3.2.17 \(\int (a+a \cos (c+d x))^{5/2} \sec ^2(c+d x) \, dx\) [117]

Optimal. Leaf size=92 \[ \frac {5 a^{5/2} \tanh ^{-1}\left (\frac {\sqrt {a} \sin (c+d x)}{\sqrt {a+a \cos (c+d x)}}\right )}{d}+\frac {a^3 \sin (c+d x)}{d \sqrt {a+a \cos (c+d x)}}+\frac {a^2 \sqrt {a+a \cos (c+d x)} \tan (c+d x)}{d} \]

[Out]

5*a^(5/2)*arctanh(sin(d*x+c)*a^(1/2)/(a+a*cos(d*x+c))^(1/2))/d+a^3*sin(d*x+c)/d/(a+a*cos(d*x+c))^(1/2)+a^2*(a+
a*cos(d*x+c))^(1/2)*tan(d*x+c)/d

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Rubi [A]
time = 0.13, antiderivative size = 92, normalized size of antiderivative = 1.00, number of steps used = 4, number of rules used = 4, integrand size = 23, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.174, Rules used = {2841, 3060, 2852, 212} \begin {gather*} \frac {5 a^{5/2} \tanh ^{-1}\left (\frac {\sqrt {a} \sin (c+d x)}{\sqrt {a \cos (c+d x)+a}}\right )}{d}+\frac {a^3 \sin (c+d x)}{d \sqrt {a \cos (c+d x)+a}}+\frac {a^2 \tan (c+d x) \sqrt {a \cos (c+d x)+a}}{d} \end {gather*}

Antiderivative was successfully verified.

[In]

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

[Out]

(5*a^(5/2)*ArcTanh[(Sqrt[a]*Sin[c + d*x])/Sqrt[a + a*Cos[c + d*x]]])/d + (a^3*Sin[c + d*x])/(d*Sqrt[a + a*Cos[
c + d*x]]) + (a^2*Sqrt[a + a*Cos[c + d*x]]*Tan[c + d*x])/d

Rule 212

Int[((a_) + (b_.)*(x_)^2)^(-1), x_Symbol] :> Simp[(1/(Rt[a, 2]*Rt[-b, 2]))*ArcTanh[Rt[-b, 2]*(x/Rt[a, 2])], x]
/; FreeQ[{a, b}, x] && NegQ[a/b] && (GtQ[a, 0] || LtQ[b, 0])

Rule 2841

Int[((a_) + (b_.)*sin[(e_.) + (f_.)*(x_)])^(m_)*((c_.) + (d_.)*sin[(e_.) + (f_.)*(x_)])^(n_), x_Symbol] :> Sim
p[(-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 2852

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 3060

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

Rubi steps

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

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Mathematica [C] Result contains complex when optimal does not.
time = 34.77, size = 1547, normalized size = 16.82 \begin {gather*} -\frac {\left (\frac {5}{32}-\frac {5 i}{32}\right ) \left (1+e^{i c}\right ) \left (\sqrt {2}-(1-i) e^{\frac {i c}{2}}+(16-16 i) e^{\frac {3 i c}{2}+i d x}+(20+20 i) \sqrt {2} e^{2 i c+\frac {3 i d x}{2}}-(34-34 i) e^{\frac {5 i c}{2}+2 i d x}-(20+20 i) \sqrt {2} e^{3 i c+\frac {5 i d x}{2}}+(16-16 i) e^{\frac {7 i c}{2}+3 i d x}+(4+4 i) \sqrt {2} e^{4 i c+\frac {7 i d x}{2}}-(1-i) e^{\frac {9 i c}{2}+4 i d x}+8 i e^{\frac {1}{2} i (c+d x)}-16 \sqrt {2} e^{i (c+d x)}-40 i e^{\frac {3}{2} i (c+d x)}+34 \sqrt {2} e^{2 i (c+d x)}+40 i e^{\frac {5}{2} i (c+d x)}-16 \sqrt {2} e^{3 i (c+d x)}-8 i e^{\frac {7}{2} i (c+d x)}+\sqrt {2} e^{4 i (c+d x)}-(4+4 i) \sqrt {2} e^{\frac {1}{2} i (2 c+d x)}\right ) x (a (1+\cos (c+d x)))^{5/2} \sec ^5\left (\frac {c}{2}+\frac {d x}{2}\right )}{\left ((-1-i)+\sqrt {2} e^{\frac {i c}{2}}\right ) \left (-1+e^{i c}\right ) \left (i-2 \sqrt {2} e^{\frac {1}{2} i (c+d x)}-4 i e^{i (c+d x)}+2 \sqrt {2} e^{\frac {3}{2} i (c+d x)}+i e^{2 i (c+d x)}\right )^2}-\frac {5 i \text {ArcTan}\left (\frac {\cos \left (\frac {c}{4}+\frac {d x}{4}\right )-\sin \left (\frac {c}{4}+\frac {d x}{4}\right )-\sqrt {2} \sin \left (\frac {c}{4}+\frac {d x}{4}\right )}{-\cos \left (\frac {c}{4}+\frac {d x}{4}\right )+\sqrt {2} \cos \left (\frac {c}{4}+\frac {d x}{4}\right )-\sin \left (\frac {c}{4}+\frac {d x}{4}\right )}\right ) (a (1+\cos (c+d x)))^{5/2} \sec ^5\left (\frac {c}{2}+\frac {d x}{2}\right )}{8 \sqrt {2} d}-\frac {5 i \text {ArcTan}\left (\frac {\cos \left (\frac {c}{4}+\frac {d x}{4}\right )+\sin \left (\frac {c}{4}+\frac {d x}{4}\right )-\sqrt {2} \sin \left (\frac {c}{4}+\frac {d x}{4}\right )}{\cos \left (\frac {c}{4}+\frac {d x}{4}\right )+\sqrt {2} \cos \left (\frac {c}{4}+\frac {d x}{4}\right )-\sin \left (\frac {c}{4}+\frac {d x}{4}\right )}\right ) (a (1+\cos (c+d x)))^{5/2} \sec ^5\left (\frac {c}{2}+\frac {d x}{2}\right )}{8 \sqrt {2} d}-\frac {5 (a (1+\cos (c+d x)))^{5/2} \log \left (2-\sqrt {2} \cos \left (\frac {c}{2}+\frac {d x}{2}\right )-\sqrt {2} \sin \left (\frac {c}{2}+\frac {d x}{2}\right )\right ) \sec ^5\left (\frac {c}{2}+\frac {d x}{2}\right )}{16 \sqrt {2} d}-\frac {5 (a (1+\cos (c+d x)))^{5/2} \log \left (2+\sqrt {2} \cos \left (\frac {c}{2}+\frac {d x}{2}\right )-\sqrt {2} \sin \left (\frac {c}{2}+\frac {d x}{2}\right )\right ) \sec ^5\left (\frac {c}{2}+\frac {d x}{2}\right )}{16 \sqrt {2} d}+\frac {\cos \left (\frac {d x}{2}\right ) (a (1+\cos (c+d x)))^{5/2} \sec ^5\left (\frac {c}{2}+\frac {d x}{2}\right ) \sin \left (\frac {c}{2}\right )}{2 d}-\frac {5 i \text {ArcTan}\left (\frac {2 i \cos \left (\frac {c}{2}\right )-i \left (-\sqrt {2}+2 \sin \left (\frac {c}{2}\right )\right ) \tan \left (\frac {d x}{4}\right )}{\sqrt {-2+4 \cos ^2\left (\frac {c}{2}\right )+4 \sin ^2\left (\frac {c}{2}\right )}}\right ) (a (1+\cos (c+d x)))^{5/2} \cot \left (\frac {c}{2}\right ) \sec ^5\left (\frac {c}{2}+\frac {d x}{2}\right )}{4 d \sqrt {-2+4 \cos ^2\left (\frac {c}{2}\right )+4 \sin ^2\left (\frac {c}{2}\right )}}+\frac {5 (a (1+\cos (c+d x)))^{5/2} \csc \left (\frac {c}{2}\right ) \sec ^5\left (\frac {c}{2}+\frac {d x}{2}\right ) \left (-d x \cos \left (\frac {c}{2}\right )+2 \log \left (\sqrt {2}+2 \cos \left (\frac {d x}{2}\right ) \sin \left (\frac {c}{2}\right )+2 \cos \left (\frac {c}{2}\right ) \sin \left (\frac {d x}{2}\right )\right ) \sin \left (\frac {c}{2}\right )+\frac {4 i \sqrt {2} \text {ArcTan}\left (\frac {2 i \cos \left (\frac {c}{2}\right )-i \left (-\sqrt {2}+2 \sin \left (\frac {c}{2}\right )\right ) \tan \left (\frac {d x}{4}\right )}{\sqrt {-2+4 \cos ^2\left (\frac {c}{2}\right )+4 \sin ^2\left (\frac {c}{2}\right )}}\right ) \cos \left (\frac {c}{2}\right )}{\sqrt {-2+4 \cos ^2\left (\frac {c}{2}\right )+4 \sin ^2\left (\frac {c}{2}\right )}}\right )}{4 \sqrt {2} d \left (4 \cos ^2\left (\frac {c}{2}\right )+4 \sin ^2\left (\frac {c}{2}\right )\right )}+\frac {\cos \left (\frac {c}{2}\right ) (a (1+\cos (c+d x)))^{5/2} \sec ^5\left (\frac {c}{2}+\frac {d x}{2}\right ) \sin \left (\frac {d x}{2}\right )}{2 d}+\frac {(a (1+\cos (c+d x)))^{5/2} \sec ^5\left (\frac {c}{2}+\frac {d x}{2}\right )}{8 d \left (\cos \left (\frac {c}{2}+\frac {d x}{2}\right )-\sin \left (\frac {c}{2}+\frac {d x}{2}\right )\right )}-\frac {(a (1+\cos (c+d x)))^{5/2} \sec ^5\left (\frac {c}{2}+\frac {d x}{2}\right )}{8 d \left (\cos \left (\frac {c}{2}+\frac {d x}{2}\right )+\sin \left (\frac {c}{2}+\frac {d x}{2}\right )\right )} \end {gather*}

Warning: Unable to verify antiderivative.

[In]

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

[Out]

((-5/32 + (5*I)/32)*(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) - (((5*I)/8)*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) - (((5*I)/8)*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) - (5*(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)/(16*Sqrt[2]*d) - (5*(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)/(16*Sqrt[2]*d) + (Cos[(d*x)/2]*(a*(1 + Cos[c + d
*x]))^(5/2)*Sec[c/2 + (d*x)/2]^5*Sin[c/2])/(2*d) - (((5*I)/4)*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]) + (5*(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)*Sq
rt[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]))/(4*Sqrt[2]*d*(4*Cos[c/2]^2 + 4*Sin[c/2]^2)) + (Cos[c/2]*(a
*(1 + Cos[c + d*x]))^(5/2)*Sec[c/2 + (d*x)/2]^5*Sin[(d*x)/2])/(2*d) + ((a*(1 + Cos[c + d*x]))^(5/2)*Sec[c/2 +
(d*x)/2]^5)/(8*d*(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)/(8*d*(Cos[c/2 + (d*x)/2] + Sin[c/2 + (d*x)/2]))

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Maple [B] Leaf count of result is larger than twice the leaf count of optimal. \(431\) vs. \(2(82)=164\).
time = 0.15, size = 432, normalized size = 4.70

method	result	size
default	\(\frac {a^{\frac {3}{2}} \cos \left (\frac {d x}{2}+\frac {c}{2}\right ) \sqrt {a \left (\sin ^{2}\left (\frac {d x}{2}+\frac {c}{2}\right )\right )}\, \left (-8 \sqrt {2}\, \sqrt {a \left (\sin ^{2}\left (\frac {d x}{2}+\frac {c}{2}\right )\right )}\, \sqrt {a}\, \left (\sin ^{2}\left (\frac {d x}{2}+\frac {c}{2}\right )\right )-10 \ln \left (-\frac {4 \left (a \sqrt {2}\, \cos \left (\frac {d x}{2}+\frac {c}{2}\right )-\sqrt {a}\, \sqrt {2}\, \sqrt {a \left (\sin ^{2}\left (\frac {d x}{2}+\frac {c}{2}\right )\right )}-2 a \right )}{2 \cos \left (\frac {d x}{2}+\frac {c}{2}\right )-\sqrt {2}}\right ) \left (\sin ^{2}\left (\frac {d x}{2}+\frac {c}{2}\right )\right ) a -10 \ln \left (\frac {4 a \sqrt {2}\, \cos \left (\frac {d x}{2}+\frac {c}{2}\right )+4 \sqrt {a}\, \sqrt {2}\, \sqrt {a \left (\sin ^{2}\left (\frac {d x}{2}+\frac {c}{2}\right )\right )}+8 a}{2 \cos \left (\frac {d x}{2}+\frac {c}{2}\right )+\sqrt {2}}\right ) \left (\sin ^{2}\left (\frac {d x}{2}+\frac {c}{2}\right )\right ) a +6 \sqrt {a}\, \sqrt {2}\, \sqrt {a \left (\sin ^{2}\left (\frac {d x}{2}+\frac {c}{2}\right )\right )}+5 \ln \left (-\frac {4 \left (a \sqrt {2}\, \cos \left (\frac {d x}{2}+\frac {c}{2}\right )-\sqrt {a}\, \sqrt {2}\, \sqrt {a \left (\sin ^{2}\left (\frac {d x}{2}+\frac {c}{2}\right )\right )}-2 a \right )}{2 \cos \left (\frac {d x}{2}+\frac {c}{2}\right )-\sqrt {2}}\right ) a +5 \ln \left (\frac {4 a \sqrt {2}\, \cos \left (\frac {d x}{2}+\frac {c}{2}\right )+4 \sqrt {a}\, \sqrt {2}\, \sqrt {a \left (\sin ^{2}\left (\frac {d x}{2}+\frac {c}{2}\right )\right )}+8 a}{2 \cos \left (\frac {d x}{2}+\frac {c}{2}\right )+\sqrt {2}}\right ) a \right )}{\left (2 \cos \left (\frac {d x}{2}+\frac {c}{2}\right )-\sqrt {2}\right ) \left (2 \cos \left (\frac {d x}{2}+\frac {c}{2}\right )+\sqrt {2}\right ) \sin \left (\frac {d x}{2}+\frac {c}{2}\right ) \sqrt {a \left (\cos ^{2}\left (\frac {d x}{2}+\frac {c}{2}\right )\right )}\, d}\)	\(432\)

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

a^(3/2)*cos(1/2*d*x+1/2*c)*(sin(1/2*d*x+1/2*c)^2*a)^(1/2)*(-8*2^(1/2)*(sin(1/2*d*x+1/2*c)^2*a)^(1/2)*a^(1/2)*s
in(1/2*d*x+1/2*c)^2-10*ln(-4/(2*cos(1/2*d*x+1/2*c)-2^(1/2))*(a*2^(1/2)*cos(1/2*d*x+1/2*c)-a^(1/2)*2^(1/2)*(sin
(1/2*d*x+1/2*c)^2*a)^(1/2)-2*a))*sin(1/2*d*x+1/2*c)^2*a-10*ln(4/(2*cos(1/2*d*x+1/2*c)+2^(1/2))*(a*2^(1/2)*cos(
1/2*d*x+1/2*c)+a^(1/2)*2^(1/2)*(sin(1/2*d*x+1/2*c)^2*a)^(1/2)+2*a))*sin(1/2*d*x+1/2*c)^2*a+6*a^(1/2)*2^(1/2)*(
sin(1/2*d*x+1/2*c)^2*a)^(1/2)+5*ln(-4/(2*cos(1/2*d*x+1/2*c)-2^(1/2))*(a*2^(1/2)*cos(1/2*d*x+1/2*c)-a^(1/2)*2^(
1/2)*(sin(1/2*d*x+1/2*c)^2*a)^(1/2)-2*a))*a+5*ln(4/(2*cos(1/2*d*x+1/2*c)+2^(1/2))*(a*2^(1/2)*cos(1/2*d*x+1/2*c
)+a^(1/2)*2^(1/2)*(sin(1/2*d*x+1/2*c)^2*a)^(1/2)+2*a))*a)/(2*cos(1/2*d*x+1/2*c)-2^(1/2))/(2*cos(1/2*d*x+1/2*c)
+2^(1/2))/sin(1/2*d*x+1/2*c)/(a*cos(1/2*d*x+1/2*c)^2)^(1/2)/d

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Maxima [B] Leaf count of result is larger than twice the leaf count of optimal. 10847 vs. \(2 (82) = 164\).
time = 0.84, size = 10847, normalized size = 117.90 \begin {gather*} \text {Too large to display} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

-1/252*(1449*sqrt(2)*a^2*cos(5/2*d*x + 5/2*c)^3*sin(2*d*x + 2*c) - 63*(sqrt(2)*a^2*cos(2*d*x + 2*c)^2 + sqrt(2
)*a^2*sin(2*d*x + 2*c)^2 + 25*sqrt(2)*a^2*cos(2*d*x + 2*c) + 24*sqrt(2)*a^2)*sin(5/2*d*x + 5/2*c)^3 - 252*sqrt
(2)*a^2*sin(1/2*d*x + 1/2*c) + 21*(5*(5*sqrt(2)*a^2*sin(3/2*d*x + 3/2*c) + 12*sqrt(2)*a^2*sin(1/2*d*x + 1/2*c)
- 3*a^2*log(2*cos(1/2*d*x + 1/2*c)^2 + 2*sin(1/2*d*x + 1/2*c)^2 + 2*sqrt(2)*cos(1/2*d*x + 1/2*c) + 2*sqrt(2)*
sin(1/2*d*x + 1/2*c) + 2) + 3*a^2*log(2*cos(1/2*d*x + 1/2*c)^2 + 2*sin(1/2*d*x + 1/2*c)^2 + 2*sqrt(2)*cos(1/2*
d*x + 1/2*c) - 2*sqrt(2)*sin(1/2*d*x + 1/2*c) + 2) - 3*a^2*log(2*cos(1/2*d*x + 1/2*c)^2 + 2*sin(1/2*d*x + 1/2*
c)^2 - 2*sqrt(2)*cos(1/2*d*x + 1/2*c) + 2*sqrt(2)*sin(1/2*d*x + 1/2*c) + 2) + 3*a^2*log(2*cos(1/2*d*x + 1/2*c)
^2 + 2*sin(1/2*d*x + 1/2*c)^2 - 2*sqrt(2)*cos(1/2*d*x + 1/2*c) - 2*sqrt(2)*sin(1/2*d*x + 1/2*c) + 2))*cos(2*d*
x + 2*c)^2 - 15*a^2*log(2*cos(1/2*d*x + 1/2*c)^2 + 2*sin(1/2*d*x + 1/2*c)^2 + 2*sqrt(2)*cos(1/2*d*x + 1/2*c) +
2*sqrt(2)*sin(1/2*d*x + 1/2*c) + 2) + 15*a^2*log(2*cos(1/2*d*x + 1/2*c)^2 + 2*sin(1/2*d*x + 1/2*c)^2 + 2*sqrt
(2)*cos(1/2*d*x + 1/2*c) - 2*sqrt(2)*sin(1/2*d*x + 1/2*c) + 2) - 15*a^2*log(2*cos(1/2*d*x + 1/2*c)^2 + 2*sin(1
/2*d*x + 1/2*c)^2 - 2*sqrt(2)*cos(1/2*d*x + 1/2*c) + 2*sqrt(2)*sin(1/2*d*x + 1/2*c) + 2) + 15*a^2*log(2*cos(1/
2*d*x + 1/2*c)^2 + 2*sin(1/2*d*x + 1/2*c)^2 - 2*sqrt(2)*cos(1/2*d*x + 1/2*c) - 2*sqrt(2)*sin(1/2*d*x + 1/2*c)
+ 2) + 5*(5*sqrt(2)*a^2*sin(3/2*d*x + 3/2*c) + 12*sqrt(2)*a^2*sin(1/2*d*x + 1/2*c) - 3*a^2*log(2*cos(1/2*d*x +
1/2*c)^2 + 2*sin(1/2*d*x + 1/2*c)^2 + 2*sqrt(2)*cos(1/2*d*x + 1/2*c) + 2*sqrt(2)*sin(1/2*d*x + 1/2*c) + 2) +
3*a^2*log(2*cos(1/2*d*x + 1/2*c)^2 + 2*sin(1/2*d*x + 1/2*c)^2 + 2*sqrt(2)*cos(1/2*d*x + 1/2*c) - 2*sqrt(2)*sin
(1/2*d*x + 1/2*c) + 2) - 3*a^2*log(2*cos(1/2*d*x + 1/2*c)^2 + 2*sin(1/2*d*x + 1/2*c)^2 - 2*sqrt(2)*cos(1/2*d*x
+ 1/2*c) + 2*sqrt(2)*sin(1/2*d*x + 1/2*c) + 2) + 3*a^2*log(2*cos(1/2*d*x + 1/2*c)^2 + 2*sin(1/2*d*x + 1/2*c)^
2 - 2*sqrt(2)*cos(1/2*d*x + 1/2*c) - 2*sqrt(2)*sin(1/2*d*x + 1/2*c) + 2))*sin(2*d*x + 2*c)^2 + 5*(5*sqrt(2)*a^
2*sin(3/2*d*x + 3/2*c) + 12*sqrt(2)*a^2*sin(1/2*d*x + 1/2*c) - 6*a^2*log(2*cos(1/2*d*x + 1/2*c)^2 + 2*sin(1/2*
d*x + 1/2*c)^2 + 2*sqrt(2)*cos(1/2*d*x + 1/2*c) + 2*sqrt(2)*sin(1/2*d*x + 1/2*c) + 2) + 6*a^2*log(2*cos(1/2*d*
x + 1/2*c)^2 + 2*sin(1/2*d*x + 1/2*c)^2 + 2*sqrt(2)*cos(1/2*d*x + 1/2*c) - 2*sqrt(2)*sin(1/2*d*x + 1/2*c) + 2)
- 6*a^2*log(2*cos(1/2*d*x + 1/2*c)^2 + 2*sin(1/2*d*x + 1/2*c)^2 - 2*sqrt(2)*cos(1/2*d*x + 1/2*c) + 2*sqrt(2)*
sin(1/2*d*x + 1/2*c) + 2) + 6*a^2*log(2*cos(1/2*d*x + 1/2*c)^2 + 2*sin(1/2*d*x + 1/2*c)^2 - 2*sqrt(2)*cos(1/2*
d*x + 1/2*c) - 2*sqrt(2)*sin(1/2*d*x + 1/2*c) + 2))*cos(2*d*x + 2*c) + (25*sqrt(2)*a^2*cos(3/2*d*x + 3/2*c) +
198*sqrt(2)*a^2*cos(1/2*d*x + 1/2*c))*sin(2*d*x + 2*c))*cos(5/2*d*x + 5/2*c)^2 + 21*(60*sqrt(2)*a^2*sin(1/2*d*
x + 1/2*c)^3 - 15*(a^2*log(2*cos(1/2*d*x + 1/2*c)^2 + 2*sin(1/2*d*x + 1/2*c)^2 + 2*sqrt(2)*cos(1/2*d*x + 1/2*c
) + 2*sqrt(2)*sin(1/2*d*x + 1/2*c) + 2) - a^2*log(2*cos(1/2*d*x + 1/2*c)^2 + 2*sin(1/2*d*x + 1/2*c)^2 + 2*sqrt
(2)*cos(1/2*d*x + 1/2*c) - 2*sqrt(2)*sin(1/2*d*x + 1/2*c) + 2) + a^2*log(2*cos(1/2*d*x + 1/2*c)^2 + 2*sin(1/2*
d*x + 1/2*c)^2 - 2*sqrt(2)*cos(1/2*d*x + 1/2*c) + 2*sqrt(2)*sin(1/2*d*x + 1/2*c) + 2) - a^2*log(2*cos(1/2*d*x
+ 1/2*c)^2 + 2*sin(1/2*d*x + 1/2*c)^2 - 2*sqrt(2)*cos(1/2*d*x + 1/2*c) - 2*sqrt(2)*sin(1/2*d*x + 1/2*c) + 2))*
cos(1/2*d*x + 1/2*c)^2 - 15*(a^2*log(2*cos(1/2*d*x + 1/2*c)^2 + 2*sin(1/2*d*x + 1/2*c)^2 + 2*sqrt(2)*cos(1/2*d
*x + 1/2*c) + 2*sqrt(2)*sin(1/2*d*x + 1/2*c) + 2) - a^2*log(2*cos(1/2*d*x + 1/2*c)^2 + 2*sin(1/2*d*x + 1/2*c)^
2 + 2*sqrt(2)*cos(1/2*d*x + 1/2*c) - 2*sqrt(2)*sin(1/2*d*x + 1/2*c) + 2) + a^2*log(2*cos(1/2*d*x + 1/2*c)^2 +
2*sin(1/2*d*x + 1/2*c)^2 - 2*sqrt(2)*cos(1/2*d*x + 1/2*c) + 2*sqrt(2)*sin(1/2*d*x + 1/2*c) + 2) - a^2*log(2*co
s(1/2*d*x + 1/2*c)^2 + 2*sin(1/2*d*x + 1/2*c)^2 - 2*sqrt(2)*cos(1/2*d*x + 1/2*c) - 2*sqrt(2)*sin(1/2*d*x + 1/2
*c) + 2))*sin(1/2*d*x + 1/2*c)^2 + 25*(sqrt(2)*a^2*cos(1/2*d*x + 1/2*c)^2 + sqrt(2)*a^2*sin(1/2*d*x + 1/2*c)^2
)*sin(3/2*d*x + 3/2*c) + 12*(5*sqrt(2)*a^2*cos(1/2*d*x + 1/2*c)^2 - sqrt(2)*a^2)*sin(1/2*d*x + 1/2*c))*cos(2*d
*x + 2*c)^2 - 315*(a^2*log(2*cos(1/2*d*x + 1/2*c)^2 + 2*sin(1/2*d*x + 1/2*c)^2 + 2*sqrt(2)*cos(1/2*d*x + 1/2*c
) + 2*sqrt(2)*sin(1/2*d*x + 1/2*c) + 2) - a^2*log(2*cos(1/2*d*x + 1/2*c)^2 + 2*sin(1/2*d*x + 1/2*c)^2 + 2*sqrt
(2)*cos(1/2*d*x + 1/2*c) - 2*sqrt(2)*sin(1/2*d*x + 1/2*c) + 2) + a^2*log(2*cos(1/2*d*x + 1/2*c)^2 + 2*sin(1/2*
d*x + 1/2*c)^2 - 2*sqrt(2)*cos(1/2*d*x + 1/2*c) + 2*sqrt(2)*sin(1/2*d*x + 1/2*c) + 2) - a^2*log(2*cos(1/2*d*x
+ 1/2*c)^2 + 2*sin(1/2*d*x + 1/2*c)^2 - 2*sqrt(2)*cos(1/2*d*x + 1/2*c) - 2*sqrt(2)*sin(1/2*d*x + 1/2*c) + 2))*
cos(1/2*d*x + 1/2*c)^2 + 21*(69*sqrt(2)*a^2*cos(5/2*d*x + 5/2*c)*sin(2*d*x + 2*c) - 144*sqrt(2)*a^2*sin(1/2*d*
x + 1/2*c) + (25*sqrt(2)*a^2*sin(3/2*d*x + 3/2*c) + 54*sqrt(2)*a^2*sin(1/2*d*x + 1/2*c) - 15*a^2*log(2*cos(1/2
*d*x + 1/2*c)^2 + 2*sin(1/2*d*x + 1/2*c)^2 + 2*...

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Fricas [A]
time = 0.39, size = 164, normalized size = 1.78 \begin {gather*} \frac {5 \, {\left (a^{2} \cos \left (d x + c\right )^{2} + a^{2} \cos \left (d x + c\right )\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 (2 \, a^{2} \cos \left (d x + c\right ) + a^{2}\right )} \sqrt {a \cos \left (d x + c\right ) + a} \sin \left (d x + c\right )}{4 \, {\left (d \cos \left (d x + c\right )^{2} + d \cos \left (d x + c\right )\right )}} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

1/4*(5*(a^2*cos(d*x + c)^2 + a^2*cos(d*x + c))*sqrt(a)*log((a*cos(d*x + c)^3 - 7*a*cos(d*x + c)^2 - 4*sqrt(a*c
os(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*(2*a^2*
cos(d*x + c) + a^2)*sqrt(a*cos(d*x + c) + a)*sin(d*x + c))/(d*cos(d*x + c)^2 + d*cos(d*x + c))

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Sympy [F(-1)] Timed out
time = 0.00, size = 0, normalized size = 0.00 \begin {gather*} \text {Timed out} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

Timed out

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Giac [A]
time = 0.52, size = 135, normalized size = 1.47 \begin {gather*} -\frac {\sqrt {2} {\left (5 \, \sqrt {2} a^{2} \log \left (\frac {{\left | -2 \, \sqrt {2} + 4 \, \sin \left (\frac {1}{2} \, d x + \frac {1}{2} \, c\right ) \right |}}{{\left | 2 \, \sqrt {2} + 4 \, \sin \left (\frac {1}{2} \, d x + \frac {1}{2} \, c\right ) \right |}}\right ) \mathrm {sgn}\left (\cos \left (\frac {1}{2} \, d x + \frac {1}{2} \, c\right )\right ) - 8 \, a^{2} \mathrm {sgn}\left (\cos \left (\frac {1}{2} \, d x + \frac {1}{2} \, c\right )\right ) \sin \left (\frac {1}{2} \, d x + \frac {1}{2} \, c\right ) + \frac {4 \, a^{2} \mathrm {sgn}\left (\cos \left (\frac {1}{2} \, d x + \frac {1}{2} \, c\right )\right ) \sin \left (\frac {1}{2} \, d x + \frac {1}{2} \, c\right )}{2 \, \sin \left (\frac {1}{2} \, d x + \frac {1}{2} \, c\right )^{2} - 1}\right )} \sqrt {a}}{4 \, d} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

-1/4*sqrt(2)*(5*sqrt(2)*a^2*log(abs(-2*sqrt(2) + 4*sin(1/2*d*x + 1/2*c))/abs(2*sqrt(2) + 4*sin(1/2*d*x + 1/2*c
)))*sgn(cos(1/2*d*x + 1/2*c)) - 8*a^2*sgn(cos(1/2*d*x + 1/2*c))*sin(1/2*d*x + 1/2*c) + 4*a^2*sgn(cos(1/2*d*x +
1/2*c))*sin(1/2*d*x + 1/2*c)/(2*sin(1/2*d*x + 1/2*c)^2 - 1))*sqrt(a)/d

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

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

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

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