3.98 \(\int (a+a \cos (c+d x))^{5/2} (A+C \cos ^2(c+d x)) \sec ^4(c+d x) \, dx\)
Optimal. Leaf size=192 \[ \frac {5 a^{5/2} (5 A+8 C) \tanh ^{-1}\left (\frac {\sqrt {a} \sin (c+d x)}{\sqrt {a \cos (c+d x)+a}}\right )}{8 d}-\frac {a^3 (49 A-24 C) \sin (c+d x)}{24 d \sqrt {a \cos (c+d x)+a}}+\frac {a^2 (31 A+24 C) \tan (c+d x) \sqrt {a \cos (c+d x)+a}}{24 d}+\frac {A \tan (c+d x) \sec ^2(c+d x) (a \cos (c+d x)+a)^{5/2}}{3 d}+\frac {5 a A \tan (c+d x) \sec (c+d x) (a \cos (c+d x)+a)^{3/2}}{12 d} \]
[Out]
5/8*a^(5/2)*(5*A+8*C)*arctanh(sin(d*x+c)*a^(1/2)/(a+a*cos(d*x+c))^(1/2))/d-1/24*a^3*(49*A-24*C)*sin(d*x+c)/d/(
a+a*cos(d*x+c))^(1/2)+5/12*a*A*(a+a*cos(d*x+c))^(3/2)*sec(d*x+c)*tan(d*x+c)/d+1/3*A*(a+a*cos(d*x+c))^(5/2)*sec
(d*x+c)^2*tan(d*x+c)/d+1/24*a^2*(31*A+24*C)*(a+a*cos(d*x+c))^(1/2)*tan(d*x+c)/d
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Rubi [A] time = 0.66, antiderivative size = 192, normalized size of antiderivative = 1.00,
number of steps used = 6, number of rules used = 5, integrand size = 35, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.143, Rules used =
{3044, 2975, 2981, 2773, 206} \[ -\frac {a^3 (49 A-24 C) \sin (c+d x)}{24 d \sqrt {a \cos (c+d x)+a}}+\frac {a^2 (31 A+24 C) \tan (c+d x) \sqrt {a \cos (c+d x)+a}}{24 d}+\frac {5 a^{5/2} (5 A+8 C) \tanh ^{-1}\left (\frac {\sqrt {a} \sin (c+d x)}{\sqrt {a \cos (c+d x)+a}}\right )}{8 d}+\frac {A \tan (c+d x) \sec ^2(c+d x) (a \cos (c+d x)+a)^{5/2}}{3 d}+\frac {5 a A \tan (c+d x) \sec (c+d x) (a \cos (c+d x)+a)^{3/2}}{12 d} \]
Antiderivative was successfully verified.
[In]
Int[(a + a*Cos[c + d*x])^(5/2)*(A + C*Cos[c + d*x]^2)*Sec[c + d*x]^4,x]
[Out]
(5*a^(5/2)*(5*A + 8*C)*ArcTanh[(Sqrt[a]*Sin[c + d*x])/Sqrt[a + a*Cos[c + d*x]]])/(8*d) - (a^3*(49*A - 24*C)*Si
n[c + d*x])/(24*d*Sqrt[a + a*Cos[c + d*x]]) + (a^2*(31*A + 24*C)*Sqrt[a + a*Cos[c + d*x]]*Tan[c + d*x])/(24*d)
+ (5*a*A*(a + a*Cos[c + d*x])^(3/2)*Sec[c + d*x]*Tan[c + d*x])/(12*d) + (A*(a + a*Cos[c + d*x])^(5/2)*Sec[c +
d*x]^2*Tan[c + d*x])/(3*d)
Rule 206
Int[((a_) + (b_.)*(x_)^2)^(-1), x_Symbol] :> Simp[(1*ArcTanh[(Rt[-b, 2]*x)/Rt[a, 2]])/(Rt[a, 2]*Rt[-b, 2]), x]
/; FreeQ[{a, b}, x] && NegQ[a/b] && (GtQ[a, 0] || LtQ[b, 0])
Rule 2773
Int[Sqrt[(a_) + (b_.)*sin[(e_.) + (f_.)*(x_)]]/((c_.) + (d_.)*sin[(e_.) + (f_.)*(x_)]), x_Symbol] :> Dist[(-2*
b)/f, Subst[Int[1/(b*c + a*d - d*x^2), x], x, (b*Cos[e + f*x])/Sqrt[a + b*Sin[e + f*x]]], x] /; FreeQ[{a, b, c
, d, e, f}, x] && NeQ[b*c - a*d, 0] && EqQ[a^2 - b^2, 0] && NeQ[c^2 - d^2, 0]
Rule 2975
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^2*(B*c - A*d)*Cos[e + f*x]*(a + b*Sin[e + f*x])^(m - 1)*(c + d*S
in[e + f*x])^(n + 1))/(d*f*(n + 1)*(b*c + a*d)), x] - Dist[b/(d*(n + 1)*(b*c + a*d)), Int[(a + b*Sin[e + f*x])
^(m - 1)*(c + d*Sin[e + f*x])^(n + 1)*Simp[a*A*d*(m - n - 2) - B*(a*c*(m - 1) + b*d*(n + 1)) - (A*b*d*(m + n +
1) - B*(b*c*m - a*d*(n + 1)))*Sin[e + f*x], x], x], x] /; FreeQ[{a, b, c, d, e, f, A, B}, 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 2981
Int[Sqrt[(a_) + (b_.)*sin[(e_.) + (f_.)*(x_)]]*((A_.) + (B_.)*sin[(e_.) + (f_.)*(x_)])*((c_.) + (d_.)*sin[(e_.
) + (f_.)*(x_)])^(n_), x_Symbol] :> Simp[(-2*b*B*Cos[e + f*x]*(c + d*Sin[e + f*x])^(n + 1))/(d*f*(2*n + 3)*Sqr
t[a + b*Sin[e + f*x]]), x] + Dist[(A*b*d*(2*n + 3) - B*(b*c - 2*a*d*(n + 1)))/(b*d*(2*n + 3)), Int[Sqrt[a + b*
Sin[e + f*x]]*(c + d*Sin[e + f*x])^n, 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] && !LtQ[n, -1]
Rule 3044
Int[((a_) + (b_.)*sin[(e_.) + (f_.)*(x_)])^(m_.)*((c_.) + (d_.)*sin[(e_.) + (f_.)*(x_)])^(n_)*((A_.) + (C_.)*s
in[(e_.) + (f_.)*(x_)]^2), x_Symbol] :> -Simp[((c^2*C + A*d^2)*Cos[e + f*x]*(a + b*Sin[e + f*x])^m*(c + d*Sin[
e + f*x])^(n + 1))/(d*f*(n + 1)*(c^2 - d^2)), x] + Dist[1/(b*d*(n + 1)*(c^2 - d^2)), Int[(a + b*Sin[e + f*x])^
m*(c + d*Sin[e + f*x])^(n + 1)*Simp[A*d*(a*d*m + b*c*(n + 1)) + c*C*(a*c*m + b*d*(n + 1)) - b*(A*d^2*(m + n +
2) + C*(c^2*(m + 1) + d^2*(n + 1)))*Sin[e + f*x], x], x], x] /; FreeQ[{a, b, c, d, e, f, A, C, m}, x] && NeQ[b
*c - a*d, 0] && EqQ[a^2 - b^2, 0] && NeQ[c^2 - d^2, 0] && !LtQ[m, -2^(-1)] && (LtQ[n, -1] || EqQ[m + n + 2, 0
])
Rubi steps
\begin {align*} \int (a+a \cos (c+d x))^{5/2} \left (A+C \cos ^2(c+d x)\right ) \sec ^4(c+d x) \, dx &=\frac {A (a+a \cos (c+d x))^{5/2} \sec ^2(c+d x) \tan (c+d x)}{3 d}+\frac {\int (a+a \cos (c+d x))^{5/2} \left (\frac {5 a A}{2}-\frac {1}{2} a (A-6 C) \cos (c+d x)\right ) \sec ^3(c+d x) \, dx}{3 a}\\ &=\frac {5 a A (a+a \cos (c+d x))^{3/2} \sec (c+d x) \tan (c+d x)}{12 d}+\frac {A (a+a \cos (c+d x))^{5/2} \sec ^2(c+d x) \tan (c+d x)}{3 d}+\frac {\int (a+a \cos (c+d x))^{3/2} \left (\frac {1}{4} a^2 (31 A+24 C)-\frac {3}{4} a^2 (3 A-8 C) \cos (c+d x)\right ) \sec ^2(c+d x) \, dx}{6 a}\\ &=\frac {a^2 (31 A+24 C) \sqrt {a+a \cos (c+d x)} \tan (c+d x)}{24 d}+\frac {5 a A (a+a \cos (c+d x))^{3/2} \sec (c+d x) \tan (c+d x)}{12 d}+\frac {A (a+a \cos (c+d x))^{5/2} \sec ^2(c+d x) \tan (c+d x)}{3 d}+\frac {\int \sqrt {a+a \cos (c+d x)} \left (\frac {15}{8} a^3 (5 A+8 C)-\frac {1}{8} a^3 (49 A-24 C) \cos (c+d x)\right ) \sec (c+d x) \, dx}{6 a}\\ &=-\frac {a^3 (49 A-24 C) \sin (c+d x)}{24 d \sqrt {a+a \cos (c+d x)}}+\frac {a^2 (31 A+24 C) \sqrt {a+a \cos (c+d x)} \tan (c+d x)}{24 d}+\frac {5 a A (a+a \cos (c+d x))^{3/2} \sec (c+d x) \tan (c+d x)}{12 d}+\frac {A (a+a \cos (c+d x))^{5/2} \sec ^2(c+d x) \tan (c+d x)}{3 d}+\frac {1}{16} \left (5 a^2 (5 A+8 C)\right ) \int \sqrt {a+a \cos (c+d x)} \sec (c+d x) \, dx\\ &=-\frac {a^3 (49 A-24 C) \sin (c+d x)}{24 d \sqrt {a+a \cos (c+d x)}}+\frac {a^2 (31 A+24 C) \sqrt {a+a \cos (c+d x)} \tan (c+d x)}{24 d}+\frac {5 a A (a+a \cos (c+d x))^{3/2} \sec (c+d x) \tan (c+d x)}{12 d}+\frac {A (a+a \cos (c+d x))^{5/2} \sec ^2(c+d x) \tan (c+d x)}{3 d}-\frac {\left (5 a^3 (5 A+8 C)\right ) \operatorname {Subst}\left (\int \frac {1}{a-x^2} \, dx,x,-\frac {a \sin (c+d x)}{\sqrt {a+a \cos (c+d x)}}\right )}{8 d}\\ &=\frac {5 a^{5/2} (5 A+8 C) \tanh ^{-1}\left (\frac {\sqrt {a} \sin (c+d x)}{\sqrt {a+a \cos (c+d x)}}\right )}{8 d}-\frac {a^3 (49 A-24 C) \sin (c+d x)}{24 d \sqrt {a+a \cos (c+d x)}}+\frac {a^2 (31 A+24 C) \sqrt {a+a \cos (c+d x)} \tan (c+d x)}{24 d}+\frac {5 a A (a+a \cos (c+d x))^{3/2} \sec (c+d x) \tan (c+d x)}{12 d}+\frac {A (a+a \cos (c+d x))^{5/2} \sec ^2(c+d x) \tan (c+d x)}{3 d}\\ \end {align*}
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Mathematica [A] time = 1.13, size = 142, normalized size = 0.74 \[ \frac {a^2 \sec \left (\frac {1}{2} (c+d x)\right ) \sec ^3(c+d x) \sqrt {a (\cos (c+d x)+1)} \left (\sin \left (\frac {1}{2} (c+d x)\right ) ((68 A+72 C) \cos (c+d x)+3 (25 A+8 C) \cos (2 (c+d x))+91 A+24 C \cos (3 (c+d x))+24 C)+15 \sqrt {2} (5 A+8 C) \cos ^3(c+d x) \tanh ^{-1}\left (\sqrt {2} \sin \left (\frac {1}{2} (c+d x)\right )\right )\right )}{48 d} \]
Antiderivative was successfully verified.
[In]
Integrate[(a + a*Cos[c + d*x])^(5/2)*(A + C*Cos[c + d*x]^2)*Sec[c + d*x]^4,x]
[Out]
(a^2*Sqrt[a*(1 + Cos[c + d*x])]*Sec[(c + d*x)/2]*Sec[c + d*x]^3*(15*Sqrt[2]*(5*A + 8*C)*ArcTanh[Sqrt[2]*Sin[(c
+ d*x)/2]]*Cos[c + d*x]^3 + (91*A + 24*C + (68*A + 72*C)*Cos[c + d*x] + 3*(25*A + 8*C)*Cos[2*(c + d*x)] + 24*
C*Cos[3*(c + d*x)])*Sin[(c + d*x)/2]))/(48*d)
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fricas [A] time = 0.65, size = 220, normalized size = 1.15 \[ \frac {15 \, {\left ({\left (5 \, A + 8 \, C\right )} a^{2} \cos \left (d x + c\right )^{4} + {\left (5 \, A + 8 \, C\right )} a^{2} \cos \left (d x + c\right )^{3}\right )} \sqrt {a} \log \left (\frac {a \cos \left (d x + c\right )^{3} - 7 \, a \cos \left (d x + c\right )^{2} - 4 \, \sqrt {a \cos \left (d x + c\right ) + a} \sqrt {a} {\left (\cos \left (d x + c\right ) - 2\right )} \sin \left (d x + c\right ) + 8 \, a}{\cos \left (d x + c\right )^{3} + \cos \left (d x + c\right )^{2}}\right ) + 4 \, {\left (48 \, C a^{2} \cos \left (d x + c\right )^{3} + 3 \, {\left (25 \, A + 8 \, C\right )} a^{2} \cos \left (d x + c\right )^{2} + 34 \, A a^{2} \cos \left (d x + c\right ) + 8 \, A a^{2}\right )} \sqrt {a \cos \left (d x + c\right ) + a} \sin \left (d x + c\right )}{96 \, {\left (d \cos \left (d x + c\right )^{4} + d \cos \left (d x + c\right )^{3}\right )}} \]
Verification of antiderivative is not currently implemented for this CAS.
[In]
integrate((a+a*cos(d*x+c))^(5/2)*(A+C*cos(d*x+c)^2)*sec(d*x+c)^4,x, algorithm="fricas")
[Out]
1/96*(15*((5*A + 8*C)*a^2*cos(d*x + c)^4 + (5*A + 8*C)*a^2*cos(d*x + c)^3)*sqrt(a)*log((a*cos(d*x + c)^3 - 7*a
*cos(d*x + c)^2 - 4*sqrt(a*cos(d*x + c) + a)*sqrt(a)*(cos(d*x + c) - 2)*sin(d*x + c) + 8*a)/(cos(d*x + c)^3 +
cos(d*x + c)^2)) + 4*(48*C*a^2*cos(d*x + c)^3 + 3*(25*A + 8*C)*a^2*cos(d*x + c)^2 + 34*A*a^2*cos(d*x + c) + 8*
A*a^2)*sqrt(a*cos(d*x + c) + a)*sin(d*x + c))/(d*cos(d*x + c)^4 + d*cos(d*x + c)^3)
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giac [F(-1)] time = 0.00, size = 0, normalized size = 0.00 \[ \text {Timed out} \]
Verification of antiderivative is not currently implemented for this CAS.
[In]
integrate((a+a*cos(d*x+c))^(5/2)*(A+C*cos(d*x+c)^2)*sec(d*x+c)^4,x, algorithm="giac")
[Out]
Timed out
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maple [B] time = 2.39, size = 1337, normalized size = 6.96 \[ \text {result too large to display} \]
Verification of antiderivative is not currently implemented for this CAS.
[In]
int((a+a*cos(d*x+c))^(5/2)*(A+C*cos(d*x+c)^2)*sec(d*x+c)^4,x)
[Out]
1/6*a^(3/2)*cos(1/2*d*x+1/2*c)*(a*sin(1/2*d*x+1/2*c)^2)^(1/2)*(-24*(32*C*2^(1/2)*(a*sin(1/2*d*x+1/2*c)^2)^(1/2
)*a^(1/2)+25*A*ln(-4/(-2*cos(1/2*d*x+1/2*c)+2^(1/2))*(2^(1/2)*(a*sin(1/2*d*x+1/2*c)^2)^(1/2)*a^(1/2)-a*2^(1/2)
*cos(1/2*d*x+1/2*c)+2*a))*a+25*A*ln(4/(2*cos(1/2*d*x+1/2*c)+2^(1/2))*(2^(1/2)*(a*sin(1/2*d*x+1/2*c)^2)^(1/2)*a
^(1/2)+a*2^(1/2)*cos(1/2*d*x+1/2*c)+2*a))*a+40*C*ln(-4/(-2*cos(1/2*d*x+1/2*c)+2^(1/2))*(2^(1/2)*(a*sin(1/2*d*x
+1/2*c)^2)^(1/2)*a^(1/2)-a*2^(1/2)*cos(1/2*d*x+1/2*c)+2*a))*a+40*C*ln(4/(2*cos(1/2*d*x+1/2*c)+2^(1/2))*(2^(1/2
)*(a*sin(1/2*d*x+1/2*c)^2)^(1/2)*a^(1/2)+a*2^(1/2)*cos(1/2*d*x+1/2*c)+2*a))*a)*sin(1/2*d*x+1/2*c)^6+12*(50*A*2
^(1/2)*(a*sin(1/2*d*x+1/2*c)^2)^(1/2)*a^(1/2)+112*C*2^(1/2)*(a*sin(1/2*d*x+1/2*c)^2)^(1/2)*a^(1/2)+75*A*ln(-4/
(-2*cos(1/2*d*x+1/2*c)+2^(1/2))*(2^(1/2)*(a*sin(1/2*d*x+1/2*c)^2)^(1/2)*a^(1/2)-a*2^(1/2)*cos(1/2*d*x+1/2*c)+2
*a))*a+75*A*ln(4/(2*cos(1/2*d*x+1/2*c)+2^(1/2))*(2^(1/2)*(a*sin(1/2*d*x+1/2*c)^2)^(1/2)*a^(1/2)+a*2^(1/2)*cos(
1/2*d*x+1/2*c)+2*a))*a+120*C*ln(-4/(-2*cos(1/2*d*x+1/2*c)+2^(1/2))*(2^(1/2)*(a*sin(1/2*d*x+1/2*c)^2)^(1/2)*a^(
1/2)-a*2^(1/2)*cos(1/2*d*x+1/2*c)+2*a))*a+120*C*ln(4/(2*cos(1/2*d*x+1/2*c)+2^(1/2))*(2^(1/2)*(a*sin(1/2*d*x+1/
2*c)^2)^(1/2)*a^(1/2)+a*2^(1/2)*cos(1/2*d*x+1/2*c)+2*a))*a)*sin(1/2*d*x+1/2*c)^4+(-736*A*2^(1/2)*(a*sin(1/2*d*
x+1/2*c)^2)^(1/2)*a^(1/2)-450*A*ln(-4/(-2*cos(1/2*d*x+1/2*c)+2^(1/2))*(2^(1/2)*(a*sin(1/2*d*x+1/2*c)^2)^(1/2)*
a^(1/2)-a*2^(1/2)*cos(1/2*d*x+1/2*c)+2*a))*a-450*A*ln(4/(2*cos(1/2*d*x+1/2*c)+2^(1/2))*(2^(1/2)*(a*sin(1/2*d*x
+1/2*c)^2)^(1/2)*a^(1/2)+a*2^(1/2)*cos(1/2*d*x+1/2*c)+2*a))*a-768*C*2^(1/2)*(a*sin(1/2*d*x+1/2*c)^2)^(1/2)*a^(
1/2)-720*C*ln(-4/(-2*cos(1/2*d*x+1/2*c)+2^(1/2))*(2^(1/2)*(a*sin(1/2*d*x+1/2*c)^2)^(1/2)*a^(1/2)-a*2^(1/2)*cos
(1/2*d*x+1/2*c)+2*a))*a-720*C*ln(4/(2*cos(1/2*d*x+1/2*c)+2^(1/2))*(2^(1/2)*(a*sin(1/2*d*x+1/2*c)^2)^(1/2)*a^(1
/2)+a*2^(1/2)*cos(1/2*d*x+1/2*c)+2*a))*a)*sin(1/2*d*x+1/2*c)^2+234*A*2^(1/2)*(a*sin(1/2*d*x+1/2*c)^2)^(1/2)*a^
(1/2)+75*A*ln(-4/(-2*cos(1/2*d*x+1/2*c)+2^(1/2))*(2^(1/2)*(a*sin(1/2*d*x+1/2*c)^2)^(1/2)*a^(1/2)-a*2^(1/2)*cos
(1/2*d*x+1/2*c)+2*a))*a+75*A*ln(4/(2*cos(1/2*d*x+1/2*c)+2^(1/2))*(2^(1/2)*(a*sin(1/2*d*x+1/2*c)^2)^(1/2)*a^(1/
2)+a*2^(1/2)*cos(1/2*d*x+1/2*c)+2*a))*a+144*C*2^(1/2)*(a*sin(1/2*d*x+1/2*c)^2)^(1/2)*a^(1/2)+120*C*ln(-4/(-2*c
os(1/2*d*x+1/2*c)+2^(1/2))*(2^(1/2)*(a*sin(1/2*d*x+1/2*c)^2)^(1/2)*a^(1/2)-a*2^(1/2)*cos(1/2*d*x+1/2*c)+2*a))*
a+120*C*ln(4/(2*cos(1/2*d*x+1/2*c)+2^(1/2))*(2^(1/2)*(a*sin(1/2*d*x+1/2*c)^2)^(1/2)*a^(1/2)+a*2^(1/2)*cos(1/2*
d*x+1/2*c)+2*a))*a)/(2*cos(1/2*d*x+1/2*c)-2^(1/2))^3/(2*cos(1/2*d*x+1/2*c)+2^(1/2))^3/sin(1/2*d*x+1/2*c)/(a*co
s(1/2*d*x+1/2*c)^2)^(1/2)/d
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maxima [F(-1)] time = 0.00, size = 0, normalized size = 0.00 \[ \text {Timed out} \]
Verification of antiderivative is not currently implemented for this CAS.
[In]
integrate((a+a*cos(d*x+c))^(5/2)*(A+C*cos(d*x+c)^2)*sec(d*x+c)^4,x, algorithm="maxima")
[Out]
Timed out
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mupad [F] time = 0.00, size = -1, normalized size = -0.01 \[ \int \frac {\left (C\,{\cos \left (c+d\,x\right )}^2+A\right )\,{\left (a+a\,\cos \left (c+d\,x\right )\right )}^{5/2}}{{\cos \left (c+d\,x\right )}^4} \,d x \]
Verification of antiderivative is not currently implemented for this CAS.
[In]
int(((A + C*cos(c + d*x)^2)*(a + a*cos(c + d*x))^(5/2))/cos(c + d*x)^4,x)
[Out]
int(((A + C*cos(c + d*x)^2)*(a + a*cos(c + d*x))^(5/2))/cos(c + d*x)^4, x)
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sympy [F(-1)] time = 0.00, size = 0, normalized size = 0.00 \[ \text {Timed out} \]
Verification of antiderivative is not currently implemented for this CAS.
[In]
integrate((a+a*cos(d*x+c))**(5/2)*(A+C*cos(d*x+c)**2)*sec(d*x+c)**4,x)
[Out]
Timed out
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