3.3.49 \(\int \frac {1}{\cos ^{\frac {3}{2}}(c+d x) (a+a \cos (c+d x))^{5/2}} \, dx\) [249]

Optimal. Leaf size=177 \[ -\frac {75 \text {ArcTan}\left (\frac {\sqrt {a} \sin (c+d x)}{\sqrt {2} \sqrt {\cos (c+d x)} \sqrt {a+a \cos (c+d x)}}\right )}{16 \sqrt {2} a^{5/2} d}-\frac {\sin (c+d x)}{4 d \sqrt {\cos (c+d x)} (a+a \cos (c+d x))^{5/2}}-\frac {13 \sin (c+d x)}{16 a d \sqrt {\cos (c+d x)} (a+a \cos (c+d x))^{3/2}}+\frac {49 \sin (c+d x)}{16 a^2 d \sqrt {\cos (c+d x)} \sqrt {a+a \cos (c+d x)}} \]

[Out]

-75/32*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))/a^(5/2)/d*2^(1/2)-1/4*si
n(d*x+c)/d/(a+a*cos(d*x+c))^(5/2)/cos(d*x+c)^(1/2)-13/16*sin(d*x+c)/a/d/(a+a*cos(d*x+c))^(3/2)/cos(d*x+c)^(1/2
)+49/16*sin(d*x+c)/a^2/d/cos(d*x+c)^(1/2)/(a+a*cos(d*x+c))^(1/2)

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Rubi [A]
time = 0.26, antiderivative size = 177, normalized size of antiderivative = 1.00, number of steps used = 6, number of rules used = 6, integrand size = 25, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.240, Rules used = {2845, 3057, 3063, 12, 2861, 211} \begin {gather*} -\frac {75 \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 )}{16 \sqrt {2} a^{5/2} d}+\frac {49 \sin (c+d x)}{16 a^2 d \sqrt {\cos (c+d x)} \sqrt {a \cos (c+d x)+a}}-\frac {13 \sin (c+d x)}{16 a d \sqrt {\cos (c+d x)} (a \cos (c+d x)+a)^{3/2}}-\frac {\sin (c+d x)}{4 d \sqrt {\cos (c+d x)} (a \cos (c+d x)+a)^{5/2}} \end {gather*}

Antiderivative was successfully verified.

[In]

Int[1/(Cos[c + d*x]^(3/2)*(a + a*Cos[c + d*x])^(5/2)),x]

[Out]

(-75*ArcTan[(Sqrt[a]*Sin[c + d*x])/(Sqrt[2]*Sqrt[Cos[c + d*x]]*Sqrt[a + a*Cos[c + d*x]])])/(16*Sqrt[2]*a^(5/2)
*d) - Sin[c + d*x]/(4*d*Sqrt[Cos[c + d*x]]*(a + a*Cos[c + d*x])^(5/2)) - (13*Sin[c + d*x])/(16*a*d*Sqrt[Cos[c
+ d*x]]*(a + a*Cos[c + d*x])^(3/2)) + (49*Sin[c + d*x])/(16*a^2*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 2845

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

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 3057

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*(A*b - a*B)*Cos[e + f*x]*(a + b*Sin[e + f*x])^m*((c + d*Sin[e + f*
x])^(n + 1)/(a*f*(2*m + 1)*(b*c - a*d))), x] + Dist[1/(a*(2*m + 1)*(b*c - a*d)), Int[(a + b*Sin[e + f*x])^(m +
1)*(c + d*Sin[e + f*x])^n*Simp[B*(a*c*m + b*d*(n + 1)) + A*(b*c*(m + 1) - a*d*(2*m + n + 2)) + d*(A*b - a*B)*
(m + n + 2)*Sin[e + f*x], x], 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[m, -2^(-1)] &&  !GtQ[n, 0] && IntegerQ[2*m] && (IntegerQ[2*n] || EqQ[c,
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 {3}{2}}(c+d x) (a+a \cos (c+d x))^{5/2}} \, dx &=-\frac {\sin (c+d x)}{4 d \sqrt {\cos (c+d x)} (a+a \cos (c+d x))^{5/2}}+\frac {\int \frac {\frac {9 a}{2}-2 a \cos (c+d x)}{\cos ^{\frac {3}{2}}(c+d x) (a+a \cos (c+d x))^{3/2}} \, dx}{4 a^2}\\ &=-\frac {\sin (c+d x)}{4 d \sqrt {\cos (c+d x)} (a+a \cos (c+d x))^{5/2}}-\frac {13 \sin (c+d x)}{16 a d \sqrt {\cos (c+d x)} (a+a \cos (c+d x))^{3/2}}+\frac {\int \frac {\frac {49 a^2}{4}-\frac {13}{2} a^2 \cos (c+d x)}{\cos ^{\frac {3}{2}}(c+d x) \sqrt {a+a \cos (c+d x)}} \, dx}{8 a^4}\\ &=-\frac {\sin (c+d x)}{4 d \sqrt {\cos (c+d x)} (a+a \cos (c+d x))^{5/2}}-\frac {13 \sin (c+d x)}{16 a d \sqrt {\cos (c+d x)} (a+a \cos (c+d x))^{3/2}}+\frac {49 \sin (c+d x)}{16 a^2 d \sqrt {\cos (c+d x)} \sqrt {a+a \cos (c+d x)}}+\frac {\int -\frac {75 a^3}{8 \sqrt {\cos (c+d x)} \sqrt {a+a \cos (c+d x)}} \, dx}{4 a^5}\\ &=-\frac {\sin (c+d x)}{4 d \sqrt {\cos (c+d x)} (a+a \cos (c+d x))^{5/2}}-\frac {13 \sin (c+d x)}{16 a d \sqrt {\cos (c+d x)} (a+a \cos (c+d x))^{3/2}}+\frac {49 \sin (c+d x)}{16 a^2 d \sqrt {\cos (c+d x)} \sqrt {a+a \cos (c+d x)}}-\frac {75 \int \frac {1}{\sqrt {\cos (c+d x)} \sqrt {a+a \cos (c+d x)}} \, dx}{32 a^2}\\ &=-\frac {\sin (c+d x)}{4 d \sqrt {\cos (c+d x)} (a+a \cos (c+d x))^{5/2}}-\frac {13 \sin (c+d x)}{16 a d \sqrt {\cos (c+d x)} (a+a \cos (c+d x))^{3/2}}+\frac {49 \sin (c+d x)}{16 a^2 d \sqrt {\cos (c+d x)} \sqrt {a+a \cos (c+d x)}}+\frac {75 \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 )}{16 a d}\\ &=-\frac {75 \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 )}{16 \sqrt {2} a^{5/2} d}-\frac {\sin (c+d x)}{4 d \sqrt {\cos (c+d x)} (a+a \cos (c+d x))^{5/2}}-\frac {13 \sin (c+d x)}{16 a d \sqrt {\cos (c+d x)} (a+a \cos (c+d x))^{3/2}}+\frac {49 \sin (c+d x)}{16 a^2 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 = 7.94, size = 506, normalized size = 2.86 \begin {gather*} \frac {2 \cos ^5\left (\frac {c}{2}+\frac {d x}{2}\right ) \sec ^4\left (\frac {1}{2} (c+d x)\right ) \sin \left (\frac {c}{2}+\frac {d x}{2}\right ) \left (\frac {8 \cos ^6\left (\frac {1}{2} (c+d x)\right ) \, _4F_3\left (2,2,2,\frac {5}{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 ^2\left (\frac {c}{2}+\frac {d x}{2}\right )}{315 \left (-1+2 \sin ^2\left (\frac {c}{2}+\frac {d x}{2}\right )\right )}+\frac {1}{120} \csc ^8\left (\frac {c}{2}+\frac {d x}{2}\right ) \left (1-2 \sin ^2\left (\frac {c}{2}+\frac {d x}{2}\right )\right )^2 \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 )}} \left (-15 \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 ) \cos ^4\left (\frac {1}{2} (c+d x)\right ) \left (-343+1465 \sin ^2\left (\frac {c}{2}+\frac {d x}{2}\right )-2021 \sin ^4\left (\frac {c}{2}+\frac {d x}{2}\right )+824 \sin ^6\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 )}} \left (-5145+33980 \sin ^2\left (\frac {c}{2}+\frac {d x}{2}\right )-87764 \sin ^4\left (\frac {c}{2}+\frac {d x}{2}\right )+109737 \sin ^6\left (\frac {c}{2}+\frac {d x}{2}\right )-66122 \sin ^8\left (\frac {c}{2}+\frac {d x}{2}\right )+15344 \sin ^{10}\left (\frac {c}{2}+\frac {d x}{2}\right )\right )\right )\right )}{d (a (1+\cos (c+d x)))^{5/2} \left (1-2 \sin ^2\left (\frac {c}{2}+\frac {d x}{2}\right )\right )^{3/2}} \end {gather*}

Warning: Unable to verify antiderivative.

[In]

Integrate[1/(Cos[c + d*x]^(3/2)*(a + a*Cos[c + d*x])^(5/2)),x]

[Out]

(2*Cos[c/2 + (d*x)/2]^5*Sec[(c + d*x)/2]^4*Sin[c/2 + (d*x)/2]*((8*Cos[(c + d*x)/2]^6*HypergeometricPFQ[{2, 2,
2, 5/2}, {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]^2)/(315*(-1 + 2*
Sin[c/2 + (d*x)/2]^2)) + (Csc[c/2 + (d*x)/2]^8*(1 - 2*Sin[c/2 + (d*x)/2]^2)^2*Sqrt[Sin[c/2 + (d*x)/2]^2/(-1 +
2*Sin[c/2 + (d*x)/2]^2)]*(-15*ArcTanh[Sqrt[Sin[c/2 + (d*x)/2]^2/(-1 + 2*Sin[c/2 + (d*x)/2]^2)]]*Cos[(c + d*x)/
2]^4*(-343 + 1465*Sin[c/2 + (d*x)/2]^2 - 2021*Sin[c/2 + (d*x)/2]^4 + 824*Sin[c/2 + (d*x)/2]^6) + Sqrt[Sin[c/2
+ (d*x)/2]^2/(-1 + 2*Sin[c/2 + (d*x)/2]^2)]*(-5145 + 33980*Sin[c/2 + (d*x)/2]^2 - 87764*Sin[c/2 + (d*x)/2]^4 +
109737*Sin[c/2 + (d*x)/2]^6 - 66122*Sin[c/2 + (d*x)/2]^8 + 15344*Sin[c/2 + (d*x)/2]^10)))/120))/(d*(a*(1 + Co
s[c + d*x]))^(5/2)*(1 - 2*Sin[c/2 + (d*x)/2]^2)^(3/2))

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Maple [B] Leaf count of result is larger than twice the leaf count of optimal. \(302\) vs. \(2(146)=292\).
time = 0.21, size = 303, normalized size = 1.71

method	result	size
default	\(\frac {\left (75 \arcsin \left (\frac {-1+\cos \left (d x +c \right )}{\sin \left (d x +c \right )}\right ) \sin \left (d x +c \right ) \left (\cos ^{3}\left (d x +c \right )\right ) \left (\frac {\cos \left (d x +c \right )}{1+\cos \left (d x +c \right )}\right )^{\frac {3}{2}}+225 \sin \left (d x +c \right ) \left (\cos ^{2}\left (d x +c \right )\right ) \arcsin \left (\frac {-1+\cos \left (d x +c \right )}{\sin \left (d x +c \right )}\right ) \left (\frac {\cos \left (d x +c \right )}{1+\cos \left (d x +c \right )}\right )^{\frac {3}{2}}+225 \sin \left (d x +c \right ) \cos \left (d x +c \right ) \arcsin \left (\frac {-1+\cos \left (d x +c \right )}{\sin \left (d x +c \right )}\right ) \left (\frac {\cos \left (d x +c \right )}{1+\cos \left (d x +c \right )}\right )^{\frac {3}{2}}+75 \sin \left (d x +c \right ) \arcsin \left (\frac {-1+\cos \left (d x +c \right )}{\sin \left (d x +c \right )}\right ) \left (\frac {\cos \left (d x +c \right )}{1+\cos \left (d x +c \right )}\right )^{\frac {3}{2}}-49 \left (\cos ^{4}\left (d x +c \right )\right ) \sqrt {2}-36 \left (\cos ^{3}\left (d x +c \right )\right ) \sqrt {2}+53 \left (\cos ^{2}\left (d x +c \right )\right ) \sqrt {2}+32 \cos \left (d x +c \right ) \sqrt {2}\right ) \sqrt {a \left (1+\cos \left (d x +c \right )\right )}\, \sqrt {2}}{32 d \sin \left (d x +c \right ) \left (1+\cos \left (d x +c \right )\right )^{2} \cos \left (d x +c \right )^{\frac {3}{2}} a^{3}}\)	\(303\)

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

1/32/d*(75*arcsin((-1+cos(d*x+c))/sin(d*x+c))*sin(d*x+c)*cos(d*x+c)^3*(cos(d*x+c)/(1+cos(d*x+c)))^(3/2)+225*si
n(d*x+c)*cos(d*x+c)^2*arcsin((-1+cos(d*x+c))/sin(d*x+c))*(cos(d*x+c)/(1+cos(d*x+c)))^(3/2)+225*sin(d*x+c)*cos(
d*x+c)*arcsin((-1+cos(d*x+c))/sin(d*x+c))*(cos(d*x+c)/(1+cos(d*x+c)))^(3/2)+75*sin(d*x+c)*arcsin((-1+cos(d*x+c
))/sin(d*x+c))*(cos(d*x+c)/(1+cos(d*x+c)))^(3/2)-49*cos(d*x+c)^4*2^(1/2)-36*cos(d*x+c)^3*2^(1/2)+53*cos(d*x+c)
^2*2^(1/2)+32*cos(d*x+c)*2^(1/2))*(a*(1+cos(d*x+c)))^(1/2)/sin(d*x+c)/(1+cos(d*x+c))^2/cos(d*x+c)^(3/2)*2^(1/2
)/a^3

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Maxima [F]
time = 0.00, size = 0, normalized size = 0.00 \begin {gather*} \text {Failed to integrate} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate(1/cos(d*x+c)^(3/2)/(a+a*cos(d*x+c))^(5/2),x, algorithm="maxima")

[Out]

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

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Fricas [A]
time = 0.47, size = 205, normalized size = 1.16 \begin {gather*} -\frac {75 \, \sqrt {2} {\left (\cos \left (d x + c\right )^{4} + 3 \, \cos \left (d x + c\right )^{3} + 3 \, \cos \left (d x + c\right )^{2} + \cos \left (d x + c\right )\right )} \sqrt {a} \arctan \left (\frac {\sqrt {2} \sqrt {a \cos \left (d x + c\right ) + a} \sqrt {a} \sqrt {\cos \left (d x + c\right )} \sin \left (d x + c\right )}{2 \, {\left (a \cos \left (d x + c\right )^{2} + a \cos \left (d x + c\right )\right )}}\right ) - 2 \, \sqrt {a \cos \left (d x + c\right ) + a} {\left (49 \, \cos \left (d x + c\right )^{2} + 85 \, \cos \left (d x + c\right ) + 32\right )} \sqrt {\cos \left (d x + c\right )} \sin \left (d x + c\right )}{32 \, {\left (a^{3} d \cos \left (d x + c\right )^{4} + 3 \, a^{3} d \cos \left (d x + c\right )^{3} + 3 \, a^{3} d \cos \left (d x + c\right )^{2} + a^{3} d \cos \left (d x + c\right )\right )}} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate(1/cos(d*x+c)^(3/2)/(a+a*cos(d*x+c))^(5/2),x, algorithm="fricas")

[Out]

-1/32*(75*sqrt(2)*(cos(d*x + c)^4 + 3*cos(d*x + c)^3 + 3*cos(d*x + c)^2 + cos(d*x + c))*sqrt(a)*arctan(1/2*sqr
t(2)*sqrt(a*cos(d*x + c) + a)*sqrt(a)*sqrt(cos(d*x + c))*sin(d*x + c)/(a*cos(d*x + c)^2 + a*cos(d*x + c))) - 2
*sqrt(a*cos(d*x + c) + a)*(49*cos(d*x + c)^2 + 85*cos(d*x + c) + 32)*sqrt(cos(d*x + c))*sin(d*x + c))/(a^3*d*c
os(d*x + c)^4 + 3*a^3*d*cos(d*x + c)^3 + 3*a^3*d*cos(d*x + c)^2 + a^3*d*cos(d*x + c))

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

[Out]

Exception raised: SystemError >> excessive stack use: stack is 3005 deep

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Giac [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(1/cos(d*x+c)^(3/2)/(a+a*cos(d*x+c))^(5/2),x, algorithm="giac")

[Out]

Timed out

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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 )}^{3/2}\,{\left (a+a\,\cos \left (c+d\,x\right )\right )}^{5/2}} \,d x \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

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

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