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

Optimal. Leaf size=160 \[ \frac {25 a^{5/2} \sinh ^{-1}\left (\frac {\sqrt {a} \tan (c+d x)}{\sqrt {a \sec (c+d x)+a}}\right )}{8 d}+\frac {13 a^3 \sin (c+d x) \sec ^{\frac {5}{2}}(c+d x)}{12 d \sqrt {a \sec (c+d x)+a}}+\frac {25 a^3 \sin (c+d x) \sec ^{\frac {3}{2}}(c+d x)}{8 d \sqrt {a \sec (c+d x)+a}}+\frac {a^2 \sin (c+d x) \sec ^{\frac {5}{2}}(c+d x) \sqrt {a \sec (c+d x)+a}}{3 d} \]

[Out]

25/8*a^(5/2)*arcsinh(a^(1/2)*tan(d*x+c)/(a+a*sec(d*x+c))^(1/2))/d+25/8*a^3*sec(d*x+c)^(3/2)*sin(d*x+c)/d/(a+a*
sec(d*x+c))^(1/2)+13/12*a^3*sec(d*x+c)^(5/2)*sin(d*x+c)/d/(a+a*sec(d*x+c))^(1/2)+1/3*a^2*sec(d*x+c)^(5/2)*sin(
d*x+c)*(a+a*sec(d*x+c))^(1/2)/d

________________________________________________________________________________________

Rubi [A]  time = 0.27, antiderivative size = 160, normalized size of antiderivative = 1.00, number of steps used = 5, number of rules used = 5, integrand size = 25, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.200, Rules used = {3814, 4016, 3803, 3801, 215} \[ \frac {13 a^3 \sin (c+d x) \sec ^{\frac {5}{2}}(c+d x)}{12 d \sqrt {a \sec (c+d x)+a}}+\frac {25 a^3 \sin (c+d x) \sec ^{\frac {3}{2}}(c+d x)}{8 d \sqrt {a \sec (c+d x)+a}}+\frac {a^2 \sin (c+d x) \sec ^{\frac {5}{2}}(c+d x) \sqrt {a \sec (c+d x)+a}}{3 d}+\frac {25 a^{5/2} \sinh ^{-1}\left (\frac {\sqrt {a} \tan (c+d x)}{\sqrt {a \sec (c+d x)+a}}\right )}{8 d} \]

Antiderivative was successfully verified.

[In]

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

[Out]

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

Rule 215

Int[1/Sqrt[(a_) + (b_.)*(x_)^2], x_Symbol] :> Simp[ArcSinh[(Rt[b, 2]*x)/Sqrt[a]]/Rt[b, 2], x] /; FreeQ[{a, b},
x] && GtQ[a, 0] && PosQ[b]

Rule 3801

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

Rule 3803

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

Rule 3814

Int[(csc[(e_.) + (f_.)*(x_)]*(d_.))^(n_)*(csc[(e_.) + (f_.)*(x_)]*(b_.) + (a_))^(m_), x_Symbol] :> -Simp[(b^2*
Cot[e + f*x]*(a + b*Csc[e + f*x])^(m - 2)*(d*Csc[e + f*x])^n)/(f*(m + n - 1)), x] + Dist[b/(m + n - 1), Int[(a
+ b*Csc[e + f*x])^(m - 2)*(d*Csc[e + f*x])^n*(b*(m + 2*n - 1) + a*(3*m + 2*n - 4)*Csc[e + f*x]), x], x] /; Fr
eeQ[{a, b, d, e, f, n}, x] && EqQ[a^2 - b^2, 0] && GtQ[m, 1] && NeQ[m + n - 1, 0] && IntegerQ[2*m]

Rule 4016

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

Rubi steps

\begin {align*} \int \sec ^{\frac {3}{2}}(c+d x) (a+a \sec (c+d x))^{5/2} \, dx &=\frac {a^2 \sec ^{\frac {5}{2}}(c+d x) \sqrt {a+a \sec (c+d x)} \sin (c+d x)}{3 d}+\frac {1}{3} a \int \sec ^{\frac {3}{2}}(c+d x) \sqrt {a+a \sec (c+d x)} \left (\frac {9 a}{2}+\frac {13}{2} a \sec (c+d x)\right ) \, dx\\ &=\frac {13 a^3 \sec ^{\frac {5}{2}}(c+d x) \sin (c+d x)}{12 d \sqrt {a+a \sec (c+d x)}}+\frac {a^2 \sec ^{\frac {5}{2}}(c+d x) \sqrt {a+a \sec (c+d x)} \sin (c+d x)}{3 d}+\frac {1}{8} \left (25 a^2\right ) \int \sec ^{\frac {3}{2}}(c+d x) \sqrt {a+a \sec (c+d x)} \, dx\\ &=\frac {25 a^3 \sec ^{\frac {3}{2}}(c+d x) \sin (c+d x)}{8 d \sqrt {a+a \sec (c+d x)}}+\frac {13 a^3 \sec ^{\frac {5}{2}}(c+d x) \sin (c+d x)}{12 d \sqrt {a+a \sec (c+d x)}}+\frac {a^2 \sec ^{\frac {5}{2}}(c+d x) \sqrt {a+a \sec (c+d x)} \sin (c+d x)}{3 d}+\frac {1}{16} \left (25 a^2\right ) \int \sqrt {\sec (c+d x)} \sqrt {a+a \sec (c+d x)} \, dx\\ &=\frac {25 a^3 \sec ^{\frac {3}{2}}(c+d x) \sin (c+d x)}{8 d \sqrt {a+a \sec (c+d x)}}+\frac {13 a^3 \sec ^{\frac {5}{2}}(c+d x) \sin (c+d x)}{12 d \sqrt {a+a \sec (c+d x)}}+\frac {a^2 \sec ^{\frac {5}{2}}(c+d x) \sqrt {a+a \sec (c+d x)} \sin (c+d x)}{3 d}-\frac {\left (25 a^2\right ) \operatorname {Subst}\left (\int \frac {1}{\sqrt {1+\frac {x^2}{a}}} \, dx,x,-\frac {a \tan (c+d x)}{\sqrt {a+a \sec (c+d x)}}\right )}{8 d}\\ &=\frac {25 a^{5/2} \sinh ^{-1}\left (\frac {\sqrt {a} \tan (c+d x)}{\sqrt {a+a \sec (c+d x)}}\right )}{8 d}+\frac {25 a^3 \sec ^{\frac {3}{2}}(c+d x) \sin (c+d x)}{8 d \sqrt {a+a \sec (c+d x)}}+\frac {13 a^3 \sec ^{\frac {5}{2}}(c+d x) \sin (c+d x)}{12 d \sqrt {a+a \sec (c+d x)}}+\frac {a^2 \sec ^{\frac {5}{2}}(c+d x) \sqrt {a+a \sec (c+d x)} \sin (c+d x)}{3 d}\\ \end {align*}

________________________________________________________________________________________

Mathematica [C]  time = 8.06, size = 458, normalized size = 2.86 \[ \frac {a^2 \sec \left (\frac {1}{2} (c+d x)\right ) \sqrt {a (\sec (c+d x)+1)} \left (-600 i \tan ^{-1}\left (\frac {\cos \left (\frac {1}{4} (c+d x)\right )-\left (\sqrt {2}-1\right ) \sin \left (\frac {1}{4} (c+d x)\right )}{\left (1+\sqrt {2}\right ) \cos \left (\frac {1}{4} (c+d x)\right )-\sin \left (\frac {1}{4} (c+d x)\right )}\right )-600 i \tan ^{-1}\left (\frac {\cos \left (\frac {1}{4} (c+d x)\right )-\left (1+\sqrt {2}\right ) \sin \left (\frac {1}{4} (c+d x)\right )}{\left (\sqrt {2}-1\right ) \cos \left (\frac {1}{4} (c+d x)\right )-\sin \left (\frac {1}{4} (c+d x)\right )}\right )+\sec ^3(c+d x) \left (4 \sqrt {2} \left (114 \sin \left (\frac {1}{2} (c+d x)\right )-7 \sin \left (\frac {3}{2} (c+d x)\right )+75 \sin \left (\frac {5}{2} (c+d x)\right )\right )+225 \cos (c+d x) \left (2 \log \left (2 \sin \left (\frac {1}{2} (c+d x)\right )+\sqrt {2}\right )-\log \left (-\sqrt {2} \sin \left (\frac {1}{2} (c+d x)\right )-\sqrt {2} \cos \left (\frac {1}{2} (c+d x)\right )+2\right )-\log \left (-\sqrt {2} \sin \left (\frac {1}{2} (c+d x)\right )+\sqrt {2} \cos \left (\frac {1}{2} (c+d x)\right )+2\right )\right )+75 \cos (3 (c+d x)) \left (2 \log \left (2 \sin \left (\frac {1}{2} (c+d x)\right )+\sqrt {2}\right )-\log \left (-\sqrt {2} \sin \left (\frac {1}{2} (c+d x)\right )-\sqrt {2} \cos \left (\frac {1}{2} (c+d x)\right )+2\right )-\log \left (-\sqrt {2} \sin \left (\frac {1}{2} (c+d x)\right )+\sqrt {2} \cos \left (\frac {1}{2} (c+d x)\right )+2\right )\right )\right )\right )}{384 \sqrt {2} d \sqrt {\sec (c+d x)}} \]

Antiderivative was successfully verified.

[In]

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

[Out]

(a^2*Sec[(c + d*x)/2]*Sqrt[a*(1 + Sec[c + d*x])]*((-600*I)*ArcTan[(Cos[(c + d*x)/4] - (-1 + Sqrt[2])*Sin[(c +
d*x)/4])/((1 + Sqrt[2])*Cos[(c + d*x)/4] - Sin[(c + d*x)/4])] - (600*I)*ArcTan[(Cos[(c + d*x)/4] - (1 + Sqrt[2
])*Sin[(c + d*x)/4])/((-1 + Sqrt[2])*Cos[(c + d*x)/4] - Sin[(c + d*x)/4])] + Sec[c + d*x]^3*(225*Cos[c + d*x]*
(2*Log[Sqrt[2] + 2*Sin[(c + d*x)/2]] - Log[2 - Sqrt[2]*Cos[(c + d*x)/2] - Sqrt[2]*Sin[(c + d*x)/2]] - Log[2 +
Sqrt[2]*Cos[(c + d*x)/2] - Sqrt[2]*Sin[(c + d*x)/2]]) + 75*Cos[3*(c + d*x)]*(2*Log[Sqrt[2] + 2*Sin[(c + d*x)/2
]] - Log[2 - Sqrt[2]*Cos[(c + d*x)/2] - Sqrt[2]*Sin[(c + d*x)/2]] - Log[2 + Sqrt[2]*Cos[(c + d*x)/2] - Sqrt[2]
*Sin[(c + d*x)/2]]) + 4*Sqrt[2]*(114*Sin[(c + d*x)/2] - 7*Sin[(3*(c + d*x))/2] + 75*Sin[(5*(c + d*x))/2]))))/(
384*Sqrt[2]*d*Sqrt[Sec[c + d*x]])

________________________________________________________________________________________

fricas [A]  time = 1.60, size = 420, normalized size = 2.62 \[ \left [\frac {75 \, {\left (a^{2} \cos \left (d x + c\right )^{3} + a^{2} \cos \left (d x + c\right )^{2}\right )} \sqrt {a} \log \left (\frac {a \cos \left (d x + c\right )^{3} - 7 \, a \cos \left (d x + c\right )^{2} - \frac {4 \, {\left (\cos \left (d x + c\right )^{2} - 2 \, \cos \left (d x + c\right )\right )} \sqrt {a} \sqrt {\frac {a \cos \left (d x + c\right ) + a}{\cos \left (d x + c\right )}} \sin \left (d x + c\right )}{\sqrt {\cos \left (d x + c\right )}} + 8 \, a}{\cos \left (d x + c\right )^{3} + \cos \left (d x + c\right )^{2}}\right ) + \frac {4 \, {\left (75 \, a^{2} \cos \left (d x + c\right )^{2} + 34 \, a^{2} \cos \left (d x + c\right ) + 8 \, a^{2}\right )} \sqrt {\frac {a \cos \left (d x + c\right ) + a}{\cos \left (d x + c\right )}} \sin \left (d x + c\right )}{\sqrt {\cos \left (d x + c\right )}}}{96 \, {\left (d \cos \left (d x + c\right )^{3} + d \cos \left (d x + c\right )^{2}\right )}}, \frac {75 \, {\left (a^{2} \cos \left (d x + c\right )^{3} + a^{2} \cos \left (d x + c\right )^{2}\right )} \sqrt {-a} \arctan \left (\frac {2 \, \sqrt {-a} \sqrt {\frac {a \cos \left (d x + c\right ) + a}{\cos \left (d x + c\right )}} \sqrt {\cos \left (d x + c\right )} \sin \left (d x + c\right )}{a \cos \left (d x + c\right )^{2} - a \cos \left (d x + c\right ) - 2 \, a}\right ) + \frac {2 \, {\left (75 \, a^{2} \cos \left (d x + c\right )^{2} + 34 \, a^{2} \cos \left (d x + c\right ) + 8 \, a^{2}\right )} \sqrt {\frac {a \cos \left (d x + c\right ) + a}{\cos \left (d x + c\right )}} \sin \left (d x + c\right )}{\sqrt {\cos \left (d x + c\right )}}}{48 \, {\left (d \cos \left (d x + c\right )^{3} + d \cos \left (d x + c\right )^{2}\right )}}\right ] \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

[1/96*(75*(a^2*cos(d*x + c)^3 + a^2*cos(d*x + c)^2)*sqrt(a)*log((a*cos(d*x + c)^3 - 7*a*cos(d*x + c)^2 - 4*(co
s(d*x + c)^2 - 2*cos(d*x + c))*sqrt(a)*sqrt((a*cos(d*x + c) + a)/cos(d*x + c))*sin(d*x + c)/sqrt(cos(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*c
os(d*x + c) + a)/cos(d*x + c))*sin(d*x + c)/sqrt(cos(d*x + c)))/(d*cos(d*x + c)^3 + d*cos(d*x + c)^2), 1/48*(7
5*(a^2*cos(d*x + c)^3 + a^2*cos(d*x + c)^2)*sqrt(-a)*arctan(2*sqrt(-a)*sqrt((a*cos(d*x + c) + a)/cos(d*x + c))
*sqrt(cos(d*x + c))*sin(d*x + c)/(a*cos(d*x + c)^2 - a*cos(d*x + c) - 2*a)) + 2*(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)/cos(d*x + c))*sin(d*x + c)/sqrt(cos(d*x + c)))/(d*cos(d*x +
c)^3 + d*cos(d*x + c)^2)]

________________________________________________________________________________________

giac [F]  time = 0.00, size = 0, normalized size = 0.00 \[ \int {\left (a \sec \left (d x + c\right ) + a\right )}^{\frac {5}{2}} \sec \left (d x + c\right )^{\frac {3}{2}}\,{d x} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

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

________________________________________________________________________________________

maple [A]  time = 1.65, size = 254, normalized size = 1.59 \[ -\frac {\left (-1+\cos \left (d x +c \right )\right ) \left (75 \sqrt {2}\, \left (\cos ^{3}\left (d x +c \right )\right ) \arctan \left (\frac {\sqrt {-\frac {2}{1+\cos \left (d x +c \right )}}\, \left (\cos \left (d x +c \right )+1+\sin \left (d x +c \right )\right ) \sqrt {2}}{4}\right )-75 \sqrt {2}\, \left (\cos ^{3}\left (d x +c \right )\right ) \arctan \left (\frac {\sqrt {-\frac {2}{1+\cos \left (d x +c \right )}}\, \left (\cos \left (d x +c \right )+1-\sin \left (d x +c \right )\right ) \sqrt {2}}{4}\right )+150 \left (\cos ^{2}\left (d x +c \right )\right ) \sin \left (d x +c \right ) \sqrt {-\frac {2}{1+\cos \left (d x +c \right )}}+68 \sqrt {-\frac {2}{1+\cos \left (d x +c \right )}}\, \cos \left (d x +c \right ) \sin \left (d x +c \right )+16 \sin \left (d x +c \right ) \sqrt {-\frac {2}{1+\cos \left (d x +c \right )}}\right ) \sqrt {\frac {a \left (1+\cos \left (d x +c \right )\right )}{\cos \left (d x +c \right )}}\, \left (\frac {1}{\cos \left (d x +c \right )}\right )^{\frac {3}{2}} a^{2}}{48 d \sqrt {-\frac {2}{1+\cos \left (d x +c \right )}}\, \cos \left (d x +c \right ) \sin \left (d x +c \right )^{2}} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

-1/48/d*(-1+cos(d*x+c))*(75*2^(1/2)*cos(d*x+c)^3*arctan(1/4*(-2/(1+cos(d*x+c)))^(1/2)*(cos(d*x+c)+1+sin(d*x+c)
)*2^(1/2))-75*2^(1/2)*cos(d*x+c)^3*arctan(1/4*(-2/(1+cos(d*x+c)))^(1/2)*(cos(d*x+c)+1-sin(d*x+c))*2^(1/2))+150
*cos(d*x+c)^2*sin(d*x+c)*(-2/(1+cos(d*x+c)))^(1/2)+68*(-2/(1+cos(d*x+c)))^(1/2)*cos(d*x+c)*sin(d*x+c)+16*sin(d
*x+c)*(-2/(1+cos(d*x+c)))^(1/2))*(a*(1+cos(d*x+c))/cos(d*x+c))^(1/2)*(1/cos(d*x+c))^(3/2)/(-2/(1+cos(d*x+c)))^
(1/2)/cos(d*x+c)/sin(d*x+c)^2*a^2

________________________________________________________________________________________

maxima [B]  time = 0.99, size = 3469, normalized size = 21.68 \[ \text {result too large to display} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

1/96*(300*sqrt(2)*a^2*cos(1/3*arctan2(sin(3/2*d*x + 3/2*c), cos(3/2*d*x + 3/2*c)))*sin(6*d*x + 6*c) - 28*sqrt(
2)*a^2*sin(9/2*d*x + 9/2*c) + 28*sqrt(2)*a^2*sin(3/2*d*x + 3/2*c) - 28*(sqrt(2)*a^2*sin(9/2*d*x + 9/2*c) - sqr
t(2)*a^2*sin(3/2*d*x + 3/2*c))*cos(6*d*x + 6*c) - 300*(sqrt(2)*a^2*sin(6*d*x + 6*c) + 3*sqrt(2)*a^2*sin(8/3*ar
ctan2(sin(3/2*d*x + 3/2*c), cos(3/2*d*x + 3/2*c))) + 3*sqrt(2)*a^2*sin(4/3*arctan2(sin(3/2*d*x + 3/2*c), cos(3
/2*d*x + 3/2*c))))*cos(11/3*arctan2(sin(3/2*d*x + 3/2*c), cos(3/2*d*x + 3/2*c))) - 12*(7*sqrt(2)*a^2*sin(9/2*d
*x + 9/2*c) - 7*sqrt(2)*a^2*sin(3/2*d*x + 3/2*c) - 114*sqrt(2)*a^2*sin(7/3*arctan2(sin(3/2*d*x + 3/2*c), cos(3
/2*d*x + 3/2*c))) + 114*sqrt(2)*a^2*sin(5/3*arctan2(sin(3/2*d*x + 3/2*c), cos(3/2*d*x + 3/2*c))) + 75*sqrt(2)*
a^2*sin(1/3*arctan2(sin(3/2*d*x + 3/2*c), cos(3/2*d*x + 3/2*c))))*cos(8/3*arctan2(sin(3/2*d*x + 3/2*c), cos(3/
2*d*x + 3/2*c))) - 456*(sqrt(2)*a^2*sin(6*d*x + 6*c) + 3*sqrt(2)*a^2*sin(4/3*arctan2(sin(3/2*d*x + 3/2*c), cos
(3/2*d*x + 3/2*c))))*cos(7/3*arctan2(sin(3/2*d*x + 3/2*c), cos(3/2*d*x + 3/2*c))) + 456*(sqrt(2)*a^2*sin(6*d*x
+ 6*c) + 3*sqrt(2)*a^2*sin(4/3*arctan2(sin(3/2*d*x + 3/2*c), cos(3/2*d*x + 3/2*c))))*cos(5/3*arctan2(sin(3/2*
d*x + 3/2*c), cos(3/2*d*x + 3/2*c))) - 12*(7*sqrt(2)*a^2*sin(9/2*d*x + 9/2*c) - 7*sqrt(2)*a^2*sin(3/2*d*x + 3/
2*c) + 75*sqrt(2)*a^2*sin(1/3*arctan2(sin(3/2*d*x + 3/2*c), cos(3/2*d*x + 3/2*c))))*cos(4/3*arctan2(sin(3/2*d*
x + 3/2*c), cos(3/2*d*x + 3/2*c))) + 75*(a^2*cos(6*d*x + 6*c)^2 + 9*a^2*cos(8/3*arctan2(sin(3/2*d*x + 3/2*c),
cos(3/2*d*x + 3/2*c)))^2 + 9*a^2*cos(4/3*arctan2(sin(3/2*d*x + 3/2*c), cos(3/2*d*x + 3/2*c)))^2 + a^2*sin(6*d*
x + 6*c)^2 + 9*a^2*sin(8/3*arctan2(sin(3/2*d*x + 3/2*c), cos(3/2*d*x + 3/2*c)))^2 + 6*a^2*sin(6*d*x + 6*c)*sin
(4/3*arctan2(sin(3/2*d*x + 3/2*c), cos(3/2*d*x + 3/2*c))) + 9*a^2*sin(4/3*arctan2(sin(3/2*d*x + 3/2*c), cos(3/
2*d*x + 3/2*c)))^2 + 2*a^2*cos(6*d*x + 6*c) + a^2 + 6*(a^2*cos(6*d*x + 6*c) + 3*a^2*cos(4/3*arctan2(sin(3/2*d*
x + 3/2*c), cos(3/2*d*x + 3/2*c))) + a^2)*cos(8/3*arctan2(sin(3/2*d*x + 3/2*c), cos(3/2*d*x + 3/2*c))) + 6*(a^
2*cos(6*d*x + 6*c) + a^2)*cos(4/3*arctan2(sin(3/2*d*x + 3/2*c), cos(3/2*d*x + 3/2*c))) + 6*(a^2*sin(6*d*x + 6*
c) + 3*a^2*sin(4/3*arctan2(sin(3/2*d*x + 3/2*c), cos(3/2*d*x + 3/2*c))))*sin(8/3*arctan2(sin(3/2*d*x + 3/2*c),
cos(3/2*d*x + 3/2*c))))*log(2*cos(1/3*arctan2(sin(3/2*d*x + 3/2*c), cos(3/2*d*x + 3/2*c)))^2 + 2*sin(1/3*arct
an2(sin(3/2*d*x + 3/2*c), cos(3/2*d*x + 3/2*c)))^2 + 2*sqrt(2)*cos(1/3*arctan2(sin(3/2*d*x + 3/2*c), cos(3/2*d
*x + 3/2*c))) + 2*sqrt(2)*sin(1/3*arctan2(sin(3/2*d*x + 3/2*c), cos(3/2*d*x + 3/2*c))) + 2) - 75*(a^2*cos(6*d*
x + 6*c)^2 + 9*a^2*cos(8/3*arctan2(sin(3/2*d*x + 3/2*c), cos(3/2*d*x + 3/2*c)))^2 + 9*a^2*cos(4/3*arctan2(sin(
3/2*d*x + 3/2*c), cos(3/2*d*x + 3/2*c)))^2 + a^2*sin(6*d*x + 6*c)^2 + 9*a^2*sin(8/3*arctan2(sin(3/2*d*x + 3/2*
c), cos(3/2*d*x + 3/2*c)))^2 + 6*a^2*sin(6*d*x + 6*c)*sin(4/3*arctan2(sin(3/2*d*x + 3/2*c), cos(3/2*d*x + 3/2*
c))) + 9*a^2*sin(4/3*arctan2(sin(3/2*d*x + 3/2*c), cos(3/2*d*x + 3/2*c)))^2 + 2*a^2*cos(6*d*x + 6*c) + a^2 + 6
*(a^2*cos(6*d*x + 6*c) + 3*a^2*cos(4/3*arctan2(sin(3/2*d*x + 3/2*c), cos(3/2*d*x + 3/2*c))) + a^2)*cos(8/3*arc
tan2(sin(3/2*d*x + 3/2*c), cos(3/2*d*x + 3/2*c))) + 6*(a^2*cos(6*d*x + 6*c) + a^2)*cos(4/3*arctan2(sin(3/2*d*x
+ 3/2*c), cos(3/2*d*x + 3/2*c))) + 6*(a^2*sin(6*d*x + 6*c) + 3*a^2*sin(4/3*arctan2(sin(3/2*d*x + 3/2*c), cos(
3/2*d*x + 3/2*c))))*sin(8/3*arctan2(sin(3/2*d*x + 3/2*c), cos(3/2*d*x + 3/2*c))))*log(2*cos(1/3*arctan2(sin(3/
2*d*x + 3/2*c), cos(3/2*d*x + 3/2*c)))^2 + 2*sin(1/3*arctan2(sin(3/2*d*x + 3/2*c), cos(3/2*d*x + 3/2*c)))^2 +
2*sqrt(2)*cos(1/3*arctan2(sin(3/2*d*x + 3/2*c), cos(3/2*d*x + 3/2*c))) - 2*sqrt(2)*sin(1/3*arctan2(sin(3/2*d*x
+ 3/2*c), cos(3/2*d*x + 3/2*c))) + 2) + 75*(a^2*cos(6*d*x + 6*c)^2 + 9*a^2*cos(8/3*arctan2(sin(3/2*d*x + 3/2*
c), cos(3/2*d*x + 3/2*c)))^2 + 9*a^2*cos(4/3*arctan2(sin(3/2*d*x + 3/2*c), cos(3/2*d*x + 3/2*c)))^2 + a^2*sin(
6*d*x + 6*c)^2 + 9*a^2*sin(8/3*arctan2(sin(3/2*d*x + 3/2*c), cos(3/2*d*x + 3/2*c)))^2 + 6*a^2*sin(6*d*x + 6*c)
*sin(4/3*arctan2(sin(3/2*d*x + 3/2*c), cos(3/2*d*x + 3/2*c))) + 9*a^2*sin(4/3*arctan2(sin(3/2*d*x + 3/2*c), co
s(3/2*d*x + 3/2*c)))^2 + 2*a^2*cos(6*d*x + 6*c) + a^2 + 6*(a^2*cos(6*d*x + 6*c) + 3*a^2*cos(4/3*arctan2(sin(3/
2*d*x + 3/2*c), cos(3/2*d*x + 3/2*c))) + a^2)*cos(8/3*arctan2(sin(3/2*d*x + 3/2*c), cos(3/2*d*x + 3/2*c))) + 6
*(a^2*cos(6*d*x + 6*c) + a^2)*cos(4/3*arctan2(sin(3/2*d*x + 3/2*c), cos(3/2*d*x + 3/2*c))) + 6*(a^2*sin(6*d*x
+ 6*c) + 3*a^2*sin(4/3*arctan2(sin(3/2*d*x + 3/2*c), cos(3/2*d*x + 3/2*c))))*sin(8/3*arctan2(sin(3/2*d*x + 3/2
*c), cos(3/2*d*x + 3/2*c))))*log(2*cos(1/3*arctan2(sin(3/2*d*x + 3/2*c), cos(3/2*d*x + 3/2*c)))^2 + 2*sin(1/3*
arctan2(sin(3/2*d*x + 3/2*c), cos(3/2*d*x + 3/2*c)))^2 - 2*sqrt(2)*cos(1/3*arctan2(sin(3/2*d*x + 3/2*c), cos(3
/2*d*x + 3/2*c))) + 2*sqrt(2)*sin(1/3*arctan2(sin(3/2*d*x + 3/2*c), cos(3/2*d*x + 3/2*c))) + 2) - 75*(a^2*cos(
6*d*x + 6*c)^2 + 9*a^2*cos(8/3*arctan2(sin(3/2*d*x + 3/2*c), cos(3/2*d*x + 3/2*c)))^2 + 9*a^2*cos(4/3*arctan2(
sin(3/2*d*x + 3/2*c), cos(3/2*d*x + 3/2*c)))^2 + a^2*sin(6*d*x + 6*c)^2 + 9*a^2*sin(8/3*arctan2(sin(3/2*d*x +
3/2*c), cos(3/2*d*x + 3/2*c)))^2 + 6*a^2*sin(6*d*x + 6*c)*sin(4/3*arctan2(sin(3/2*d*x + 3/2*c), cos(3/2*d*x +
3/2*c))) + 9*a^2*sin(4/3*arctan2(sin(3/2*d*x + 3/2*c), cos(3/2*d*x + 3/2*c)))^2 + 2*a^2*cos(6*d*x + 6*c) + a^2
+ 6*(a^2*cos(6*d*x + 6*c) + 3*a^2*cos(4/3*arctan2(sin(3/2*d*x + 3/2*c), cos(3/2*d*x + 3/2*c))) + a^2)*cos(8/3
*arctan2(sin(3/2*d*x + 3/2*c), cos(3/2*d*x + 3/2*c))) + 6*(a^2*cos(6*d*x + 6*c) + a^2)*cos(4/3*arctan2(sin(3/2
*d*x + 3/2*c), cos(3/2*d*x + 3/2*c))) + 6*(a^2*sin(6*d*x + 6*c) + 3*a^2*sin(4/3*arctan2(sin(3/2*d*x + 3/2*c),
cos(3/2*d*x + 3/2*c))))*sin(8/3*arctan2(sin(3/2*d*x + 3/2*c), cos(3/2*d*x + 3/2*c))))*log(2*cos(1/3*arctan2(si
n(3/2*d*x + 3/2*c), cos(3/2*d*x + 3/2*c)))^2 + 2*sin(1/3*arctan2(sin(3/2*d*x + 3/2*c), cos(3/2*d*x + 3/2*c)))^
2 - 2*sqrt(2)*cos(1/3*arctan2(sin(3/2*d*x + 3/2*c), cos(3/2*d*x + 3/2*c))) - 2*sqrt(2)*sin(1/3*arctan2(sin(3/2
*d*x + 3/2*c), cos(3/2*d*x + 3/2*c))) + 2) + 28*(sqrt(2)*a^2*cos(9/2*d*x + 9/2*c) - sqrt(2)*a^2*cos(3/2*d*x +
3/2*c))*sin(6*d*x + 6*c) + 300*(sqrt(2)*a^2*cos(6*d*x + 6*c) + 3*sqrt(2)*a^2*cos(8/3*arctan2(sin(3/2*d*x + 3/2
*c), cos(3/2*d*x + 3/2*c))) + 3*sqrt(2)*a^2*cos(4/3*arctan2(sin(3/2*d*x + 3/2*c), cos(3/2*d*x + 3/2*c))) + sqr
t(2)*a^2)*sin(11/3*arctan2(sin(3/2*d*x + 3/2*c), cos(3/2*d*x + 3/2*c))) + 12*(7*sqrt(2)*a^2*cos(9/2*d*x + 9/2*
c) - 7*sqrt(2)*a^2*cos(3/2*d*x + 3/2*c) - 114*sqrt(2)*a^2*cos(7/3*arctan2(sin(3/2*d*x + 3/2*c), cos(3/2*d*x +
3/2*c))) + 114*sqrt(2)*a^2*cos(5/3*arctan2(sin(3/2*d*x + 3/2*c), cos(3/2*d*x + 3/2*c))) + 75*sqrt(2)*a^2*cos(1
/3*arctan2(sin(3/2*d*x + 3/2*c), cos(3/2*d*x + 3/2*c))))*sin(8/3*arctan2(sin(3/2*d*x + 3/2*c), cos(3/2*d*x + 3
/2*c))) + 456*(sqrt(2)*a^2*cos(6*d*x + 6*c) + 3*sqrt(2)*a^2*cos(4/3*arctan2(sin(3/2*d*x + 3/2*c), cos(3/2*d*x
+ 3/2*c))) + sqrt(2)*a^2)*sin(7/3*arctan2(sin(3/2*d*x + 3/2*c), cos(3/2*d*x + 3/2*c))) - 456*(sqrt(2)*a^2*cos(
6*d*x + 6*c) + 3*sqrt(2)*a^2*cos(4/3*arctan2(sin(3/2*d*x + 3/2*c), cos(3/2*d*x + 3/2*c))) + sqrt(2)*a^2)*sin(5
/3*arctan2(sin(3/2*d*x + 3/2*c), cos(3/2*d*x + 3/2*c))) + 12*(7*sqrt(2)*a^2*cos(9/2*d*x + 9/2*c) - 7*sqrt(2)*a
^2*cos(3/2*d*x + 3/2*c) + 75*sqrt(2)*a^2*cos(1/3*arctan2(sin(3/2*d*x + 3/2*c), cos(3/2*d*x + 3/2*c))))*sin(4/3
*arctan2(sin(3/2*d*x + 3/2*c), cos(3/2*d*x + 3/2*c))) - 300*(sqrt(2)*a^2*cos(6*d*x + 6*c) + sqrt(2)*a^2)*sin(1
/3*arctan2(sin(3/2*d*x + 3/2*c), cos(3/2*d*x + 3/2*c))))*sqrt(a)/((cos(6*d*x + 6*c)^2 + 6*(cos(6*d*x + 6*c) +
3*cos(4/3*arctan2(sin(3/2*d*x + 3/2*c), cos(3/2*d*x + 3/2*c))) + 1)*cos(8/3*arctan2(sin(3/2*d*x + 3/2*c), cos(
3/2*d*x + 3/2*c))) + 9*cos(8/3*arctan2(sin(3/2*d*x + 3/2*c), cos(3/2*d*x + 3/2*c)))^2 + 6*(cos(6*d*x + 6*c) +
1)*cos(4/3*arctan2(sin(3/2*d*x + 3/2*c), cos(3/2*d*x + 3/2*c))) + 9*cos(4/3*arctan2(sin(3/2*d*x + 3/2*c), cos(
3/2*d*x + 3/2*c)))^2 + sin(6*d*x + 6*c)^2 + 6*(sin(6*d*x + 6*c) + 3*sin(4/3*arctan2(sin(3/2*d*x + 3/2*c), cos(
3/2*d*x + 3/2*c))))*sin(8/3*arctan2(sin(3/2*d*x + 3/2*c), cos(3/2*d*x + 3/2*c))) + 9*sin(8/3*arctan2(sin(3/2*d
*x + 3/2*c), cos(3/2*d*x + 3/2*c)))^2 + 6*sin(6*d*x + 6*c)*sin(4/3*arctan2(sin(3/2*d*x + 3/2*c), cos(3/2*d*x +
3/2*c))) + 9*sin(4/3*arctan2(sin(3/2*d*x + 3/2*c), cos(3/2*d*x + 3/2*c)))^2 + 2*cos(6*d*x + 6*c) + 1)*d)

________________________________________________________________________________________

mupad [F]  time = 0.00, size = -1, normalized size = -0.01 \[ \int {\left (a+\frac {a}{\cos \left (c+d\,x\right )}\right )}^{5/2}\,{\left (\frac {1}{\cos \left (c+d\,x\right )}\right )}^{3/2} \,d x \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

int((a + a/cos(c + d*x))^(5/2)*(1/cos(c + d*x))^(3/2),x)

[Out]

int((a + a/cos(c + d*x))^(5/2)*(1/cos(c + d*x))^(3/2), x)

________________________________________________________________________________________

sympy [F(-1)]  time = 0.00, size = 0, normalized size = 0.00 \[ \text {Timed out} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate(sec(d*x+c)**(3/2)*(a+a*sec(d*x+c))**(5/2),x)

[Out]

Timed out

________________________________________________________________________________________