3.1287 \(\int \frac{\cos ^{\frac{3}{2}}(c+d x) (A+B \sec (c+d x)+C \sec ^2(c+d x))}{(a+a \sec (c+d x))^{5/2}} \, dx\)

Optimal. Leaf size=281 \[ \frac{(95 A-39 B+15 C) \sin (c+d x) \sqrt{\cos (c+d x)}}{48 a^2 d \sqrt{a \sec (c+d x)+a}}-\frac{(299 A-147 B+27 C) \sin (c+d x)}{48 a^2 d \sqrt{\cos (c+d x)} \sqrt{a \sec (c+d x)+a}}+\frac{(163 A-75 B+19 C) \sqrt{\cos (c+d x)} \sqrt{\sec (c+d x)} \tanh ^{-1}\left (\frac{\sqrt{a} \sin (c+d x) \sqrt{\sec (c+d x)}}{\sqrt{2} \sqrt{a \sec (c+d x)+a}}\right )}{16 \sqrt{2} a^{5/2} d}-\frac{(17 A-9 B+C) \sin (c+d x) \sqrt{\cos (c+d x)}}{16 a d (a \sec (c+d x)+a)^{3/2}}-\frac{(A-B+C) \sin (c+d x) \sqrt{\cos (c+d x)}}{4 d (a \sec (c+d x)+a)^{5/2}} \]

[Out]

((163*A - 75*B + 19*C)*ArcTanh[(Sqrt[a]*Sqrt[Sec[c + d*x]]*Sin[c + d*x])/(Sqrt[2]*Sqrt[a + a*Sec[c + d*x]])]*S
qrt[Cos[c + d*x]]*Sqrt[Sec[c + d*x]])/(16*Sqrt[2]*a^(5/2)*d) - ((A - B + C)*Sqrt[Cos[c + d*x]]*Sin[c + d*x])/(
4*d*(a + a*Sec[c + d*x])^(5/2)) - ((17*A - 9*B + C)*Sqrt[Cos[c + d*x]]*Sin[c + d*x])/(16*a*d*(a + a*Sec[c + d*
x])^(3/2)) - ((299*A - 147*B + 27*C)*Sin[c + d*x])/(48*a^2*d*Sqrt[Cos[c + d*x]]*Sqrt[a + a*Sec[c + d*x]]) + ((
95*A - 39*B + 15*C)*Sqrt[Cos[c + d*x]]*Sin[c + d*x])/(48*a^2*d*Sqrt[a + a*Sec[c + d*x]])

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Rubi [A]  time = 0.95592, antiderivative size = 281, 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 = {4265, 4084, 4020, 4022, 4013, 3808, 206} \[ \frac{(95 A-39 B+15 C) \sin (c+d x) \sqrt{\cos (c+d x)}}{48 a^2 d \sqrt{a \sec (c+d x)+a}}-\frac{(299 A-147 B+27 C) \sin (c+d x)}{48 a^2 d \sqrt{\cos (c+d x)} \sqrt{a \sec (c+d x)+a}}+\frac{(163 A-75 B+19 C) \sqrt{\cos (c+d x)} \sqrt{\sec (c+d x)} \tanh ^{-1}\left (\frac{\sqrt{a} \sin (c+d x) \sqrt{\sec (c+d x)}}{\sqrt{2} \sqrt{a \sec (c+d x)+a}}\right )}{16 \sqrt{2} a^{5/2} d}-\frac{(17 A-9 B+C) \sin (c+d x) \sqrt{\cos (c+d x)}}{16 a d (a \sec (c+d x)+a)^{3/2}}-\frac{(A-B+C) \sin (c+d x) \sqrt{\cos (c+d x)}}{4 d (a \sec (c+d x)+a)^{5/2}} \]

Antiderivative was successfully verified.

[In]

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

[Out]

((163*A - 75*B + 19*C)*ArcTanh[(Sqrt[a]*Sqrt[Sec[c + d*x]]*Sin[c + d*x])/(Sqrt[2]*Sqrt[a + a*Sec[c + d*x]])]*S
qrt[Cos[c + d*x]]*Sqrt[Sec[c + d*x]])/(16*Sqrt[2]*a^(5/2)*d) - ((A - B + C)*Sqrt[Cos[c + d*x]]*Sin[c + d*x])/(
4*d*(a + a*Sec[c + d*x])^(5/2)) - ((17*A - 9*B + C)*Sqrt[Cos[c + d*x]]*Sin[c + d*x])/(16*a*d*(a + a*Sec[c + d*
x])^(3/2)) - ((299*A - 147*B + 27*C)*Sin[c + d*x])/(48*a^2*d*Sqrt[Cos[c + d*x]]*Sqrt[a + a*Sec[c + d*x]]) + ((
95*A - 39*B + 15*C)*Sqrt[Cos[c + d*x]]*Sin[c + d*x])/(48*a^2*d*Sqrt[a + a*Sec[c + d*x]])

Rule 4265

Int[(cos[(a_.) + (b_.)*(x_)]*(c_.))^(m_.)*(u_), x_Symbol] :> Dist[(c*Cos[a + b*x])^m*(c*Sec[a + b*x])^m, Int[A
ctivateTrig[u]/(c*Sec[a + b*x])^m, x], x] /; FreeQ[{a, b, c, m}, x] &&  !IntegerQ[m] && KnownSecantIntegrandQ[
u, x]

Rule 4084

Int[((A_.) + csc[(e_.) + (f_.)*(x_)]*(B_.) + csc[(e_.) + (f_.)*(x_)]^2*(C_.))*(csc[(e_.) + (f_.)*(x_)]*(d_.))^
(n_)*(csc[(e_.) + (f_.)*(x_)]*(b_.) + (a_))^(m_), x_Symbol] :> -Simp[((a*A - b*B + a*C)*Cot[e + f*x]*(a + b*Cs
c[e + f*x])^m*(d*Csc[e + f*x])^n)/(a*f*(2*m + 1)), x] - Dist[1/(a*b*(2*m + 1)), Int[(a + b*Csc[e + f*x])^(m +
1)*(d*Csc[e + f*x])^n*Simp[a*B*n - b*C*n - A*b*(2*m + n + 1) - (b*B*(m + n + 1) - a*(A*(m + n + 1) - C*(m - n)
))*Csc[e + f*x], x], x], x] /; FreeQ[{a, b, d, e, f, A, B, C, n}, x] && EqQ[a^2 - b^2, 0] && LtQ[m, -2^(-1)]

Rule 4020

Int[(csc[(e_.) + (f_.)*(x_)]*(d_.))^(n_)*(csc[(e_.) + (f_.)*(x_)]*(b_.) + (a_))^(m_)*(csc[(e_.) + (f_.)*(x_)]*
(B_.) + (A_)), x_Symbol] :> -Simp[((A*b - a*B)*Cot[e + f*x]*(a + b*Csc[e + f*x])^m*(d*Csc[e + f*x])^n)/(b*f*(2
*m + 1)), x] - Dist[1/(a^2*(2*m + 1)), Int[(a + b*Csc[e + f*x])^(m + 1)*(d*Csc[e + f*x])^n*Simp[b*B*n - a*A*(2
*m + n + 1) + (A*b - a*B)*(m + n + 1)*Csc[e + f*x], x], x], x] /; FreeQ[{a, b, d, e, f, A, B, n}, x] && NeQ[A*
b - a*B, 0] && EqQ[a^2 - b^2, 0] && LtQ[m, -2^(-1)] &&  !GtQ[n, 0]

Rule 4022

Int[(csc[(e_.) + (f_.)*(x_)]*(d_.))^(n_)*(csc[(e_.) + (f_.)*(x_)]*(b_.) + (a_))^(m_)*(csc[(e_.) + (f_.)*(x_)]*
(B_.) + (A_)), x_Symbol] :> Simp[(A*Cot[e + f*x]*(a + b*Csc[e + f*x])^m*(d*Csc[e + f*x])^n)/(f*n), x] - Dist[1
/(b*d*n), Int[(a + b*Csc[e + f*x])^m*(d*Csc[e + f*x])^(n + 1)*Simp[a*A*m - b*B*n - A*b*(m + n + 1)*Csc[e + f*x
], x], x], x] /; FreeQ[{a, b, d, e, f, A, B, m}, x] && NeQ[A*b - a*B, 0] && EqQ[a^2 - b^2, 0] && LtQ[n, 0]

Rule 4013

Int[(csc[(e_.) + (f_.)*(x_)]*(d_.))^(n_)*(csc[(e_.) + (f_.)*(x_)]*(b_.) + (a_))^(m_)*(csc[(e_.) + (f_.)*(x_)]*
(B_.) + (A_)), x_Symbol] :> Simp[(A*Cot[e + f*x]*(a + b*Csc[e + f*x])^m*(d*Csc[e + f*x])^n)/(f*n), x] - Dist[(
a*A*m - b*B*n)/(b*d*n), Int[(a + b*Csc[e + f*x])^m*(d*Csc[e + f*x])^(n + 1), x], x] /; FreeQ[{a, b, d, e, f, A
, B, m, n}, x] && NeQ[A*b - a*B, 0] && EqQ[a^2 - b^2, 0] && EqQ[m + n + 1, 0] &&  !LeQ[m, -1]

Rule 3808

Int[Sqrt[csc[(e_.) + (f_.)*(x_)]*(d_.)]/Sqrt[csc[(e_.) + (f_.)*(x_)]*(b_.) + (a_)], x_Symbol] :> Dist[(-2*b*d)
/(a*f), Subst[Int[1/(2*b - d*x^2), x], x, (b*Cot[e + f*x])/(Sqrt[a + b*Csc[e + f*x]]*Sqrt[d*Csc[e + f*x]])], x
] /; FreeQ[{a, b, d, e, f}, x] && EqQ[a^2 - b^2, 0]

Rule 206

Int[((a_) + (b_.)*(x_)^2)^(-1), x_Symbol] :> Simp[(1*ArcTanh[(Rt[-b, 2]*x)/Rt[a, 2]])/(Rt[a, 2]*Rt[-b, 2]), x]
/; FreeQ[{a, b}, x] && NegQ[a/b] && (GtQ[a, 0] || LtQ[b, 0])

Rubi steps

\begin{align*} \int \frac{\cos ^{\frac{3}{2}}(c+d x) \left (A+B \sec (c+d x)+C \sec ^2(c+d x)\right )}{(a+a \sec (c+d x))^{5/2}} \, dx &=\left (\sqrt{\cos (c+d x)} \sqrt{\sec (c+d x)}\right ) \int \frac{A+B \sec (c+d x)+C \sec ^2(c+d x)}{\sec ^{\frac{3}{2}}(c+d x) (a+a \sec (c+d x))^{5/2}} \, dx\\ &=-\frac{(A-B+C) \sqrt{\cos (c+d x)} \sin (c+d x)}{4 d (a+a \sec (c+d x))^{5/2}}+\frac{\left (\sqrt{\cos (c+d x)} \sqrt{\sec (c+d x)}\right ) \int \frac{\frac{1}{2} a (11 A-3 B+3 C)-a (3 A-3 B-C) \sec (c+d x)}{\sec ^{\frac{3}{2}}(c+d x) (a+a \sec (c+d x))^{3/2}} \, dx}{4 a^2}\\ &=-\frac{(A-B+C) \sqrt{\cos (c+d x)} \sin (c+d x)}{4 d (a+a \sec (c+d x))^{5/2}}-\frac{(17 A-9 B+C) \sqrt{\cos (c+d x)} \sin (c+d x)}{16 a d (a+a \sec (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 (95 A-39 B+15 C)-a^2 (17 A-9 B+C) \sec (c+d x)}{\sec ^{\frac{3}{2}}(c+d x) \sqrt{a+a \sec (c+d x)}} \, dx}{8 a^4}\\ &=-\frac{(A-B+C) \sqrt{\cos (c+d x)} \sin (c+d x)}{4 d (a+a \sec (c+d x))^{5/2}}-\frac{(17 A-9 B+C) \sqrt{\cos (c+d x)} \sin (c+d x)}{16 a d (a+a \sec (c+d x))^{3/2}}+\frac{(95 A-39 B+15 C) \sqrt{\cos (c+d x)} \sin (c+d x)}{48 a^2 d \sqrt{a+a \sec (c+d x)}}+\frac{\left (\sqrt{\cos (c+d x)} \sqrt{\sec (c+d x)}\right ) \int \frac{-\frac{1}{8} a^3 (299 A-147 B+27 C)+\frac{1}{4} a^3 (95 A-39 B+15 C) \sec (c+d x)}{\sqrt{\sec (c+d x)} \sqrt{a+a \sec (c+d x)}} \, dx}{12 a^5}\\ &=-\frac{(A-B+C) \sqrt{\cos (c+d x)} \sin (c+d x)}{4 d (a+a \sec (c+d x))^{5/2}}-\frac{(17 A-9 B+C) \sqrt{\cos (c+d x)} \sin (c+d x)}{16 a d (a+a \sec (c+d x))^{3/2}}-\frac{(299 A-147 B+27 C) \sin (c+d x)}{48 a^2 d \sqrt{\cos (c+d x)} \sqrt{a+a \sec (c+d x)}}+\frac{(95 A-39 B+15 C) \sqrt{\cos (c+d x)} \sin (c+d x)}{48 a^2 d \sqrt{a+a \sec (c+d x)}}+\frac{\left ((163 A-75 B+19 C) \sqrt{\cos (c+d x)} \sqrt{\sec (c+d x)}\right ) \int \frac{\sqrt{\sec (c+d x)}}{\sqrt{a+a \sec (c+d x)}} \, dx}{32 a^2}\\ &=-\frac{(A-B+C) \sqrt{\cos (c+d x)} \sin (c+d x)}{4 d (a+a \sec (c+d x))^{5/2}}-\frac{(17 A-9 B+C) \sqrt{\cos (c+d x)} \sin (c+d x)}{16 a d (a+a \sec (c+d x))^{3/2}}-\frac{(299 A-147 B+27 C) \sin (c+d x)}{48 a^2 d \sqrt{\cos (c+d x)} \sqrt{a+a \sec (c+d x)}}+\frac{(95 A-39 B+15 C) \sqrt{\cos (c+d x)} \sin (c+d x)}{48 a^2 d \sqrt{a+a \sec (c+d x)}}-\frac{\left ((163 A-75 B+19 C) \sqrt{\cos (c+d x)} \sqrt{\sec (c+d x)}\right ) \operatorname{Subst}\left (\int \frac{1}{2 a-x^2} \, dx,x,-\frac{a \sqrt{\sec (c+d x)} \sin (c+d x)}{\sqrt{a+a \sec (c+d x)}}\right )}{16 a^2 d}\\ &=\frac{(163 A-75 B+19 C) \tanh ^{-1}\left (\frac{\sqrt{a} \sqrt{\sec (c+d x)} \sin (c+d x)}{\sqrt{2} \sqrt{a+a \sec (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{\cos (c+d x)} \sin (c+d x)}{4 d (a+a \sec (c+d x))^{5/2}}-\frac{(17 A-9 B+C) \sqrt{\cos (c+d x)} \sin (c+d x)}{16 a d (a+a \sec (c+d x))^{3/2}}-\frac{(299 A-147 B+27 C) \sin (c+d x)}{48 a^2 d \sqrt{\cos (c+d x)} \sqrt{a+a \sec (c+d x)}}+\frac{(95 A-39 B+15 C) \sqrt{\cos (c+d x)} \sin (c+d x)}{48 a^2 d \sqrt{a+a \sec (c+d x)}}\\ \end{align*}

Mathematica [A]  time = 3.36112, size = 146, normalized size = 0.52 \[ \frac{\sec \left (\frac{1}{2} (c+d x)\right ) \left (4 \sin \left (\frac{1}{2} (c+d x)\right ) ((-479 A+255 B-39 C) \cos (c+d x)+(48 B-80 A) \cos (2 (c+d x))+8 A \cos (3 (c+d x))-379 A+195 B-27 C)+24 (163 A-75 B+19 C) \cos ^4\left (\frac{1}{2} (c+d x)\right ) \tanh ^{-1}\left (\sin \left (\frac{1}{2} (c+d x)\right )\right )\right )}{192 a d \cos ^{\frac{3}{2}}(c+d x) (a (\sec (c+d x)+1))^{3/2}} \]

Antiderivative was successfully verified.

[In]

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

[Out]

(Sec[(c + d*x)/2]*(24*(163*A - 75*B + 19*C)*ArcTanh[Sin[(c + d*x)/2]]*Cos[(c + d*x)/2]^4 + 4*(-379*A + 195*B -
27*C + (-479*A + 255*B - 39*C)*Cos[c + d*x] + (-80*A + 48*B)*Cos[2*(c + d*x)] + 8*A*Cos[3*(c + d*x)])*Sin[(c
+ d*x)/2]))/(192*a*d*Cos[c + d*x]^(3/2)*(a*(1 + Sec[c + d*x]))^(3/2))

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Maple [B]  time = 0.305, size = 550, 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(cos(d*x+c)^(3/2)*(A+B*sec(d*x+c)+C*sec(d*x+c)^2)/(a+a*sec(d*x+c))^(5/2),x)

[Out]

-1/48/d*(a*(cos(d*x+c)+1)/cos(d*x+c))^(1/2)*(-1+cos(d*x+c))^2*(32*A*cos(d*x+c)^4*(-2/(cos(d*x+c)+1))^(1/2)-192
*A*(-2/(cos(d*x+c)+1))^(1/2)*cos(d*x+c)^3+96*B*(-2/(cos(d*x+c)+1))^(1/2)*cos(d*x+c)^3-343*A*cos(d*x+c)^2*(-2/(
cos(d*x+c)+1))^(1/2)-489*A*sin(d*x+c)*cos(d*x+c)*arctan(1/2*sin(d*x+c)*(-2/(cos(d*x+c)+1))^(1/2))+159*B*(-2/(c
os(d*x+c)+1))^(1/2)*cos(d*x+c)^2+225*B*sin(d*x+c)*cos(d*x+c)*arctan(1/2*sin(d*x+c)*(-2/(cos(d*x+c)+1))^(1/2))-
39*C*cos(d*x+c)^2*(-2/(cos(d*x+c)+1))^(1/2)-57*C*sin(d*x+c)*cos(d*x+c)*arctan(1/2*sin(d*x+c)*(-2/(cos(d*x+c)+1
))^(1/2))+204*A*cos(d*x+c)*(-2/(cos(d*x+c)+1))^(1/2)-489*A*sin(d*x+c)*arctan(1/2*sin(d*x+c)*(-2/(cos(d*x+c)+1)
)^(1/2))-108*B*(-2/(cos(d*x+c)+1))^(1/2)*cos(d*x+c)+225*B*sin(d*x+c)*arctan(1/2*sin(d*x+c)*(-2/(cos(d*x+c)+1))
^(1/2))+12*C*cos(d*x+c)*(-2/(cos(d*x+c)+1))^(1/2)-57*C*sin(d*x+c)*arctan(1/2*sin(d*x+c)*(-2/(cos(d*x+c)+1))^(1
/2))+299*A*(-2/(cos(d*x+c)+1))^(1/2)-147*B*(-2/(cos(d*x+c)+1))^(1/2)+27*C*(-2/(cos(d*x+c)+1))^(1/2))*cos(d*x+c
)^(1/2)/a^3/sin(d*x+c)^5/(-2/(cos(d*x+c)+1))^(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(cos(d*x+c)^(3/2)*(A+B*sec(d*x+c)+C*sec(d*x+c)^2)/(a+a*sec(d*x+c))^(5/2),x, algorithm="maxima")

[Out]

Timed out

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Fricas [A]  time = 0.573833, size = 1569, normalized size = 5.58 \begin{align*} \text{result too large to display} \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

[1/192*(3*sqrt(2)*((163*A - 75*B + 19*C)*cos(d*x + c)^3 + 3*(163*A - 75*B + 19*C)*cos(d*x + c)^2 + 3*(163*A -
75*B + 19*C)*cos(d*x + c) + 163*A - 75*B + 19*C)*sqrt(a)*log(-(a*cos(d*x + c)^2 - 2*sqrt(2)*sqrt(a)*sqrt((a*co
s(d*x + c) + a)/cos(d*x + c))*sqrt(cos(d*x + c))*sin(d*x + c) - 2*a*cos(d*x + c) - 3*a)/(cos(d*x + c)^2 + 2*co
s(d*x + c) + 1)) + 4*(32*A*cos(d*x + c)^3 - 32*(5*A - 3*B)*cos(d*x + c)^2 - (503*A - 255*B + 39*C)*cos(d*x + c
) - 299*A + 147*B - 27*C)*sqrt((a*cos(d*x + c) + a)/cos(d*x + c))*sqrt(cos(d*x + c))*sin(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), -1/96*(3*sqrt(2)*((163*A - 75*B + 19*C)*c
os(d*x + c)^3 + 3*(163*A - 75*B + 19*C)*cos(d*x + c)^2 + 3*(163*A - 75*B + 19*C)*cos(d*x + c) + 163*A - 75*B +
19*C)*sqrt(-a)*arctan(sqrt(2)*sqrt(-a)*sqrt((a*cos(d*x + c) + a)/cos(d*x + c))*sqrt(cos(d*x + c))/(a*sin(d*x
+ c))) - 2*(32*A*cos(d*x + c)^3 - 32*(5*A - 3*B)*cos(d*x + c)^2 - (503*A - 255*B + 39*C)*cos(d*x + c) - 299*A
+ 147*B - 27*C)*sqrt((a*cos(d*x + c) + a)/cos(d*x + c))*sqrt(cos(d*x + c))*sin(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(cos(d*x+c)**(3/2)*(A+B*sec(d*x+c)+C*sec(d*x+c)**2)/(a+a*sec(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 \sec \left (d x + c\right )^{2} + B \sec \left (d x + c\right ) + A\right )} \cos \left (d x + c\right )^{\frac{3}{2}}}{{\left (a \sec \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(cos(d*x+c)^(3/2)*(A+B*sec(d*x+c)+C*sec(d*x+c)^2)/(a+a*sec(d*x+c))^(5/2),x, algorithm="giac")

[Out]

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