3.1226 \(\int (a+a \cos (c+d x))^{5/2} (A+C \cos ^2(c+d x)) \sec ^{\frac{5}{2}}(c+d x) \, dx\)

Optimal. Leaf size=238 \[ \frac{a^{5/2} (8 A+19 C) \sqrt{\cos (c+d x)} \sqrt{\sec (c+d x)} \sin ^{-1}\left (\frac{\sqrt{a} \sin (c+d x)}{\sqrt{a \cos (c+d x)+a}}\right )}{4 d}-\frac{a^3 (56 A-27 C) \sin (c+d x)}{12 d \sqrt{\sec (c+d x)} \sqrt{a \cos (c+d x)+a}}-\frac{a^2 (8 A-C) \sin (c+d x) \sqrt{a \cos (c+d x)+a}}{2 d \sqrt{\sec (c+d x)}}+\frac{2 A \sin (c+d x) \sec ^{\frac{3}{2}}(c+d x) (a \cos (c+d x)+a)^{5/2}}{3 d}+\frac{10 a A \sin (c+d x) \sqrt{\sec (c+d x)} (a \cos (c+d x)+a)^{3/2}}{3 d} \]

[Out]

(a^(5/2)*(8*A + 19*C)*ArcSin[(Sqrt[a]*Sin[c + d*x])/Sqrt[a + a*Cos[c + d*x]]]*Sqrt[Cos[c + d*x]]*Sqrt[Sec[c +
d*x]])/(4*d) - (a^3*(56*A - 27*C)*Sin[c + d*x])/(12*d*Sqrt[a + a*Cos[c + d*x]]*Sqrt[Sec[c + d*x]]) - (a^2*(8*A
- C)*Sqrt[a + a*Cos[c + d*x]]*Sin[c + d*x])/(2*d*Sqrt[Sec[c + d*x]]) + (10*a*A*(a + a*Cos[c + d*x])^(3/2)*Sqr
t[Sec[c + d*x]]*Sin[c + d*x])/(3*d) + (2*A*(a + a*Cos[c + d*x])^(5/2)*Sec[c + d*x]^(3/2)*Sin[c + d*x])/(3*d)

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Rubi [A]  time = 0.852188, antiderivative size = 238, normalized size of antiderivative = 1., number of steps used = 7, number of rules used = 7, integrand size = 37, \(\frac{\text{number of rules}}{\text{integrand size}}\) = 0.189, Rules used = {4221, 3044, 2975, 2976, 2981, 2774, 216} \[ \frac{a^{5/2} (8 A+19 C) \sqrt{\cos (c+d x)} \sqrt{\sec (c+d x)} \sin ^{-1}\left (\frac{\sqrt{a} \sin (c+d x)}{\sqrt{a \cos (c+d x)+a}}\right )}{4 d}-\frac{a^3 (56 A-27 C) \sin (c+d x)}{12 d \sqrt{\sec (c+d x)} \sqrt{a \cos (c+d x)+a}}-\frac{a^2 (8 A-C) \sin (c+d x) \sqrt{a \cos (c+d x)+a}}{2 d \sqrt{\sec (c+d x)}}+\frac{2 A \sin (c+d x) \sec ^{\frac{3}{2}}(c+d x) (a \cos (c+d x)+a)^{5/2}}{3 d}+\frac{10 a A \sin (c+d x) \sqrt{\sec (c+d x)} (a \cos (c+d x)+a)^{3/2}}{3 d} \]

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]^(5/2),x]

[Out]

(a^(5/2)*(8*A + 19*C)*ArcSin[(Sqrt[a]*Sin[c + d*x])/Sqrt[a + a*Cos[c + d*x]]]*Sqrt[Cos[c + d*x]]*Sqrt[Sec[c +
d*x]])/(4*d) - (a^3*(56*A - 27*C)*Sin[c + d*x])/(12*d*Sqrt[a + a*Cos[c + d*x]]*Sqrt[Sec[c + d*x]]) - (a^2*(8*A
- C)*Sqrt[a + a*Cos[c + d*x]]*Sin[c + d*x])/(2*d*Sqrt[Sec[c + d*x]]) + (10*a*A*(a + a*Cos[c + d*x])^(3/2)*Sqr
t[Sec[c + d*x]]*Sin[c + d*x])/(3*d) + (2*A*(a + a*Cos[c + d*x])^(5/2)*Sec[c + d*x]^(3/2)*Sin[c + d*x])/(3*d)

Rule 4221

Int[(u_)*((c_.)*sec[(a_.) + (b_.)*(x_)])^(m_.), x_Symbol] :> Dist[(c*Sec[a + b*x])^m*(c*Cos[a + b*x])^m, Int[A
ctivateTrig[u]/(c*Cos[a + b*x])^m, x], x] /; FreeQ[{a, b, c, m}, x] &&  !IntegerQ[m] && KnownSineIntegrandQ[u,
x]

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
])

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 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 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 2774

Int[Sqrt[(a_) + (b_.)*sin[(e_.) + (f_.)*(x_)]]/Sqrt[(d_.)*sin[(e_.) + (f_.)*(x_)]], x_Symbol] :> Dist[-2/f, Su
bst[Int[1/Sqrt[1 - x^2/a], x], x, (b*Cos[e + f*x])/Sqrt[a + b*Sin[e + f*x]]], x] /; FreeQ[{a, b, d, e, f}, x]
&& EqQ[a^2 - b^2, 0] && EqQ[d, a/b]

Rule 216

Int[1/Sqrt[(a_) + (b_.)*(x_)^2], x_Symbol] :> Simp[ArcSin[(Rt[-b, 2]*x)/Sqrt[a]]/Rt[-b, 2], x] /; FreeQ[{a, b}
, x] && GtQ[a, 0] && NegQ[b]

Rubi steps

\begin{align*} \int (a+a \cos (c+d x))^{5/2} \left (A+C \cos ^2(c+d x)\right ) \sec ^{\frac{5}{2}}(c+d x) \, dx &=\left (\sqrt{\cos (c+d x)} \sqrt{\sec (c+d x)}\right ) \int \frac{(a+a \cos (c+d x))^{5/2} \left (A+C \cos ^2(c+d x)\right )}{\cos ^{\frac{5}{2}}(c+d x)} \, dx\\ &=\frac{2 A (a+a \cos (c+d x))^{5/2} \sec ^{\frac{3}{2}}(c+d x) \sin (c+d x)}{3 d}+\frac{\left (2 \sqrt{\cos (c+d x)} \sqrt{\sec (c+d x)}\right ) \int \frac{(a+a \cos (c+d x))^{5/2} \left (\frac{5 a A}{2}-\frac{1}{2} a (4 A-3 C) \cos (c+d x)\right )}{\cos ^{\frac{3}{2}}(c+d x)} \, dx}{3 a}\\ &=\frac{10 a A (a+a \cos (c+d x))^{3/2} \sqrt{\sec (c+d x)} \sin (c+d x)}{3 d}+\frac{2 A (a+a \cos (c+d x))^{5/2} \sec ^{\frac{3}{2}}(c+d x) \sin (c+d x)}{3 d}+\frac{\left (4 \sqrt{\cos (c+d x)} \sqrt{\sec (c+d x)}\right ) \int \frac{(a+a \cos (c+d x))^{3/2} \left (\frac{1}{4} a^2 (16 A+3 C)-\frac{3}{4} a^2 (8 A-C) \cos (c+d x)\right )}{\sqrt{\cos (c+d x)}} \, dx}{3 a}\\ &=-\frac{a^2 (8 A-C) \sqrt{a+a \cos (c+d x)} \sin (c+d x)}{2 d \sqrt{\sec (c+d x)}}+\frac{10 a A (a+a \cos (c+d x))^{3/2} \sqrt{\sec (c+d x)} \sin (c+d x)}{3 d}+\frac{2 A (a+a \cos (c+d x))^{5/2} \sec ^{\frac{3}{2}}(c+d x) \sin (c+d x)}{3 d}+\frac{\left (2 \sqrt{\cos (c+d x)} \sqrt{\sec (c+d x)}\right ) \int \frac{\sqrt{a+a \cos (c+d x)} \left (\frac{5}{8} a^3 (8 A+3 C)-\frac{1}{8} a^3 (56 A-27 C) \cos (c+d x)\right )}{\sqrt{\cos (c+d x)}} \, dx}{3 a}\\ &=-\frac{a^3 (56 A-27 C) \sin (c+d x)}{12 d \sqrt{a+a \cos (c+d x)} \sqrt{\sec (c+d x)}}-\frac{a^2 (8 A-C) \sqrt{a+a \cos (c+d x)} \sin (c+d x)}{2 d \sqrt{\sec (c+d x)}}+\frac{10 a A (a+a \cos (c+d x))^{3/2} \sqrt{\sec (c+d x)} \sin (c+d x)}{3 d}+\frac{2 A (a+a \cos (c+d x))^{5/2} \sec ^{\frac{3}{2}}(c+d x) \sin (c+d x)}{3 d}+\frac{1}{8} \left (a^2 (8 A+19 C) \sqrt{\cos (c+d x)} \sqrt{\sec (c+d x)}\right ) \int \frac{\sqrt{a+a \cos (c+d x)}}{\sqrt{\cos (c+d x)}} \, dx\\ &=-\frac{a^3 (56 A-27 C) \sin (c+d x)}{12 d \sqrt{a+a \cos (c+d x)} \sqrt{\sec (c+d x)}}-\frac{a^2 (8 A-C) \sqrt{a+a \cos (c+d x)} \sin (c+d x)}{2 d \sqrt{\sec (c+d x)}}+\frac{10 a A (a+a \cos (c+d x))^{3/2} \sqrt{\sec (c+d x)} \sin (c+d x)}{3 d}+\frac{2 A (a+a \cos (c+d x))^{5/2} \sec ^{\frac{3}{2}}(c+d x) \sin (c+d x)}{3 d}-\frac{\left (a^2 (8 A+19 C) \sqrt{\cos (c+d x)} \sqrt{\sec (c+d x)}\right ) \operatorname{Subst}\left (\int \frac{1}{\sqrt{1-\frac{x^2}{a}}} \, dx,x,-\frac{a \sin (c+d x)}{\sqrt{a+a \cos (c+d x)}}\right )}{4 d}\\ &=\frac{a^{5/2} (8 A+19 C) \sin ^{-1}\left (\frac{\sqrt{a} \sin (c+d x)}{\sqrt{a+a \cos (c+d x)}}\right ) \sqrt{\cos (c+d x)} \sqrt{\sec (c+d x)}}{4 d}-\frac{a^3 (56 A-27 C) \sin (c+d x)}{12 d \sqrt{a+a \cos (c+d x)} \sqrt{\sec (c+d x)}}-\frac{a^2 (8 A-C) \sqrt{a+a \cos (c+d x)} \sin (c+d x)}{2 d \sqrt{\sec (c+d x)}}+\frac{10 a A (a+a \cos (c+d x))^{3/2} \sqrt{\sec (c+d x)} \sin (c+d x)}{3 d}+\frac{2 A (a+a \cos (c+d x))^{5/2} \sec ^{\frac{3}{2}}(c+d x) \sin (c+d x)}{3 d}\\ \end{align*}

Mathematica [A]  time = 0.85915, size = 141, normalized size = 0.59 \[ \frac{a^2 \sec \left (\frac{1}{2} (c+d x)\right ) \sec ^{\frac{3}{2}}(c+d x) \sqrt{a (\cos (c+d x)+1)} \left (6 \sqrt{2} (8 A+19 C) \sin ^{-1}\left (\sqrt{2} \sin \left (\frac{1}{2} (c+d x)\right )\right ) \cos ^{\frac{3}{2}}(c+d x)+2 \sin \left (\frac{1}{2} (c+d x)\right ) ((128 A+9 C) \cos (c+d x)+16 A+33 C \cos (2 (c+d x))+3 C \cos (3 (c+d x))+33 C)\right )}{48 d} \]

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]^(5/2),x]

[Out]

(a^2*Sqrt[a*(1 + Cos[c + d*x])]*Sec[(c + d*x)/2]*Sec[c + d*x]^(3/2)*(6*Sqrt[2]*(8*A + 19*C)*ArcSin[Sqrt[2]*Sin
[(c + d*x)/2]]*Cos[c + d*x]^(3/2) + 2*(16*A + 33*C + (128*A + 9*C)*Cos[c + d*x] + 33*C*Cos[2*(c + d*x)] + 3*C*
Cos[3*(c + d*x)])*Sin[(c + d*x)/2]))/(48*d)

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Maple [B]  time = 0.214, size = 494, 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((a+a*cos(d*x+c))^(5/2)*(A+C*cos(d*x+c)^2)*sec(d*x+c)^(5/2),x)

[Out]

-1/12/d*a^2*(24*A*(cos(d*x+c)/(1+cos(d*x+c)))^(3/2)*arctan(sin(d*x+c)*(cos(d*x+c)/(1+cos(d*x+c)))^(1/2)/cos(d*
x+c))*cos(d*x+c)^2+57*C*(cos(d*x+c)/(1+cos(d*x+c)))^(3/2)*arctan(sin(d*x+c)*(cos(d*x+c)/(1+cos(d*x+c)))^(1/2)/
cos(d*x+c))*cos(d*x+c)^2+48*A*(cos(d*x+c)/(1+cos(d*x+c)))^(3/2)*arctan(sin(d*x+c)*(cos(d*x+c)/(1+cos(d*x+c)))^
(1/2)/cos(d*x+c))*cos(d*x+c)+114*C*(cos(d*x+c)/(1+cos(d*x+c)))^(3/2)*arctan(sin(d*x+c)*(cos(d*x+c)/(1+cos(d*x+
c)))^(1/2)/cos(d*x+c))*cos(d*x+c)+24*A*(cos(d*x+c)/(1+cos(d*x+c)))^(3/2)*arctan(sin(d*x+c)*(cos(d*x+c)/(1+cos(
d*x+c)))^(1/2)/cos(d*x+c))+57*C*(cos(d*x+c)/(1+cos(d*x+c)))^(3/2)*arctan(sin(d*x+c)*(cos(d*x+c)/(1+cos(d*x+c))
)^(1/2)/cos(d*x+c))+6*C*sin(d*x+c)*cos(d*x+c)^3+33*C*cos(d*x+c)^2*sin(d*x+c)+64*A*sin(d*x+c)*cos(d*x+c)+8*A*si
n(d*x+c))*cos(d*x+c)*(1/cos(d*x+c))^(5/2)*(a*(1+cos(d*x+c)))^(1/2)*sin(d*x+c)^2/(-1+cos(d*x+c))/(1+cos(d*x+c))
^2

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

[Out]

Timed out

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Fricas [A]  time = 2.15052, size = 459, normalized size = 1.93 \begin{align*} -\frac{3 \,{\left ({\left (8 \, A + 19 \, C\right )} a^{2} \cos \left (d x + c\right )^{2} +{\left (8 \, A + 19 \, C\right )} a^{2} \cos \left (d x + c\right )\right )} \sqrt{a} \arctan \left (\frac{\sqrt{a \cos \left (d x + c\right ) + a} \sqrt{\cos \left (d x + c\right )}}{\sqrt{a} \sin \left (d x + c\right )}\right ) - \frac{{\left (6 \, C a^{2} \cos \left (d x + c\right )^{3} + 33 \, C a^{2} \cos \left (d x + c\right )^{2} + 64 \, 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 )}{\sqrt{\cos \left (d x + c\right )}}}{12 \,{\left (d \cos \left (d x + c\right )^{2} + d \cos \left (d x + c\right )\right )}} \end{align*}

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)^(5/2),x, algorithm="fricas")

[Out]

-1/12*(3*((8*A + 19*C)*a^2*cos(d*x + c)^2 + (8*A + 19*C)*a^2*cos(d*x + c))*sqrt(a)*arctan(sqrt(a*cos(d*x + c)
+ a)*sqrt(cos(d*x + c))/(sqrt(a)*sin(d*x + c))) - (6*C*a^2*cos(d*x + c)^3 + 33*C*a^2*cos(d*x + c)^2 + 64*A*a^2
*cos(d*x + c) + 8*A*a^2)*sqrt(a*cos(d*x + c) + a)*sin(d*x + c)/sqrt(cos(d*x + c)))/(d*cos(d*x + c)^2 + d*cos(d
*x + c))

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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((a+a*cos(d*x+c))**(5/2)*(A+C*cos(d*x+c)**2)*sec(d*x+c)**(5/2),x)

[Out]

Timed out

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

[Out]

Timed out