3.489 \(\int \cos ^{\frac{9}{2}}(c+d x) (a+a \sec (c+d x))^2 (A+B \sec (c+d x)) \, dx\)
Optimal. Leaf size=194 \[ \frac{4 a^2 (5 A+6 B) \text{EllipticF}\left (\frac{1}{2} (c+d x),2\right )}{21 d}+\frac{4 a^2 (8 A+9 B) E\left (\left .\frac{1}{2} (c+d x)\right |2\right )}{15 d}+\frac{2 a^2 (11 A+9 B) \sin (c+d x) \cos ^{\frac{5}{2}}(c+d x)}{63 d}+\frac{4 a^2 (8 A+9 B) \sin (c+d x) \cos ^{\frac{3}{2}}(c+d x)}{45 d}+\frac{4 a^2 (5 A+6 B) \sin (c+d x) \sqrt{\cos (c+d x)}}{21 d}+\frac{2 A \sin (c+d x) \cos ^{\frac{5}{2}}(c+d x) \left (a^2 \cos (c+d x)+a^2\right )}{9 d} \]
[Out]
(4*a^2*(8*A + 9*B)*EllipticE[(c + d*x)/2, 2])/(15*d) + (4*a^2*(5*A + 6*B)*EllipticF[(c + d*x)/2, 2])/(21*d) +
(4*a^2*(5*A + 6*B)*Sqrt[Cos[c + d*x]]*Sin[c + d*x])/(21*d) + (4*a^2*(8*A + 9*B)*Cos[c + d*x]^(3/2)*Sin[c + d*x
])/(45*d) + (2*a^2*(11*A + 9*B)*Cos[c + d*x]^(5/2)*Sin[c + d*x])/(63*d) + (2*A*Cos[c + d*x]^(5/2)*(a^2 + a^2*C
os[c + d*x])*Sin[c + d*x])/(9*d)
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Rubi [A] time = 0.403935, antiderivative size = 194, normalized size of antiderivative = 1.,
number of steps used = 9, number of rules used = 8, integrand size = 33, \(\frac{\text{number of rules}}{\text{integrand size}}\) = 0.242, Rules used =
{2954, 2976, 2968, 3023, 2748, 2635, 2641, 2639} \[ \frac{4 a^2 (5 A+6 B) F\left (\left .\frac{1}{2} (c+d x)\right |2\right )}{21 d}+\frac{4 a^2 (8 A+9 B) E\left (\left .\frac{1}{2} (c+d x)\right |2\right )}{15 d}+\frac{2 a^2 (11 A+9 B) \sin (c+d x) \cos ^{\frac{5}{2}}(c+d x)}{63 d}+\frac{4 a^2 (8 A+9 B) \sin (c+d x) \cos ^{\frac{3}{2}}(c+d x)}{45 d}+\frac{4 a^2 (5 A+6 B) \sin (c+d x) \sqrt{\cos (c+d x)}}{21 d}+\frac{2 A \sin (c+d x) \cos ^{\frac{5}{2}}(c+d x) \left (a^2 \cos (c+d x)+a^2\right )}{9 d} \]
Antiderivative was successfully verified.
[In]
Int[Cos[c + d*x]^(9/2)*(a + a*Sec[c + d*x])^2*(A + B*Sec[c + d*x]),x]
[Out]
(4*a^2*(8*A + 9*B)*EllipticE[(c + d*x)/2, 2])/(15*d) + (4*a^2*(5*A + 6*B)*EllipticF[(c + d*x)/2, 2])/(21*d) +
(4*a^2*(5*A + 6*B)*Sqrt[Cos[c + d*x]]*Sin[c + d*x])/(21*d) + (4*a^2*(8*A + 9*B)*Cos[c + d*x]^(3/2)*Sin[c + d*x
])/(45*d) + (2*a^2*(11*A + 9*B)*Cos[c + d*x]^(5/2)*Sin[c + d*x])/(63*d) + (2*A*Cos[c + d*x]^(5/2)*(a^2 + a^2*C
os[c + d*x])*Sin[c + d*x])/(9*d)
Rule 2954
Int[((a_.) + csc[(e_.) + (f_.)*(x_)]*(b_.))^(m_.)*(csc[(e_.) + (f_.)*(x_)]*(d_.) + (c_))^(n_.)*((g_.)*sin[(e_.
) + (f_.)*(x_)])^(p_.), x_Symbol] :> Dist[g^(m + n), Int[(g*Sin[e + f*x])^(p - m - n)*(b + a*Sin[e + f*x])^m*(
d + c*Sin[e + f*x])^n, x], x] /; FreeQ[{a, b, c, d, e, f, g, p}, x] && NeQ[b*c - a*d, 0] && !IntegerQ[p] && I
ntegerQ[m] && IntegerQ[n]
Rule 2976
Int[((a_) + (b_.)*sin[(e_.) + (f_.)*(x_)])^(m_)*((A_.) + (B_.)*sin[(e_.) + (f_.)*(x_)])*((c_.) + (d_.)*sin[(e_
.) + (f_.)*(x_)])^(n_), x_Symbol] :> -Simp[(b*B*Cos[e + f*x]*(a + b*Sin[e + f*x])^(m - 1)*(c + d*Sin[e + f*x])
^(n + 1))/(d*f*(m + n + 1)), x] + Dist[1/(d*(m + n + 1)), Int[(a + b*Sin[e + f*x])^(m - 1)*(c + d*Sin[e + f*x]
)^n*Simp[a*A*d*(m + n + 1) + B*(a*c*(m - 1) + b*d*(n + 1)) + (A*b*d*(m + n + 1) - B*(b*c*m - a*d*(2*m + n)))*S
in[e + f*x], x], x], x] /; FreeQ[{a, b, c, d, e, f, A, B, n}, x] && NeQ[b*c - a*d, 0] && EqQ[a^2 - b^2, 0] &&
NeQ[c^2 - d^2, 0] && GtQ[m, 1/2] && !LtQ[n, -1] && IntegerQ[2*m] && (IntegerQ[2*n] || EqQ[c, 0])
Rule 2968
Int[((a_.) + (b_.)*sin[(e_.) + (f_.)*(x_)])^(m_.)*((A_.) + (B_.)*sin[(e_.) + (f_.)*(x_)])*((c_.) + (d_.)*sin[(
e_.) + (f_.)*(x_)]), x_Symbol] :> Int[(a + b*Sin[e + f*x])^m*(A*c + (B*c + A*d)*Sin[e + f*x] + B*d*Sin[e + f*x
]^2), x] /; FreeQ[{a, b, c, d, e, f, A, B, m}, x] && NeQ[b*c - a*d, 0]
Rule 3023
Int[((a_.) + (b_.)*sin[(e_.) + (f_.)*(x_)])^(m_.)*((A_.) + (B_.)*sin[(e_.) + (f_.)*(x_)] + (C_.)*sin[(e_.) + (
f_.)*(x_)]^2), x_Symbol] :> -Simp[(C*Cos[e + f*x]*(a + b*Sin[e + f*x])^(m + 1))/(b*f*(m + 2)), x] + Dist[1/(b*
(m + 2)), Int[(a + b*Sin[e + f*x])^m*Simp[A*b*(m + 2) + b*C*(m + 1) + (b*B*(m + 2) - a*C)*Sin[e + f*x], x], x]
, x] /; FreeQ[{a, b, e, f, A, B, C, m}, x] && !LtQ[m, -1]
Rule 2748
Int[((b_.)*sin[(e_.) + (f_.)*(x_)])^(m_)*((c_) + (d_.)*sin[(e_.) + (f_.)*(x_)]), x_Symbol] :> Dist[c, Int[(b*S
in[e + f*x])^m, x], x] + Dist[d/b, Int[(b*Sin[e + f*x])^(m + 1), x], x] /; FreeQ[{b, c, d, e, f, m}, x]
Rule 2635
Int[((b_.)*sin[(c_.) + (d_.)*(x_)])^(n_), x_Symbol] :> -Simp[(b*Cos[c + d*x]*(b*Sin[c + d*x])^(n - 1))/(d*n),
x] + Dist[(b^2*(n - 1))/n, Int[(b*Sin[c + d*x])^(n - 2), x], x] /; FreeQ[{b, c, d}, x] && GtQ[n, 1] && Integer
Q[2*n]
Rule 2641
Int[1/Sqrt[sin[(c_.) + (d_.)*(x_)]], x_Symbol] :> Simp[(2*EllipticF[(1*(c - Pi/2 + d*x))/2, 2])/d, x] /; FreeQ
[{c, d}, x]
Rule 2639
Int[Sqrt[sin[(c_.) + (d_.)*(x_)]], x_Symbol] :> Simp[(2*EllipticE[(1*(c - Pi/2 + d*x))/2, 2])/d, x] /; FreeQ[{
c, d}, x]
Rubi steps
\begin{align*} \int \cos ^{\frac{9}{2}}(c+d x) (a+a \sec (c+d x))^2 (A+B \sec (c+d x)) \, dx &=\int \cos ^{\frac{3}{2}}(c+d x) (a+a \cos (c+d x))^2 (B+A \cos (c+d x)) \, dx\\ &=\frac{2 A \cos ^{\frac{5}{2}}(c+d x) \left (a^2+a^2 \cos (c+d x)\right ) \sin (c+d x)}{9 d}+\frac{2}{9} \int \cos ^{\frac{3}{2}}(c+d x) (a+a \cos (c+d x)) \left (\frac{1}{2} a (5 A+9 B)+\frac{1}{2} a (11 A+9 B) \cos (c+d x)\right ) \, dx\\ &=\frac{2 A \cos ^{\frac{5}{2}}(c+d x) \left (a^2+a^2 \cos (c+d x)\right ) \sin (c+d x)}{9 d}+\frac{2}{9} \int \cos ^{\frac{3}{2}}(c+d x) \left (\frac{1}{2} a^2 (5 A+9 B)+\left (\frac{1}{2} a^2 (5 A+9 B)+\frac{1}{2} a^2 (11 A+9 B)\right ) \cos (c+d x)+\frac{1}{2} a^2 (11 A+9 B) \cos ^2(c+d x)\right ) \, dx\\ &=\frac{2 a^2 (11 A+9 B) \cos ^{\frac{5}{2}}(c+d x) \sin (c+d x)}{63 d}+\frac{2 A \cos ^{\frac{5}{2}}(c+d x) \left (a^2+a^2 \cos (c+d x)\right ) \sin (c+d x)}{9 d}+\frac{4}{63} \int \cos ^{\frac{3}{2}}(c+d x) \left (\frac{9}{2} a^2 (5 A+6 B)+\frac{7}{2} a^2 (8 A+9 B) \cos (c+d x)\right ) \, dx\\ &=\frac{2 a^2 (11 A+9 B) \cos ^{\frac{5}{2}}(c+d x) \sin (c+d x)}{63 d}+\frac{2 A \cos ^{\frac{5}{2}}(c+d x) \left (a^2+a^2 \cos (c+d x)\right ) \sin (c+d x)}{9 d}+\frac{1}{7} \left (2 a^2 (5 A+6 B)\right ) \int \cos ^{\frac{3}{2}}(c+d x) \, dx+\frac{1}{9} \left (2 a^2 (8 A+9 B)\right ) \int \cos ^{\frac{5}{2}}(c+d x) \, dx\\ &=\frac{4 a^2 (5 A+6 B) \sqrt{\cos (c+d x)} \sin (c+d x)}{21 d}+\frac{4 a^2 (8 A+9 B) \cos ^{\frac{3}{2}}(c+d x) \sin (c+d x)}{45 d}+\frac{2 a^2 (11 A+9 B) \cos ^{\frac{5}{2}}(c+d x) \sin (c+d x)}{63 d}+\frac{2 A \cos ^{\frac{5}{2}}(c+d x) \left (a^2+a^2 \cos (c+d x)\right ) \sin (c+d x)}{9 d}+\frac{1}{21} \left (2 a^2 (5 A+6 B)\right ) \int \frac{1}{\sqrt{\cos (c+d x)}} \, dx+\frac{1}{15} \left (2 a^2 (8 A+9 B)\right ) \int \sqrt{\cos (c+d x)} \, dx\\ &=\frac{4 a^2 (8 A+9 B) E\left (\left .\frac{1}{2} (c+d x)\right |2\right )}{15 d}+\frac{4 a^2 (5 A+6 B) F\left (\left .\frac{1}{2} (c+d x)\right |2\right )}{21 d}+\frac{4 a^2 (5 A+6 B) \sqrt{\cos (c+d x)} \sin (c+d x)}{21 d}+\frac{4 a^2 (8 A+9 B) \cos ^{\frac{3}{2}}(c+d x) \sin (c+d x)}{45 d}+\frac{2 a^2 (11 A+9 B) \cos ^{\frac{5}{2}}(c+d x) \sin (c+d x)}{63 d}+\frac{2 A \cos ^{\frac{5}{2}}(c+d x) \left (a^2+a^2 \cos (c+d x)\right ) \sin (c+d x)}{9 d}\\ \end{align*}
Mathematica [C] time = 6.34127, size = 1086, normalized size = 5.6 \[ \text{result too large to display} \]
Warning: Unable to verify antiderivative.
[In]
Integrate[Cos[c + d*x]^(9/2)*(a + a*Sec[c + d*x])^2*(A + B*Sec[c + d*x]),x]
[Out]
(Cos[c + d*x]^(7/2)*Sec[c/2 + (d*x)/2]^4*(a + a*Sec[c + d*x])^2*(A + B*Sec[c + d*x])*(-((8*A + 9*B)*Cot[c])/(1
5*d) + ((46*A + 51*B)*Cos[d*x]*Sin[c])/(168*d) + ((37*A + 36*B)*Cos[2*d*x]*Sin[2*c])/(360*d) + ((2*A + B)*Cos[
3*d*x]*Sin[3*c])/(56*d) + (A*Cos[4*d*x]*Sin[4*c])/(144*d) + ((46*A + 51*B)*Cos[c]*Sin[d*x])/(168*d) + ((37*A +
36*B)*Cos[2*c]*Sin[2*d*x])/(360*d) + ((2*A + B)*Cos[3*c]*Sin[3*d*x])/(56*d) + (A*Cos[4*c]*Sin[4*d*x])/(144*d)
))/(B + A*Cos[c + d*x]) - (5*A*Cos[c + d*x]^3*Csc[c]*HypergeometricPFQ[{1/4, 1/2}, {5/4}, Sin[d*x - ArcTan[Cot
[c]]]^2]*Sec[c/2 + (d*x)/2]^4*(a + a*Sec[c + d*x])^2*(A + B*Sec[c + d*x])*Sec[d*x - ArcTan[Cot[c]]]*Sqrt[1 - S
in[d*x - ArcTan[Cot[c]]]]*Sqrt[-(Sqrt[1 + Cot[c]^2]*Sin[c]*Sin[d*x - ArcTan[Cot[c]]])]*Sqrt[1 + Sin[d*x - ArcT
an[Cot[c]]]])/(21*d*(B + A*Cos[c + d*x])*Sqrt[1 + Cot[c]^2]) - (2*B*Cos[c + d*x]^3*Csc[c]*HypergeometricPFQ[{1
/4, 1/2}, {5/4}, Sin[d*x - ArcTan[Cot[c]]]^2]*Sec[c/2 + (d*x)/2]^4*(a + a*Sec[c + d*x])^2*(A + B*Sec[c + d*x])
*Sec[d*x - ArcTan[Cot[c]]]*Sqrt[1 - Sin[d*x - ArcTan[Cot[c]]]]*Sqrt[-(Sqrt[1 + Cot[c]^2]*Sin[c]*Sin[d*x - ArcT
an[Cot[c]]])]*Sqrt[1 + Sin[d*x - ArcTan[Cot[c]]]])/(7*d*(B + A*Cos[c + d*x])*Sqrt[1 + Cot[c]^2]) - (4*A*Cos[c
+ d*x]^3*Csc[c]*Sec[c/2 + (d*x)/2]^4*(a + a*Sec[c + d*x])^2*(A + B*Sec[c + d*x])*((HypergeometricPFQ[{-1/2, -1
/4}, {3/4}, Cos[d*x + ArcTan[Tan[c]]]^2]*Sin[d*x + ArcTan[Tan[c]]]*Tan[c])/(Sqrt[1 - Cos[d*x + ArcTan[Tan[c]]]
]*Sqrt[1 + Cos[d*x + ArcTan[Tan[c]]]]*Sqrt[Cos[c]*Cos[d*x + ArcTan[Tan[c]]]*Sqrt[1 + Tan[c]^2]]*Sqrt[1 + Tan[c
]^2]) - ((Sin[d*x + ArcTan[Tan[c]]]*Tan[c])/Sqrt[1 + Tan[c]^2] + (2*Cos[c]^2*Cos[d*x + ArcTan[Tan[c]]]*Sqrt[1
+ Tan[c]^2])/(Cos[c]^2 + Sin[c]^2))/Sqrt[Cos[c]*Cos[d*x + ArcTan[Tan[c]]]*Sqrt[1 + Tan[c]^2]]))/(15*d*(B + A*C
os[c + d*x])) - (3*B*Cos[c + d*x]^3*Csc[c]*Sec[c/2 + (d*x)/2]^4*(a + a*Sec[c + d*x])^2*(A + B*Sec[c + d*x])*((
HypergeometricPFQ[{-1/2, -1/4}, {3/4}, Cos[d*x + ArcTan[Tan[c]]]^2]*Sin[d*x + ArcTan[Tan[c]]]*Tan[c])/(Sqrt[1
- Cos[d*x + ArcTan[Tan[c]]]]*Sqrt[1 + Cos[d*x + ArcTan[Tan[c]]]]*Sqrt[Cos[c]*Cos[d*x + ArcTan[Tan[c]]]*Sqrt[1
+ Tan[c]^2]]*Sqrt[1 + Tan[c]^2]) - ((Sin[d*x + ArcTan[Tan[c]]]*Tan[c])/Sqrt[1 + Tan[c]^2] + (2*Cos[c]^2*Cos[d*
x + ArcTan[Tan[c]]]*Sqrt[1 + Tan[c]^2])/(Cos[c]^2 + Sin[c]^2))/Sqrt[Cos[c]*Cos[d*x + ArcTan[Tan[c]]]*Sqrt[1 +
Tan[c]^2]]))/(10*d*(B + A*Cos[c + d*x]))
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Maple [A] time = 1.789, size = 413, normalized size = 2.1 \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)^(9/2)*(a+a*sec(d*x+c))^2*(A+B*sec(d*x+c)),x)
[Out]
-4/315*((2*cos(1/2*d*x+1/2*c)^2-1)*sin(1/2*d*x+1/2*c)^2)^(1/2)*a^2*(-560*A*cos(1/2*d*x+1/2*c)*sin(1/2*d*x+1/2*
c)^10+(1840*A+360*B)*sin(1/2*d*x+1/2*c)^8*cos(1/2*d*x+1/2*c)+(-2368*A-1044*B)*sin(1/2*d*x+1/2*c)^6*cos(1/2*d*x
+1/2*c)+(1568*A+1134*B)*sin(1/2*d*x+1/2*c)^4*cos(1/2*d*x+1/2*c)+(-387*A-351*B)*sin(1/2*d*x+1/2*c)^2*cos(1/2*d*
x+1/2*c)+75*A*(sin(1/2*d*x+1/2*c)^2)^(1/2)*(2*sin(1/2*d*x+1/2*c)^2-1)^(1/2)*EllipticF(cos(1/2*d*x+1/2*c),2^(1/
2))-168*A*(sin(1/2*d*x+1/2*c)^2)^(1/2)*(2*sin(1/2*d*x+1/2*c)^2-1)^(1/2)*EllipticE(cos(1/2*d*x+1/2*c),2^(1/2))+
90*B*(sin(1/2*d*x+1/2*c)^2)^(1/2)*(2*sin(1/2*d*x+1/2*c)^2-1)^(1/2)*EllipticF(cos(1/2*d*x+1/2*c),2^(1/2))-189*B
*(sin(1/2*d*x+1/2*c)^2)^(1/2)*(2*sin(1/2*d*x+1/2*c)^2-1)^(1/2)*EllipticE(cos(1/2*d*x+1/2*c),2^(1/2)))/(-2*sin(
1/2*d*x+1/2*c)^4+sin(1/2*d*x+1/2*c)^2)^(1/2)/sin(1/2*d*x+1/2*c)/(2*cos(1/2*d*x+1/2*c)^2-1)^(1/2)/d
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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)^(9/2)*(a+a*sec(d*x+c))^2*(A+B*sec(d*x+c)),x, algorithm="maxima")
[Out]
Timed out
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Fricas [F] time = 0., size = 0, normalized size = 0. \begin{align*}{\rm integral}\left ({\left (B a^{2} \cos \left (d x + c\right )^{4} \sec \left (d x + c\right )^{3} +{\left (A + 2 \, B\right )} a^{2} \cos \left (d x + c\right )^{4} \sec \left (d x + c\right )^{2} +{\left (2 \, A + B\right )} a^{2} \cos \left (d x + c\right )^{4} \sec \left (d x + c\right ) + A a^{2} \cos \left (d x + c\right )^{4}\right )} \sqrt{\cos \left (d x + c\right )}, x\right ) \end{align*}
Verification of antiderivative is not currently implemented for this CAS.
[In]
integrate(cos(d*x+c)^(9/2)*(a+a*sec(d*x+c))^2*(A+B*sec(d*x+c)),x, algorithm="fricas")
[Out]
integral((B*a^2*cos(d*x + c)^4*sec(d*x + c)^3 + (A + 2*B)*a^2*cos(d*x + c)^4*sec(d*x + c)^2 + (2*A + B)*a^2*co
s(d*x + c)^4*sec(d*x + c) + A*a^2*cos(d*x + c)^4)*sqrt(cos(d*x + c)), x)
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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)**(9/2)*(a+a*sec(d*x+c))**2*(A+B*sec(d*x+c)),x)
[Out]
Timed out
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Giac [F] time = 0., size = 0, normalized size = 0. \begin{align*} \int{\left (B \sec \left (d x + c\right ) + A\right )}{\left (a \sec \left (d x + c\right ) + a\right )}^{2} \cos \left (d x + c\right )^{\frac{9}{2}}\,{d x} \end{align*}
Verification of antiderivative is not currently implemented for this CAS.
[In]
integrate(cos(d*x+c)^(9/2)*(a+a*sec(d*x+c))^2*(A+B*sec(d*x+c)),x, algorithm="giac")
[Out]
integrate((B*sec(d*x + c) + A)*(a*sec(d*x + c) + a)^2*cos(d*x + c)^(9/2), x)