3.1280 \(\int \frac{\cos ^{\frac{5}{2}}(c+d x) (A+B \sec (c+d x)+C \sec ^2(c+d x))}{(a+a \sec (c+d x))^{3/2}} \, dx\)

Optimal. Leaf size=283 \[ -\frac{(15 A-11 B+7 C) \sqrt{\cos (c+d x)} \sqrt{\sec (c+d x)} \tanh ^{-1}\left (\frac{\sqrt{a} \sin (c+d x) \sqrt{\sec (c+d x)}}{\sqrt{2} \sqrt{a \sec (c+d x)+a}}\right )}{2 \sqrt{2} a^{3/2} d}+\frac{(9 A-5 B+5 C) \sin (c+d x) \cos ^{\frac{3}{2}}(c+d x)}{10 a d \sqrt{a \sec (c+d x)+a}}-\frac{(A-B+C) \sin (c+d x) \cos ^{\frac{3}{2}}(c+d x)}{2 d (a \sec (c+d x)+a)^{3/2}}-\frac{(39 A-35 B+15 C) \sin (c+d x) \sqrt{\cos (c+d x)}}{30 a d \sqrt{a \sec (c+d x)+a}}+\frac{(147 A-95 B+75 C) \sin (c+d x)}{30 a d \sqrt{\cos (c+d x)} \sqrt{a \sec (c+d x)+a}} \]

[Out]

-((15*A - 11*B + 7*C)*ArcTanh[(Sqrt[a]*Sqrt[Sec[c + d*x]]*Sin[c + d*x])/(Sqrt[2]*Sqrt[a + a*Sec[c + d*x]])]*Sq
rt[Cos[c + d*x]]*Sqrt[Sec[c + d*x]])/(2*Sqrt[2]*a^(3/2)*d) - ((A - B + C)*Cos[c + d*x]^(3/2)*Sin[c + d*x])/(2*
d*(a + a*Sec[c + d*x])^(3/2)) + ((147*A - 95*B + 75*C)*Sin[c + d*x])/(30*a*d*Sqrt[Cos[c + d*x]]*Sqrt[a + a*Sec
[c + d*x]]) - ((39*A - 35*B + 15*C)*Sqrt[Cos[c + d*x]]*Sin[c + d*x])/(30*a*d*Sqrt[a + a*Sec[c + d*x]]) + ((9*A
- 5*B + 5*C)*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 = 0.917505, antiderivative size = 283, normalized size of antiderivative = 1., number of steps used = 7, number of rules used = 6, integrand size = 45, \(\frac{\text{number of rules}}{\text{integrand size}}\) = 0.133, Rules used = {4265, 4084, 4022, 4013, 3808, 206} \[ -\frac{(15 A-11 B+7 C) \sqrt{\cos (c+d x)} \sqrt{\sec (c+d x)} \tanh ^{-1}\left (\frac{\sqrt{a} \sin (c+d x) \sqrt{\sec (c+d x)}}{\sqrt{2} \sqrt{a \sec (c+d x)+a}}\right )}{2 \sqrt{2} a^{3/2} d}+\frac{(9 A-5 B+5 C) \sin (c+d x) \cos ^{\frac{3}{2}}(c+d x)}{10 a d \sqrt{a \sec (c+d x)+a}}-\frac{(A-B+C) \sin (c+d x) \cos ^{\frac{3}{2}}(c+d x)}{2 d (a \sec (c+d x)+a)^{3/2}}-\frac{(39 A-35 B+15 C) \sin (c+d x) \sqrt{\cos (c+d x)}}{30 a d \sqrt{a \sec (c+d x)+a}}+\frac{(147 A-95 B+75 C) \sin (c+d x)}{30 a d \sqrt{\cos (c+d x)} \sqrt{a \sec (c+d x)+a}} \]

Antiderivative was successfully verified.

[In]

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

[Out]

-((15*A - 11*B + 7*C)*ArcTanh[(Sqrt[a]*Sqrt[Sec[c + d*x]]*Sin[c + d*x])/(Sqrt[2]*Sqrt[a + a*Sec[c + d*x]])]*Sq
rt[Cos[c + d*x]]*Sqrt[Sec[c + d*x]])/(2*Sqrt[2]*a^(3/2)*d) - ((A - B + C)*Cos[c + d*x]^(3/2)*Sin[c + d*x])/(2*
d*(a + a*Sec[c + d*x])^(3/2)) + ((147*A - 95*B + 75*C)*Sin[c + d*x])/(30*a*d*Sqrt[Cos[c + d*x]]*Sqrt[a + a*Sec
[c + d*x]]) - ((39*A - 35*B + 15*C)*Sqrt[Cos[c + d*x]]*Sin[c + d*x])/(30*a*d*Sqrt[a + a*Sec[c + d*x]]) + ((9*A
- 5*B + 5*C)*Cos[c + d*x]^(3/2)*Sin[c + d*x])/(10*a*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 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) \left (A+B \sec (c+d x)+C \sec ^2(c+d x)\right )}{(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)+C \sec ^2(c+d x)}{\sec ^{\frac{5}{2}}(c+d x) (a+a \sec (c+d x))^{3/2}} \, dx\\ &=-\frac{(A-B+C) \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+5 C)-a (3 A-3 B+C) \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+C) \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+5 C) \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+15 C)+a^2 (9 A-5 B+5 C) \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+C) \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+15 C) \sqrt{\cos (c+d x)} \sin (c+d x)}{30 a d \sqrt{a+a \sec (c+d x)}}+\frac{(9 A-5 B+5 C) \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+75 C)-\frac{1}{4} a^3 (39 A-35 B+15 C) \sec (c+d x)}{\sqrt{\sec (c+d x)} \sqrt{a+a \sec (c+d x)}} \, dx}{15 a^4}\\ &=-\frac{(A-B+C) \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+75 C) \sin (c+d x)}{30 a d \sqrt{\cos (c+d x)} \sqrt{a+a \sec (c+d x)}}-\frac{(39 A-35 B+15 C) \sqrt{\cos (c+d x)} \sin (c+d x)}{30 a d \sqrt{a+a \sec (c+d x)}}+\frac{(9 A-5 B+5 C) \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+7 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}{4 a}\\ &=-\frac{(A-B+C) \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+75 C) \sin (c+d x)}{30 a d \sqrt{\cos (c+d x)} \sqrt{a+a \sec (c+d x)}}-\frac{(39 A-35 B+15 C) \sqrt{\cos (c+d x)} \sin (c+d x)}{30 a d \sqrt{a+a \sec (c+d x)}}+\frac{(9 A-5 B+5 C) \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+7 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 )}{2 a d}\\ &=-\frac{(15 A-11 B+7 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)}}{2 \sqrt{2} a^{3/2} d}-\frac{(A-B+C) \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+75 C) \sin (c+d x)}{30 a d \sqrt{\cos (c+d x)} \sqrt{a+a \sec (c+d x)}}-\frac{(39 A-35 B+15 C) \sqrt{\cos (c+d x)} \sin (c+d x)}{30 a d \sqrt{a+a \sec (c+d x)}}+\frac{(9 A-5 B+5 C) \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 = 3.06527, size = 135, normalized size = 0.48 \[ \frac{\tan \left (\frac{1}{2} (c+d x)\right ) (3 (39 A+20 (C-B)) \cos (c+d x)+(10 B-6 A) \cos (2 (c+d x))+3 A \cos (3 (c+d x))+141 A-85 B+75 C)-15 (15 A-11 B+7 C) \cos \left (\frac{1}{2} (c+d x)\right ) \tanh ^{-1}\left (\sin \left (\frac{1}{2} (c+d x)\right )\right )}{30 a d \sqrt{\cos (c+d x)} \sqrt{a (\sec (c+d x)+1)}} \]

Antiderivative was successfully verified.

[In]

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

[Out]

(-15*(15*A - 11*B + 7*C)*ArcTanh[Sin[(c + d*x)/2]]*Cos[(c + d*x)/2] + (141*A - 85*B + 75*C + 3*(39*A + 20*(-B
+ C))*Cos[c + d*x] + (-6*A + 10*B)*Cos[2*(c + d*x)] + 3*A*Cos[3*(c + d*x)])*Tan[(c + d*x)/2])/(30*a*d*Sqrt[Cos
[c + d*x]]*Sqrt[a*(1 + Sec[c + d*x])])

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Maple [A]  time = 0.29, size = 450, normalized size = 1.6 \begin{align*} -{\frac{-1+\cos \left ( dx+c \right ) }{60\,d{a}^{2} \left ( \sin \left ( dx+c \right ) \right ) ^{3}}\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}}+105\,C\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}+105\,C\sqrt{-2\, \left ( \cos \left ( dx+c \right ) +1 \right ) ^{-1}}\arctan \left ( 1/2\,\sin \left ( dx+c \right ) \sqrt{-2\, \left ( \cos \left ( dx+c \right ) +1 \right ) ^{-1}} \right ) \sin \left ( dx+c \right ) -240\,A \left ( \cos \left ( dx+c \right ) \right ) ^{2}+160\,B \left ( \cos \left ( dx+c \right ) \right ) ^{2}-120\,C \left ( \cos \left ( dx+c \right ) \right ) ^{2}-78\,A\cos \left ( dx+c \right ) +70\,B\cos \left ( dx+c \right ) -30\,C\cos \left ( dx+c \right ) +294\,A-190\,B+150\,C \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)+C*sec(d*x+c)^2)/(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)+105*C*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+105*C*(-2/(cos(d*x+c)+1))^(1/2)*arc
tan(1/2*sin(d*x+c)*(-2/(cos(d*x+c)+1))^(1/2))*sin(d*x+c)-240*A*cos(d*x+c)^2+160*B*cos(d*x+c)^2-120*C*cos(d*x+c
)^2-78*A*cos(d*x+c)+70*B*cos(d*x+c)-30*C*cos(d*x+c)+294*A-190*B+150*C)/a^2/sin(d*x+c)^3

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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)+C*sec(d*x+c)^2)/(a+a*sec(d*x+c))^(3/2),x, algorithm="maxima")

[Out]

Exception raised: RuntimeError

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Fricas [A]  time = 0.565991, size = 1365, normalized size = 4.82 \begin{align*} \left [\frac{15 \, \sqrt{2}{\left ({\left (15 \, A - 11 \, B + 7 \, C\right )} \cos \left (d x + c\right )^{2} + 2 \,{\left (15 \, A - 11 \, B + 7 \, C\right )} \cos \left (d x + c\right ) + 15 \, A - 11 \, B + 7 \, C\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 + 5 \, C\right )} \cos \left (d x + c\right ) + 147 \, A - 95 \, B + 75 \, C\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 + 7 \, C\right )} \cos \left (d x + c\right )^{2} + 2 \,{\left (15 \, A - 11 \, B + 7 \, C\right )} \cos \left (d x + c\right ) + 15 \, A - 11 \, B + 7 \, C\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 + 5 \, C\right )} \cos \left (d x + c\right ) + 147 \, A - 95 \, B + 75 \, C\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)+C*sec(d*x+c)^2)/(a+a*sec(d*x+c))^(3/2),x, algorithm="fricas")

[Out]

[1/120*(15*sqrt(2)*((15*A - 11*B + 7*C)*cos(d*x + c)^2 + 2*(15*A - 11*B + 7*C)*cos(d*x + c) + 15*A - 11*B + 7*
C)*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 + 5*C)*cos(d*x + c) + 147*A - 95*B + 75*C)*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 + 7*C)*cos(d*x + c)^2 + 2*(15*A - 11*B + 7*C)*cos(d*x + c) + 15*A - 11*B + 7*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*(
12*A*cos(d*x + c)^3 - 4*(3*A - 5*B)*cos(d*x + c)^2 + 12*(9*A - 5*B + 5*C)*cos(d*x + c) + 147*A - 95*B + 75*C)*
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)]

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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)+C*sec(d*x+c)**2)/(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 (C \sec \left (d x + c\right )^{2} + 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)+C*sec(d*x+c)^2)/(a+a*sec(d*x+c))^(3/2),x, algorithm="giac")

[Out]

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