3.2.95 \(\int \frac {1}{\sqrt {\cos (c+d x)} (a+a \cos (c+d x))^3} \, dx\) [195]

Optimal. Leaf size=155 \[ \frac {9 E\left (\left .\frac {1}{2} (c+d x)\right |2\right )}{10 a^3 d}+\frac {F\left (\left .\frac {1}{2} (c+d x)\right |2\right )}{2 a^3 d}-\frac {\sqrt {\cos (c+d x)} \sin (c+d x)}{5 d (a+a \cos (c+d x))^3}-\frac {2 \sqrt {\cos (c+d x)} \sin (c+d x)}{5 a d (a+a \cos (c+d x))^2}-\frac {9 \sqrt {\cos (c+d x)} \sin (c+d x)}{10 d \left (a^3+a^3 \cos (c+d x)\right )} \]

[Out]

9/10*(cos(1/2*d*x+1/2*c)^2)^(1/2)/cos(1/2*d*x+1/2*c)*EllipticE(sin(1/2*d*x+1/2*c),2^(1/2))/a^3/d+1/2*(cos(1/2*
d*x+1/2*c)^2)^(1/2)/cos(1/2*d*x+1/2*c)*EllipticF(sin(1/2*d*x+1/2*c),2^(1/2))/a^3/d-1/5*sin(d*x+c)*cos(d*x+c)^(
1/2)/d/(a+a*cos(d*x+c))^3-2/5*sin(d*x+c)*cos(d*x+c)^(1/2)/a/d/(a+a*cos(d*x+c))^2-9/10*sin(d*x+c)*cos(d*x+c)^(1
/2)/d/(a^3+a^3*cos(d*x+c))

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Rubi [A]
time = 0.21, antiderivative size = 155, normalized size of antiderivative = 1.00, number of steps used = 6, number of rules used = 5, integrand size = 23, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.217, Rules used = {2845, 3057, 2827, 2720, 2719} \begin {gather*} \frac {F\left (\left .\frac {1}{2} (c+d x)\right |2\right )}{2 a^3 d}+\frac {9 E\left (\left .\frac {1}{2} (c+d x)\right |2\right )}{10 a^3 d}-\frac {9 \sin (c+d x) \sqrt {\cos (c+d x)}}{10 d \left (a^3 \cos (c+d x)+a^3\right )}-\frac {2 \sin (c+d x) \sqrt {\cos (c+d x)}}{5 a d (a \cos (c+d x)+a)^2}-\frac {\sin (c+d x) \sqrt {\cos (c+d x)}}{5 d (a \cos (c+d x)+a)^3} \end {gather*}

Antiderivative was successfully verified.

[In]

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

[Out]

(9*EllipticE[(c + d*x)/2, 2])/(10*a^3*d) + EllipticF[(c + d*x)/2, 2]/(2*a^3*d) - (Sqrt[Cos[c + d*x]]*Sin[c + d
*x])/(5*d*(a + a*Cos[c + d*x])^3) - (2*Sqrt[Cos[c + d*x]]*Sin[c + d*x])/(5*a*d*(a + a*Cos[c + d*x])^2) - (9*Sq
rt[Cos[c + d*x]]*Sin[c + d*x])/(10*d*(a^3 + a^3*Cos[c + d*x]))

Rule 2719

Int[Sqrt[sin[(c_.) + (d_.)*(x_)]], x_Symbol] :> Simp[(2/d)*EllipticE[(1/2)*(c - Pi/2 + d*x), 2], x] /; FreeQ[{
c, d}, x]

Rule 2720

Int[1/Sqrt[sin[(c_.) + (d_.)*(x_)]], x_Symbol] :> Simp[(2/d)*EllipticF[(1/2)*(c - Pi/2 + d*x), 2], x] /; FreeQ
[{c, d}, x]

Rule 2827

Int[((b_.)*sin[(e_.) + (f_.)*(x_)])^(m_)*((c_) + (d_.)*sin[(e_.) + (f_.)*(x_)]), x_Symbol] :> Dist[c, Int[(b*S
in[e + f*x])^m, x], x] + Dist[d/b, Int[(b*Sin[e + f*x])^(m + 1), x], x] /; FreeQ[{b, c, d, e, f, m}, x]

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 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])

Rubi steps

\begin {align*} \int \frac {1}{\sqrt {\cos (c+d x)} (a+a \cos (c+d x))^3} \, dx &=-\frac {\sqrt {\cos (c+d x)} \sin (c+d x)}{5 d (a+a \cos (c+d x))^3}+\frac {\int \frac {\frac {9 a}{2}-\frac {3}{2} a \cos (c+d x)}{\sqrt {\cos (c+d x)} (a+a \cos (c+d x))^2} \, dx}{5 a^2}\\ &=-\frac {\sqrt {\cos (c+d x)} \sin (c+d x)}{5 d (a+a \cos (c+d x))^3}-\frac {2 \sqrt {\cos (c+d x)} \sin (c+d x)}{5 a d (a+a \cos (c+d x))^2}+\frac {\int \frac {\frac {21 a^2}{2}-3 a^2 \cos (c+d x)}{\sqrt {\cos (c+d x)} (a+a \cos (c+d x))} \, dx}{15 a^4}\\ &=-\frac {\sqrt {\cos (c+d x)} \sin (c+d x)}{5 d (a+a \cos (c+d x))^3}-\frac {2 \sqrt {\cos (c+d x)} \sin (c+d x)}{5 a d (a+a \cos (c+d x))^2}-\frac {9 \sqrt {\cos (c+d x)} \sin (c+d x)}{10 d \left (a^3+a^3 \cos (c+d x)\right )}+\frac {\int \frac {\frac {15 a^3}{4}+\frac {27}{4} a^3 \cos (c+d x)}{\sqrt {\cos (c+d x)}} \, dx}{15 a^6}\\ &=-\frac {\sqrt {\cos (c+d x)} \sin (c+d x)}{5 d (a+a \cos (c+d x))^3}-\frac {2 \sqrt {\cos (c+d x)} \sin (c+d x)}{5 a d (a+a \cos (c+d x))^2}-\frac {9 \sqrt {\cos (c+d x)} \sin (c+d x)}{10 d \left (a^3+a^3 \cos (c+d x)\right )}+\frac {\int \frac {1}{\sqrt {\cos (c+d x)}} \, dx}{4 a^3}+\frac {9 \int \sqrt {\cos (c+d x)} \, dx}{20 a^3}\\ &=\frac {9 E\left (\left .\frac {1}{2} (c+d x)\right |2\right )}{10 a^3 d}+\frac {F\left (\left .\frac {1}{2} (c+d x)\right |2\right )}{2 a^3 d}-\frac {\sqrt {\cos (c+d x)} \sin (c+d x)}{5 d (a+a \cos (c+d x))^3}-\frac {2 \sqrt {\cos (c+d x)} \sin (c+d x)}{5 a d (a+a \cos (c+d x))^2}-\frac {9 \sqrt {\cos (c+d x)} \sin (c+d x)}{10 d \left (a^3+a^3 \cos (c+d x)\right )}\\ \end {align*}

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Mathematica [C] Result contains higher order function than in optimal. Order 5 vs. order 4 in optimal.
time = 6.49, size = 705, normalized size = 4.55 \begin {gather*} \frac {9 i \cos ^6\left (\frac {c}{2}+\frac {d x}{2}\right ) \csc \left (\frac {c}{2}\right ) \sec \left (\frac {c}{2}\right ) \left (\frac {2 e^{2 i d x} \, _2F_1\left (\frac {1}{2},\frac {3}{4};\frac {7}{4};-e^{2 i d x} (\cos (c)+i \sin (c))^2\right ) \sqrt {e^{-i d x} \left (2 \left (1+e^{2 i d x}\right ) \cos (c)+2 i \left (-1+e^{2 i d x}\right ) \sin (c)\right )} \sqrt {1+e^{2 i d x} \cos (2 c)+i e^{2 i d x} \sin (2 c)}}{3 i d \left (1+e^{2 i d x}\right ) \cos (c)-3 d \left (-1+e^{2 i d x}\right ) \sin (c)}-\frac {2 \, _2F_1\left (-\frac {1}{4},\frac {1}{2};\frac {3}{4};-e^{2 i d x} (\cos (c)+i \sin (c))^2\right ) \sqrt {e^{-i d x} \left (2 \left (1+e^{2 i d x}\right ) \cos (c)+2 i \left (-1+e^{2 i d x}\right ) \sin (c)\right )} \sqrt {1+e^{2 i d x} \cos (2 c)+i e^{2 i d x} \sin (2 c)}}{-i d \left (1+e^{2 i d x}\right ) \cos (c)+d \left (-1+e^{2 i d x}\right ) \sin (c)}\right )}{10 (a+a \cos (c+d x))^3}-\frac {2 \cos ^6\left (\frac {c}{2}+\frac {d x}{2}\right ) \csc \left (\frac {c}{2}\right ) \, _2F_1\left (\frac {1}{4},\frac {1}{2};\frac {5}{4};\sin ^2(d x-\text {ArcTan}(\cot (c)))\right ) \sec \left (\frac {c}{2}\right ) \sec (d x-\text {ArcTan}(\cot (c))) \sqrt {1-\sin (d x-\text {ArcTan}(\cot (c)))} \sqrt {-\sqrt {1+\cot ^2(c)} \sin (c) \sin (d x-\text {ArcTan}(\cot (c)))} \sqrt {1+\sin (d x-\text {ArcTan}(\cot (c)))}}{d (a+a \cos (c+d x))^3 \sqrt {1+\cot ^2(c)}}+\frac {\cos ^6\left (\frac {c}{2}+\frac {d x}{2}\right ) \sqrt {\cos (c+d x)} \left (-\frac {36 \csc (c)}{5 d}-\frac {36 \sec \left (\frac {c}{2}\right ) \sec \left (\frac {c}{2}+\frac {d x}{2}\right ) \sin \left (\frac {d x}{2}\right )}{5 d}-\frac {8 \sec \left (\frac {c}{2}\right ) \sec ^3\left (\frac {c}{2}+\frac {d x}{2}\right ) \sin \left (\frac {d x}{2}\right )}{5 d}-\frac {2 \sec \left (\frac {c}{2}\right ) \sec ^5\left (\frac {c}{2}+\frac {d x}{2}\right ) \sin \left (\frac {d x}{2}\right )}{5 d}-\frac {8 \sec ^2\left (\frac {c}{2}+\frac {d x}{2}\right ) \tan \left (\frac {c}{2}\right )}{5 d}-\frac {2 \sec ^4\left (\frac {c}{2}+\frac {d x}{2}\right ) \tan \left (\frac {c}{2}\right )}{5 d}\right )}{(a+a \cos (c+d x))^3} \end {gather*}

Warning: Unable to verify antiderivative.

[In]

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

[Out]

(((9*I)/10)*Cos[c/2 + (d*x)/2]^6*Csc[c/2]*Sec[c/2]*((2*E^((2*I)*d*x)*Hypergeometric2F1[1/2, 3/4, 7/4, -(E^((2*
I)*d*x)*(Cos[c] + I*Sin[c])^2)]*Sqrt[(2*(1 + E^((2*I)*d*x))*Cos[c] + (2*I)*(-1 + E^((2*I)*d*x))*Sin[c])/E^(I*d
*x)]*Sqrt[1 + E^((2*I)*d*x)*Cos[2*c] + I*E^((2*I)*d*x)*Sin[2*c]])/((3*I)*d*(1 + E^((2*I)*d*x))*Cos[c] - 3*d*(-
1 + E^((2*I)*d*x))*Sin[c]) - (2*Hypergeometric2F1[-1/4, 1/2, 3/4, -(E^((2*I)*d*x)*(Cos[c] + I*Sin[c])^2)]*Sqrt
[(2*(1 + E^((2*I)*d*x))*Cos[c] + (2*I)*(-1 + E^((2*I)*d*x))*Sin[c])/E^(I*d*x)]*Sqrt[1 + E^((2*I)*d*x)*Cos[2*c]
+ I*E^((2*I)*d*x)*Sin[2*c]])/((-I)*d*(1 + E^((2*I)*d*x))*Cos[c] + d*(-1 + E^((2*I)*d*x))*Sin[c])))/(a + a*Cos
[c + d*x])^3 - (2*Cos[c/2 + (d*x)/2]^6*Csc[c/2]*HypergeometricPFQ[{1/4, 1/2}, {5/4}, Sin[d*x - ArcTan[Cot[c]]]
^2]*Sec[c/2]*Sec[d*x - ArcTan[Cot[c]]]*Sqrt[1 - Sin[d*x - ArcTan[Cot[c]]]]*Sqrt[-(Sqrt[1 + Cot[c]^2]*Sin[c]*Si
n[d*x - ArcTan[Cot[c]]])]*Sqrt[1 + Sin[d*x - ArcTan[Cot[c]]]])/(d*(a + a*Cos[c + d*x])^3*Sqrt[1 + Cot[c]^2]) +
(Cos[c/2 + (d*x)/2]^6*Sqrt[Cos[c + d*x]]*((-36*Csc[c])/(5*d) - (36*Sec[c/2]*Sec[c/2 + (d*x)/2]*Sin[(d*x)/2])/
(5*d) - (8*Sec[c/2]*Sec[c/2 + (d*x)/2]^3*Sin[(d*x)/2])/(5*d) - (2*Sec[c/2]*Sec[c/2 + (d*x)/2]^5*Sin[(d*x)/2])/
(5*d) - (8*Sec[c/2 + (d*x)/2]^2*Tan[c/2])/(5*d) - (2*Sec[c/2 + (d*x)/2]^4*Tan[c/2])/(5*d)))/(a + a*Cos[c + d*x
])^3

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Maple [A]
time = 0.16, size = 268, normalized size = 1.73

method	result	size
default	\(\frac {\sqrt {\left (2 \left (\cos ^{2}\left (\frac {d x}{2}+\frac {c}{2}\right )\right )-1\right ) \left (\sin ^{2}\left (\frac {d x}{2}+\frac {c}{2}\right )\right )}\, \left (36 \left (\cos ^{8}\left (\frac {d x}{2}+\frac {c}{2}\right )\right )-10 \sqrt {\frac {1}{2}-\frac {\cos \left (d x +c \right )}{2}}\, \sqrt {-2 \left (\cos ^{2}\left (\frac {d x}{2}+\frac {c}{2}\right )\right )+1}\, \EllipticF \left (\cos \left (\frac {d x}{2}+\frac {c}{2}\right ), \sqrt {2}\right ) \left (\cos ^{5}\left (\frac {d x}{2}+\frac {c}{2}\right )\right )+18 \sqrt {\frac {1}{2}-\frac {\cos \left (d x +c \right )}{2}}\, \sqrt {-2 \left (\cos ^{2}\left (\frac {d x}{2}+\frac {c}{2}\right )\right )+1}\, \left (\cos ^{5}\left (\frac {d x}{2}+\frac {c}{2}\right )\right ) \EllipticE \left (\cos \left (\frac {d x}{2}+\frac {c}{2}\right ), \sqrt {2}\right )-46 \left (\cos ^{6}\left (\frac {d x}{2}+\frac {c}{2}\right )\right )+8 \left (\cos ^{4}\left (\frac {d x}{2}+\frac {c}{2}\right )\right )+\cos ^{2}\left (\frac {d x}{2}+\frac {c}{2}\right )+1\right )}{20 a^{3} \cos \left (\frac {d x}{2}+\frac {c}{2}\right )^{5} \sqrt {-2 \left (\sin ^{4}\left (\frac {d x}{2}+\frac {c}{2}\right )\right )+\sin ^{2}\left (\frac {d x}{2}+\frac {c}{2}\right )}\, \sin \left (\frac {d x}{2}+\frac {c}{2}\right ) \sqrt {2 \left (\cos ^{2}\left (\frac {d x}{2}+\frac {c}{2}\right )\right )-1}\, d}\)	\(268\)

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

1/20*((2*cos(1/2*d*x+1/2*c)^2-1)*sin(1/2*d*x+1/2*c)^2)^(1/2)*(36*cos(1/2*d*x+1/2*c)^8-10*(sin(1/2*d*x+1/2*c)^2
)^(1/2)*(-2*cos(1/2*d*x+1/2*c)^2+1)^(1/2)*EllipticF(cos(1/2*d*x+1/2*c),2^(1/2))*cos(1/2*d*x+1/2*c)^5+18*(sin(1
/2*d*x+1/2*c)^2)^(1/2)*(-2*cos(1/2*d*x+1/2*c)^2+1)^(1/2)*cos(1/2*d*x+1/2*c)^5*EllipticE(cos(1/2*d*x+1/2*c),2^(
1/2))-46*cos(1/2*d*x+1/2*c)^6+8*cos(1/2*d*x+1/2*c)^4+cos(1/2*d*x+1/2*c)^2+1)/a^3/cos(1/2*d*x+1/2*c)^5/(-2*sin(
1/2*d*x+1/2*c)^4+sin(1/2*d*x+1/2*c)^2)^(1/2)/sin(1/2*d*x+1/2*c)/(2*cos(1/2*d*x+1/2*c)^2-1)^(1/2)/d

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

[Out]

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

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Fricas [C] Result contains higher order function than in optimal. Order 9 vs. order 4.
time = 0.12, size = 344, normalized size = 2.22 \begin {gather*} -\frac {2 \, {\left (9 \, \cos \left (d x + c\right )^{2} + 22 \, \cos \left (d x + c\right ) + 15\right )} \sqrt {\cos \left (d x + c\right )} \sin \left (d x + c\right ) + 5 \, {\left (i \, \sqrt {2} \cos \left (d x + c\right )^{3} + 3 i \, \sqrt {2} \cos \left (d x + c\right )^{2} + 3 i \, \sqrt {2} \cos \left (d x + c\right ) + i \, \sqrt {2}\right )} {\rm weierstrassPInverse}\left (-4, 0, \cos \left (d x + c\right ) + i \, \sin \left (d x + c\right )\right ) + 5 \, {\left (-i \, \sqrt {2} \cos \left (d x + c\right )^{3} - 3 i \, \sqrt {2} \cos \left (d x + c\right )^{2} - 3 i \, \sqrt {2} \cos \left (d x + c\right ) - i \, \sqrt {2}\right )} {\rm weierstrassPInverse}\left (-4, 0, \cos \left (d x + c\right ) - i \, \sin \left (d x + c\right )\right ) + 9 \, {\left (-i \, \sqrt {2} \cos \left (d x + c\right )^{3} - 3 i \, \sqrt {2} \cos \left (d x + c\right )^{2} - 3 i \, \sqrt {2} \cos \left (d x + c\right ) - i \, \sqrt {2}\right )} {\rm weierstrassZeta}\left (-4, 0, {\rm weierstrassPInverse}\left (-4, 0, \cos \left (d x + c\right ) + i \, \sin \left (d x + c\right )\right )\right ) + 9 \, {\left (i \, \sqrt {2} \cos \left (d x + c\right )^{3} + 3 i \, \sqrt {2} \cos \left (d x + c\right )^{2} + 3 i \, \sqrt {2} \cos \left (d x + c\right ) + i \, \sqrt {2}\right )} {\rm weierstrassZeta}\left (-4, 0, {\rm weierstrassPInverse}\left (-4, 0, \cos \left (d x + c\right ) - i \, \sin \left (d x + c\right )\right )\right )}{20 \, {\left (a^{3} d \cos \left (d x + c\right )^{3} + 3 \, a^{3} d \cos \left (d x + c\right )^{2} + 3 \, a^{3} d \cos \left (d x + c\right ) + a^{3} d\right )}} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

-1/20*(2*(9*cos(d*x + c)^2 + 22*cos(d*x + c) + 15)*sqrt(cos(d*x + c))*sin(d*x + c) + 5*(I*sqrt(2)*cos(d*x + c)
^3 + 3*I*sqrt(2)*cos(d*x + c)^2 + 3*I*sqrt(2)*cos(d*x + c) + I*sqrt(2))*weierstrassPInverse(-4, 0, cos(d*x + c
) + I*sin(d*x + c)) + 5*(-I*sqrt(2)*cos(d*x + c)^3 - 3*I*sqrt(2)*cos(d*x + c)^2 - 3*I*sqrt(2)*cos(d*x + c) - I
*sqrt(2))*weierstrassPInverse(-4, 0, cos(d*x + c) - I*sin(d*x + c)) + 9*(-I*sqrt(2)*cos(d*x + c)^3 - 3*I*sqrt(
2)*cos(d*x + c)^2 - 3*I*sqrt(2)*cos(d*x + c) - I*sqrt(2))*weierstrassZeta(-4, 0, weierstrassPInverse(-4, 0, co
s(d*x + c) + I*sin(d*x + c))) + 9*(I*sqrt(2)*cos(d*x + c)^3 + 3*I*sqrt(2)*cos(d*x + c)^2 + 3*I*sqrt(2)*cos(d*x
+ c) + I*sqrt(2))*weierstrassZeta(-4, 0, weierstrassPInverse(-4, 0, cos(d*x + c) - I*sin(d*x + c))))/(a^3*d*c
os(d*x + c)^3 + 3*a^3*d*cos(d*x + c)^2 + 3*a^3*d*cos(d*x + c) + a^3*d)

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Sympy [F]
time = 0.00, size = 0, normalized size = 0.00 \begin {gather*} \frac {\int \frac {1}{\cos ^{\frac {7}{2}}{\left (c + d x \right )} + 3 \cos ^{\frac {5}{2}}{\left (c + d x \right )} + 3 \cos ^{\frac {3}{2}}{\left (c + d x \right )} + \sqrt {\cos {\left (c + d x \right )}}}\, dx}{a^{3}} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

Integral(1/(cos(c + d*x)**(7/2) + 3*cos(c + d*x)**(5/2) + 3*cos(c + d*x)**(3/2) + sqrt(cos(c + d*x))), x)/a**3

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

[Out]

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

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Mupad [F]
time = 0.00, size = -1, normalized size = -0.01 \begin {gather*} \int \frac {1}{\sqrt {\cos \left (c+d\,x\right )}\,{\left (a+a\,\cos \left (c+d\,x\right )\right )}^3} \,d x \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

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

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