3.1333 \(\int (a+a \cos (c+d x))^{5/2} (A+B \cos (c+d x)+C \cos ^2(c+d x)) \sec ^{\frac{3}{2}}(c+d x) \, dx\)

Optimal. Leaf size=251 \[ \frac{a^{5/2} (40 A+38 B+25 C) \sqrt{\cos (c+d x)} \sqrt{\sec (c+d x)} \sin ^{-1}\left (\frac{\sqrt{a} \sin (c+d x)}{\sqrt{a \cos (c+d x)+a}}\right )}{8 d}-\frac{a^3 (24 A-54 B-49 C) \sin (c+d x)}{24 d \sqrt{\sec (c+d x)} \sqrt{a \cos (c+d x)+a}}-\frac{a^2 (8 A-2 B-3 C) \sin (c+d x) \sqrt{a \cos (c+d x)+a}}{4 d \sqrt{\sec (c+d x)}}-\frac{a (6 A-C) \sin (c+d x) (a \cos (c+d x)+a)^{3/2}}{3 d \sqrt{\sec (c+d x)}}+\frac{2 A \sin (c+d x) \sqrt{\sec (c+d x)} (a \cos (c+d x)+a)^{5/2}}{d} \]

[Out]

(a^(5/2)*(40*A + 38*B + 25*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]])/(8*d) - (a^3*(24*A - 54*B - 49*C)*Sin[c + d*x])/(24*d*Sqrt[a + a*Cos[c + d*x]]*Sqrt[Sec[c + d*x
]]) - (a^2*(8*A - 2*B - 3*C)*Sqrt[a + a*Cos[c + d*x]]*Sin[c + d*x])/(4*d*Sqrt[Sec[c + d*x]]) - (a*(6*A - C)*(a
+ a*Cos[c + d*x])^(3/2)*Sin[c + d*x])/(3*d*Sqrt[Sec[c + d*x]]) + (2*A*(a + a*Cos[c + d*x])^(5/2)*Sqrt[Sec[c +
d*x]]*Sin[c + d*x])/d

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Rubi [A]  time = 0.954069, antiderivative size = 251, normalized size of antiderivative = 1., number of steps used = 7, number of rules used = 6, integrand size = 45, \(\frac{\text{number of rules}}{\text{integrand size}}\) = 0.133, Rules used = {4221, 3043, 2976, 2981, 2774, 216} \[ \frac{a^{5/2} (40 A+38 B+25 C) \sqrt{\cos (c+d x)} \sqrt{\sec (c+d x)} \sin ^{-1}\left (\frac{\sqrt{a} \sin (c+d x)}{\sqrt{a \cos (c+d x)+a}}\right )}{8 d}-\frac{a^3 (24 A-54 B-49 C) \sin (c+d x)}{24 d \sqrt{\sec (c+d x)} \sqrt{a \cos (c+d x)+a}}-\frac{a^2 (8 A-2 B-3 C) \sin (c+d x) \sqrt{a \cos (c+d x)+a}}{4 d \sqrt{\sec (c+d x)}}-\frac{a (6 A-C) \sin (c+d x) (a \cos (c+d x)+a)^{3/2}}{3 d \sqrt{\sec (c+d x)}}+\frac{2 A \sin (c+d x) \sqrt{\sec (c+d x)} (a \cos (c+d x)+a)^{5/2}}{d} \]

Antiderivative was successfully verified.

[In]

Int[(a + a*Cos[c + d*x])^(5/2)*(A + B*Cos[c + d*x] + C*Cos[c + d*x]^2)*Sec[c + d*x]^(3/2),x]

[Out]

(a^(5/2)*(40*A + 38*B + 25*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]])/(8*d) - (a^3*(24*A - 54*B - 49*C)*Sin[c + d*x])/(24*d*Sqrt[a + a*Cos[c + d*x]]*Sqrt[Sec[c + d*x
]]) - (a^2*(8*A - 2*B - 3*C)*Sqrt[a + a*Cos[c + d*x]]*Sin[c + d*x])/(4*d*Sqrt[Sec[c + d*x]]) - (a*(6*A - C)*(a
+ a*Cos[c + d*x])^(3/2)*Sin[c + d*x])/(3*d*Sqrt[Sec[c + d*x]]) + (2*A*(a + a*Cos[c + d*x])^(5/2)*Sqrt[Sec[c +
d*x]]*Sin[c + d*x])/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 3043

Int[((a_) + (b_.)*sin[(e_.) + (f_.)*(x_)])^(m_.)*((c_.) + (d_.)*sin[(e_.) + (f_.)*(x_)])^(n_)*((A_.) + (B_.)*s
in[(e_.) + (f_.)*(x_)] + (C_.)*sin[(e_.) + (f_.)*(x_)]^2), x_Symbol] :> -Simp[((c^2*C - B*c*d + 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 -
B*d)*(a*c*m + b*d*(n + 1)) + b*(d*(B*c - A*d)*(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, B, 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 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+B \cos (c+d x)+C \cos ^2(c+d x)\right ) \sec ^{\frac{3}{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+B \cos (c+d x)+C \cos ^2(c+d x)\right )}{\cos ^{\frac{3}{2}}(c+d x)} \, dx\\ &=\frac{2 A (a+a \cos (c+d x))^{5/2} \sqrt{\sec (c+d x)} \sin (c+d x)}{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{1}{2} a (5 A+B)-\frac{1}{2} a (6 A-C) \cos (c+d x)\right )}{\sqrt{\cos (c+d x)}} \, dx}{a}\\ &=-\frac{a (6 A-C) (a+a \cos (c+d x))^{3/2} \sin (c+d x)}{3 d \sqrt{\sec (c+d x)}}+\frac{2 A (a+a \cos (c+d x))^{5/2} \sqrt{\sec (c+d x)} \sin (c+d x)}{d}+\frac{\left (2 \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 (24 A+6 B+C)-\frac{3}{4} a^2 (8 A-2 B-3 C) \cos (c+d x)\right )}{\sqrt{\cos (c+d x)}} \, dx}{3 a}\\ &=-\frac{a^2 (8 A-2 B-3 C) \sqrt{a+a \cos (c+d x)} \sin (c+d x)}{4 d \sqrt{\sec (c+d x)}}-\frac{a (6 A-C) (a+a \cos (c+d x))^{3/2} \sin (c+d x)}{3 d \sqrt{\sec (c+d x)}}+\frac{2 A (a+a \cos (c+d x))^{5/2} \sqrt{\sec (c+d x)} \sin (c+d x)}{d}+\frac{\left (\sqrt{\cos (c+d x)} \sqrt{\sec (c+d x)}\right ) \int \frac{\sqrt{a+a \cos (c+d x)} \left (\frac{1}{8} a^3 (72 A+30 B+13 C)-\frac{1}{8} a^3 (24 A-54 B-49 C) \cos (c+d x)\right )}{\sqrt{\cos (c+d x)}} \, dx}{3 a}\\ &=-\frac{a^3 (24 A-54 B-49 C) \sin (c+d x)}{24 d \sqrt{a+a \cos (c+d x)} \sqrt{\sec (c+d x)}}-\frac{a^2 (8 A-2 B-3 C) \sqrt{a+a \cos (c+d x)} \sin (c+d x)}{4 d \sqrt{\sec (c+d x)}}-\frac{a (6 A-C) (a+a \cos (c+d x))^{3/2} \sin (c+d x)}{3 d \sqrt{\sec (c+d x)}}+\frac{2 A (a+a \cos (c+d x))^{5/2} \sqrt{\sec (c+d x)} \sin (c+d x)}{d}+\frac{1}{16} \left (a^2 (40 A+38 B+25 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 (24 A-54 B-49 C) \sin (c+d x)}{24 d \sqrt{a+a \cos (c+d x)} \sqrt{\sec (c+d x)}}-\frac{a^2 (8 A-2 B-3 C) \sqrt{a+a \cos (c+d x)} \sin (c+d x)}{4 d \sqrt{\sec (c+d x)}}-\frac{a (6 A-C) (a+a \cos (c+d x))^{3/2} \sin (c+d x)}{3 d \sqrt{\sec (c+d x)}}+\frac{2 A (a+a \cos (c+d x))^{5/2} \sqrt{\sec (c+d x)} \sin (c+d x)}{d}-\frac{\left (a^2 (40 A+38 B+25 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 )}{8 d}\\ &=\frac{a^{5/2} (40 A+38 B+25 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)}}{8 d}-\frac{a^3 (24 A-54 B-49 C) \sin (c+d x)}{24 d \sqrt{a+a \cos (c+d x)} \sqrt{\sec (c+d x)}}-\frac{a^2 (8 A-2 B-3 C) \sqrt{a+a \cos (c+d x)} \sin (c+d x)}{4 d \sqrt{\sec (c+d x)}}-\frac{a (6 A-C) (a+a \cos (c+d x))^{3/2} \sin (c+d x)}{3 d \sqrt{\sec (c+d x)}}+\frac{2 A (a+a \cos (c+d x))^{5/2} \sqrt{\sec (c+d x)} \sin (c+d x)}{d}\\ \end{align*}

Mathematica [A]  time = 0.995426, size = 156, normalized size = 0.62 \[ \frac{a^2 \sec \left (\frac{1}{2} (c+d x)\right ) \sqrt{\sec (c+d x)} \sqrt{a (\cos (c+d x)+1)} \left (3 \sqrt{2} (40 A+38 B+25 C) \sin ^{-1}\left (\sqrt{2} \sin \left (\frac{1}{2} (c+d x)\right )\right ) \sqrt{\cos (c+d x)}+2 \sin \left (\frac{1}{2} (c+d x)\right ) (3 (8 A+22 B+27 C) \cos (c+d x)+48 A+(6 B+17 C) \cos (2 (c+d x))+6 B+2 C \cos (3 (c+d x))+17 C)\right )}{48 d} \]

Antiderivative was successfully verified.

[In]

Integrate[(a + a*Cos[c + d*x])^(5/2)*(A + B*Cos[c + d*x] + C*Cos[c + d*x]^2)*Sec[c + d*x]^(3/2),x]

[Out]

(a^2*Sqrt[a*(1 + Cos[c + d*x])]*Sec[(c + d*x)/2]*Sqrt[Sec[c + d*x]]*(3*Sqrt[2]*(40*A + 38*B + 25*C)*ArcSin[Sqr
t[2]*Sin[(c + d*x)/2]]*Sqrt[Cos[c + d*x]] + 2*(48*A + 6*B + 17*C + 3*(8*A + 22*B + 27*C)*Cos[c + d*x] + (6*B +
17*C)*Cos[2*(c + d*x)] + 2*C*Cos[3*(c + d*x)])*Sin[(c + d*x)/2]))/(48*d)

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Maple [B]  time = 0.185, size = 513, normalized size = 2. \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+B*cos(d*x+c)+C*cos(d*x+c)^2)*sec(d*x+c)^(3/2),x)

[Out]

1/24/d*a^2*(8*C*sin(d*x+c)*cos(d*x+c)^3+120*A*(cos(d*x+c)/(1+cos(d*x+c)))^(1/2)*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*B*(cos(d*x+c)/(1+cos(d*x+c)))^(1/2)*arctan(sin(d*x+c)*(cos(d*
x+c)/(1+cos(d*x+c)))^(1/2)/cos(d*x+c))*cos(d*x+c)+12*B*cos(d*x+c)^2*sin(d*x+c)+75*C*(cos(d*x+c)/(1+cos(d*x+c))
)^(1/2)*arctan(sin(d*x+c)*(cos(d*x+c)/(1+cos(d*x+c)))^(1/2)/cos(d*x+c))*cos(d*x+c)+34*C*cos(d*x+c)^2*sin(d*x+c
)+120*A*(cos(d*x+c)/(1+cos(d*x+c)))^(1/2)*arctan(sin(d*x+c)*(cos(d*x+c)/(1+cos(d*x+c)))^(1/2)/cos(d*x+c))+24*A
*sin(d*x+c)*cos(d*x+c)+114*B*(cos(d*x+c)/(1+cos(d*x+c)))^(1/2)*arctan(sin(d*x+c)*(cos(d*x+c)/(1+cos(d*x+c)))^(
1/2)/cos(d*x+c))+66*B*cos(d*x+c)*sin(d*x+c)+75*C*(cos(d*x+c)/(1+cos(d*x+c)))^(1/2)*arctan(sin(d*x+c)*(cos(d*x+
c)/(1+cos(d*x+c)))^(1/2)/cos(d*x+c))+75*C*cos(d*x+c)*sin(d*x+c)+48*A*sin(d*x+c))*cos(d*x+c)*(1/cos(d*x+c))^(3/
2)*(a*(1+cos(d*x+c)))^(1/2)/(1+cos(d*x+c))

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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+B*cos(d*x+c)+C*cos(d*x+c)^2)*sec(d*x+c)^(3/2),x, algorithm="maxima")

[Out]

Timed out

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Fricas [A]  time = 3.26127, size = 478, normalized size = 1.9 \begin{align*} -\frac{3 \,{\left ({\left (40 \, A + 38 \, B + 25 \, C\right )} a^{2} \cos \left (d x + c\right ) +{\left (40 \, A + 38 \, B + 25 \, C\right )} a^{2}\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 (8 \, C a^{2} \cos \left (d x + c\right )^{3} + 2 \,{\left (6 \, B + 17 \, C\right )} a^{2} \cos \left (d x + c\right )^{2} + 3 \,{\left (8 \, A + 22 \, B + 25 \, C\right )} a^{2} \cos \left (d x + c\right ) + 48 \, 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 )}}}{24 \,{\left (d \cos \left (d x + c\right ) + d\right )}} \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate((a+a*cos(d*x+c))^(5/2)*(A+B*cos(d*x+c)+C*cos(d*x+c)^2)*sec(d*x+c)^(3/2),x, algorithm="fricas")

[Out]

-1/24*(3*((40*A + 38*B + 25*C)*a^2*cos(d*x + c) + (40*A + 38*B + 25*C)*a^2)*sqrt(a)*arctan(sqrt(a*cos(d*x + c)
+ a)*sqrt(cos(d*x + c))/(sqrt(a)*sin(d*x + c))) - (8*C*a^2*cos(d*x + c)^3 + 2*(6*B + 17*C)*a^2*cos(d*x + c)^2
+ 3*(8*A + 22*B + 25*C)*a^2*cos(d*x + c) + 48*A*a^2)*sqrt(a*cos(d*x + c) + a)*sin(d*x + c)/sqrt(cos(d*x + c))
)/(d*cos(d*x + c) + 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((a+a*cos(d*x+c))**(5/2)*(A+B*cos(d*x+c)+C*cos(d*x+c)**2)*sec(d*x+c)**(3/2),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+B*cos(d*x+c)+C*cos(d*x+c)^2)*sec(d*x+c)^(3/2),x, algorithm="giac")

[Out]

Timed out