3.3.30 \(\int \frac {1}{\cos ^{\frac {7}{2}}(c+d x) \sqrt {a+a \cos (c+d x)}} \, dx\) [230]

Optimal. Leaf size=169 \[ -\frac {\sqrt {2} \text {ArcTan}\left (\frac {\sqrt {a} \sin (c+d x)}{\sqrt {2} \sqrt {\cos (c+d x)} \sqrt {a+a \cos (c+d x)}}\right )}{\sqrt {a} d}+\frac {2 \sin (c+d x)}{5 d \cos ^{\frac {5}{2}}(c+d x) \sqrt {a+a \cos (c+d x)}}-\frac {2 \sin (c+d x)}{15 d \cos ^{\frac {3}{2}}(c+d x) \sqrt {a+a \cos (c+d x)}}+\frac {26 \sin (c+d x)}{15 d \sqrt {\cos (c+d x)} \sqrt {a+a \cos (c+d x)}} \]

[Out]

-arctan(1/2*sin(d*x+c)*a^(1/2)*2^(1/2)/cos(d*x+c)^(1/2)/(a+a*cos(d*x+c))^(1/2))*2^(1/2)/d/a^(1/2)+2/5*sin(d*x+
c)/d/cos(d*x+c)^(5/2)/(a+a*cos(d*x+c))^(1/2)-2/15*sin(d*x+c)/d/cos(d*x+c)^(3/2)/(a+a*cos(d*x+c))^(1/2)+26/15*s
in(d*x+c)/d/cos(d*x+c)^(1/2)/(a+a*cos(d*x+c))^(1/2)

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Rubi [A]
time = 0.24, antiderivative size = 169, normalized size of antiderivative = 1.00, number of steps used = 6, number of rules used = 5, integrand size = 25, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.200, Rules used = {2858, 3063, 12, 2861, 211} \begin {gather*} -\frac {\sqrt {2} \text {ArcTan}\left (\frac {\sqrt {a} \sin (c+d x)}{\sqrt {2} \sqrt {\cos (c+d x)} \sqrt {a \cos (c+d x)+a}}\right )}{\sqrt {a} d}-\frac {2 \sin (c+d x)}{15 d \cos ^{\frac {3}{2}}(c+d x) \sqrt {a \cos (c+d x)+a}}+\frac {2 \sin (c+d x)}{5 d \cos ^{\frac {5}{2}}(c+d x) \sqrt {a \cos (c+d x)+a}}+\frac {26 \sin (c+d x)}{15 d \sqrt {\cos (c+d x)} \sqrt {a \cos (c+d x)+a}} \end {gather*}

Antiderivative was successfully verified.

[In]

Int[1/(Cos[c + d*x]^(7/2)*Sqrt[a + a*Cos[c + d*x]]),x]

[Out]

-((Sqrt[2]*ArcTan[(Sqrt[a]*Sin[c + d*x])/(Sqrt[2]*Sqrt[Cos[c + d*x]]*Sqrt[a + a*Cos[c + d*x]])])/(Sqrt[a]*d))
+ (2*Sin[c + d*x])/(5*d*Cos[c + d*x]^(5/2)*Sqrt[a + a*Cos[c + d*x]]) - (2*Sin[c + d*x])/(15*d*Cos[c + d*x]^(3/
2)*Sqrt[a + a*Cos[c + d*x]]) + (26*Sin[c + d*x])/(15*d*Sqrt[Cos[c + d*x]]*Sqrt[a + a*Cos[c + d*x]])

Rule 12

Int[(a_)*(u_), x_Symbol] :> Dist[a, Int[u, x], x] /; FreeQ[a, x] &&  !MatchQ[u, (b_)*(v_) /; FreeQ[b, x]]

Rule 211

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

Rule 2858

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

Rule 2861

Int[1/(Sqrt[(a_) + (b_.)*sin[(e_.) + (f_.)*(x_)]]*Sqrt[(c_.) + (d_.)*sin[(e_.) + (f_.)*(x_)]]), x_Symbol] :> D
ist[-2*(a/f), Subst[Int[1/(2*b^2 - (a*c - b*d)*x^2), x], x, b*(Cos[e + f*x]/(Sqrt[a + b*Sin[e + f*x]]*Sqrt[c +
d*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 3063

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

Rubi steps

\begin {align*} \int \frac {1}{\cos ^{\frac {7}{2}}(c+d x) \sqrt {a+a \cos (c+d x)}} \, dx &=\frac {2 \sin (c+d x)}{5 d \cos ^{\frac {5}{2}}(c+d x) \sqrt {a+a \cos (c+d x)}}-\frac {\int \frac {a-4 a \cos (c+d x)}{\cos ^{\frac {5}{2}}(c+d x) \sqrt {a+a \cos (c+d x)}} \, dx}{5 a}\\ &=\frac {2 \sin (c+d x)}{5 d \cos ^{\frac {5}{2}}(c+d x) \sqrt {a+a \cos (c+d x)}}-\frac {2 \sin (c+d x)}{15 d \cos ^{\frac {3}{2}}(c+d x) \sqrt {a+a \cos (c+d x)}}-\frac {2 \int \frac {-\frac {13 a^2}{2}+a^2 \cos (c+d x)}{\cos ^{\frac {3}{2}}(c+d x) \sqrt {a+a \cos (c+d x)}} \, dx}{15 a^2}\\ &=\frac {2 \sin (c+d x)}{5 d \cos ^{\frac {5}{2}}(c+d x) \sqrt {a+a \cos (c+d x)}}-\frac {2 \sin (c+d x)}{15 d \cos ^{\frac {3}{2}}(c+d x) \sqrt {a+a \cos (c+d x)}}+\frac {26 \sin (c+d x)}{15 d \sqrt {\cos (c+d x)} \sqrt {a+a \cos (c+d x)}}-\frac {4 \int \frac {15 a^3}{4 \sqrt {\cos (c+d x)} \sqrt {a+a \cos (c+d x)}} \, dx}{15 a^3}\\ &=\frac {2 \sin (c+d x)}{5 d \cos ^{\frac {5}{2}}(c+d x) \sqrt {a+a \cos (c+d x)}}-\frac {2 \sin (c+d x)}{15 d \cos ^{\frac {3}{2}}(c+d x) \sqrt {a+a \cos (c+d x)}}+\frac {26 \sin (c+d x)}{15 d \sqrt {\cos (c+d x)} \sqrt {a+a \cos (c+d x)}}-\int \frac {1}{\sqrt {\cos (c+d x)} \sqrt {a+a \cos (c+d x)}} \, dx\\ &=\frac {2 \sin (c+d x)}{5 d \cos ^{\frac {5}{2}}(c+d x) \sqrt {a+a \cos (c+d x)}}-\frac {2 \sin (c+d x)}{15 d \cos ^{\frac {3}{2}}(c+d x) \sqrt {a+a \cos (c+d x)}}+\frac {26 \sin (c+d x)}{15 d \sqrt {\cos (c+d x)} \sqrt {a+a \cos (c+d x)}}+\frac {(2 a) \text {Subst}\left (\int \frac {1}{2 a^2+a x^2} \, dx,x,-\frac {a \sin (c+d x)}{\sqrt {\cos (c+d x)} \sqrt {a+a \cos (c+d x)}}\right )}{d}\\ &=-\frac {\sqrt {2} \tan ^{-1}\left (\frac {\sqrt {a} \sin (c+d x)}{\sqrt {2} \sqrt {\cos (c+d x)} \sqrt {a+a \cos (c+d x)}}\right )}{\sqrt {a} d}+\frac {2 \sin (c+d x)}{5 d \cos ^{\frac {5}{2}}(c+d x) \sqrt {a+a \cos (c+d x)}}-\frac {2 \sin (c+d x)}{15 d \cos ^{\frac {3}{2}}(c+d x) \sqrt {a+a \cos (c+d x)}}+\frac {26 \sin (c+d x)}{15 d \sqrt {\cos (c+d x)} \sqrt {a+a \cos (c+d x)}}\\ \end {align*}

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Mathematica [C] Result contains higher order function than in optimal. Order 5 vs. order 3 in optimal.
time = 10.18, size = 1540, normalized size = 9.11 \begin {gather*} -\frac {2 \cot \left (\frac {c}{2}+\frac {d x}{2}\right ) \csc ^6\left (\frac {c}{2}+\frac {d x}{2}\right ) \left (4725 \sin ^2\left (\frac {c}{2}+\frac {d x}{2}\right )-48825 \sin ^4\left (\frac {c}{2}+\frac {d x}{2}\right )+210105 \sin ^6\left (\frac {c}{2}+\frac {d x}{2}\right )-486630 \sin ^8\left (\frac {c}{2}+\frac {d x}{2}\right )+655812 \sin ^{10}\left (\frac {c}{2}+\frac {d x}{2}\right )-710 \, _2F_1\left (2,\frac {9}{2};\frac {11}{2};\frac {\sin ^2\left (\frac {c}{2}+\frac {d x}{2}\right )}{-1+2 \sin ^2\left (\frac {c}{2}+\frac {d x}{2}\right )}\right ) \sin ^{10}\left (\frac {c}{2}+\frac {d x}{2}\right )-40 \cos ^6\left (\frac {1}{2} (c+d x)\right ) \, _4F_3\left (2,2,2,\frac {9}{2};1,1,\frac {11}{2};\frac {\sin ^2\left (\frac {c}{2}+\frac {d x}{2}\right )}{-1+2 \sin ^2\left (\frac {c}{2}+\frac {d x}{2}\right )}\right ) \sin ^{10}\left (\frac {c}{2}+\frac {d x}{2}\right )-518760 \sin ^{12}\left (\frac {c}{2}+\frac {d x}{2}\right )+1770 \, _2F_1\left (2,\frac {9}{2};\frac {11}{2};\frac {\sin ^2\left (\frac {c}{2}+\frac {d x}{2}\right )}{-1+2 \sin ^2\left (\frac {c}{2}+\frac {d x}{2}\right )}\right ) \sin ^{12}\left (\frac {c}{2}+\frac {d x}{2}\right )+226656 \sin ^{14}\left (\frac {c}{2}+\frac {d x}{2}\right )-1500 \, _2F_1\left (2,\frac {9}{2};\frac {11}{2};\frac {\sin ^2\left (\frac {c}{2}+\frac {d x}{2}\right )}{-1+2 \sin ^2\left (\frac {c}{2}+\frac {d x}{2}\right )}\right ) \sin ^{14}\left (\frac {c}{2}+\frac {d x}{2}\right )-42048 \sin ^{16}\left (\frac {c}{2}+\frac {d x}{2}\right )+440 \, _2F_1\left (2,\frac {9}{2};\frac {11}{2};\frac {\sin ^2\left (\frac {c}{2}+\frac {d x}{2}\right )}{-1+2 \sin ^2\left (\frac {c}{2}+\frac {d x}{2}\right )}\right ) \sin ^{16}\left (\frac {c}{2}+\frac {d x}{2}\right )+4725 \tanh ^{-1}\left (\sqrt {\frac {\sin ^2\left (\frac {c}{2}+\frac {d x}{2}\right )}{-1+2 \sin ^2\left (\frac {c}{2}+\frac {d x}{2}\right )}}\right ) \sqrt {\frac {\sin ^2\left (\frac {c}{2}+\frac {d x}{2}\right )}{-1+2 \sin ^2\left (\frac {c}{2}+\frac {d x}{2}\right )}}-56700 \tanh ^{-1}\left (\sqrt {\frac {\sin ^2\left (\frac {c}{2}+\frac {d x}{2}\right )}{-1+2 \sin ^2\left (\frac {c}{2}+\frac {d x}{2}\right )}}\right ) \sin ^2\left (\frac {c}{2}+\frac {d x}{2}\right ) \sqrt {\frac {\sin ^2\left (\frac {c}{2}+\frac {d x}{2}\right )}{-1+2 \sin ^2\left (\frac {c}{2}+\frac {d x}{2}\right )}}+291060 \tanh ^{-1}\left (\sqrt {\frac {\sin ^2\left (\frac {c}{2}+\frac {d x}{2}\right )}{-1+2 \sin ^2\left (\frac {c}{2}+\frac {d x}{2}\right )}}\right ) \sin ^4\left (\frac {c}{2}+\frac {d x}{2}\right ) \sqrt {\frac {\sin ^2\left (\frac {c}{2}+\frac {d x}{2}\right )}{-1+2 \sin ^2\left (\frac {c}{2}+\frac {d x}{2}\right )}}-833760 \tanh ^{-1}\left (\sqrt {\frac {\sin ^2\left (\frac {c}{2}+\frac {d x}{2}\right )}{-1+2 \sin ^2\left (\frac {c}{2}+\frac {d x}{2}\right )}}\right ) \sin ^6\left (\frac {c}{2}+\frac {d x}{2}\right ) \sqrt {\frac {\sin ^2\left (\frac {c}{2}+\frac {d x}{2}\right )}{-1+2 \sin ^2\left (\frac {c}{2}+\frac {d x}{2}\right )}}+1458000 \tanh ^{-1}\left (\sqrt {\frac {\sin ^2\left (\frac {c}{2}+\frac {d x}{2}\right )}{-1+2 \sin ^2\left (\frac {c}{2}+\frac {d x}{2}\right )}}\right ) \sin ^8\left (\frac {c}{2}+\frac {d x}{2}\right ) \sqrt {\frac {\sin ^2\left (\frac {c}{2}+\frac {d x}{2}\right )}{-1+2 \sin ^2\left (\frac {c}{2}+\frac {d x}{2}\right )}}-1598400 \tanh ^{-1}\left (\sqrt {\frac {\sin ^2\left (\frac {c}{2}+\frac {d x}{2}\right )}{-1+2 \sin ^2\left (\frac {c}{2}+\frac {d x}{2}\right )}}\right ) \sin ^{10}\left (\frac {c}{2}+\frac {d x}{2}\right ) \sqrt {\frac {\sin ^2\left (\frac {c}{2}+\frac {d x}{2}\right )}{-1+2 \sin ^2\left (\frac {c}{2}+\frac {d x}{2}\right )}}+1080000 \tanh ^{-1}\left (\sqrt {\frac {\sin ^2\left (\frac {c}{2}+\frac {d x}{2}\right )}{-1+2 \sin ^2\left (\frac {c}{2}+\frac {d x}{2}\right )}}\right ) \sin ^{12}\left (\frac {c}{2}+\frac {d x}{2}\right ) \sqrt {\frac {\sin ^2\left (\frac {c}{2}+\frac {d x}{2}\right )}{-1+2 \sin ^2\left (\frac {c}{2}+\frac {d x}{2}\right )}}-414720 \tanh ^{-1}\left (\sqrt {\frac {\sin ^2\left (\frac {c}{2}+\frac {d x}{2}\right )}{-1+2 \sin ^2\left (\frac {c}{2}+\frac {d x}{2}\right )}}\right ) \sin ^{14}\left (\frac {c}{2}+\frac {d x}{2}\right ) \sqrt {\frac {\sin ^2\left (\frac {c}{2}+\frac {d x}{2}\right )}{-1+2 \sin ^2\left (\frac {c}{2}+\frac {d x}{2}\right )}}+69120 \tanh ^{-1}\left (\sqrt {\frac {\sin ^2\left (\frac {c}{2}+\frac {d x}{2}\right )}{-1+2 \sin ^2\left (\frac {c}{2}+\frac {d x}{2}\right )}}\right ) \sin ^{16}\left (\frac {c}{2}+\frac {d x}{2}\right ) \sqrt {\frac {\sin ^2\left (\frac {c}{2}+\frac {d x}{2}\right )}{-1+2 \sin ^2\left (\frac {c}{2}+\frac {d x}{2}\right )}}+60 \cos ^4\left (\frac {1}{2} (c+d x)\right ) \, _3F_2\left (2,2,\frac {9}{2};1,\frac {11}{2};\frac {\sin ^2\left (\frac {c}{2}+\frac {d x}{2}\right )}{-1+2 \sin ^2\left (\frac {c}{2}+\frac {d x}{2}\right )}\right ) \sin ^{10}\left (\frac {c}{2}+\frac {d x}{2}\right ) \left (-5+4 \sin ^2\left (\frac {c}{2}+\frac {d x}{2}\right )\right )\right )}{675 d \sqrt {a (1+\cos (c+d x))} \left (1-2 \sin ^2\left (\frac {c}{2}+\frac {d x}{2}\right )\right )^{7/2} \left (-1+2 \sin ^2\left (\frac {c}{2}+\frac {d x}{2}\right )\right )} \end {gather*}

Warning: Unable to verify antiderivative.

[In]

Integrate[1/(Cos[c + d*x]^(7/2)*Sqrt[a + a*Cos[c + d*x]]),x]

[Out]

(-2*Cot[c/2 + (d*x)/2]*Csc[c/2 + (d*x)/2]^6*(4725*Sin[c/2 + (d*x)/2]^2 - 48825*Sin[c/2 + (d*x)/2]^4 + 210105*S
in[c/2 + (d*x)/2]^6 - 486630*Sin[c/2 + (d*x)/2]^8 + 655812*Sin[c/2 + (d*x)/2]^10 - 710*Hypergeometric2F1[2, 9/
2, 11/2, Sin[c/2 + (d*x)/2]^2/(-1 + 2*Sin[c/2 + (d*x)/2]^2)]*Sin[c/2 + (d*x)/2]^10 - 40*Cos[(c + d*x)/2]^6*Hyp
ergeometricPFQ[{2, 2, 2, 9/2}, {1, 1, 11/2}, Sin[c/2 + (d*x)/2]^2/(-1 + 2*Sin[c/2 + (d*x)/2]^2)]*Sin[c/2 + (d*
x)/2]^10 - 518760*Sin[c/2 + (d*x)/2]^12 + 1770*Hypergeometric2F1[2, 9/2, 11/2, Sin[c/2 + (d*x)/2]^2/(-1 + 2*Si
n[c/2 + (d*x)/2]^2)]*Sin[c/2 + (d*x)/2]^12 + 226656*Sin[c/2 + (d*x)/2]^14 - 1500*Hypergeometric2F1[2, 9/2, 11/
2, Sin[c/2 + (d*x)/2]^2/(-1 + 2*Sin[c/2 + (d*x)/2]^2)]*Sin[c/2 + (d*x)/2]^14 - 42048*Sin[c/2 + (d*x)/2]^16 + 4
40*Hypergeometric2F1[2, 9/2, 11/2, Sin[c/2 + (d*x)/2]^2/(-1 + 2*Sin[c/2 + (d*x)/2]^2)]*Sin[c/2 + (d*x)/2]^16 +
4725*ArcTanh[Sqrt[Sin[c/2 + (d*x)/2]^2/(-1 + 2*Sin[c/2 + (d*x)/2]^2)]]*Sqrt[Sin[c/2 + (d*x)/2]^2/(-1 + 2*Sin[
c/2 + (d*x)/2]^2)] - 56700*ArcTanh[Sqrt[Sin[c/2 + (d*x)/2]^2/(-1 + 2*Sin[c/2 + (d*x)/2]^2)]]*Sin[c/2 + (d*x)/2
]^2*Sqrt[Sin[c/2 + (d*x)/2]^2/(-1 + 2*Sin[c/2 + (d*x)/2]^2)] + 291060*ArcTanh[Sqrt[Sin[c/2 + (d*x)/2]^2/(-1 +
2*Sin[c/2 + (d*x)/2]^2)]]*Sin[c/2 + (d*x)/2]^4*Sqrt[Sin[c/2 + (d*x)/2]^2/(-1 + 2*Sin[c/2 + (d*x)/2]^2)] - 8337
60*ArcTanh[Sqrt[Sin[c/2 + (d*x)/2]^2/(-1 + 2*Sin[c/2 + (d*x)/2]^2)]]*Sin[c/2 + (d*x)/2]^6*Sqrt[Sin[c/2 + (d*x)
/2]^2/(-1 + 2*Sin[c/2 + (d*x)/2]^2)] + 1458000*ArcTanh[Sqrt[Sin[c/2 + (d*x)/2]^2/(-1 + 2*Sin[c/2 + (d*x)/2]^2)
]]*Sin[c/2 + (d*x)/2]^8*Sqrt[Sin[c/2 + (d*x)/2]^2/(-1 + 2*Sin[c/2 + (d*x)/2]^2)] - 1598400*ArcTanh[Sqrt[Sin[c/
2 + (d*x)/2]^2/(-1 + 2*Sin[c/2 + (d*x)/2]^2)]]*Sin[c/2 + (d*x)/2]^10*Sqrt[Sin[c/2 + (d*x)/2]^2/(-1 + 2*Sin[c/2
+ (d*x)/2]^2)] + 1080000*ArcTanh[Sqrt[Sin[c/2 + (d*x)/2]^2/(-1 + 2*Sin[c/2 + (d*x)/2]^2)]]*Sin[c/2 + (d*x)/2]
^12*Sqrt[Sin[c/2 + (d*x)/2]^2/(-1 + 2*Sin[c/2 + (d*x)/2]^2)] - 414720*ArcTanh[Sqrt[Sin[c/2 + (d*x)/2]^2/(-1 +
2*Sin[c/2 + (d*x)/2]^2)]]*Sin[c/2 + (d*x)/2]^14*Sqrt[Sin[c/2 + (d*x)/2]^2/(-1 + 2*Sin[c/2 + (d*x)/2]^2)] + 691
20*ArcTanh[Sqrt[Sin[c/2 + (d*x)/2]^2/(-1 + 2*Sin[c/2 + (d*x)/2]^2)]]*Sin[c/2 + (d*x)/2]^16*Sqrt[Sin[c/2 + (d*x
)/2]^2/(-1 + 2*Sin[c/2 + (d*x)/2]^2)] + 60*Cos[(c + d*x)/2]^4*HypergeometricPFQ[{2, 2, 9/2}, {1, 11/2}, Sin[c/
2 + (d*x)/2]^2/(-1 + 2*Sin[c/2 + (d*x)/2]^2)]*Sin[c/2 + (d*x)/2]^10*(-5 + 4*Sin[c/2 + (d*x)/2]^2)))/(675*d*Sqr
t[a*(1 + Cos[c + d*x])]*(1 - 2*Sin[c/2 + (d*x)/2]^2)^(7/2)*(-1 + 2*Sin[c/2 + (d*x)/2]^2))

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Maple [B] Leaf count of result is larger than twice the leaf count of optimal. \(340\) vs. \(2(140)=280\).
time = 0.23, size = 341, normalized size = 2.02

method	result	size
default	\(-\frac {\left (15 \left (\frac {\cos \left (d x +c \right )}{1+\cos \left (d x +c \right )}\right )^{\frac {7}{2}} \arcsin \left (\frac {-1+\cos \left (d x +c \right )}{\sin \left (d x +c \right )}\right ) \left (\cos ^{4}\left (d x +c \right )\right )+60 \left (\frac {\cos \left (d x +c \right )}{1+\cos \left (d x +c \right )}\right )^{\frac {7}{2}} \arcsin \left (\frac {-1+\cos \left (d x +c \right )}{\sin \left (d x +c \right )}\right ) \left (\cos ^{3}\left (d x +c \right )\right )+90 \left (\frac {\cos \left (d x +c \right )}{1+\cos \left (d x +c \right )}\right )^{\frac {7}{2}} \arcsin \left (\frac {-1+\cos \left (d x +c \right )}{\sin \left (d x +c \right )}\right ) \left (\cos ^{2}\left (d x +c \right )\right )+60 \left (\frac {\cos \left (d x +c \right )}{1+\cos \left (d x +c \right )}\right )^{\frac {7}{2}} \arcsin \left (\frac {-1+\cos \left (d x +c \right )}{\sin \left (d x +c \right )}\right ) \cos \left (d x +c \right )+15 \left (\frac {\cos \left (d x +c \right )}{1+\cos \left (d x +c \right )}\right )^{\frac {7}{2}} \arcsin \left (\frac {-1+\cos \left (d x +c \right )}{\sin \left (d x +c \right )}\right )+13 \sqrt {2}\, \left (\cos ^{3}\left (d x +c \right )\right ) \sin \left (d x +c \right )-\sqrt {2}\, \left (\cos ^{2}\left (d x +c \right )\right ) \sin \left (d x +c \right )+3 \sqrt {2}\, \cos \left (d x +c \right ) \sin \left (d x +c \right )\right ) \sqrt {a \left (1+\cos \left (d x +c \right )\right )}\, \left (\sin ^{6}\left (d x +c \right )\right ) \sqrt {2}}{15 d \left (-1+\cos \left (d x +c \right )\right )^{3} \left (1+\cos \left (d x +c \right )\right )^{4} \cos \left (d x +c \right )^{\frac {7}{2}} a}\)	\(341\)

Verification of antiderivative is not currently implemented for this CAS.

[In]

int(1/cos(d*x+c)^(7/2)/(a+a*cos(d*x+c))^(1/2),x,method=_RETURNVERBOSE)

[Out]

-1/15/d*(15*(cos(d*x+c)/(1+cos(d*x+c)))^(7/2)*arcsin((-1+cos(d*x+c))/sin(d*x+c))*cos(d*x+c)^4+60*(cos(d*x+c)/(
1+cos(d*x+c)))^(7/2)*arcsin((-1+cos(d*x+c))/sin(d*x+c))*cos(d*x+c)^3+90*(cos(d*x+c)/(1+cos(d*x+c)))^(7/2)*arcs
in((-1+cos(d*x+c))/sin(d*x+c))*cos(d*x+c)^2+60*(cos(d*x+c)/(1+cos(d*x+c)))^(7/2)*arcsin((-1+cos(d*x+c))/sin(d*
x+c))*cos(d*x+c)+15*(cos(d*x+c)/(1+cos(d*x+c)))^(7/2)*arcsin((-1+cos(d*x+c))/sin(d*x+c))+13*2^(1/2)*cos(d*x+c)
^3*sin(d*x+c)-2^(1/2)*cos(d*x+c)^2*sin(d*x+c)+3*2^(1/2)*cos(d*x+c)*sin(d*x+c))*(a*(1+cos(d*x+c)))^(1/2)*sin(d*
x+c)^6/(-1+cos(d*x+c))^3/(1+cos(d*x+c))^4/cos(d*x+c)^(7/2)*2^(1/2)/a

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Maxima [F(-2)]
time = 0.00, size = 0, normalized size = 0.00 \begin {gather*} \text {Exception raised: RuntimeError} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate(1/cos(d*x+c)^(7/2)/(a+a*cos(d*x+c))^(1/2),x, algorithm="maxima")

[Out]

Exception raised: RuntimeError >> ECL says: Error executing code in Maxima: sign: argument cannot be imaginary
; found %i

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Fricas [A]
time = 0.47, size = 157, normalized size = 0.93 \begin {gather*} \frac {2 \, \sqrt {a \cos \left (d x + c\right ) + a} {\left (13 \, \cos \left (d x + c\right )^{2} - \cos \left (d x + c\right ) + 3\right )} \sqrt {\cos \left (d x + c\right )} \sin \left (d x + c\right ) - \frac {15 \, \sqrt {2} {\left (a \cos \left (d x + c\right )^{4} + a \cos \left (d x + c\right )^{3}\right )} \arctan \left (\frac {\sqrt {2} \sqrt {a \cos \left (d x + c\right ) + a} \sqrt {\cos \left (d x + c\right )} \sin \left (d x + c\right )}{2 \, {\left (\cos \left (d x + c\right )^{2} + \cos \left (d x + c\right )\right )} \sqrt {a}}\right )}{\sqrt {a}}}{15 \, {\left (a d \cos \left (d x + c\right )^{4} + a d \cos \left (d x + c\right )^{3}\right )}} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate(1/cos(d*x+c)^(7/2)/(a+a*cos(d*x+c))^(1/2),x, algorithm="fricas")

[Out]

1/15*(2*sqrt(a*cos(d*x + c) + a)*(13*cos(d*x + c)^2 - cos(d*x + c) + 3)*sqrt(cos(d*x + c))*sin(d*x + c) - 15*s
qrt(2)*(a*cos(d*x + c)^4 + a*cos(d*x + c)^3)*arctan(1/2*sqrt(2)*sqrt(a*cos(d*x + c) + a)*sqrt(cos(d*x + c))*si
n(d*x + c)/((cos(d*x + c)^2 + cos(d*x + c))*sqrt(a)))/sqrt(a))/(a*d*cos(d*x + c)^4 + a*d*cos(d*x + c)^3)

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Sympy [F(-2)]
time = 0.00, size = 0, normalized size = 0.00 \begin {gather*} \text {Exception raised: SystemError} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate(1/cos(d*x+c)**(7/2)/(a+a*cos(d*x+c))**(1/2),x)

[Out]

Exception raised: SystemError >> excessive stack use: stack is 6190 deep

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Giac [F]
time = 0.00, size = 0, normalized size = 0.00 \begin {gather*} \text {could not integrate} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate(1/cos(d*x+c)^(7/2)/(a+a*cos(d*x+c))^(1/2),x, algorithm="giac")

[Out]

integrate(1/(sqrt(a*cos(d*x + c) + a)*cos(d*x + c)^(7/2)), x)

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Mupad [F]
time = 0.00, size = -1, normalized size = -0.01 \begin {gather*} \int \frac {1}{{\cos \left (c+d\,x\right )}^{7/2}\,\sqrt {a+a\,\cos \left (c+d\,x\right )}} \,d x \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

int(1/(cos(c + d*x)^(7/2)*(a + a*cos(c + d*x))^(1/2)),x)

[Out]

int(1/(cos(c + d*x)^(7/2)*(a + a*cos(c + d*x))^(1/2)), x)

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