3.119 \(\int (a+a \cos (c+d x))^{5/2} \sec ^4(c+d x) \, dx\)

Optimal. Leaf size=144 \[ \frac{25 a^3 \tan (c+d x)}{8 d \sqrt{a \cos (c+d x)+a}}+\frac{25 a^{5/2} \tanh ^{-1}\left (\frac{\sqrt{a} \sin (c+d x)}{\sqrt{a \cos (c+d x)+a}}\right )}{8 d}+\frac{a^2 \tan (c+d x) \sec ^2(c+d x) \sqrt{a \cos (c+d x)+a}}{3 d}+\frac{13 a^3 \tan (c+d x) \sec (c+d x)}{12 d \sqrt{a \cos (c+d x)+a}} \]

[Out]

(25*a^(5/2)*ArcTanh[(Sqrt[a]*Sin[c + d*x])/Sqrt[a + a*Cos[c + d*x]]])/(8*d) + (25*a^3*Tan[c + d*x])/(8*d*Sqrt[
a + a*Cos[c + d*x]]) + (13*a^3*Sec[c + d*x]*Tan[c + d*x])/(12*d*Sqrt[a + a*Cos[c + d*x]]) + (a^2*Sqrt[a + a*Co
s[c + d*x]]*Sec[c + d*x]^2*Tan[c + d*x])/(3*d)

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Rubi [A]  time = 0.282293, antiderivative size = 144, normalized size of antiderivative = 1., number of steps used = 5, number of rules used = 5, integrand size = 23, \(\frac{\text{number of rules}}{\text{integrand size}}\) = 0.217, Rules used = {2762, 2980, 2772, 2773, 206} \[ \frac{25 a^3 \tan (c+d x)}{8 d \sqrt{a \cos (c+d x)+a}}+\frac{25 a^{5/2} \tanh ^{-1}\left (\frac{\sqrt{a} \sin (c+d x)}{\sqrt{a \cos (c+d x)+a}}\right )}{8 d}+\frac{a^2 \tan (c+d x) \sec ^2(c+d x) \sqrt{a \cos (c+d x)+a}}{3 d}+\frac{13 a^3 \tan (c+d x) \sec (c+d x)}{12 d \sqrt{a \cos (c+d x)+a}} \]

Antiderivative was successfully verified.

[In]

Int[(a + a*Cos[c + d*x])^(5/2)*Sec[c + d*x]^4,x]

[Out]

(25*a^(5/2)*ArcTanh[(Sqrt[a]*Sin[c + d*x])/Sqrt[a + a*Cos[c + d*x]]])/(8*d) + (25*a^3*Tan[c + d*x])/(8*d*Sqrt[
a + a*Cos[c + d*x]]) + (13*a^3*Sec[c + d*x]*Tan[c + d*x])/(12*d*Sqrt[a + a*Cos[c + d*x]]) + (a^2*Sqrt[a + a*Co
s[c + d*x]]*Sec[c + d*x]^2*Tan[c + d*x])/(3*d)

Rule 2762

Int[((a_) + (b_.)*sin[(e_.) + (f_.)*(x_)])^(m_)*((c_.) + (d_.)*sin[(e_.) + (f_.)*(x_)])^(n_), x_Symbol] :> -Si
mp[(b^2*(b*c - a*d)*Cos[e + f*x]*(a + b*Sin[e + f*x])^(m - 2)*(c + d*Sin[e + f*x])^(n + 1))/(d*f*(n + 1)*(b*c
+ a*d)), x] + Dist[b^2/(d*(n + 1)*(b*c + a*d)), Int[(a + b*Sin[e + f*x])^(m - 2)*(c + d*Sin[e + f*x])^(n + 1)*
Simp[a*c*(m - 2) - b*d*(m - 2*n - 4) - (b*c*(m - 1) - a*d*(m + 2*n + 1))*Sin[e + f*x], x], 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] && GtQ[m, 1] && LtQ[n, -1]
&& (IntegersQ[2*m, 2*n] || IntegerQ[m + 1/2] || (IntegerQ[m] && EqQ[c, 0]))

Rule 2980

Int[Sqrt[(a_) + (b_.)*sin[(e_.) + (f_.)*(x_)]]*((A_.) + (B_.)*sin[(e_.) + (f_.)*(x_)])*((c_.) + (d_.)*sin[(e_.
) + (f_.)*(x_)])^(n_), x_Symbol] :> -Simp[(b^2*(B*c - A*d)*Cos[e + f*x]*(c + d*Sin[e + f*x])^(n + 1))/(d*f*(n
+ 1)*(b*c + a*d)*Sqrt[a + b*Sin[e + f*x]]), x] + Dist[(A*b*d*(2*n + 3) - B*(b*c - 2*a*d*(n + 1)))/(2*d*(n + 1)
*(b*c + a*d)), Int[Sqrt[a + b*Sin[e + f*x]]*(c + d*Sin[e + f*x])^(n + 1), x], x] /; FreeQ[{a, b, c, d, e, f, A
, B}, x] && NeQ[b*c - a*d, 0] && EqQ[a^2 - b^2, 0] && NeQ[c^2 - d^2, 0] && LtQ[n, -1]

Rule 2772

Int[Sqrt[(a_) + (b_.)*sin[(e_.) + (f_.)*(x_)]]*((c_.) + (d_.)*sin[(e_.) + (f_.)*(x_)])^(n_), x_Symbol] :> Simp
[((b*c - a*d)*Cos[e + f*x]*(c + d*Sin[e + f*x])^(n + 1))/(f*(n + 1)*(c^2 - d^2)*Sqrt[a + b*Sin[e + f*x]]), x]
+ Dist[((2*n + 3)*(b*c - a*d))/(2*b*(n + 1)*(c^2 - d^2)), Int[Sqrt[a + b*Sin[e + f*x]]*(c + d*Sin[e + f*x])^(n
+ 1), 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] &
& LtQ[n, -1] && NeQ[2*n + 3, 0] && IntegerQ[2*n]

Rule 2773

Int[Sqrt[(a_) + (b_.)*sin[(e_.) + (f_.)*(x_)]]/((c_.) + (d_.)*sin[(e_.) + (f_.)*(x_)]), x_Symbol] :> Dist[(-2*
b)/f, Subst[Int[1/(b*c + a*d - d*x^2), x], x, (b*Cos[e + f*x])/Sqrt[a + b*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 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 (a+a \cos (c+d x))^{5/2} \sec ^4(c+d x) \, dx &=\frac{a^2 \sqrt{a+a \cos (c+d x)} \sec ^2(c+d x) \tan (c+d x)}{3 d}-\frac{1}{3} a \int \left (-\frac{13 a}{2}-\frac{9}{2} a \cos (c+d x)\right ) \sqrt{a+a \cos (c+d x)} \sec ^3(c+d x) \, dx\\ &=\frac{13 a^3 \sec (c+d x) \tan (c+d x)}{12 d \sqrt{a+a \cos (c+d x)}}+\frac{a^2 \sqrt{a+a \cos (c+d x)} \sec ^2(c+d x) \tan (c+d x)}{3 d}+\frac{1}{8} \left (25 a^2\right ) \int \sqrt{a+a \cos (c+d x)} \sec ^2(c+d x) \, dx\\ &=\frac{25 a^3 \tan (c+d x)}{8 d \sqrt{a+a \cos (c+d x)}}+\frac{13 a^3 \sec (c+d x) \tan (c+d x)}{12 d \sqrt{a+a \cos (c+d x)}}+\frac{a^2 \sqrt{a+a \cos (c+d x)} \sec ^2(c+d x) \tan (c+d x)}{3 d}+\frac{1}{16} \left (25 a^2\right ) \int \sqrt{a+a \cos (c+d x)} \sec (c+d x) \, dx\\ &=\frac{25 a^3 \tan (c+d x)}{8 d \sqrt{a+a \cos (c+d x)}}+\frac{13 a^3 \sec (c+d x) \tan (c+d x)}{12 d \sqrt{a+a \cos (c+d x)}}+\frac{a^2 \sqrt{a+a \cos (c+d x)} \sec ^2(c+d x) \tan (c+d x)}{3 d}-\frac{\left (25 a^3\right ) \operatorname{Subst}\left (\int \frac{1}{a-x^2} \, dx,x,-\frac{a \sin (c+d x)}{\sqrt{a+a \cos (c+d x)}}\right )}{8 d}\\ &=\frac{25 a^{5/2} \tanh ^{-1}\left (\frac{\sqrt{a} \sin (c+d x)}{\sqrt{a+a \cos (c+d x)}}\right )}{8 d}+\frac{25 a^3 \tan (c+d x)}{8 d \sqrt{a+a \cos (c+d x)}}+\frac{13 a^3 \sec (c+d x) \tan (c+d x)}{12 d \sqrt{a+a \cos (c+d x)}}+\frac{a^2 \sqrt{a+a \cos (c+d x)} \sec ^2(c+d x) \tan (c+d x)}{3 d}\\ \end{align*}

Mathematica [C]  time = 33.5679, size = 1825, normalized size = 12.67 \[ \text{result too large to display} \]

Warning: Unable to verify antiderivative.

[In]

Integrate[(a + a*Cos[c + d*x])^(5/2)*Sec[c + d*x]^4,x]

[Out]

((-25/256 + (25*I)/256)*(1 + E^(I*c))*(Sqrt[2] - (1 - I)*E^((I/2)*c) + (16 - 16*I)*E^(((3*I)/2)*c + I*d*x) + (
20 + 20*I)*Sqrt[2]*E^((2*I)*c + ((3*I)/2)*d*x) - (34 - 34*I)*E^(((5*I)/2)*c + (2*I)*d*x) - (20 + 20*I)*Sqrt[2]
*E^((3*I)*c + ((5*I)/2)*d*x) + (16 - 16*I)*E^(((7*I)/2)*c + (3*I)*d*x) + (4 + 4*I)*Sqrt[2]*E^((4*I)*c + ((7*I)
/2)*d*x) - (1 - I)*E^(((9*I)/2)*c + (4*I)*d*x) + (8*I)*E^((I/2)*(c + d*x)) - 16*Sqrt[2]*E^(I*(c + d*x)) - (40*
I)*E^(((3*I)/2)*(c + d*x)) + 34*Sqrt[2]*E^((2*I)*(c + d*x)) + (40*I)*E^(((5*I)/2)*(c + d*x)) - 16*Sqrt[2]*E^((
3*I)*(c + d*x)) - (8*I)*E^(((7*I)/2)*(c + d*x)) + Sqrt[2]*E^((4*I)*(c + d*x)) - (4 + 4*I)*Sqrt[2]*E^((I/2)*(2*
c + d*x)))*x*(a*(1 + Cos[c + d*x]))^(5/2)*Sec[c/2 + (d*x)/2]^5)/(((-1 - I) + Sqrt[2]*E^((I/2)*c))*(-1 + E^(I*c
))*(I - 2*Sqrt[2]*E^((I/2)*(c + d*x)) - (4*I)*E^(I*(c + d*x)) + 2*Sqrt[2]*E^(((3*I)/2)*(c + d*x)) + I*E^((2*I)
*(c + d*x)))^2) - (((25*I)/64)*ArcTan[(Cos[c/4 + (d*x)/4] - Sin[c/4 + (d*x)/4] - Sqrt[2]*Sin[c/4 + (d*x)/4])/(
-Cos[c/4 + (d*x)/4] + Sqrt[2]*Cos[c/4 + (d*x)/4] - Sin[c/4 + (d*x)/4])]*(a*(1 + Cos[c + d*x]))^(5/2)*Sec[c/2 +
(d*x)/2]^5)/(Sqrt[2]*d) - (((25*I)/64)*ArcTan[(Cos[c/4 + (d*x)/4] + Sin[c/4 + (d*x)/4] - Sqrt[2]*Sin[c/4 + (d
*x)/4])/(Cos[c/4 + (d*x)/4] + Sqrt[2]*Cos[c/4 + (d*x)/4] - Sin[c/4 + (d*x)/4])]*(a*(1 + Cos[c + d*x]))^(5/2)*S
ec[c/2 + (d*x)/2]^5)/(Sqrt[2]*d) - (25*(a*(1 + Cos[c + d*x]))^(5/2)*Log[2 - Sqrt[2]*Cos[c/2 + (d*x)/2] - Sqrt[
2]*Sin[c/2 + (d*x)/2]]*Sec[c/2 + (d*x)/2]^5)/(128*Sqrt[2]*d) - (25*(a*(1 + Cos[c + d*x]))^(5/2)*Log[2 + Sqrt[2
]*Cos[c/2 + (d*x)/2] - Sqrt[2]*Sin[c/2 + (d*x)/2]]*Sec[c/2 + (d*x)/2]^5)/(128*Sqrt[2]*d) - (((25*I)/32)*ArcTan
[((2*I)*Cos[c/2] - I*(-Sqrt[2] + 2*Sin[c/2])*Tan[(d*x)/4])/Sqrt[-2 + 4*Cos[c/2]^2 + 4*Sin[c/2]^2]]*(a*(1 + Cos
[c + d*x]))^(5/2)*Cot[c/2]*Sec[c/2 + (d*x)/2]^5)/(d*Sqrt[-2 + 4*Cos[c/2]^2 + 4*Sin[c/2]^2]) + (25*(a*(1 + Cos[
c + d*x]))^(5/2)*Csc[c/2]*Sec[c/2 + (d*x)/2]^5*(-(d*x*Cos[c/2]) + 2*Log[Sqrt[2] + 2*Cos[(d*x)/2]*Sin[c/2] + 2*
Cos[c/2]*Sin[(d*x)/2]]*Sin[c/2] + ((4*I)*Sqrt[2]*ArcTan[((2*I)*Cos[c/2] - I*(-Sqrt[2] + 2*Sin[c/2])*Tan[(d*x)/
4])/Sqrt[-2 + 4*Cos[c/2]^2 + 4*Sin[c/2]^2]]*Cos[c/2])/Sqrt[-2 + 4*Cos[c/2]^2 + 4*Sin[c/2]^2]))/(32*Sqrt[2]*d*(
4*Cos[c/2]^2 + 4*Sin[c/2]^2)) + ((a*(1 + Cos[c + d*x]))^(5/2)*Sec[c/2 + (d*x)/2]^5)/(48*d*(Cos[c/2 + (d*x)/2]
- Sin[c/2 + (d*x)/2])^3) + (5*(a*(1 + Cos[c + d*x]))^(5/2)*Sec[c/2 + (d*x)/2]^5*Sin[(d*x)/2])/(32*d*(Cos[c/2]
- Sin[c/2])*(Cos[c/2 + (d*x)/2] - Sin[c/2 + (d*x)/2])^2) + (5*(a*(1 + Cos[c + d*x]))^(5/2)*Sec[c/2 + (d*x)/2]^
5*(5*Cos[c/2] - 3*Sin[c/2]))/(64*d*(Cos[c/2] - Sin[c/2])*(Cos[c/2 + (d*x)/2] - Sin[c/2 + (d*x)/2])) - ((a*(1 +
Cos[c + d*x]))^(5/2)*Sec[c/2 + (d*x)/2]^5)/(48*d*(Cos[c/2 + (d*x)/2] + Sin[c/2 + (d*x)/2])^3) + (5*(a*(1 + Co
s[c + d*x]))^(5/2)*Sec[c/2 + (d*x)/2]^5*Sin[(d*x)/2])/(32*d*(Cos[c/2] + Sin[c/2])*(Cos[c/2 + (d*x)/2] + Sin[c/
2 + (d*x)/2])^2) - (5*(a*(1 + Cos[c + d*x]))^(5/2)*Sec[c/2 + (d*x)/2]^5*(5*Cos[c/2] + 3*Sin[c/2]))/(64*d*(Cos[
c/2] + Sin[c/2])*(Cos[c/2 + (d*x)/2] + Sin[c/2 + (d*x)/2]))

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Maple [B]  time = 2.53, size = 709, normalized size = 4.9 \begin{align*} \text{result too large to display} \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

int((a+cos(d*x+c)*a)^(5/2)*sec(d*x+c)^4,x)

[Out]

1/6*a^(3/2)*cos(1/2*d*x+1/2*c)*(a*sin(1/2*d*x+1/2*c)^2)^(1/2)*(-600*a*(ln(-4/(-2*cos(1/2*d*x+1/2*c)+2^(1/2))*(
a^(1/2)*2^(1/2)*(a*sin(1/2*d*x+1/2*c)^2)^(1/2)-a*2^(1/2)*cos(1/2*d*x+1/2*c)+2*a))+ln(4/(2*cos(1/2*d*x+1/2*c)+2
^(1/2))*(a^(1/2)*2^(1/2)*(a*sin(1/2*d*x+1/2*c)^2)^(1/2)+a*2^(1/2)*cos(1/2*d*x+1/2*c)+2*a)))*sin(1/2*d*x+1/2*c)
^6+300*(2*a^(1/2)*2^(1/2)*(a*sin(1/2*d*x+1/2*c)^2)^(1/2)+3*ln(4/(2*cos(1/2*d*x+1/2*c)+2^(1/2))*(a^(1/2)*2^(1/2
)*(a*sin(1/2*d*x+1/2*c)^2)^(1/2)+a*2^(1/2)*cos(1/2*d*x+1/2*c)+2*a))*a+3*ln(-4/(-2*cos(1/2*d*x+1/2*c)+2^(1/2))*
(a^(1/2)*2^(1/2)*(a*sin(1/2*d*x+1/2*c)^2)^(1/2)-a*2^(1/2)*cos(1/2*d*x+1/2*c)+2*a))*a)*sin(1/2*d*x+1/2*c)^4+(-4
50*ln(4/(2*cos(1/2*d*x+1/2*c)+2^(1/2))*(a^(1/2)*2^(1/2)*(a*sin(1/2*d*x+1/2*c)^2)^(1/2)+a*2^(1/2)*cos(1/2*d*x+1
/2*c)+2*a))*a-450*ln(-4/(-2*cos(1/2*d*x+1/2*c)+2^(1/2))*(a^(1/2)*2^(1/2)*(a*sin(1/2*d*x+1/2*c)^2)^(1/2)-a*2^(1
/2)*cos(1/2*d*x+1/2*c)+2*a))*a-736*a^(1/2)*2^(1/2)*(a*sin(1/2*d*x+1/2*c)^2)^(1/2))*sin(1/2*d*x+1/2*c)^2+75*ln(
4/(2*cos(1/2*d*x+1/2*c)+2^(1/2))*(a^(1/2)*2^(1/2)*(a*sin(1/2*d*x+1/2*c)^2)^(1/2)+a*2^(1/2)*cos(1/2*d*x+1/2*c)+
2*a))*a+75*ln(-4/(-2*cos(1/2*d*x+1/2*c)+2^(1/2))*(a^(1/2)*2^(1/2)*(a*sin(1/2*d*x+1/2*c)^2)^(1/2)-a*2^(1/2)*cos
(1/2*d*x+1/2*c)+2*a))*a+234*a^(1/2)*2^(1/2)*(a*sin(1/2*d*x+1/2*c)^2)^(1/2))/(2*cos(1/2*d*x+1/2*c)+2^(1/2))^3/(
2*cos(1/2*d*x+1/2*c)-2^(1/2))^3/sin(1/2*d*x+1/2*c)/(cos(1/2*d*x+1/2*c)^2*a)^(1/2)/d

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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+a*cos(d*x+c))^(5/2)*sec(d*x+c)^4,x, algorithm="maxima")

[Out]

Timed out

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Fricas [A]  time = 1.73438, size = 471, normalized size = 3.27 \begin{align*} \frac{75 \,{\left (a^{2} \cos \left (d x + c\right )^{4} + a^{2} \cos \left (d x + c\right )^{3}\right )} \sqrt{a} \log \left (\frac{a \cos \left (d x + c\right )^{3} - 7 \, a \cos \left (d x + c\right )^{2} - 4 \, \sqrt{a \cos \left (d x + c\right ) + a} \sqrt{a}{\left (\cos \left (d x + c\right ) - 2\right )} \sin \left (d x + c\right ) + 8 \, a}{\cos \left (d x + c\right )^{3} + \cos \left (d x + c\right )^{2}}\right ) + 4 \,{\left (75 \, a^{2} \cos \left (d x + c\right )^{2} + 34 \, a^{2} \cos \left (d x + c\right ) + 8 \, a^{2}\right )} \sqrt{a \cos \left (d x + c\right ) + a} \sin \left (d x + c\right )}{96 \,{\left (d \cos \left (d x + c\right )^{4} + d \cos \left (d x + c\right )^{3}\right )}} \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate((a+a*cos(d*x+c))^(5/2)*sec(d*x+c)^4,x, algorithm="fricas")

[Out]

1/96*(75*(a^2*cos(d*x + c)^4 + a^2*cos(d*x + c)^3)*sqrt(a)*log((a*cos(d*x + c)^3 - 7*a*cos(d*x + c)^2 - 4*sqrt
(a*cos(d*x + c) + a)*sqrt(a)*(cos(d*x + c) - 2)*sin(d*x + c) + 8*a)/(cos(d*x + c)^3 + cos(d*x + c)^2)) + 4*(75
*a^2*cos(d*x + c)^2 + 34*a^2*cos(d*x + c) + 8*a^2)*sqrt(a*cos(d*x + c) + a)*sin(d*x + c))/(d*cos(d*x + c)^4 +
d*cos(d*x + c)^3)

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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+a*cos(d*x+c))**(5/2)*sec(d*x+c)**4,x)

[Out]

Timed out

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Giac [B]  time = 3.45762, size = 541, normalized size = 3.76 \begin{align*} \frac{75 \, a^{\frac{5}{2}} \log \left ({\left |{\left (\sqrt{a} \tan \left (\frac{1}{2} \, d x + \frac{1}{2} \, c\right ) - \sqrt{a \tan \left (\frac{1}{2} \, d x + \frac{1}{2} \, c\right )^{2} + a}\right )}^{2} - a{\left (2 \, \sqrt{2} + 3\right )} \right |}\right ) - 75 \, a^{\frac{5}{2}} \log \left ({\left |{\left (\sqrt{a} \tan \left (\frac{1}{2} \, d x + \frac{1}{2} \, c\right ) - \sqrt{a \tan \left (\frac{1}{2} \, d x + \frac{1}{2} \, c\right )^{2} + a}\right )}^{2} + a{\left (2 \, \sqrt{2} - 3\right )} \right |}\right ) + \frac{4 \, \sqrt{2}{\left (75 \,{\left (\sqrt{a} \tan \left (\frac{1}{2} \, d x + \frac{1}{2} \, c\right ) - \sqrt{a \tan \left (\frac{1}{2} \, d x + \frac{1}{2} \, c\right )^{2} + a}\right )}^{10} a^{\frac{7}{2}} - 1125 \,{\left (\sqrt{a} \tan \left (\frac{1}{2} \, d x + \frac{1}{2} \, c\right ) - \sqrt{a \tan \left (\frac{1}{2} \, d x + \frac{1}{2} \, c\right )^{2} + a}\right )}^{8} a^{\frac{9}{2}} + 6174 \,{\left (\sqrt{a} \tan \left (\frac{1}{2} \, d x + \frac{1}{2} \, c\right ) - \sqrt{a \tan \left (\frac{1}{2} \, d x + \frac{1}{2} \, c\right )^{2} + a}\right )}^{6} a^{\frac{11}{2}} - 4314 \,{\left (\sqrt{a} \tan \left (\frac{1}{2} \, d x + \frac{1}{2} \, c\right ) - \sqrt{a \tan \left (\frac{1}{2} \, d x + \frac{1}{2} \, c\right )^{2} + a}\right )}^{4} a^{\frac{13}{2}} + 807 \,{\left (\sqrt{a} \tan \left (\frac{1}{2} \, d x + \frac{1}{2} \, c\right ) - \sqrt{a \tan \left (\frac{1}{2} \, d x + \frac{1}{2} \, c\right )^{2} + a}\right )}^{2} a^{\frac{15}{2}} - 49 \, a^{\frac{17}{2}}\right )}}{{\left ({\left (\sqrt{a} \tan \left (\frac{1}{2} \, d x + \frac{1}{2} \, c\right ) - \sqrt{a \tan \left (\frac{1}{2} \, d x + \frac{1}{2} \, c\right )^{2} + a}\right )}^{4} - 6 \,{\left (\sqrt{a} \tan \left (\frac{1}{2} \, d x + \frac{1}{2} \, c\right ) - \sqrt{a \tan \left (\frac{1}{2} \, d x + \frac{1}{2} \, c\right )^{2} + a}\right )}^{2} a + a^{2}\right )}^{3}}}{48 \, d} \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate((a+a*cos(d*x+c))^(5/2)*sec(d*x+c)^4,x, algorithm="giac")

[Out]

1/48*(75*a^(5/2)*log(abs((sqrt(a)*tan(1/2*d*x + 1/2*c) - sqrt(a*tan(1/2*d*x + 1/2*c)^2 + a))^2 - a*(2*sqrt(2)
+ 3))) - 75*a^(5/2)*log(abs((sqrt(a)*tan(1/2*d*x + 1/2*c) - sqrt(a*tan(1/2*d*x + 1/2*c)^2 + a))^2 + a*(2*sqrt(
2) - 3))) + 4*sqrt(2)*(75*(sqrt(a)*tan(1/2*d*x + 1/2*c) - sqrt(a*tan(1/2*d*x + 1/2*c)^2 + a))^10*a^(7/2) - 112
5*(sqrt(a)*tan(1/2*d*x + 1/2*c) - sqrt(a*tan(1/2*d*x + 1/2*c)^2 + a))^8*a^(9/2) + 6174*(sqrt(a)*tan(1/2*d*x +
1/2*c) - sqrt(a*tan(1/2*d*x + 1/2*c)^2 + a))^6*a^(11/2) - 4314*(sqrt(a)*tan(1/2*d*x + 1/2*c) - sqrt(a*tan(1/2*
d*x + 1/2*c)^2 + a))^4*a^(13/2) + 807*(sqrt(a)*tan(1/2*d*x + 1/2*c) - sqrt(a*tan(1/2*d*x + 1/2*c)^2 + a))^2*a^
(15/2) - 49*a^(17/2))/((sqrt(a)*tan(1/2*d*x + 1/2*c) - sqrt(a*tan(1/2*d*x + 1/2*c)^2 + a))^4 - 6*(sqrt(a)*tan(
1/2*d*x + 1/2*c) - sqrt(a*tan(1/2*d*x + 1/2*c)^2 + a))^2*a + a^2)^3)/d