) + a d\right )}}\right ] \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate(cos(d*x+c)^3*(B*sec(d*x+c)+C*sec(d*x+c)^2)/(a+a*sec(d*x+c))^(1/2),x, algorithm="fricas")

[Out]

[-1/8*(4*sqrt(2)*((B - C)*a*cos(d*x + c) + (B - C)*a)*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*c
os(d*x + c) + 1)) - ((7*B - 4*C)*cos(d*x + c) + 7*B - 4*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*(2*
B*cos(d*x + c)^2 - (B - 4*C)*cos(d*x + c))*sqrt((a*cos(d*x + c) + a)/cos(d*x + c))*sin(d*x + c))/(a*d*cos(d*x
+ c) + a*d), -1/4*(((7*B - 4*C)*cos(d*x + c) + 7*B - 4*C)*sqrt(a)*arctan(sqrt((a*cos(d*x + c) + a)/cos(d*x + c
))*cos(d*x + c)/(sqrt(a)*sin(d*x + c))) - (2*B*cos(d*x + c)^2 - (B - 4*C)*cos(d*x + c))*sqrt((a*cos(d*x + c) +
a)/cos(d*x + c))*sin(d*x + c) - 4*sqrt(2)*((B - C)*a*cos(d*x + c) + (B - C)*a)*arctan(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) + a*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)**3*(B*sec(d*x+c)+C*sec(d*x+c)**2)/(a+a*sec(d*x+c))**(1/2),x)

[Out]

Timed out

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Giac [B]  time = 11.5572, size = 876, normalized size = 5.31 \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)^3*(B*sec(d*x+c)+C*sec(d*x+c)^2)/(a+a*sec(d*x+c))^(1/2),x, algorithm="giac")

[Out]

-1/8*(4*sqrt(2)*(B - C)*log((sqrt(-a)*tan(1/2*d*x + 1/2*c) - sqrt(-a*tan(1/2*d*x + 1/2*c)^2 + a))^2)/(sqrt(-a)
*sgn(tan(1/2*d*x + 1/2*c)^2 - 1)) + (7*B - 4*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)))/(sqrt(-a)*sgn(tan(1/2*d*x + 1/2*c)^2 - 1)) - (7*B - 4*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)))/(sqrt(-a)*sgn(tan(1/2
*d*x + 1/2*c)^2 - 1)) + 4*sqrt(2)*(17*(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) - 12*(sqrt(-a)*tan(1/2*d*x + 1/2*c) - sqrt(-a*tan(1/2*d*x + 1/2*c)^2 + a))^6*C*sqrt(-a) - 57*(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 + 76*(sqrt(-a)*tan(1/2*d*x + 1/
2*c) - sqrt(-a*tan(1/2*d*x + 1/2*c)^2 + a))^4*C*sqrt(-a)*a + 19*(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^2 - 36*(sqrt(-a)*tan(1/2*d*x + 1/2*c) - sqrt(-a*tan(1/2*d*x + 1/2*c)^2
+ a))^2*C*sqrt(-a)*a^2 - 3*B*sqrt(-a)*a^3 + 4*C*sqrt(-a)*a^3)/(((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*sgn(tan(1/2*d*x + 1/2*c)^2 - 1)))/d