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

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

[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 = 0.20, 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 = {2762, 2981, 2773, 206} \[ \frac {a^3 \sin (c+d x)}{d \sqrt {a \cos (c+d x)+a}}+\frac {a^2 \tan (c+d x) \sqrt {a \cos (c+d x)+a}}{d}+\frac {5 a^{5/2} \tanh ^{-1}\left (\frac {\sqrt {a} \sin (c+d x)}{\sqrt {a \cos (c+d x)+a}}\right )}{d} \]

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 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])

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 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 2981

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)*Sqr
t[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 ) \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 )}{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]  time = 36.23, size = 1547, normalized size = 16.82 \[ \text {result too large to display} \]

Warning: Unable to verify antiderivative.

[In]

Integrate[(a + a*Cos[c + d*x])^(5/2)*Sec[c + d*x]^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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fricas [A]  time = 1.20, size = 164, normalized size = 1.78 \[ \frac {5 \, {\left (a^{2} \cos \left (d x + c\right )^{2} + a^{2} \cos \left (d x + c\right )\right )} \sqrt {a} \log \left (\frac {a \cos \left (d x + c\right )^{3} - 7 \, a \cos \left (d x + c\right )^{2} - 4 \, \sqrt {a \cos \left (d x + c\right ) + a} \sqrt {a} {\left (\cos \left (d x + c\right ) - 2\right )} \sin \left (d x + c\right ) + 8 \, a}{\cos \left (d x + c\right )^{3} + \cos \left (d x + c\right )^{2}}\right ) + 4 \, {\left (2 \, a^{2} \cos \left (d x + c\right ) + a^{2}\right )} \sqrt {a \cos \left (d x + c\right ) + a} \sin \left (d x + c\right )}{4 \, {\left (d \cos \left (d x + c\right )^{2} + d \cos \left (d x + c\right )\right )}} \]

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

[Out]

Timed out

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maple [B]  time = 0.48, size = 408, normalized size = 4.43 \[ \frac {a^{\frac {3}{2}} \cos \left (\frac {d x}{2}+\frac {c}{2}\right ) \sqrt {a \left (\sin ^{2}\left (\frac {d x}{2}+\frac {c}{2}\right )\right )}\, \left (\left (-8 \sqrt {2}\, \sqrt {a \left (\sin ^{2}\left (\frac {d x}{2}+\frac {c}{2}\right )\right )}\, \sqrt {a}-10 \ln \left (-\frac {4 \left (\sqrt {2}\, \sqrt {a \left (\sin ^{2}\left (\frac {d x}{2}+\frac {c}{2}\right )\right )}\, \sqrt {a}-a \sqrt {2}\, \cos \left (\frac {d x}{2}+\frac {c}{2}\right )+2 a \right )}{-2 \cos \left (\frac {d x}{2}+\frac {c}{2}\right )+\sqrt {2}}\right ) a -10 \ln \left (\frac {4 \sqrt {2}\, \sqrt {a \left (\sin ^{2}\left (\frac {d x}{2}+\frac {c}{2}\right )\right )}\, \sqrt {a}+4 a \sqrt {2}\, \cos \left (\frac {d x}{2}+\frac {c}{2}\right )+8 a}{2 \cos \left (\frac {d x}{2}+\frac {c}{2}\right )+\sqrt {2}}\right ) a \right ) \left (\sin ^{2}\left (\frac {d x}{2}+\frac {c}{2}\right )\right )+6 \sqrt {2}\, \sqrt {a \left (\sin ^{2}\left (\frac {d x}{2}+\frac {c}{2}\right )\right )}\, \sqrt {a}+5 \ln \left (-\frac {4 \left (\sqrt {2}\, \sqrt {a \left (\sin ^{2}\left (\frac {d x}{2}+\frac {c}{2}\right )\right )}\, \sqrt {a}-a \sqrt {2}\, \cos \left (\frac {d x}{2}+\frac {c}{2}\right )+2 a \right )}{-2 \cos \left (\frac {d x}{2}+\frac {c}{2}\right )+\sqrt {2}}\right ) a +5 \ln \left (\frac {4 \sqrt {2}\, \sqrt {a \left (\sin ^{2}\left (\frac {d x}{2}+\frac {c}{2}\right )\right )}\, \sqrt {a}+4 a \sqrt {2}\, \cos \left (\frac {d x}{2}+\frac {c}{2}\right )+8 a}{2 \cos \left (\frac {d x}{2}+\frac {c}{2}\right )+\sqrt {2}}\right ) a \right )}{\left (2 \cos \left (\frac {d x}{2}+\frac {c}{2}\right )-\sqrt {2}\right ) \left (2 \cos \left (\frac {d x}{2}+\frac {c}{2}\right )+\sqrt {2}\right ) \sin \left (\frac {d x}{2}+\frac {c}{2}\right ) \sqrt {a \left (\cos ^{2}\left (\frac {d x}{2}+\frac {c}{2}\right )\right )}\, d} \]

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)

[Out]

a^(3/2)*cos(1/2*d*x+1/2*c)*(a*sin(1/2*d*x+1/2*c)^2)^(1/2)*((-8*2^(1/2)*(a*sin(1/2*d*x+1/2*c)^2)^(1/2)*a^(1/2)-
10*ln(-4/(-2*cos(1/2*d*x+1/2*c)+2^(1/2))*(2^(1/2)*(a*sin(1/2*d*x+1/2*c)^2)^(1/2)*a^(1/2)-a*2^(1/2)*cos(1/2*d*x
+1/2*c)+2*a))*a-10*ln(4/(2*cos(1/2*d*x+1/2*c)+2^(1/2))*(2^(1/2)*(a*sin(1/2*d*x+1/2*c)^2)^(1/2)*a^(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)^2+6*2^(1/2)*(a*sin(1/2*d*x+1/2*c)^2)^(1/2)*a^(1/2)+5*ln(-4/(
-2*cos(1/2*d*x+1/2*c)+2^(1/2))*(2^(1/2)*(a*sin(1/2*d*x+1/2*c)^2)^(1/2)*a^(1/2)-a*2^(1/2)*cos(1/2*d*x+1/2*c)+2*
a))*a+5*ln(4/(2*cos(1/2*d*x+1/2*c)+2^(1/2))*(2^(1/2)*(a*sin(1/2*d*x+1/2*c)^2)^(1/2)*a^(1/2)+a*2^(1/2)*cos(1/2*
d*x+1/2*c)+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 [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((a+a*cos(d*x+c))^(5/2)*sec(d*x+c)^2,x, algorithm="maxima")

[Out]

Timed out

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

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

[Out]

Timed out

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