3.1355 \(\int \frac{(A+B \cos (c+d x)+C \cos ^2(c+d x)) \sec ^{\frac{3}{2}}(c+d x)}{(a+a \cos (c+d x))^{5/2}} \, dx\)

Optimal. Leaf size=231 \[ \frac{(49 A-9 B+C) \sin (c+d x) \sqrt{\sec (c+d x)}}{16 a^2 d \sqrt{a \cos (c+d x)+a}}-\frac{(75 A-19 B-5 C) \sqrt{\cos (c+d x)} \sqrt{\sec (c+d x)} \tan ^{-1}\left (\frac{\sqrt{a} \sin (c+d x)}{\sqrt{2} \sqrt{\cos (c+d x)} \sqrt{a \cos (c+d x)+a}}\right )}{16 \sqrt{2} a^{5/2} d}-\frac{(13 A-5 B-3 C) \sin (c+d x) \sqrt{\sec (c+d x)}}{16 a d (a \cos (c+d x)+a)^{3/2}}-\frac{(A-B+C) \sin (c+d x) \sqrt{\sec (c+d x)}}{4 d (a \cos (c+d x)+a)^{5/2}} \]

[Out]

-((75*A - 19*B - 5*C)*ArcTan[(Sqrt[a]*Sin[c + d*x])/(Sqrt[2]*Sqrt[Cos[c + d*x]]*Sqrt[a + a*Cos[c + d*x]])]*Sqr
t[Cos[c + d*x]]*Sqrt[Sec[c + d*x]])/(16*Sqrt[2]*a^(5/2)*d) - ((A - B + C)*Sqrt[Sec[c + d*x]]*Sin[c + d*x])/(4*
d*(a + a*Cos[c + d*x])^(5/2)) - ((13*A - 5*B - 3*C)*Sqrt[Sec[c + d*x]]*Sin[c + d*x])/(16*a*d*(a + a*Cos[c + d*
x])^(3/2)) + ((49*A - 9*B + C)*Sqrt[Sec[c + d*x]]*Sin[c + d*x])/(16*a^2*d*Sqrt[a + a*Cos[c + d*x]])

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Rubi [A]  time = 0.804608, antiderivative size = 231, normalized size of antiderivative = 1., number of steps used = 7, number of rules used = 7, integrand size = 45, \(\frac{\text{number of rules}}{\text{integrand size}}\) = 0.156, Rules used = {4221, 3041, 2978, 2984, 12, 2782, 205} \[ \frac{(49 A-9 B+C) \sin (c+d x) \sqrt{\sec (c+d x)}}{16 a^2 d \sqrt{a \cos (c+d x)+a}}-\frac{(75 A-19 B-5 C) \sqrt{\cos (c+d x)} \sqrt{\sec (c+d x)} \tan ^{-1}\left (\frac{\sqrt{a} \sin (c+d x)}{\sqrt{2} \sqrt{\cos (c+d x)} \sqrt{a \cos (c+d x)+a}}\right )}{16 \sqrt{2} a^{5/2} d}-\frac{(13 A-5 B-3 C) \sin (c+d x) \sqrt{\sec (c+d x)}}{16 a d (a \cos (c+d x)+a)^{3/2}}-\frac{(A-B+C) \sin (c+d x) \sqrt{\sec (c+d x)}}{4 d (a \cos (c+d x)+a)^{5/2}} \]

Antiderivative was successfully verified.

[In]

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

[Out]

-((75*A - 19*B - 5*C)*ArcTan[(Sqrt[a]*Sin[c + d*x])/(Sqrt[2]*Sqrt[Cos[c + d*x]]*Sqrt[a + a*Cos[c + d*x]])]*Sqr
t[Cos[c + d*x]]*Sqrt[Sec[c + d*x]])/(16*Sqrt[2]*a^(5/2)*d) - ((A - B + C)*Sqrt[Sec[c + d*x]]*Sin[c + d*x])/(4*
d*(a + a*Cos[c + d*x])^(5/2)) - ((13*A - 5*B - 3*C)*Sqrt[Sec[c + d*x]]*Sin[c + d*x])/(16*a*d*(a + a*Cos[c + d*
x])^(3/2)) + ((49*A - 9*B + C)*Sqrt[Sec[c + d*x]]*Sin[c + d*x])/(16*a^2*d*Sqrt[a + a*Cos[c + d*x]])

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 3041

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[((a*A - b*B + a*C)*Cos[e + f*x]*(
a + b*Sin[e + f*x])^m*(c + d*Sin[e + f*x])^(n + 1))/(f*(b*c - a*d)*(2*m + 1)), x] + Dist[1/(b*(b*c - a*d)*(2*m
+ 1)), Int[(a + b*Sin[e + f*x])^(m + 1)*(c + d*Sin[e + f*x])^n*Simp[A*(a*c*(m + 1) - b*d*(2*m + n + 2)) + B*(
b*c*m + a*d*(n + 1)) - C*(a*c*m + b*d*(n + 1)) + (d*(a*A - b*B)*(m + n + 2) + C*(b*c*(2*m + 1) - a*d*(m - n -
1)))*Sin[e + f*x], x], x], x] /; FreeQ[{a, b, c, d, e, f, A, B, C, n}, x] && NeQ[b*c - a*d, 0] && EqQ[a^2 - b^
2, 0] && NeQ[c^2 - d^2, 0] && LtQ[m, -2^(-1)]

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 2984

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*c - A*d)*Cos[e + f*x]*(a + b*Sin[e + f*x])^m*(c + d*Sin[e + f*x]
)^(n + 1))/(f*(n + 1)*(c^2 - d^2)), x] + Dist[1/(b*(n + 1)*(c^2 - d^2)), Int[(a + b*Sin[e + f*x])^m*(c + d*Sin
[e + f*x])^(n + 1)*Simp[A*(a*d*m + b*c*(n + 1)) - B*(a*c*m + b*d*(n + 1)) + b*(B*c - A*d)*(m + n + 2)*Sin[e +
f*x], x], x], x] /; FreeQ[{a, b, c, d, e, f, A, B, m}, x] && NeQ[b*c - a*d, 0] && EqQ[a^2 - b^2, 0] && NeQ[c^2
- d^2, 0] && LtQ[n, -1] && (IntegerQ[n] || EqQ[m + 1/2, 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{\left (A+B \cos (c+d x)+C \cos ^2(c+d x)\right ) \sec ^{\frac{3}{2}}(c+d x)}{(a+a \cos (c+d x))^{5/2}} \, dx &=\left (\sqrt{\cos (c+d x)} \sqrt{\sec (c+d x)}\right ) \int \frac{A+B \cos (c+d x)+C \cos ^2(c+d x)}{\cos ^{\frac{3}{2}}(c+d x) (a+a \cos (c+d x))^{5/2}} \, dx\\ &=-\frac{(A-B+C) \sqrt{\sec (c+d x)} \sin (c+d x)}{4 d (a+a \cos (c+d x))^{5/2}}+\frac{\left (\sqrt{\cos (c+d x)} \sqrt{\sec (c+d x)}\right ) \int \frac{\frac{1}{2} a (9 A-B+C)-2 a (A-B-C) \cos (c+d x)}{\cos ^{\frac{3}{2}}(c+d x) (a+a \cos (c+d x))^{3/2}} \, dx}{4 a^2}\\ &=-\frac{(A-B+C) \sqrt{\sec (c+d x)} \sin (c+d x)}{4 d (a+a \cos (c+d x))^{5/2}}-\frac{(13 A-5 B-3 C) \sqrt{\sec (c+d x)} \sin (c+d x)}{16 a d (a+a \cos (c+d x))^{3/2}}+\frac{\left (\sqrt{\cos (c+d x)} \sqrt{\sec (c+d x)}\right ) \int \frac{\frac{1}{4} a^2 (49 A-9 B+C)-\frac{1}{2} a^2 (13 A-5 B-3 C) \cos (c+d x)}{\cos ^{\frac{3}{2}}(c+d x) \sqrt{a+a \cos (c+d x)}} \, dx}{8 a^4}\\ &=-\frac{(A-B+C) \sqrt{\sec (c+d x)} \sin (c+d x)}{4 d (a+a \cos (c+d x))^{5/2}}-\frac{(13 A-5 B-3 C) \sqrt{\sec (c+d x)} \sin (c+d x)}{16 a d (a+a \cos (c+d x))^{3/2}}+\frac{(49 A-9 B+C) \sqrt{\sec (c+d x)} \sin (c+d x)}{16 a^2 d \sqrt{a+a \cos (c+d x)}}+\frac{\left (\sqrt{\cos (c+d x)} \sqrt{\sec (c+d x)}\right ) \int -\frac{a^3 (75 A-19 B-5 C)}{8 \sqrt{\cos (c+d x)} \sqrt{a+a \cos (c+d x)}} \, dx}{4 a^5}\\ &=-\frac{(A-B+C) \sqrt{\sec (c+d x)} \sin (c+d x)}{4 d (a+a \cos (c+d x))^{5/2}}-\frac{(13 A-5 B-3 C) \sqrt{\sec (c+d x)} \sin (c+d x)}{16 a d (a+a \cos (c+d x))^{3/2}}+\frac{(49 A-9 B+C) \sqrt{\sec (c+d x)} \sin (c+d x)}{16 a^2 d \sqrt{a+a \cos (c+d x)}}-\frac{\left ((75 A-19 B-5 C) \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{(A-B+C) \sqrt{\sec (c+d x)} \sin (c+d x)}{4 d (a+a \cos (c+d x))^{5/2}}-\frac{(13 A-5 B-3 C) \sqrt{\sec (c+d x)} \sin (c+d x)}{16 a d (a+a \cos (c+d x))^{3/2}}+\frac{(49 A-9 B+C) \sqrt{\sec (c+d x)} \sin (c+d x)}{16 a^2 d \sqrt{a+a \cos (c+d x)}}+\frac{\left ((75 A-19 B-5 C) \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{(75 A-19 B-5 C) \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+C) \sqrt{\sec (c+d x)} \sin (c+d x)}{4 d (a+a \cos (c+d x))^{5/2}}-\frac{(13 A-5 B-3 C) \sqrt{\sec (c+d x)} \sin (c+d x)}{16 a d (a+a \cos (c+d x))^{3/2}}+\frac{(49 A-9 B+C) \sqrt{\sec (c+d x)} \sin (c+d x)}{16 a^2 d \sqrt{a+a \cos (c+d x)}}\\ \end{align*}

Mathematica [C]  time = 25.0702, size = 7100, normalized size = 30.74 \[ \text{Result too large to show} \]

Warning: Unable to verify antiderivative.

[In]

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

[Out]

Result too large to show

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Maple [B]  time = 0.204, size = 649, normalized size = 2.8 \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)+C*cos(d*x+c)^2)*sec(d*x+c)^(3/2)/(a+a*cos(d*x+c))^(5/2),x)

[Out]

-1/32/d*2^(1/2)/a^3*(-1+cos(d*x+c))*(75*A*sin(d*x+c)*cos(d*x+c)^2*arcsin((-1+cos(d*x+c))/sin(d*x+c))*(cos(d*x+
c)/(1+cos(d*x+c)))^(1/2)-19*B*(cos(d*x+c)/(1+cos(d*x+c)))^(1/2)*sin(d*x+c)*cos(d*x+c)^2*arcsin((-1+cos(d*x+c))
/sin(d*x+c))-5*C*sin(d*x+c)*cos(d*x+c)^2*arcsin((-1+cos(d*x+c))/sin(d*x+c))*(cos(d*x+c)/(1+cos(d*x+c)))^(1/2)+
150*A*arcsin((-1+cos(d*x+c))/sin(d*x+c))*cos(d*x+c)*sin(d*x+c)*(cos(d*x+c)/(1+cos(d*x+c)))^(1/2)-49*A*cos(d*x+
c)^3*2^(1/2)-38*B*arcsin((-1+cos(d*x+c))/sin(d*x+c))*sin(d*x+c)*(cos(d*x+c)/(1+cos(d*x+c)))^(1/2)*cos(d*x+c)+9
*B*2^(1/2)*cos(d*x+c)^3-10*C*arcsin((-1+cos(d*x+c))/sin(d*x+c))*cos(d*x+c)*sin(d*x+c)*(cos(d*x+c)/(1+cos(d*x+c
)))^(1/2)-C*cos(d*x+c)^3*2^(1/2)+75*A*arcsin((-1+cos(d*x+c))/sin(d*x+c))*sin(d*x+c)*(cos(d*x+c)/(1+cos(d*x+c))
)^(1/2)-36*A*cos(d*x+c)^2*2^(1/2)-19*B*arcsin((-1+cos(d*x+c))/sin(d*x+c))*sin(d*x+c)*(cos(d*x+c)/(1+cos(d*x+c)
))^(1/2)+4*B*2^(1/2)*cos(d*x+c)^2-5*C*arcsin((-1+cos(d*x+c))/sin(d*x+c))*sin(d*x+c)*(cos(d*x+c)/(1+cos(d*x+c))
)^(1/2)-4*C*cos(d*x+c)^2*2^(1/2)+53*A*cos(d*x+c)*2^(1/2)-13*B*2^(1/2)*cos(d*x+c)+5*C*2^(1/2)*cos(d*x+c)+32*A*2
^(1/2))*cos(d*x+c)*(1/cos(d*x+c))^(3/2)*(a*(1+cos(d*x+c)))^(1/2)/sin(d*x+c)^3/(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+B*cos(d*x+c)+C*cos(d*x+c)^2)*sec(d*x+c)^(3/2)/(a+a*cos(d*x+c))^(5/2),x, algorithm="maxima")

[Out]

Timed out

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Fricas [A]  time = 1.69506, size = 616, normalized size = 2.67 \begin{align*} \frac{\sqrt{2}{\left ({\left (75 \, A - 19 \, B - 5 \, C\right )} \cos \left (d x + c\right )^{3} + 3 \,{\left (75 \, A - 19 \, B - 5 \, C\right )} \cos \left (d x + c\right )^{2} + 3 \,{\left (75 \, A - 19 \, B - 5 \, C\right )} \cos \left (d x + c\right ) + 75 \, A - 19 \, B - 5 \, C\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 (49 \, A - 9 \, B + C\right )} \cos \left (d x + c\right )^{2} +{\left (85 \, A - 13 \, B + 5 \, C\right )} \cos \left (d x + c\right ) + 32 \, A\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)+C*cos(d*x+c)^2)*sec(d*x+c)^(3/2)/(a+a*cos(d*x+c))^(5/2),x, algorithm="fricas")

[Out]

1/32*(sqrt(2)*((75*A - 19*B - 5*C)*cos(d*x + c)^3 + 3*(75*A - 19*B - 5*C)*cos(d*x + c)^2 + 3*(75*A - 19*B - 5*
C)*cos(d*x + c) + 75*A - 19*B - 5*C)*sqrt(a)*arctan(sqrt(2)*sqrt(a*cos(d*x + c) + a)*sqrt(cos(d*x + c))/(sqrt(
a)*sin(d*x + c))) + 2*((49*A - 9*B + C)*cos(d*x + c)^2 + (85*A - 13*B + 5*C)*cos(d*x + c) + 32*A)*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)+C*cos(d*x+c)**2)*sec(d*x+c)**(3/2)/(a+a*cos(d*x+c))**(5/2),x)

[Out]

Timed out

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

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

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