3.545 \(\int \frac{(A+B \cos (c+d x)) \sqrt{\sec (c+d x)}}{(a+a \cos (c+d x))^{7/2}} \, dx\)

Optimal. Leaf size=223 \[ -\frac{(103 A+5 B) \sin (c+d x)}{192 a^2 d \sqrt{\sec (c+d x)} (a \cos (c+d x)+a)^{3/2}}+\frac{(63 A+13 B) \sqrt{\cos (c+d x)} \sqrt{\sec (c+d x)} \tan ^{-1}\left (\frac{\sqrt{a} \sin (c+d x)}{\sqrt{2} \sqrt{\cos (c+d x)} \sqrt{a \cos (c+d x)+a}}\right )}{64 \sqrt{2} a^{7/2} d}-\frac{(5 A-B) \sin (c+d x)}{16 a d \sqrt{\sec (c+d x)} (a \cos (c+d x)+a)^{5/2}}-\frac{(A-B) \sin (c+d x)}{6 d \sqrt{\sec (c+d x)} (a \cos (c+d x)+a)^{7/2}} \]

[Out]

((63*A + 13*B)*ArcTan[(Sqrt[a]*Sin[c + d*x])/(Sqrt[2]*Sqrt[Cos[c + d*x]]*Sqrt[a + a*Cos[c + d*x]])]*Sqrt[Cos[c
+ d*x]]*Sqrt[Sec[c + d*x]])/(64*Sqrt[2]*a^(7/2)*d) - ((A - B)*Sin[c + d*x])/(6*d*(a + a*Cos[c + d*x])^(7/2)*S
qrt[Sec[c + d*x]]) - ((5*A - B)*Sin[c + d*x])/(16*a*d*(a + a*Cos[c + d*x])^(5/2)*Sqrt[Sec[c + d*x]]) - ((103*A
+ 5*B)*Sin[c + d*x])/(192*a^2*d*(a + a*Cos[c + d*x])^(3/2)*Sqrt[Sec[c + d*x]])

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Rubi [A]  time = 0.727955, antiderivative size = 223, normalized size of antiderivative = 1., number of steps used = 7, number of rules used = 5, integrand size = 35, \(\frac{\text{number of rules}}{\text{integrand size}}\) = 0.143, Rules used = {2961, 2978, 12, 2782, 205} \[ -\frac{(103 A+5 B) \sin (c+d x)}{192 a^2 d \sqrt{\sec (c+d x)} (a \cos (c+d x)+a)^{3/2}}+\frac{(63 A+13 B) \sqrt{\cos (c+d x)} \sqrt{\sec (c+d x)} \tan ^{-1}\left (\frac{\sqrt{a} \sin (c+d x)}{\sqrt{2} \sqrt{\cos (c+d x)} \sqrt{a \cos (c+d x)+a}}\right )}{64 \sqrt{2} a^{7/2} d}-\frac{(5 A-B) \sin (c+d x)}{16 a d \sqrt{\sec (c+d x)} (a \cos (c+d x)+a)^{5/2}}-\frac{(A-B) \sin (c+d x)}{6 d \sqrt{\sec (c+d x)} (a \cos (c+d x)+a)^{7/2}} \]

Antiderivative was successfully verified.

[In]

Int[((A + B*Cos[c + d*x])*Sqrt[Sec[c + d*x]])/(a + a*Cos[c + d*x])^(7/2),x]

[Out]

((63*A + 13*B)*ArcTan[(Sqrt[a]*Sin[c + d*x])/(Sqrt[2]*Sqrt[Cos[c + d*x]]*Sqrt[a + a*Cos[c + d*x]])]*Sqrt[Cos[c
+ d*x]]*Sqrt[Sec[c + d*x]])/(64*Sqrt[2]*a^(7/2)*d) - ((A - B)*Sin[c + d*x])/(6*d*(a + a*Cos[c + d*x])^(7/2)*S
qrt[Sec[c + d*x]]) - ((5*A - B)*Sin[c + d*x])/(16*a*d*(a + a*Cos[c + d*x])^(5/2)*Sqrt[Sec[c + d*x]]) - ((103*A
+ 5*B)*Sin[c + d*x])/(192*a^2*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 2978

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*(A*b - a*B)*Cos[e + f*x]*(a + b*Sin[e + f*x])^m*(c + d*Sin[e + f*
x])^(n + 1))/(a*f*(2*m + 1)*(b*c - a*d)), x] + Dist[1/(a*(2*m + 1)*(b*c - a*d)), Int[(a + b*Sin[e + f*x])^(m +
1)*(c + d*Sin[e + f*x])^n*Simp[B*(a*c*m + b*d*(n + 1)) + A*(b*c*(m + 1) - a*d*(2*m + n + 2)) + d*(A*b - a*B)*
(m + n + 2)*Sin[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] && LtQ[m, -2^(-1)] &&  !GtQ[n, 0] && IntegerQ[2*m] && (IntegerQ[2*n] || EqQ[c,
0])

Rule 12

Int[(a_)*(u_), x_Symbol] :> Dist[a, Int[u, x], x] /; FreeQ[a, x] &&  !MatchQ[u, (b_)*(v_) /; FreeQ[b, x]]

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]

Rubi steps

\begin{align*} \int \frac{(A+B \cos (c+d x)) \sqrt{\sec (c+d x)}}{(a+a \cos (c+d x))^{7/2}} \, dx &=\left (\sqrt{\cos (c+d x)} \sqrt{\sec (c+d x)}\right ) \int \frac{A+B \cos (c+d x)}{\sqrt{\cos (c+d x)} (a+a \cos (c+d x))^{7/2}} \, dx\\ &=-\frac{(A-B) \sin (c+d x)}{6 d (a+a \cos (c+d x))^{7/2} \sqrt{\sec (c+d x)}}+\frac{\left (\sqrt{\cos (c+d x)} \sqrt{\sec (c+d x)}\right ) \int \frac{\frac{1}{2} a (11 A+B)-2 a (A-B) \cos (c+d x)}{\sqrt{\cos (c+d x)} (a+a \cos (c+d x))^{5/2}} \, dx}{6 a^2}\\ &=-\frac{(A-B) \sin (c+d x)}{6 d (a+a \cos (c+d x))^{7/2} \sqrt{\sec (c+d x)}}-\frac{(5 A-B) \sin (c+d x)}{16 a d (a+a \cos (c+d x))^{5/2} \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 (73 A+11 B)-\frac{3}{2} a^2 (5 A-B) \cos (c+d x)}{\sqrt{\cos (c+d x)} (a+a \cos (c+d x))^{3/2}} \, dx}{24 a^4}\\ &=-\frac{(A-B) \sin (c+d x)}{6 d (a+a \cos (c+d x))^{7/2} \sqrt{\sec (c+d x)}}-\frac{(5 A-B) \sin (c+d x)}{16 a d (a+a \cos (c+d x))^{5/2} \sqrt{\sec (c+d x)}}-\frac{(103 A+5 B) \sin (c+d x)}{192 a^2 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{3 a^3 (63 A+13 B)}{8 \sqrt{\cos (c+d x)} \sqrt{a+a \cos (c+d x)}} \, dx}{48 a^6}\\ &=-\frac{(A-B) \sin (c+d x)}{6 d (a+a \cos (c+d x))^{7/2} \sqrt{\sec (c+d x)}}-\frac{(5 A-B) \sin (c+d x)}{16 a d (a+a \cos (c+d x))^{5/2} \sqrt{\sec (c+d x)}}-\frac{(103 A+5 B) \sin (c+d x)}{192 a^2 d (a+a \cos (c+d x))^{3/2} \sqrt{\sec (c+d x)}}+\frac{\left ((63 A+13 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}{128 a^3}\\ &=-\frac{(A-B) \sin (c+d x)}{6 d (a+a \cos (c+d x))^{7/2} \sqrt{\sec (c+d x)}}-\frac{(5 A-B) \sin (c+d x)}{16 a d (a+a \cos (c+d x))^{5/2} \sqrt{\sec (c+d x)}}-\frac{(103 A+5 B) \sin (c+d x)}{192 a^2 d (a+a \cos (c+d x))^{3/2} \sqrt{\sec (c+d x)}}-\frac{\left ((63 A+13 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 )}{64 a^2 d}\\ &=\frac{(63 A+13 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)}}{64 \sqrt{2} a^{7/2} d}-\frac{(A-B) \sin (c+d x)}{6 d (a+a \cos (c+d x))^{7/2} \sqrt{\sec (c+d x)}}-\frac{(5 A-B) \sin (c+d x)}{16 a d (a+a \cos (c+d x))^{5/2} \sqrt{\sec (c+d x)}}-\frac{(103 A+5 B) \sin (c+d x)}{192 a^2 d (a+a \cos (c+d x))^{3/2} \sqrt{\sec (c+d x)}}\\ \end{align*}

Mathematica [C]  time = 3.02116, size = 228, normalized size = 1.02 \[ \frac{\cos ^7\left (\frac{1}{2} (c+d x)\right ) \left (\left (\sin \left (\frac{1}{2} (c+d x)\right )-\sin \left (\frac{3}{2} (c+d x)\right )\right ) \sqrt{\sec (c+d x)} \sec ^6\left (\frac{1}{2} (c+d x)\right ) ((532 A-4 B) \cos (c+d x)+(103 A+5 B) \cos (2 (c+d x))+493 A-73 B)+48 i (63 A+13 B) e^{-\frac{1}{2} i (c+d x)} \sqrt{\frac{e^{i (c+d x)}}{1+e^{2 i (c+d x)}}} \sqrt{1+e^{2 i (c+d x)}} \tanh ^{-1}\left (\frac{1-e^{i (c+d x)}}{\sqrt{2} \sqrt{1+e^{2 i (c+d x)}}}\right )\right )}{384 d (a (\cos (c+d x)+1))^{7/2}} \]

Antiderivative was successfully verified.

[In]

Integrate[((A + B*Cos[c + d*x])*Sqrt[Sec[c + d*x]])/(a + a*Cos[c + d*x])^(7/2),x]

[Out]

(Cos[(c + d*x)/2]^7*(((48*I)*(63*A + 13*B)*Sqrt[E^(I*(c + d*x))/(1 + E^((2*I)*(c + d*x)))]*Sqrt[1 + E^((2*I)*(
c + d*x))]*ArcTanh[(1 - E^(I*(c + d*x)))/(Sqrt[2]*Sqrt[1 + E^((2*I)*(c + d*x))])])/E^((I/2)*(c + d*x)) + (493*
A - 73*B + (532*A - 4*B)*Cos[c + d*x] + (103*A + 5*B)*Cos[2*(c + d*x)])*Sec[(c + d*x)/2]^6*Sqrt[Sec[c + d*x]]*
(Sin[(c + d*x)/2] - Sin[(3*(c + d*x))/2])))/(384*d*(a*(1 + Cos[c + d*x]))^(7/2))

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Maple [B]  time = 0.707, size = 512, normalized size = 2.3 \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))*sec(d*x+c)^(1/2)/(a+cos(d*x+c)*a)^(7/2),x)

[Out]

-1/384/d*2^(1/2)/a^4*(1/cos(d*x+c))^(1/2)*(a*(1+cos(d*x+c)))^(1/2)*cos(d*x+c)*(-1+cos(d*x+c))^3*(103*A*2^(1/2)
*(cos(d*x+c)/(1+cos(d*x+c)))^(1/2)*cos(d*x+c)^3+5*B*cos(d*x+c)^3*2^(1/2)*(cos(d*x+c)/(1+cos(d*x+c)))^(1/2)+163
*A*2^(1/2)*(cos(d*x+c)/(1+cos(d*x+c)))^(1/2)*cos(d*x+c)^2-189*A*arcsin((-1+cos(d*x+c))/sin(d*x+c))*cos(d*x+c)^
2*sin(d*x+c)-7*B*2^(1/2)*(cos(d*x+c)/(1+cos(d*x+c)))^(1/2)*cos(d*x+c)^2-39*B*arcsin((-1+cos(d*x+c))/sin(d*x+c)
)*cos(d*x+c)^2*sin(d*x+c)-71*A*2^(1/2)*(cos(d*x+c)/(1+cos(d*x+c)))^(1/2)*cos(d*x+c)-378*A*arcsin((-1+cos(d*x+c
))/sin(d*x+c))*sin(d*x+c)*cos(d*x+c)-37*B*2^(1/2)*(cos(d*x+c)/(1+cos(d*x+c)))^(1/2)*cos(d*x+c)-78*B*arcsin((-1
+cos(d*x+c))/sin(d*x+c))*sin(d*x+c)*cos(d*x+c)-195*A*2^(1/2)*(cos(d*x+c)/(1+cos(d*x+c)))^(1/2)-189*A*arcsin((-
1+cos(d*x+c))/sin(d*x+c))*sin(d*x+c)+39*B*2^(1/2)*(cos(d*x+c)/(1+cos(d*x+c)))^(1/2)-39*B*arcsin((-1+cos(d*x+c)
)/sin(d*x+c))*sin(d*x+c))/sin(d*x+c)^7/(cos(d*x+c)/(1+cos(d*x+c)))^(1/2)

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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+B*cos(d*x+c))*sec(d*x+c)^(1/2)/(a+a*cos(d*x+c))^(7/2),x, algorithm="maxima")

[Out]

Timed out

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Fricas [A]  time = 1.73606, size = 686, normalized size = 3.08 \begin{align*} -\frac{3 \, \sqrt{2}{\left ({\left (63 \, A + 13 \, B\right )} \cos \left (d x + c\right )^{4} + 4 \,{\left (63 \, A + 13 \, B\right )} \cos \left (d x + c\right )^{3} + 6 \,{\left (63 \, A + 13 \, B\right )} \cos \left (d x + c\right )^{2} + 4 \,{\left (63 \, A + 13 \, B\right )} \cos \left (d x + c\right ) + 63 \, A + 13 \, 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 ) + \frac{2 \,{\left ({\left (103 \, A + 5 \, B\right )} \cos \left (d x + c\right )^{3} + 2 \,{\left (133 \, A - B\right )} \cos \left (d x + c\right )^{2} + 39 \,{\left (5 \, A - 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 )}}}{384 \,{\left (a^{4} d \cos \left (d x + c\right )^{4} + 4 \, a^{4} d \cos \left (d x + c\right )^{3} + 6 \, a^{4} d \cos \left (d x + c\right )^{2} + 4 \, a^{4} d \cos \left (d x + c\right ) + a^{4} d\right )}} \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate((A+B*cos(d*x+c))*sec(d*x+c)^(1/2)/(a+a*cos(d*x+c))^(7/2),x, algorithm="fricas")

[Out]

-1/384*(3*sqrt(2)*((63*A + 13*B)*cos(d*x + c)^4 + 4*(63*A + 13*B)*cos(d*x + c)^3 + 6*(63*A + 13*B)*cos(d*x + c
)^2 + 4*(63*A + 13*B)*cos(d*x + c) + 63*A + 13*B)*sqrt(a)*arctan(sqrt(2)*sqrt(a*cos(d*x + c) + a)*sqrt(cos(d*x
+ c))/(sqrt(a)*sin(d*x + c))) + 2*((103*A + 5*B)*cos(d*x + c)^3 + 2*(133*A - B)*cos(d*x + c)^2 + 39*(5*A - B)
*cos(d*x + c))*sqrt(a*cos(d*x + c) + a)*sin(d*x + c)/sqrt(cos(d*x + c)))/(a^4*d*cos(d*x + c)^4 + 4*a^4*d*cos(d
*x + c)^3 + 6*a^4*d*cos(d*x + c)^2 + 4*a^4*d*cos(d*x + c) + a^4*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))*sec(d*x+c)**(1/2)/(a+a*cos(d*x+c))**(7/2),x)

[Out]

Timed out

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Giac [F]  time = 0., size = 0, normalized size = 0. \begin{align*} \int \frac{{\left (B \cos \left (d x + c\right ) + A\right )} \sqrt{\sec \left (d x + c\right )}}{{\left (a \cos \left (d x + c\right ) + a\right )}^{\frac{7}{2}}}\,{d x} \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate((A+B*cos(d*x+c))*sec(d*x+c)^(1/2)/(a+a*cos(d*x+c))^(7/2),x, algorithm="giac")

[Out]

integrate((B*cos(d*x + c) + A)*sqrt(sec(d*x + c))/(a*cos(d*x + c) + a)^(7/2), x)