3.407 \(\int \frac{\cos ^2(c+d x) (B \sec (c+d x)+C \sec ^2(c+d x))}{(a+a \sec (c+d x))^{5/2}} \, dx\)

Optimal. Leaf size=207 \[ \frac{(35 B-11 C) \sin (c+d x)}{16 a^2 d \sqrt{a \sec (c+d x)+a}}-\frac{(5 B-2 C) \tan ^{-1}\left (\frac{\sqrt{a} \tan (c+d x)}{\sqrt{a \sec (c+d x)+a}}\right )}{a^{5/2} d}+\frac{(115 B-43 C) \tan ^{-1}\left (\frac{\sqrt{a} \tan (c+d x)}{\sqrt{2} \sqrt{a \sec (c+d x)+a}}\right )}{16 \sqrt{2} a^{5/2} d}-\frac{(15 B-7 C) \sin (c+d x)}{16 a d (a \sec (c+d x)+a)^{3/2}}-\frac{(B-C) \sin (c+d x)}{4 d (a \sec (c+d x)+a)^{5/2}} \]

[Out]

-(((5*B - 2*C)*ArcTan[(Sqrt[a]*Tan[c + d*x])/Sqrt[a + a*Sec[c + d*x]]])/(a^(5/2)*d)) + ((115*B - 43*C)*ArcTan[
(Sqrt[a]*Tan[c + d*x])/(Sqrt[2]*Sqrt[a + a*Sec[c + d*x]])])/(16*Sqrt[2]*a^(5/2)*d) - ((B - C)*Sin[c + d*x])/(4
*d*(a + a*Sec[c + d*x])^(5/2)) - ((15*B - 7*C)*Sin[c + d*x])/(16*a*d*(a + a*Sec[c + d*x])^(3/2)) + ((35*B - 11
*C)*Sin[c + d*x])/(16*a^2*d*Sqrt[a + a*Sec[c + d*x]])

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Rubi [A]  time = 0.67209, antiderivative size = 207, normalized size of antiderivative = 1., number of steps used = 9, number of rules used = 7, integrand size = 42, \(\frac{\text{number of rules}}{\text{integrand size}}\) = 0.167, Rules used = {4072, 4020, 4022, 3920, 3774, 203, 3795} \[ \frac{(35 B-11 C) \sin (c+d x)}{16 a^2 d \sqrt{a \sec (c+d x)+a}}-\frac{(5 B-2 C) \tan ^{-1}\left (\frac{\sqrt{a} \tan (c+d x)}{\sqrt{a \sec (c+d x)+a}}\right )}{a^{5/2} d}+\frac{(115 B-43 C) \tan ^{-1}\left (\frac{\sqrt{a} \tan (c+d x)}{\sqrt{2} \sqrt{a \sec (c+d x)+a}}\right )}{16 \sqrt{2} a^{5/2} d}-\frac{(15 B-7 C) \sin (c+d x)}{16 a d (a \sec (c+d x)+a)^{3/2}}-\frac{(B-C) \sin (c+d x)}{4 d (a \sec (c+d x)+a)^{5/2}} \]

Antiderivative was successfully verified.

[In]

Int[(Cos[c + d*x]^2*(B*Sec[c + d*x] + C*Sec[c + d*x]^2))/(a + a*Sec[c + d*x])^(5/2),x]

[Out]

-(((5*B - 2*C)*ArcTan[(Sqrt[a]*Tan[c + d*x])/Sqrt[a + a*Sec[c + d*x]]])/(a^(5/2)*d)) + ((115*B - 43*C)*ArcTan[
(Sqrt[a]*Tan[c + d*x])/(Sqrt[2]*Sqrt[a + a*Sec[c + d*x]])])/(16*Sqrt[2]*a^(5/2)*d) - ((B - C)*Sin[c + d*x])/(4
*d*(a + a*Sec[c + d*x])^(5/2)) - ((15*B - 7*C)*Sin[c + d*x])/(16*a*d*(a + a*Sec[c + d*x])^(3/2)) + ((35*B - 11
*C)*Sin[c + d*x])/(16*a^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 4020

Int[(csc[(e_.) + (f_.)*(x_)]*(d_.))^(n_)*(csc[(e_.) + (f_.)*(x_)]*(b_.) + (a_))^(m_)*(csc[(e_.) + (f_.)*(x_)]*
(B_.) + (A_)), x_Symbol] :> -Simp[((A*b - a*B)*Cot[e + f*x]*(a + b*Csc[e + f*x])^m*(d*Csc[e + f*x])^n)/(b*f*(2
*m + 1)), x] - Dist[1/(a^2*(2*m + 1)), Int[(a + b*Csc[e + f*x])^(m + 1)*(d*Csc[e + f*x])^n*Simp[b*B*n - a*A*(2
*m + n + 1) + (A*b - a*B)*(m + n + 1)*Csc[e + f*x], x], x], x] /; FreeQ[{a, b, d, e, f, A, B, n}, x] && NeQ[A*
b - a*B, 0] && EqQ[a^2 - b^2, 0] && LtQ[m, -2^(-1)] &&  !GtQ[n, 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 ^2(c+d x) \left (B \sec (c+d x)+C \sec ^2(c+d x)\right )}{(a+a \sec (c+d x))^{5/2}} \, dx &=\int \frac{\cos (c+d x) (B+C \sec (c+d x))}{(a+a \sec (c+d x))^{5/2}} \, dx\\ &=-\frac{(B-C) \sin (c+d x)}{4 d (a+a \sec (c+d x))^{5/2}}+\frac{\int \frac{\cos (c+d x) \left (a (5 B-C)-\frac{5}{2} a (B-C) \sec (c+d x)\right )}{(a+a \sec (c+d x))^{3/2}} \, dx}{4 a^2}\\ &=-\frac{(B-C) \sin (c+d x)}{4 d (a+a \sec (c+d x))^{5/2}}-\frac{(15 B-7 C) \sin (c+d x)}{16 a d (a+a \sec (c+d x))^{3/2}}+\frac{\int \frac{\cos (c+d x) \left (\frac{1}{2} a^2 (35 B-11 C)-\frac{3}{4} a^2 (15 B-7 C) \sec (c+d x)\right )}{\sqrt{a+a \sec (c+d x)}} \, dx}{8 a^4}\\ &=-\frac{(B-C) \sin (c+d x)}{4 d (a+a \sec (c+d x))^{5/2}}-\frac{(15 B-7 C) \sin (c+d x)}{16 a d (a+a \sec (c+d x))^{3/2}}+\frac{(35 B-11 C) \sin (c+d x)}{16 a^2 d \sqrt{a+a \sec (c+d x)}}+\frac{\int \frac{-4 a^3 (5 B-2 C)+\frac{1}{4} a^3 (35 B-11 C) \sec (c+d x)}{\sqrt{a+a \sec (c+d x)}} \, dx}{8 a^5}\\ &=-\frac{(B-C) \sin (c+d x)}{4 d (a+a \sec (c+d x))^{5/2}}-\frac{(15 B-7 C) \sin (c+d x)}{16 a d (a+a \sec (c+d x))^{3/2}}+\frac{(35 B-11 C) \sin (c+d x)}{16 a^2 d \sqrt{a+a \sec (c+d x)}}+\frac{(115 B-43 C) \int \frac{\sec (c+d x)}{\sqrt{a+a \sec (c+d x)}} \, dx}{32 a^2}-\frac{(5 B-2 C) \int \sqrt{a+a \sec (c+d x)} \, dx}{2 a^3}\\ &=-\frac{(B-C) \sin (c+d x)}{4 d (a+a \sec (c+d x))^{5/2}}-\frac{(15 B-7 C) \sin (c+d x)}{16 a d (a+a \sec (c+d x))^{3/2}}+\frac{(35 B-11 C) \sin (c+d x)}{16 a^2 d \sqrt{a+a \sec (c+d x)}}-\frac{(115 B-43 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 )}{16 a^2 d}+\frac{(5 B-2 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 )}{a^2 d}\\ &=-\frac{(5 B-2 C) \tan ^{-1}\left (\frac{\sqrt{a} \tan (c+d x)}{\sqrt{a+a \sec (c+d x)}}\right )}{a^{5/2} d}+\frac{(115 B-43 C) \tan ^{-1}\left (\frac{\sqrt{a} \tan (c+d x)}{\sqrt{2} \sqrt{a+a \sec (c+d x)}}\right )}{16 \sqrt{2} a^{5/2} d}-\frac{(B-C) \sin (c+d x)}{4 d (a+a \sec (c+d x))^{5/2}}-\frac{(15 B-7 C) \sin (c+d x)}{16 a d (a+a \sec (c+d x))^{3/2}}+\frac{(35 B-11 C) \sin (c+d x)}{16 a^2 d \sqrt{a+a \sec (c+d x)}}\\ \end{align*}

Mathematica [C]  time = 26.9155, size = 10956, normalized size = 52.93 \[ \text{Result too large to show} \]

Warning: Unable to verify antiderivative.

[In]

Integrate[(Cos[c + d*x]^2*(B*Sec[c + d*x] + C*Sec[c + d*x]^2))/(a + a*Sec[c + d*x])^(5/2),x]

[Out]

Result too large to show

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Maple [B]  time = 0.348, size = 1065, normalized size = 5.1 \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*(B*sec(d*x+c)+C*sec(d*x+c)^2)/(a+a*sec(d*x+c))^(5/2),x)

[Out]

1/32/d/a^3*(-1+cos(d*x+c))^2*(80*B*cos(d*x+c)^2*sin(d*x+c)*(-2*cos(d*x+c)/(cos(d*x+c)+1))^(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^(1/2)-32*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))^(1/2)*sin(d*x+c)*cos(d*x+c)^2*
2^(1/2)+115*B*cos(d*x+c)^2*sin(d*x+c)*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))*(-2*cos(d*x+c)/(cos(d*x+c)+1))^(1/2)+160*B*cos(d*x+c)*2^(1/2)*sin(d*x+c)*(-2*cos(d*x+c)/(cos(d*x+c)+1))^
(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))-43*C*cos(d*x+c)^2*sin(d*
x+c)*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))*(-2*cos(d*x+c)/(cos(d*x+c)+
1))^(1/2)-64*C*cos(d*x+c)*2^(1/2)*sin(d*x+c)*(-2*cos(d*x+c)/(cos(d*x+c)+1))^(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))+230*B*sin(d*x+c)*cos(d*x+c)*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))*(-2*cos(d*x+c)/(cos(d*x+c)+1))^(1/2)+80*B*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))^(1/2)*sin(d
*x+c)-32*B*cos(d*x+c)^4-86*C*sin(d*x+c)*cos(d*x+c)*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))*(-2*cos(d*x+c)/(cos(d*x+c)+1))^(1/2)-32*C*(-2*cos(d*x+c)/(cos(d*x+c)+1))^(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^(1/2)*sin(d*x+c)+115*B*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))*(-2*cos(d*x+c)/(cos(d*x+c)+1))^(1/2)*sin(d*x+c)-7
8*B*cos(d*x+c)^3-43*C*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))*(-2*cos(d*
x+c)/(cos(d*x+c)+1))^(1/2)*sin(d*x+c)+30*C*cos(d*x+c)^3+40*B*cos(d*x+c)^2-8*C*cos(d*x+c)^2+70*B*cos(d*x+c)-22*
C*cos(d*x+c))*(a*(cos(d*x+c)+1)/cos(d*x+c))^(1/2)/sin(d*x+c)^5

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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 )^{2}}{{\left (a \sec \left (d x + c\right ) + a\right )}^{\frac{5}{2}}}\,{d x} \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate(cos(d*x+c)^2*(B*sec(d*x+c)+C*sec(d*x+c)^2)/(a+a*sec(d*x+c))^(5/2),x, algorithm="maxima")

[Out]

integrate((C*sec(d*x + c)^2 + B*sec(d*x + c))*cos(d*x + c)^2/(a*sec(d*x + c) + a)^(5/2), x)

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Fricas [A]  time = 22.2534, size = 1948, normalized size = 9.41 \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*(B*sec(d*x+c)+C*sec(d*x+c)^2)/(a+a*sec(d*x+c))^(5/2),x, algorithm="fricas")

[Out]

[1/64*(sqrt(2)*((115*B - 43*C)*cos(d*x + c)^3 + 3*(115*B - 43*C)*cos(d*x + c)^2 + 3*(115*B - 43*C)*cos(d*x + c
) + 115*B - 43*C)*sqrt(-a)*log(-(2*sqrt(2)*sqrt(-a)*sqrt((a*cos(d*x + c) + a)/cos(d*x + c))*cos(d*x + c)*sin(d
*x + c) - 3*a*cos(d*x + c)^2 - 2*a*cos(d*x + c) + a)/(cos(d*x + c)^2 + 2*cos(d*x + c) + 1)) + 32*((5*B - 2*C)*
cos(d*x + c)^3 + 3*(5*B - 2*C)*cos(d*x + c)^2 + 3*(5*B - 2*C)*cos(d*x + c) + 5*B - 2*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)) + 4*(16*B*cos(d*x + c)^3 + 5*(11*B - 3*C)*cos(d*x + c)^2 + (35*B - 11*C)*cos(d*x + c))*s
qrt((a*cos(d*x + c) + a)/cos(d*x + c))*sin(d*x + c))/(a^3*d*cos(d*x + c)^3 + 3*a^3*d*cos(d*x + c)^2 + 3*a^3*d*
cos(d*x + c) + a^3*d), -1/32*(sqrt(2)*((115*B - 43*C)*cos(d*x + c)^3 + 3*(115*B - 43*C)*cos(d*x + c)^2 + 3*(11
5*B - 43*C)*cos(d*x + c) + 115*B - 43*C)*sqrt(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))) - 32*((5*B - 2*C)*cos(d*x + c)^3 + 3*(5*B - 2*C)*cos(d*x + c)^2 + 3*(5*B - 2*C)
*cos(d*x + c) + 5*B - 2*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*(16*B*cos(d*x + c)^3 + 5*(11*B - 3*C)*cos(d*x + c)^2 + (35*B - 11*C)*cos(d*x + c))*sqrt((a*cos(d*
x + c) + a)/cos(d*x + c))*sin(d*x + c))/(a^3*d*cos(d*x + c)^3 + 3*a^3*d*cos(d*x + c)^2 + 3*a^3*d*cos(d*x + c)
+ a^3*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*(B*sec(d*x+c)+C*sec(d*x+c)**2)/(a+a*sec(d*x+c))**(5/2),x)

[Out]

Timed out

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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(cos(d*x+c)^2*(B*sec(d*x+c)+C*sec(d*x+c)^2)/(a+a*sec(d*x+c))^(5/2),x, algorithm="giac")

[Out]

Exception raised: NotImplementedError