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

Optimal. Leaf size=270 \[ -\frac{(15 A-11 B) \sqrt{\cos (c+d x)} \sqrt{\sec (c+d x)} \tanh ^{-1}\left (\frac{\sqrt{a} \sin (c+d x) \sqrt{\sec (c+d x)}}{\sqrt{2} \sqrt{a \sec (c+d x)+a}}\right )}{2 \sqrt{2} a^{3/2} d}+\frac{(9 A-5 B) \sin (c+d x) \cos ^{\frac{3}{2}}(c+d x)}{10 a d \sqrt{a \sec (c+d x)+a}}-\frac{(A-B) \sin (c+d x) \cos ^{\frac{3}{2}}(c+d x)}{2 d (a \sec (c+d x)+a)^{3/2}}-\frac{(39 A-35 B) \sin (c+d x) \sqrt{\cos (c+d x)}}{30 a d \sqrt{a \sec (c+d x)+a}}+\frac{(147 A-95 B) \sin (c+d x)}{30 a d \sqrt{\cos (c+d x)} \sqrt{a \sec (c+d x)+a}} \]

[Out]

-((15*A - 11*B)*ArcTanh[(Sqrt[a]*Sqrt[Sec[c + d*x]]*Sin[c + d*x])/(Sqrt[2]*Sqrt[a + a*Sec[c + d*x]])]*Sqrt[Cos
[c + d*x]]*Sqrt[Sec[c + d*x]])/(2*Sqrt[2]*a^(3/2)*d) - ((A - B)*Cos[c + d*x]^(3/2)*Sin[c + d*x])/(2*d*(a + a*S
ec[c + d*x])^(3/2)) + ((147*A - 95*B)*Sin[c + d*x])/(30*a*d*Sqrt[Cos[c + d*x]]*Sqrt[a + a*Sec[c + d*x]]) - ((3
9*A - 35*B)*Sqrt[Cos[c + d*x]]*Sin[c + d*x])/(30*a*d*Sqrt[a + a*Sec[c + d*x]]) + ((9*A - 5*B)*Cos[c + d*x]^(3/
2)*Sin[c + d*x])/(10*a*d*Sqrt[a + a*Sec[c + d*x]])

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Rubi [A]  time = 0.866346, antiderivative size = 270, normalized size of antiderivative = 1., number of steps used = 7, number of rules used = 6, integrand size = 35, \(\frac{\text{number of rules}}{\text{integrand size}}\) = 0.171, Rules used = {2955, 4020, 4022, 4013, 3808, 206} \[ -\frac{(15 A-11 B) \sqrt{\cos (c+d x)} \sqrt{\sec (c+d x)} \tanh ^{-1}\left (\frac{\sqrt{a} \sin (c+d x) \sqrt{\sec (c+d x)}}{\sqrt{2} \sqrt{a \sec (c+d x)+a}}\right )}{2 \sqrt{2} a^{3/2} d}+\frac{(9 A-5 B) \sin (c+d x) \cos ^{\frac{3}{2}}(c+d x)}{10 a d \sqrt{a \sec (c+d x)+a}}-\frac{(A-B) \sin (c+d x) \cos ^{\frac{3}{2}}(c+d x)}{2 d (a \sec (c+d x)+a)^{3/2}}-\frac{(39 A-35 B) \sin (c+d x) \sqrt{\cos (c+d x)}}{30 a d \sqrt{a \sec (c+d x)+a}}+\frac{(147 A-95 B) \sin (c+d x)}{30 a d \sqrt{\cos (c+d x)} \sqrt{a \sec (c+d x)+a}} \]

Antiderivative was successfully verified.

[In]

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

[Out]

-((15*A - 11*B)*ArcTanh[(Sqrt[a]*Sqrt[Sec[c + d*x]]*Sin[c + d*x])/(Sqrt[2]*Sqrt[a + a*Sec[c + d*x]])]*Sqrt[Cos
[c + d*x]]*Sqrt[Sec[c + d*x]])/(2*Sqrt[2]*a^(3/2)*d) - ((A - B)*Cos[c + d*x]^(3/2)*Sin[c + d*x])/(2*d*(a + a*S
ec[c + d*x])^(3/2)) + ((147*A - 95*B)*Sin[c + d*x])/(30*a*d*Sqrt[Cos[c + d*x]]*Sqrt[a + a*Sec[c + d*x]]) - ((3
9*A - 35*B)*Sqrt[Cos[c + d*x]]*Sin[c + d*x])/(30*a*d*Sqrt[a + a*Sec[c + d*x]]) + ((9*A - 5*B)*Cos[c + d*x]^(3/
2)*Sin[c + d*x])/(10*a*d*Sqrt[a + a*Sec[c + d*x]])

Rule 2955

Int[((a_.) + csc[(e_.) + (f_.)*(x_)]*(b_.))^(m_.)*(csc[(e_.) + (f_.)*(x_)]*(d_.) + (c_))^(n_.)*((g_.)*sin[(e_.
) + (f_.)*(x_)])^(p_.), x_Symbol] :> Dist[(g*Csc[e + f*x])^p*(g*Sin[e + f*x])^p, Int[((a + b*Csc[e + f*x])^m*(
c + d*Csc[e + f*x])^n)/(g*Csc[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 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{5}{2}}(c+d x) (A+B \sec (c+d x))}{(a+a \sec (c+d x))^{3/2}} \, dx &=\left (\sqrt{\cos (c+d x)} \sqrt{\sec (c+d x)}\right ) \int \frac{A+B \sec (c+d x)}{\sec ^{\frac{5}{2}}(c+d x) (a+a \sec (c+d x))^{3/2}} \, dx\\ &=-\frac{(A-B) \cos ^{\frac{3}{2}}(c+d x) \sin (c+d x)}{2 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}{2} a (9 A-5 B)-3 a (A-B) \sec (c+d x)}{\sec ^{\frac{5}{2}}(c+d x) \sqrt{a+a \sec (c+d x)}} \, dx}{2 a^2}\\ &=-\frac{(A-B) \cos ^{\frac{3}{2}}(c+d x) \sin (c+d x)}{2 d (a+a \sec (c+d x))^{3/2}}+\frac{(9 A-5 B) \cos ^{\frac{3}{2}}(c+d x) \sin (c+d x)}{10 a 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}{4} a^2 (39 A-35 B)+a^2 (9 A-5 B) \sec (c+d x)}{\sec ^{\frac{3}{2}}(c+d x) \sqrt{a+a \sec (c+d x)}} \, dx}{5 a^3}\\ &=-\frac{(A-B) \cos ^{\frac{3}{2}}(c+d x) \sin (c+d x)}{2 d (a+a \sec (c+d x))^{3/2}}-\frac{(39 A-35 B) \sqrt{\cos (c+d x)} \sin (c+d x)}{30 a d \sqrt{a+a \sec (c+d x)}}+\frac{(9 A-5 B) \cos ^{\frac{3}{2}}(c+d x) \sin (c+d x)}{10 a d \sqrt{a+a \sec (c+d x)}}+\frac{\left (2 \sqrt{\cos (c+d x)} \sqrt{\sec (c+d x)}\right ) \int \frac{\frac{1}{8} a^3 (147 A-95 B)-\frac{1}{4} a^3 (39 A-35 B) \sec (c+d x)}{\sqrt{\sec (c+d x)} \sqrt{a+a \sec (c+d x)}} \, dx}{15 a^4}\\ &=-\frac{(A-B) \cos ^{\frac{3}{2}}(c+d x) \sin (c+d x)}{2 d (a+a \sec (c+d x))^{3/2}}+\frac{(147 A-95 B) \sin (c+d x)}{30 a d \sqrt{\cos (c+d x)} \sqrt{a+a \sec (c+d x)}}-\frac{(39 A-35 B) \sqrt{\cos (c+d x)} \sin (c+d x)}{30 a d \sqrt{a+a \sec (c+d x)}}+\frac{(9 A-5 B) \cos ^{\frac{3}{2}}(c+d x) \sin (c+d x)}{10 a d \sqrt{a+a \sec (c+d x)}}-\frac{\left ((15 A-11 B) \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}{4 a}\\ &=-\frac{(A-B) \cos ^{\frac{3}{2}}(c+d x) \sin (c+d x)}{2 d (a+a \sec (c+d x))^{3/2}}+\frac{(147 A-95 B) \sin (c+d x)}{30 a d \sqrt{\cos (c+d x)} \sqrt{a+a \sec (c+d x)}}-\frac{(39 A-35 B) \sqrt{\cos (c+d x)} \sin (c+d x)}{30 a d \sqrt{a+a \sec (c+d x)}}+\frac{(9 A-5 B) \cos ^{\frac{3}{2}}(c+d x) \sin (c+d x)}{10 a d \sqrt{a+a \sec (c+d x)}}+\frac{\left ((15 A-11 B) \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 )}{2 a d}\\ &=-\frac{(15 A-11 B) \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)}}{2 \sqrt{2} a^{3/2} d}-\frac{(A-B) \cos ^{\frac{3}{2}}(c+d x) \sin (c+d x)}{2 d (a+a \sec (c+d x))^{3/2}}+\frac{(147 A-95 B) \sin (c+d x)}{30 a d \sqrt{\cos (c+d x)} \sqrt{a+a \sec (c+d x)}}-\frac{(39 A-35 B) \sqrt{\cos (c+d x)} \sin (c+d x)}{30 a d \sqrt{a+a \sec (c+d x)}}+\frac{(9 A-5 B) \cos ^{\frac{3}{2}}(c+d x) \sin (c+d x)}{10 a d \sqrt{a+a \sec (c+d x)}}\\ \end{align*}

Mathematica [A]  time = 1.29232, size = 178, normalized size = 0.66 \[ \frac{2 \tan (c+d x) \sqrt{1-\sec (c+d x)} (3 (39 A-20 B) \cos (c+d x)+(10 B-6 A) \cos (2 (c+d x))+3 A \cos (3 (c+d x))+141 A-85 B)+30 \sqrt{2} (15 A-11 B) \sin (c+d x) \cos ^2\left (\frac{1}{2} (c+d x)\right ) \sec ^{\frac{3}{2}}(c+d x) \tan ^{-1}\left (\frac{\sqrt{2} \sqrt{\sec (c+d x)}}{\sqrt{1-\sec (c+d x)}}\right )}{60 d \sqrt{\cos (c+d x)-1} (a (\sec (c+d x)+1))^{3/2}} \]

Warning: Unable to verify antiderivative.

[In]

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

[Out]

(30*Sqrt[2]*(15*A - 11*B)*ArcTan[(Sqrt[2]*Sqrt[Sec[c + d*x]])/Sqrt[1 - Sec[c + d*x]]]*Cos[(c + d*x)/2]^2*Sec[c
+ d*x]^(3/2)*Sin[c + d*x] + 2*(141*A - 85*B + 3*(39*A - 20*B)*Cos[c + d*x] + (-6*A + 10*B)*Cos[2*(c + d*x)] +
3*A*Cos[3*(c + d*x)])*Sqrt[1 - Sec[c + d*x]]*Tan[c + d*x])/(60*d*Sqrt[-1 + Cos[c + d*x]]*(a*(1 + Sec[c + d*x]
))^(3/2))

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Maple [A]  time = 0.403, size = 329, normalized size = 1.2 \begin{align*} -{\frac{-1+\cos \left ( dx+c \right ) }{60\,d \left ( \sin \left ( dx+c \right ) \right ) ^{3}{a}^{2}}\sqrt{\cos \left ( dx+c \right ) }\sqrt{{\frac{a \left ( \cos \left ( dx+c \right ) +1 \right ) }{\cos \left ( dx+c \right ) }}} \left ( 225\,A\sin \left ( dx+c \right ) \cos \left ( dx+c \right ) \arctan \left ( 1/2\,\sin \left ( dx+c \right ) \sqrt{-2\, \left ( \cos \left ( dx+c \right ) +1 \right ) ^{-1}} \right ) \sqrt{-2\, \left ( \cos \left ( dx+c \right ) +1 \right ) ^{-1}}-24\,A \left ( \cos \left ( dx+c \right ) \right ) ^{4}-165\,B\sin \left ( dx+c \right ) \cos \left ( dx+c \right ) \arctan \left ( 1/2\,\sin \left ( dx+c \right ) \sqrt{-2\, \left ( \cos \left ( dx+c \right ) +1 \right ) ^{-1}} \right ) \sqrt{-2\, \left ( \cos \left ( dx+c \right ) +1 \right ) ^{-1}}+225\,\arctan \left ( 1/2\,\sin \left ( dx+c \right ) \sqrt{-2\, \left ( \cos \left ( dx+c \right ) +1 \right ) ^{-1}} \right ) \sqrt{-2\, \left ( \cos \left ( dx+c \right ) +1 \right ) ^{-1}}A\sin \left ( dx+c \right ) +48\,A \left ( \cos \left ( dx+c \right ) \right ) ^{3}-165\,\arctan \left ( 1/2\,\sin \left ( dx+c \right ) \sqrt{-2\, \left ( \cos \left ( dx+c \right ) +1 \right ) ^{-1}} \right ) \sqrt{-2\, \left ( \cos \left ( dx+c \right ) +1 \right ) ^{-1}}B\sin \left ( dx+c \right ) -40\,B \left ( \cos \left ( dx+c \right ) \right ) ^{3}-240\,A \left ( \cos \left ( dx+c \right ) \right ) ^{2}+160\,B \left ( \cos \left ( dx+c \right ) \right ) ^{2}-78\,A\cos \left ( dx+c \right ) +70\,B\cos \left ( dx+c \right ) +294\,A-190\,B \right ) } \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

-1/60/d*cos(d*x+c)^(1/2)*(a*(cos(d*x+c)+1)/cos(d*x+c))^(1/2)*(-1+cos(d*x+c))*(225*A*sin(d*x+c)*cos(d*x+c)*arct
an(1/2*sin(d*x+c)*(-2/(cos(d*x+c)+1))^(1/2))*(-2/(cos(d*x+c)+1))^(1/2)-24*A*cos(d*x+c)^4-165*B*sin(d*x+c)*cos(
d*x+c)*arctan(1/2*sin(d*x+c)*(-2/(cos(d*x+c)+1))^(1/2))*(-2/(cos(d*x+c)+1))^(1/2)+225*arctan(1/2*sin(d*x+c)*(-
2/(cos(d*x+c)+1))^(1/2))*(-2/(cos(d*x+c)+1))^(1/2)*A*sin(d*x+c)+48*A*cos(d*x+c)^3-165*arctan(1/2*sin(d*x+c)*(-
2/(cos(d*x+c)+1))^(1/2))*(-2/(cos(d*x+c)+1))^(1/2)*B*sin(d*x+c)-40*B*cos(d*x+c)^3-240*A*cos(d*x+c)^2+160*B*cos
(d*x+c)^2-78*A*cos(d*x+c)+70*B*cos(d*x+c)+294*A-190*B)/sin(d*x+c)^3/a^2

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Maxima [F(-2)]  time = 0., size = 0, normalized size = 0. \begin{align*} \text{Exception raised: RuntimeError} \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

Exception raised: RuntimeError

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Fricas [A]  time = 0.547027, size = 1283, normalized size = 4.75 \begin{align*} \left [-\frac{15 \, \sqrt{2}{\left ({\left (15 \, A - 11 \, B\right )} \cos \left (d x + c\right )^{2} + 2 \,{\left (15 \, A - 11 \, B\right )} \cos \left (d x + c\right ) + 15 \, A - 11 \, B\right )} \sqrt{a} \log \left (-\frac{a \cos \left (d x + c\right )^{2} - 2 \, \sqrt{2} \sqrt{a} \sqrt{\frac{a \cos \left (d x + c\right ) + a}{\cos \left (d x + c\right )}} \sqrt{\cos \left (d x + c\right )} \sin \left (d x + c\right ) - 2 \, a \cos \left (d x + c\right ) - 3 \, a}{\cos \left (d x + c\right )^{2} + 2 \, \cos \left (d x + c\right ) + 1}\right ) - 4 \,{\left (12 \, A \cos \left (d x + c\right )^{3} - 4 \,{\left (3 \, A - 5 \, B\right )} \cos \left (d x + c\right )^{2} + 12 \,{\left (9 \, A - 5 \, B\right )} \cos \left (d x + c\right ) + 147 \, A - 95 \, B\right )} \sqrt{\frac{a \cos \left (d x + c\right ) + a}{\cos \left (d x + c\right )}} \sqrt{\cos \left (d x + c\right )} \sin \left (d x + c\right )}{120 \,{\left (a^{2} d \cos \left (d x + c\right )^{2} + 2 \, a^{2} d \cos \left (d x + c\right ) + a^{2} d\right )}}, \frac{15 \, \sqrt{2}{\left ({\left (15 \, A - 11 \, B\right )} \cos \left (d x + c\right )^{2} + 2 \,{\left (15 \, A - 11 \, B\right )} \cos \left (d x + c\right ) + 15 \, A - 11 \, B\right )} \sqrt{-a} \arctan \left (\frac{\sqrt{2} \sqrt{-a} \sqrt{\frac{a \cos \left (d x + c\right ) + a}{\cos \left (d x + c\right )}} \sqrt{\cos \left (d x + c\right )}}{a \sin \left (d x + c\right )}\right ) + 2 \,{\left (12 \, A \cos \left (d x + c\right )^{3} - 4 \,{\left (3 \, A - 5 \, B\right )} \cos \left (d x + c\right )^{2} + 12 \,{\left (9 \, A - 5 \, B\right )} \cos \left (d x + c\right ) + 147 \, A - 95 \, B\right )} \sqrt{\frac{a \cos \left (d x + c\right ) + a}{\cos \left (d x + c\right )}} \sqrt{\cos \left (d x + c\right )} \sin \left (d x + c\right )}{60 \,{\left (a^{2} d \cos \left (d x + c\right )^{2} + 2 \, a^{2} d \cos \left (d x + c\right ) + a^{2} d\right )}}\right ] \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

[-1/120*(15*sqrt(2)*((15*A - 11*B)*cos(d*x + c)^2 + 2*(15*A - 11*B)*cos(d*x + c) + 15*A - 11*B)*sqrt(a)*log(-(
a*cos(d*x + c)^2 - 2*sqrt(2)*sqrt(a)*sqrt((a*cos(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*cos(d*x + c) + 1)) - 4*(12*A*cos(d*x + c)^3 - 4*(3*A - 5*B)*cos(d
*x + c)^2 + 12*(9*A - 5*B)*cos(d*x + c) + 147*A - 95*B)*sqrt((a*cos(d*x + c) + a)/cos(d*x + c))*sqrt(cos(d*x +
c))*sin(d*x + c))/(a^2*d*cos(d*x + c)^2 + 2*a^2*d*cos(d*x + c) + a^2*d), 1/60*(15*sqrt(2)*((15*A - 11*B)*cos(
d*x + c)^2 + 2*(15*A - 11*B)*cos(d*x + c) + 15*A - 11*B)*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*(12*A*cos(d*x + c)^3 - 4*(3*A - 5*B)*cos(d*x + c)
^2 + 12*(9*A - 5*B)*cos(d*x + c) + 147*A - 95*B)*sqrt((a*cos(d*x + c) + a)/cos(d*x + c))*sqrt(cos(d*x + c))*si
n(d*x + c))/(a^2*d*cos(d*x + c)^2 + 2*a^2*d*cos(d*x + c) + a^2*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)**(5/2)*(A+B*sec(d*x+c))/(a+a*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{{\left (B \sec \left (d x + c\right ) + A\right )} \cos \left (d x + c\right )^{\frac{5}{2}}}{{\left (a \sec \left (d x + c\right ) + a\right )}^{\frac{3}{2}}}\,{d x} \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

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