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

Optimal. Leaf size=234 \[ \frac{(3 A-43 B) \sqrt{\cos (c+d x)} \sqrt{\sec (c+d x)} \tan ^{-1}\left (\frac{\sqrt{a} \sin (c+d x)}{\sqrt{2} \sqrt{\cos (c+d x)} \sqrt{a \cos (c+d x)+a}}\right )}{16 \sqrt{2} a^{5/2} d}+\frac{2 B \sqrt{\cos (c+d x)} \sqrt{\sec (c+d x)} \sin ^{-1}\left (\frac{\sqrt{a} \sin (c+d x)}{\sqrt{a \cos (c+d x)+a}}\right )}{a^{5/2} d}+\frac{(A-B) \sin (c+d x)}{4 d \sec ^{\frac{3}{2}}(c+d x) (a \cos (c+d x)+a)^{5/2}}+\frac{(3 A-11 B) \sin (c+d x)}{16 a d \sqrt{\sec (c+d x)} (a \cos (c+d x)+a)^{3/2}} \]

[Out]

(2*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]])/(a^(5/2)*d
) + ((3*A - 43*B)*ArcTan[(Sqrt[a]*Sin[c + d*x])/(Sqrt[2]*Sqrt[Cos[c + d*x]]*Sqrt[a + a*Cos[c + d*x]])]*Sqrt[Co
s[c + d*x]]*Sqrt[Sec[c + d*x]])/(16*Sqrt[2]*a^(5/2)*d) + ((A - B)*Sin[c + d*x])/(4*d*(a + a*Cos[c + d*x])^(5/2
)*Sec[c + d*x]^(3/2)) + ((3*A - 11*B)*Sin[c + d*x])/(16*a*d*(a + a*Cos[c + d*x])^(3/2)*Sqrt[Sec[c + d*x]])

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Rubi [A]  time = 0.738146, antiderivative size = 234, normalized size of antiderivative = 1., number of steps used = 8, number of rules used = 7, integrand size = 35, \(\frac{\text{number of rules}}{\text{integrand size}}\) = 0.2, Rules used = {2961, 2977, 2982, 2782, 205, 2774, 216} \[ \frac{(3 A-43 B) \sqrt{\cos (c+d x)} \sqrt{\sec (c+d x)} \tan ^{-1}\left (\frac{\sqrt{a} \sin (c+d x)}{\sqrt{2} \sqrt{\cos (c+d x)} \sqrt{a \cos (c+d x)+a}}\right )}{16 \sqrt{2} a^{5/2} d}+\frac{2 B \sqrt{\cos (c+d x)} \sqrt{\sec (c+d x)} \sin ^{-1}\left (\frac{\sqrt{a} \sin (c+d x)}{\sqrt{a \cos (c+d x)+a}}\right )}{a^{5/2} d}+\frac{(A-B) \sin (c+d x)}{4 d \sec ^{\frac{3}{2}}(c+d x) (a \cos (c+d x)+a)^{5/2}}+\frac{(3 A-11 B) \sin (c+d x)}{16 a d \sqrt{\sec (c+d x)} (a \cos (c+d x)+a)^{3/2}} \]

Antiderivative was successfully verified.

[In]

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

[Out]

(2*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]])/(a^(5/2)*d
) + ((3*A - 43*B)*ArcTan[(Sqrt[a]*Sin[c + d*x])/(Sqrt[2]*Sqrt[Cos[c + d*x]]*Sqrt[a + a*Cos[c + d*x]])]*Sqrt[Co
s[c + d*x]]*Sqrt[Sec[c + d*x]])/(16*Sqrt[2]*a^(5/2)*d) + ((A - B)*Sin[c + d*x])/(4*d*(a + a*Cos[c + d*x])^(5/2
)*Sec[c + d*x]^(3/2)) + ((3*A - 11*B)*Sin[c + d*x])/(16*a*d*(a + a*Cos[c + d*x])^(3/2)*Sqrt[Sec[c + d*x]])

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 2977

Int[((a_) + (b_.)*sin[(e_.) + (f_.)*(x_)])^(m_)*((A_.) + (B_.)*sin[(e_.) + (f_.)*(x_)])*((c_.) + (d_.)*sin[(e_
.) + (f_.)*(x_)])^(n_), x_Symbol] :> Simp[((A*b - a*B)*Cos[e + f*x]*(a + b*Sin[e + f*x])^m*(c + d*Sin[e + f*x]
)^n)/(a*f*(2*m + 1)), x] - Dist[1/(a*b*(2*m + 1)), Int[(a + b*Sin[e + f*x])^(m + 1)*(c + d*Sin[e + f*x])^(n -
1)*Simp[A*(a*d*n - b*c*(m + 1)) - B*(a*c*m + b*d*n) - d*(a*B*(m - n) + A*b*(m + 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] && LtQ
[m, -2^(-1)] && GtQ[n, 0] && IntegerQ[2*m] && (IntegerQ[2*n] || EqQ[c, 0])

Rule 2982

Int[((A_.) + (B_.)*sin[(e_.) + (f_.)*(x_)])/(Sqrt[(a_) + (b_.)*sin[(e_.) + (f_.)*(x_)]]*Sqrt[(c_.) + (d_.)*sin
[(e_.) + (f_.)*(x_)]]), x_Symbol] :> Dist[(A*b - a*B)/b, Int[1/(Sqrt[a + b*Sin[e + f*x]]*Sqrt[c + d*Sin[e + f*
x]]), x], x] + Dist[B/b, Int[Sqrt[a + b*Sin[e + f*x]]/Sqrt[c + d*Sin[e + f*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]

Rule 2782

Int[1/(Sqrt[(a_) + (b_.)*sin[(e_.) + (f_.)*(x_)]]*Sqrt[(c_.) + (d_.)*sin[(e_.) + (f_.)*(x_)]]), x_Symbol] :> D
ist[(-2*a)/f, Subst[Int[1/(2*b^2 - (a*c - b*d)*x^2), x], x, (b*Cos[e + f*x])/(Sqrt[a + b*Sin[e + f*x]]*Sqrt[c
+ d*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 205

Int[((a_) + (b_.)*(x_)^2)^(-1), x_Symbol] :> Simp[(Rt[a/b, 2]*ArcTan[x/Rt[a/b, 2]])/a, x] /; FreeQ[{a, b}, x]
&& PosQ[a/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 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 \frac{A+B \cos (c+d x)}{(a+a \cos (c+d x))^{5/2} \sec ^{\frac{3}{2}}(c+d x)} \, dx &=\left (\sqrt{\cos (c+d x)} \sqrt{\sec (c+d x)}\right ) \int \frac{\cos ^{\frac{3}{2}}(c+d x) (A+B \cos (c+d x))}{(a+a \cos (c+d x))^{5/2}} \, dx\\ &=\frac{(A-B) \sin (c+d x)}{4 d (a+a \cos (c+d x))^{5/2} \sec ^{\frac{3}{2}}(c+d x)}+\frac{\left (\sqrt{\cos (c+d x)} \sqrt{\sec (c+d x)}\right ) \int \frac{\sqrt{\cos (c+d x)} \left (\frac{3}{2} a (A-B)+4 a B \cos (c+d x)\right )}{(a+a \cos (c+d x))^{3/2}} \, dx}{4 a^2}\\ &=\frac{(A-B) \sin (c+d x)}{4 d (a+a \cos (c+d x))^{5/2} \sec ^{\frac{3}{2}}(c+d x)}+\frac{(3 A-11 B) \sin (c+d x)}{16 a d (a+a \cos (c+d x))^{3/2} \sqrt{\sec (c+d x)}}+\frac{\left (\sqrt{\cos (c+d x)} \sqrt{\sec (c+d x)}\right ) \int \frac{\frac{1}{4} a^2 (3 A-11 B)+8 a^2 B \cos (c+d x)}{\sqrt{\cos (c+d x)} \sqrt{a+a \cos (c+d x)}} \, dx}{8 a^4}\\ &=\frac{(A-B) \sin (c+d x)}{4 d (a+a \cos (c+d x))^{5/2} \sec ^{\frac{3}{2}}(c+d x)}+\frac{(3 A-11 B) \sin (c+d x)}{16 a d (a+a \cos (c+d x))^{3/2} \sqrt{\sec (c+d x)}}+\frac{\left ((3 A-43 B) \sqrt{\cos (c+d x)} \sqrt{\sec (c+d x)}\right ) \int \frac{1}{\sqrt{\cos (c+d x)} \sqrt{a+a \cos (c+d x)}} \, dx}{32 a^2}+\frac{\left (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}{a^3}\\ &=\frac{(A-B) \sin (c+d x)}{4 d (a+a \cos (c+d x))^{5/2} \sec ^{\frac{3}{2}}(c+d x)}+\frac{(3 A-11 B) \sin (c+d x)}{16 a d (a+a \cos (c+d x))^{3/2} \sqrt{\sec (c+d x)}}-\frac{\left ((3 A-43 B) \sqrt{\cos (c+d x)} \sqrt{\sec (c+d x)}\right ) \operatorname{Subst}\left (\int \frac{1}{2 a^2+a x^2} \, dx,x,-\frac{a \sin (c+d x)}{\sqrt{\cos (c+d x)} \sqrt{a+a \cos (c+d x)}}\right )}{16 a d}-\frac{\left (2 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 )}{a^3 d}\\ &=\frac{2 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)}}{a^{5/2} d}+\frac{(3 A-43 B) \tan ^{-1}\left (\frac{\sqrt{a} \sin (c+d x)}{\sqrt{2} \sqrt{\cos (c+d x)} \sqrt{a+a \cos (c+d x)}}\right ) \sqrt{\cos (c+d x)} \sqrt{\sec (c+d x)}}{16 \sqrt{2} a^{5/2} d}+\frac{(A-B) \sin (c+d x)}{4 d (a+a \cos (c+d x))^{5/2} \sec ^{\frac{3}{2}}(c+d x)}+\frac{(3 A-11 B) \sin (c+d x)}{16 a d (a+a \cos (c+d x))^{3/2} \sqrt{\sec (c+d x)}}\\ \end{align*}

Mathematica [C]  time = 2.5089, size = 264, normalized size = 1.13 \[ \frac{\cos ^5\left (\frac{1}{2} (c+d x)\right ) \left (\frac{1}{2} \left (\sin \left (\frac{3}{2} (c+d x)\right )-\sin \left (\frac{1}{2} (c+d x)\right )\right ) \sec ^4\left (\frac{1}{2} (c+d x)\right ) \sqrt{\sec (c+d x)} ((7 A-15 B) \cos (c+d x)+3 A-11 B)-i \sqrt{2} e^{-\frac{1}{2} i (c+d x)} \sqrt{\frac{e^{i (c+d x)}}{1+e^{2 i (c+d x)}}} \sqrt{1+e^{2 i (c+d x)}} \left (-\sqrt{2} (3 A-43 B) \tanh ^{-1}\left (\frac{1-e^{i (c+d x)}}{\sqrt{2} \sqrt{1+e^{2 i (c+d x)}}}\right )+32 B \sinh ^{-1}\left (e^{i (c+d x)}\right )-32 B \tanh ^{-1}\left (\sqrt{1+e^{2 i (c+d x)}}\right )\right )\right )}{8 d (a (\cos (c+d x)+1))^{5/2}} \]

Antiderivative was successfully verified.

[In]

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

[Out]

(Cos[(c + d*x)/2]^5*(((-I)*Sqrt[2]*Sqrt[E^(I*(c + d*x))/(1 + E^((2*I)*(c + d*x)))]*Sqrt[1 + E^((2*I)*(c + d*x)
)]*(32*B*ArcSinh[E^(I*(c + d*x))] - Sqrt[2]*(3*A - 43*B)*ArcTanh[(1 - E^(I*(c + d*x)))/(Sqrt[2]*Sqrt[1 + E^((2
*I)*(c + d*x))])] - 32*B*ArcTanh[Sqrt[1 + E^((2*I)*(c + d*x))]]))/E^((I/2)*(c + d*x)) + ((3*A - 11*B + (7*A -
15*B)*Cos[c + d*x])*Sec[(c + d*x)/2]^4*Sqrt[Sec[c + d*x]]*(-Sin[(c + d*x)/2] + Sin[(3*(c + d*x))/2]))/2))/(8*d
*(a*(1 + Cos[c + d*x]))^(5/2))

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Maple [B]  time = 0.605, size = 476, 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+B*cos(d*x+c))/(a+cos(d*x+c)*a)^(5/2)/sec(d*x+c)^(3/2),x)

[Out]

-1/32/d*2^(1/2)/a^3*(a*(1+cos(d*x+c)))^(1/2)*(-1+cos(d*x+c))^4*cos(d*x+c)*(7*A*2^(1/2)*(cos(d*x+c)/(1+cos(d*x+
c)))^(1/2)*cos(d*x+c)^2-15*B*2^(1/2)*(cos(d*x+c)/(1+cos(d*x+c)))^(1/2)*cos(d*x+c)^2-32*B*2^(1/2)*sin(d*x+c)*co
s(d*x+c)*arctan(sin(d*x+c)*(cos(d*x+c)/(1+cos(d*x+c)))^(1/2)/cos(d*x+c))+3*A*arcsin((-1+cos(d*x+c))/sin(d*x+c)
)*sin(d*x+c)*cos(d*x+c)-4*A*2^(1/2)*(cos(d*x+c)/(1+cos(d*x+c)))^(1/2)*cos(d*x+c)-43*B*arcsin((-1+cos(d*x+c))/s
in(d*x+c))*sin(d*x+c)*cos(d*x+c)+4*B*2^(1/2)*(cos(d*x+c)/(1+cos(d*x+c)))^(1/2)*cos(d*x+c)-32*B*2^(1/2)*arctan(
sin(d*x+c)*(cos(d*x+c)/(1+cos(d*x+c)))^(1/2)/cos(d*x+c))*sin(d*x+c)+3*A*arcsin((-1+cos(d*x+c))/sin(d*x+c))*sin
(d*x+c)-3*A*2^(1/2)*(cos(d*x+c)/(1+cos(d*x+c)))^(1/2)-43*B*arcsin((-1+cos(d*x+c))/sin(d*x+c))*sin(d*x+c)+11*B*
2^(1/2)*(cos(d*x+c)/(1+cos(d*x+c)))^(1/2))/(1/cos(d*x+c))^(3/2)/(cos(d*x+c)/(1+cos(d*x+c)))^(5/2)/sin(d*x+c)^9

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Maxima [F]  time = 0., size = 0, normalized size = 0. \begin{align*} \int \frac{B \cos \left (d x + c\right ) + A}{{\left (a \cos \left (d x + c\right ) + a\right )}^{\frac{5}{2}} \sec \left (d x + c\right )^{\frac{3}{2}}}\,{d x} \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

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

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Fricas [A]  time = 77.0789, size = 759, normalized size = 3.24 \begin{align*} -\frac{\sqrt{2}{\left ({\left (3 \, A - 43 \, B\right )} \cos \left (d x + c\right )^{3} + 3 \,{\left (3 \, A - 43 \, B\right )} \cos \left (d x + c\right )^{2} + 3 \,{\left (3 \, A - 43 \, B\right )} \cos \left (d x + c\right ) + 3 \, A - 43 \, B\right )} \sqrt{a} \arctan \left (\frac{\sqrt{2} \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 ) + 64 \,{\left (B \cos \left (d x + c\right )^{3} + 3 \, B \cos \left (d x + c\right )^{2} + 3 \, B \cos \left (d x + c\right ) + B\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{2 \,{\left ({\left (7 \, A - 15 \, B\right )} \cos \left (d x + c\right )^{2} +{\left (3 \, A - 11 \, B\right )} \cos \left (d x + c\right )\right )} \sqrt{a \cos \left (d x + c\right ) + a} \sin \left (d x + c\right )}{\sqrt{\cos \left (d x + c\right )}}}{32 \,{\left (a^{3} d \cos \left (d x + c\right )^{3} + 3 \, a^{3} d \cos \left (d x + c\right )^{2} + 3 \, a^{3} d \cos \left (d x + c\right ) + a^{3} d\right )}} \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

-1/32*(sqrt(2)*((3*A - 43*B)*cos(d*x + c)^3 + 3*(3*A - 43*B)*cos(d*x + c)^2 + 3*(3*A - 43*B)*cos(d*x + c) + 3*
A - 43*B)*sqrt(a)*arctan(sqrt(2)*sqrt(a*cos(d*x + c) + a)*sqrt(cos(d*x + c))/(sqrt(a)*sin(d*x + c))) + 64*(B*c
os(d*x + c)^3 + 3*B*cos(d*x + c)^2 + 3*B*cos(d*x + c) + B)*sqrt(a)*arctan(sqrt(a*cos(d*x + c) + a)*sqrt(cos(d*
x + c))/(sqrt(a)*sin(d*x + c))) - 2*((7*A - 15*B)*cos(d*x + c)^2 + (3*A - 11*B)*cos(d*x + c))*sqrt(a*cos(d*x +
c) + a)*sin(d*x + c)/sqrt(cos(d*x + c)))/(a^3*d*cos(d*x + c)^3 + 3*a^3*d*cos(d*x + c)^2 + 3*a^3*d*cos(d*x + c
) + a^3*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+B*cos(d*x+c))/(a+a*cos(d*x+c))**(5/2)/sec(d*x+c)**(3/2),x)

[Out]

Timed out

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Giac [F]  time = 0., size = 0, normalized size = 0. \begin{align*} \int \frac{B \cos \left (d x + c\right ) + A}{{\left (a \cos \left (d x + c\right ) + a\right )}^{\frac{5}{2}} \sec \left (d x + c\right )^{\frac{3}{2}}}\,{d x} \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

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