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

Optimal. Leaf size=198 \[ \frac {a^{5/2} (20 A+19 B) \sqrt {\cos (c+d x)} \sqrt {\sec (c+d x)} \sin ^{-1}\left (\frac {\sqrt {a} \sin (c+d x)}{\sqrt {a \cos (c+d x)+a}}\right )}{4 d}-\frac {a^3 (4 A-9 B) \sin (c+d x)}{4 d \sqrt {\sec (c+d x)} \sqrt {a \cos (c+d x)+a}}-\frac {a^2 (4 A-B) \sin (c+d x) \sqrt {a \cos (c+d x)+a}}{2 d \sqrt {\sec (c+d x)}}+\frac {2 a A \sin (c+d x) \sqrt {\sec (c+d x)} (a \cos (c+d x)+a)^{3/2}}{d} \]

[Out]

-1/4*a^3*(4*A-9*B)*sin(d*x+c)/d/(a+a*cos(d*x+c))^(1/2)/sec(d*x+c)^(1/2)-1/2*a^2*(4*A-B)*sin(d*x+c)*(a+a*cos(d*
x+c))^(1/2)/d/sec(d*x+c)^(1/2)+2*a*A*(a+a*cos(d*x+c))^(3/2)*sin(d*x+c)*sec(d*x+c)^(1/2)/d+1/4*a^(5/2)*(20*A+19
*B)*arcsin(sin(d*x+c)*a^(1/2)/(a+a*cos(d*x+c))^(1/2))*cos(d*x+c)^(1/2)*sec(d*x+c)^(1/2)/d

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Rubi [A]  time = 0.66, antiderivative size = 198, normalized size of antiderivative = 1.00, number of steps used = 6, number of rules used = 6, integrand size = 35, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.171, Rules used = {2961, 2975, 2976, 2981, 2774, 216} \[ \frac {a^{5/2} (20 A+19 B) \sqrt {\cos (c+d x)} \sqrt {\sec (c+d x)} \sin ^{-1}\left (\frac {\sqrt {a} \sin (c+d x)}{\sqrt {a \cos (c+d x)+a}}\right )}{4 d}-\frac {a^3 (4 A-9 B) \sin (c+d x)}{4 d \sqrt {\sec (c+d x)} \sqrt {a \cos (c+d x)+a}}-\frac {a^2 (4 A-B) \sin (c+d x) \sqrt {a \cos (c+d x)+a}}{2 d \sqrt {\sec (c+d x)}}+\frac {2 a A \sin (c+d x) \sqrt {\sec (c+d x)} (a \cos (c+d x)+a)^{3/2}}{d} \]

Antiderivative was successfully verified.

[In]

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

[Out]

(a^(5/2)*(20*A + 19*B)*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*(4*A - 9*B)*Sin[c + d*x])/(4*d*Sqrt[a + a*Cos[c + d*x]]*Sqrt[Sec[c + d*x]]) - (a^2*(4*A -
B)*Sqrt[a + a*Cos[c + d*x]]*Sin[c + d*x])/(2*d*Sqrt[Sec[c + d*x]]) + (2*a*A*(a + a*Cos[c + d*x])^(3/2)*Sqrt[S
ec[c + d*x]]*Sin[c + d*x])/d

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]

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 2961

Int[(csc[(e_.) + (f_.)*(x_)]*(g_.))^(p_.)*((a_.) + (b_.)*sin[(e_.) + (f_.)*(x_)])^(m_.)*((c_) + (d_.)*sin[(e_.
) + (f_.)*(x_)])^(n_.), x_Symbol] :> Dist[(g*Csc[e + f*x])^p*(g*Sin[e + f*x])^p, Int[((a + b*Sin[e + f*x])^m*(
c + d*Sin[e + f*x])^n)/(g*Sin[e + f*x])^p, x], x] /; FreeQ[{a, b, c, d, e, f, g, m, n, p}, x] && NeQ[b*c - a*d
, 0] &&  !IntegerQ[p] &&  !(IntegerQ[m] && IntegerQ[n])

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]

Rubi steps

\begin {align*} \int (a+a \cos (c+d x))^{5/2} (A+B \cos (c+d x)) \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} (A+B \cos (c+d x))}{\cos ^{\frac {3}{2}}(c+d x)} \, dx\\ &=\frac {2 a A (a+a \cos (c+d x))^{3/2} \sqrt {\sec (c+d x)} \sin (c+d x)}{d}+\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}{2} a (4 A+B)-\frac {1}{2} a (4 A-B) \cos (c+d x)\right )}{\sqrt {\cos (c+d x)}} \, dx\\ &=-\frac {a^2 (4 A-B) \sqrt {a+a \cos (c+d x)} \sin (c+d x)}{2 d \sqrt {\sec (c+d x)}}+\frac {2 a A (a+a \cos (c+d x))^{3/2} \sqrt {\sec (c+d x)} \sin (c+d x)}{d}+\left (\sqrt {\cos (c+d x)} \sqrt {\sec (c+d x)}\right ) \int \frac {\sqrt {a+a \cos (c+d x)} \left (\frac {1}{4} a^2 (12 A+5 B)-\frac {1}{4} a^2 (4 A-9 B) \cos (c+d x)\right )}{\sqrt {\cos (c+d x)}} \, dx\\ &=-\frac {a^3 (4 A-9 B) \sin (c+d x)}{4 d \sqrt {a+a \cos (c+d x)} \sqrt {\sec (c+d x)}}-\frac {a^2 (4 A-B) \sqrt {a+a \cos (c+d x)} \sin (c+d x)}{2 d \sqrt {\sec (c+d x)}}+\frac {2 a A (a+a \cos (c+d x))^{3/2} \sqrt {\sec (c+d x)} \sin (c+d x)}{d}+\frac {1}{8} \left (a^2 (20 A+19 B) \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 (4 A-9 B) \sin (c+d x)}{4 d \sqrt {a+a \cos (c+d x)} \sqrt {\sec (c+d x)}}-\frac {a^2 (4 A-B) \sqrt {a+a \cos (c+d x)} \sin (c+d x)}{2 d \sqrt {\sec (c+d x)}}+\frac {2 a A (a+a \cos (c+d x))^{3/2} \sqrt {\sec (c+d x)} \sin (c+d x)}{d}-\frac {\left (a^2 (20 A+19 B) \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} (20 A+19 B) \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 (4 A-9 B) \sin (c+d x)}{4 d \sqrt {a+a \cos (c+d x)} \sqrt {\sec (c+d x)}}-\frac {a^2 (4 A-B) \sqrt {a+a \cos (c+d x)} \sin (c+d x)}{2 d \sqrt {\sec (c+d x)}}+\frac {2 a A (a+a \cos (c+d x))^{3/2} \sqrt {\sec (c+d x)} \sin (c+d x)}{d}\\ \end {align*}

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Mathematica [A]  time = 0.78, size = 126, normalized size = 0.64 \[ \frac {a^2 \sec \left (\frac {1}{2} (c+d x)\right ) \sqrt {\sec (c+d x)} \sqrt {a (\cos (c+d x)+1)} \left (\sqrt {2} (20 A+19 B) \sin ^{-1}\left (\sqrt {2} \sin \left (\frac {1}{2} (c+d x)\right )\right ) \sqrt {\cos (c+d x)}+2 \sin \left (\frac {1}{2} (c+d x)\right ) ((4 A+11 B) \cos (c+d x)+8 A+B \cos (2 (c+d x))+B)\right )}{8 d} \]

Antiderivative was successfully verified.

[In]

Integrate[(a + a*Cos[c + d*x])^(5/2)*(A + B*Cos[c + d*x])*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]]*(Sqrt[2]*(20*A + 19*B)*ArcSin[Sqrt[2]*Sin[
(c + d*x)/2]]*Sqrt[Cos[c + d*x]] + 2*(8*A + B + (4*A + 11*B)*Cos[c + d*x] + B*Cos[2*(c + d*x)])*Sin[(c + d*x)/
2]))/(8*d)

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fricas [A]  time = 0.62, size = 147, normalized size = 0.74 \[ -\frac {{\left ({\left (20 \, A + 19 \, B\right )} a^{2} \cos \left (d x + c\right ) + {\left (20 \, A + 19 \, B\right )} a^{2}\right )} \sqrt {a} \arctan \left (\frac {\sqrt {a \cos \left (d x + c\right ) + a} \sqrt {\cos \left (d x + c\right )}}{\sqrt {a} \sin \left (d x + c\right )}\right ) - \frac {{\left (2 \, B a^{2} \cos \left (d x + c\right )^{2} + {\left (4 \, A + 11 \, B\right )} a^{2} \cos \left (d x + c\right ) + 8 \, A a^{2}\right )} \sqrt {a \cos \left (d x + c\right ) + a} \sin \left (d x + c\right )}{\sqrt {\cos \left (d x + c\right )}}}{4 \, {\left (d \cos \left (d x + c\right ) + d\right )}} \]

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))*sec(d*x+c)^(3/2),x, algorithm="fricas")

[Out]

-1/4*(((20*A + 19*B)*a^2*cos(d*x + c) + (20*A + 19*B)*a^2)*sqrt(a)*arctan(sqrt(a*cos(d*x + c) + a)*sqrt(cos(d*
x + c))/(sqrt(a)*sin(d*x + c))) - (2*B*a^2*cos(d*x + c)^2 + (4*A + 11*B)*a^2*cos(d*x + c) + 8*A*a^2)*sqrt(a*co
s(d*x + c) + a)*sin(d*x + c)/sqrt(cos(d*x + c)))/(d*cos(d*x + c) + d)

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

[Out]

Timed out

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maple [B]  time = 0.44, size = 344, normalized size = 1.74 \[ \frac {\left (20 A \cos \left (d x +c \right ) \sqrt {\frac {\cos \left (d x +c \right )}{1+\cos \left (d x +c \right )}}\, \arctan \left (\frac {\sin \left (d x +c \right ) \sqrt {\frac {\cos \left (d x +c \right )}{1+\cos \left (d x +c \right )}}}{\cos \left (d x +c \right )}\right )+19 B \cos \left (d x +c \right ) \sqrt {\frac {\cos \left (d x +c \right )}{1+\cos \left (d x +c \right )}}\, \arctan \left (\frac {\sin \left (d x +c \right ) \sqrt {\frac {\cos \left (d x +c \right )}{1+\cos \left (d x +c \right )}}}{\cos \left (d x +c \right )}\right )+2 B \sin \left (d x +c \right ) \left (\cos ^{2}\left (d x +c \right )\right )+20 A \sqrt {\frac {\cos \left (d x +c \right )}{1+\cos \left (d x +c \right )}}\, \arctan \left (\frac {\sin \left (d x +c \right ) \sqrt {\frac {\cos \left (d x +c \right )}{1+\cos \left (d x +c \right )}}}{\cos \left (d x +c \right )}\right )+4 A \cos \left (d x +c \right ) \sin \left (d x +c \right )+19 B \sqrt {\frac {\cos \left (d x +c \right )}{1+\cos \left (d x +c \right )}}\, \arctan \left (\frac {\sin \left (d x +c \right ) \sqrt {\frac {\cos \left (d x +c \right )}{1+\cos \left (d x +c \right )}}}{\cos \left (d x +c \right )}\right )+11 B \cos \left (d x +c \right ) \sin \left (d x +c \right )+8 A \sin \left (d x +c \right )\right ) \cos \left (d x +c \right ) \left (\frac {1}{\cos \left (d x +c \right )}\right )^{\frac {3}{2}} \sqrt {a \left (1+\cos \left (d x +c \right )\right )}\, a^{2}}{4 d \left (1+\cos \left (d x +c \right )\right )} \]

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))*sec(d*x+c)^(3/2),x)

[Out]

1/4/d*(20*A*cos(d*x+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)/c
os(d*x+c))+19*B*cos(d*x+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))+2*B*sin(d*x+c)*cos(d*x+c)^2+20*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))+4*A*cos(d*x+c)*sin(d*x+c)+19*B*(cos(d*x+c)/(1+cos(d*x+c)))^(1/2)*arctan(si
n(d*x+c)*(cos(d*x+c)/(1+cos(d*x+c)))^(1/2)/cos(d*x+c))+11*B*cos(d*x+c)*sin(d*x+c)+8*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))*a^2

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

[Out]

Timed out

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mupad [F]  time = 0.00, size = -1, normalized size = -0.01 \[ \int \left (A+B\,\cos \left (c+d\,x\right )\right )\,{\left (\frac {1}{\cos \left (c+d\,x\right )}\right )}^{3/2}\,{\left (a+a\,\cos \left (c+d\,x\right )\right )}^{5/2} \,d x \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

int((A + B*cos(c + d*x))*(1/cos(c + d*x))^(3/2)*(a + a*cos(c + d*x))^(5/2),x)

[Out]

int((A + B*cos(c + d*x))*(1/cos(c + d*x))^(3/2)*(a + a*cos(c + d*x))^(5/2), 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+B*cos(d*x+c))*sec(d*x+c)**(3/2),x)

[Out]

Timed out

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