3.406 \(\int \frac{\cos (c+d x) (B \sec (c+d x)+C \sec ^2(c+d x))}{(a+a \sec (c+d x))^{5/2}} \, dx\)
Optimal. Leaf size=164 \[ -\frac{(43 B-3 C) \tan ^{-1}\left (\frac{\sqrt{a} \tan (c+d x)}{\sqrt{2} \sqrt{a \sec (c+d x)+a}}\right )}{16 \sqrt{2} a^{5/2} d}+\frac{2 B \tan ^{-1}\left (\frac{\sqrt{a} \tan (c+d x)}{\sqrt{a \sec (c+d x)+a}}\right )}{a^{5/2} d}-\frac{(11 B-3 C) \tan (c+d x)}{16 a d (a \sec (c+d x)+a)^{3/2}}-\frac{(B-C) \tan (c+d x)}{4 d (a \sec (c+d x)+a)^{5/2}} \]
[Out]
(2*B*ArcTan[(Sqrt[a]*Tan[c + d*x])/Sqrt[a + a*Sec[c + d*x]]])/(a^(5/2)*d) - ((43*B - 3*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)*Tan[c + d*x])/(4*d*(a + a*Sec
[c + d*x])^(5/2)) - ((11*B - 3*C)*Tan[c + d*x])/(16*a*d*(a + a*Sec[c + d*x])^(3/2))
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Rubi [A] time = 0.346006, antiderivative size = 164, normalized size of antiderivative = 1.,
number of steps used = 8, number of rules used = 6, integrand size = 40, \(\frac{\text{number of rules}}{\text{integrand size}}\) = 0.15, Rules used =
{4072, 3922, 3920, 3774, 203, 3795} \[ -\frac{(43 B-3 C) \tan ^{-1}\left (\frac{\sqrt{a} \tan (c+d x)}{\sqrt{2} \sqrt{a \sec (c+d x)+a}}\right )}{16 \sqrt{2} a^{5/2} d}+\frac{2 B \tan ^{-1}\left (\frac{\sqrt{a} \tan (c+d x)}{\sqrt{a \sec (c+d x)+a}}\right )}{a^{5/2} d}-\frac{(11 B-3 C) \tan (c+d x)}{16 a d (a \sec (c+d x)+a)^{3/2}}-\frac{(B-C) \tan (c+d x)}{4 d (a \sec (c+d x)+a)^{5/2}} \]
Antiderivative was successfully verified.
[In]
Int[(Cos[c + d*x]*(B*Sec[c + d*x] + C*Sec[c + d*x]^2))/(a + a*Sec[c + d*x])^(5/2),x]
[Out]
(2*B*ArcTan[(Sqrt[a]*Tan[c + d*x])/Sqrt[a + a*Sec[c + d*x]]])/(a^(5/2)*d) - ((43*B - 3*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)*Tan[c + d*x])/(4*d*(a + a*Sec
[c + d*x])^(5/2)) - ((11*B - 3*C)*Tan[c + d*x])/(16*a*d*(a + a*Sec[c + d*x])^(3/2))
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 3922
Int[(csc[(e_.) + (f_.)*(x_)]*(b_.) + (a_))^(m_)*(csc[(e_.) + (f_.)*(x_)]*(d_.) + (c_)), x_Symbol] :> -Simp[((b
*c - a*d)*Cot[e + f*x]*(a + b*Csc[e + f*x])^m)/(b*f*(2*m + 1)), x] + Dist[1/(a^2*(2*m + 1)), Int[(a + b*Csc[e
+ f*x])^(m + 1)*Simp[a*c*(2*m + 1) - (b*c - a*d)*(m + 1)*Csc[e + f*x], x], x], x] /; FreeQ[{a, b, c, d, e, f},
x] && NeQ[b*c - a*d, 0] && LtQ[m, -1] && EqQ[a^2 - b^2, 0] && IntegerQ[2*m]
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 (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{B+C \sec (c+d x)}{(a+a \sec (c+d x))^{5/2}} \, dx\\ &=-\frac{(B-C) \tan (c+d x)}{4 d (a+a \sec (c+d x))^{5/2}}-\frac{\int \frac{-4 a B+\frac{3}{2} a (B-C) \sec (c+d x)}{(a+a \sec (c+d x))^{3/2}} \, dx}{4 a^2}\\ &=-\frac{(B-C) \tan (c+d x)}{4 d (a+a \sec (c+d x))^{5/2}}-\frac{(11 B-3 C) \tan (c+d x)}{16 a d (a+a \sec (c+d x))^{3/2}}+\frac{\int \frac{8 a^2 B-\frac{1}{4} a^2 (11 B-3 C) \sec (c+d x)}{\sqrt{a+a \sec (c+d x)}} \, dx}{8 a^4}\\ &=-\frac{(B-C) \tan (c+d x)}{4 d (a+a \sec (c+d x))^{5/2}}-\frac{(11 B-3 C) \tan (c+d x)}{16 a d (a+a \sec (c+d x))^{3/2}}+\frac{B \int \sqrt{a+a \sec (c+d x)} \, dx}{a^3}-\frac{(43 B-3 C) \int \frac{\sec (c+d x)}{\sqrt{a+a \sec (c+d x)}} \, dx}{32 a^2}\\ &=-\frac{(B-C) \tan (c+d x)}{4 d (a+a \sec (c+d x))^{5/2}}-\frac{(11 B-3 C) \tan (c+d x)}{16 a d (a+a \sec (c+d x))^{3/2}}-\frac{(2 B) \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{(43 B-3 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{2 B \tan ^{-1}\left (\frac{\sqrt{a} \tan (c+d x)}{\sqrt{a+a \sec (c+d x)}}\right )}{a^{5/2} d}-\frac{(43 B-3 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) \tan (c+d x)}{4 d (a+a \sec (c+d x))^{5/2}}-\frac{(11 B-3 C) \tan (c+d x)}{16 a d (a+a \sec (c+d x))^{3/2}}\\ \end{align*}
Mathematica [C] time = 26.6628, size = 10133, normalized size = 61.79 \[ \text{Result too large to show} \]
Warning: Unable to verify antiderivative.
[In]
Integrate[(Cos[c + d*x]*(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 = 0.24, size = 824, normalized size = 5. \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)*(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*(a*(cos(d*x+c)+1)/cos(d*x+c))^(1/2)*(-32*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)-43*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)-64*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))+3*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)-86*B*sin(d*x+c)*co
s(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*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)+6*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)+30*B*cos(d*x+c)^3-43*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)-14*C*cos(d*x+c)^3+3*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*c
os(d*x+c)/(cos(d*x+c)+1))^(1/2)*sin(d*x+c)-8*B*cos(d*x+c)^2+8*C*cos(d*x+c)^2-22*B*cos(d*x+c)+6*C*cos(d*x+c))/(
cos(d*x+c)+1)^2/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 )}{{\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)*(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)/(a*sec(d*x + c) + a)^(5/2), x)
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Fricas [B] time = 16.8055, size = 1754, normalized size = 10.7 \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)*(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)*((43*B - 3*C)*cos(d*x + c)^3 + 3*(43*B - 3*C)*cos(d*x + c)^2 + 3*(43*B - 3*C)*cos(d*x + c) + 43
*B - 3*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)) - 64*(B*cos(d*x + c)^3 + 3*
B*cos(d*x + c)^2 + 3*B*cos(d*x + c) + B)*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*((15*B - 7*C)*cos(d*x
+ c)^2 + (11*B - 3*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), 1/32*(sqrt(2)*((43*B - 3*C)*cos(d*x + c)^3 + 3*(4
3*B - 3*C)*cos(d*x + c)^2 + 3*(43*B - 3*C)*cos(d*x + c) + 43*B - 3*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))) - 64*(B*cos(d*x + c)^3 + 3*B*cos(d*x + c)^2 + 3*B*
cos(d*x + c) + B)*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*((15*B - 7*C)*cos(d*x + c)^2 + (11*B - 3*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)*(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 [B] time = 11.7508, size = 471, normalized size = 2.87 \begin{align*} -\frac{2 \, \sqrt{-a \tan \left (\frac{1}{2} \, d x + \frac{1}{2} \, c\right )^{2} + a}{\left (\frac{2 \, \sqrt{2}{\left (B a^{5} - C a^{5}\right )} \tan \left (\frac{1}{2} \, d x + \frac{1}{2} \, c\right )^{2}}{a^{8} \mathrm{sgn}\left (\tan \left (\frac{1}{2} \, d x + \frac{1}{2} \, c\right )^{2} - 1\right )} - \frac{\sqrt{2}{\left (13 \, B a^{5} - 5 \, C a^{5}\right )}}{a^{8} \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 ) + \frac{\sqrt{2}{\left (43 \, B - 3 \, 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 )}^{2}\right )}{\sqrt{-a} a^{2} \mathrm{sgn}\left (\tan \left (\frac{1}{2} \, d x + \frac{1}{2} \, c\right )^{2} - 1\right )} + \frac{64 \, B \log \left ({\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 )}^{2} - a{\left (2 \, \sqrt{2} + 3\right )} \right |}\right )}{\sqrt{-a} a^{2} \mathrm{sgn}\left (\tan \left (\frac{1}{2} \, d x + \frac{1}{2} \, c\right )^{2} - 1\right )} - \frac{64 \, B \log \left ({\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 )}^{2} + a{\left (2 \, \sqrt{2} - 3\right )} \right |}\right )}{\sqrt{-a} a^{2} \mathrm{sgn}\left (\tan \left (\frac{1}{2} \, d x + \frac{1}{2} \, c\right )^{2} - 1\right )}}{64 \, d} \end{align*}
Verification of antiderivative is not currently implemented for this CAS.
[In]
integrate(cos(d*x+c)*(B*sec(d*x+c)+C*sec(d*x+c)^2)/(a+a*sec(d*x+c))^(5/2),x, algorithm="giac")
[Out]
-1/64*(2*sqrt(-a*tan(1/2*d*x + 1/2*c)^2 + a)*(2*sqrt(2)*(B*a^5 - C*a^5)*tan(1/2*d*x + 1/2*c)^2/(a^8*sgn(tan(1/
2*d*x + 1/2*c)^2 - 1)) - sqrt(2)*(13*B*a^5 - 5*C*a^5)/(a^8*sgn(tan(1/2*d*x + 1/2*c)^2 - 1)))*tan(1/2*d*x + 1/2
*c) + sqrt(2)*(43*B - 3*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)*a^2*sgn(tan(1/2*d*x + 1/2*c)^2 - 1)) + 64*B*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)*a^2*sgn(tan(1/2*d*x + 1/2*c)^2 - 1)) - 64*B*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)*a^2*sgn(tan(1/
2*d*x + 1/2*c)^2 - 1)))/d