3.390 \(\int \frac {1}{(a+a \cos (c+d x))^{7/2} \sec ^{\frac {3}{2}}(c+d x)} \, dx\)

Optimal. Leaf size=197 \[ \frac {7 \sqrt {\cos (c+d x)} \sqrt {\sec (c+d x)} \tan ^{-1}\left (\frac {\sqrt {a} \sin (c+d x)}{\sqrt {2} \sqrt {\cos (c+d x)} \sqrt {a \cos (c+d x)+a}}\right )}{64 \sqrt {2} a^{7/2} d}+\frac {17 \sin (c+d x)}{192 a^2 d \sqrt {\sec (c+d x)} (a \cos (c+d x)+a)^{3/2}}+\frac {3 \sin (c+d x)}{16 a d \sqrt {\sec (c+d x)} (a \cos (c+d x)+a)^{5/2}}-\frac {\sin (c+d x)}{6 d \sqrt {\sec (c+d x)} (a \cos (c+d x)+a)^{7/2}} \]

[Out]

-1/6*sin(d*x+c)/d/(a+a*cos(d*x+c))^(7/2)/sec(d*x+c)^(1/2)+3/16*sin(d*x+c)/a/d/(a+a*cos(d*x+c))^(5/2)/sec(d*x+c
)^(1/2)+17/192*sin(d*x+c)/a^2/d/(a+a*cos(d*x+c))^(3/2)/sec(d*x+c)^(1/2)+7/128*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))*cos(d*x+c)^(1/2)*sec(d*x+c)^(1/2)/a^(7/2)/d*2^(1/2)

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Rubi [A]  time = 0.50, antiderivative size = 197, normalized size of antiderivative = 1.00, number of steps used = 7, number of rules used = 6, integrand size = 25, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.240, Rules used = {4222, 2765, 2978, 12, 2782, 205} \[ \frac {17 \sin (c+d x)}{192 a^2 d \sqrt {\sec (c+d x)} (a \cos (c+d x)+a)^{3/2}}+\frac {7 \sqrt {\cos (c+d x)} \sqrt {\sec (c+d x)} \tan ^{-1}\left (\frac {\sqrt {a} \sin (c+d x)}{\sqrt {2} \sqrt {\cos (c+d x)} \sqrt {a \cos (c+d x)+a}}\right )}{64 \sqrt {2} a^{7/2} d}+\frac {3 \sin (c+d x)}{16 a d \sqrt {\sec (c+d x)} (a \cos (c+d x)+a)^{5/2}}-\frac {\sin (c+d x)}{6 d \sqrt {\sec (c+d x)} (a \cos (c+d x)+a)^{7/2}} \]

Antiderivative was successfully verified.

[In]

Int[1/((a + a*Cos[c + d*x])^(7/2)*Sec[c + d*x]^(3/2)),x]

[Out]

(7*ArcTan[(Sqrt[a]*Sin[c + d*x])/(Sqrt[2]*Sqrt[Cos[c + d*x]]*Sqrt[a + a*Cos[c + d*x]])]*Sqrt[Cos[c + d*x]]*Sqr
t[Sec[c + d*x]])/(64*Sqrt[2]*a^(7/2)*d) - Sin[c + d*x]/(6*d*(a + a*Cos[c + d*x])^(7/2)*Sqrt[Sec[c + d*x]]) + (
3*Sin[c + d*x])/(16*a*d*(a + a*Cos[c + d*x])^(5/2)*Sqrt[Sec[c + d*x]]) + (17*Sin[c + d*x])/(192*a^2*d*(a + a*C
os[c + d*x])^(3/2)*Sqrt[Sec[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 205

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

Rule 2765

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

Rule 2782

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 2978

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 4222

Int[(csc[(a_.) + (b_.)*(x_)]*(c_.))^(m_.)*(u_), x_Symbol] :> Dist[(c*Csc[a + b*x])^m*(c*Sin[a + b*x])^m, Int[A
ctivateTrig[u]/(c*Sin[a + b*x])^m, x], x] /; FreeQ[{a, b, c, m}, x] &&  !IntegerQ[m] && KnownSineIntegrandQ[u,
x]

Rubi steps

\begin {align*} \int \frac {1}{(a+a \cos (c+d x))^{7/2} \sec ^{\frac {3}{2}}(c+d x)} \, dx &=\left (\sqrt {\cos (c+d x)} \sqrt {\sec (c+d x)}\right ) \int \frac {\cos ^{\frac {3}{2}}(c+d x)}{(a+a \cos (c+d x))^{7/2}} \, dx\\ &=-\frac {\sin (c+d x)}{6 d (a+a \cos (c+d x))^{7/2} \sqrt {\sec (c+d x)}}-\frac {\left (\sqrt {\cos (c+d x)} \sqrt {\sec (c+d x)}\right ) \int \frac {\frac {a}{2}-4 a \cos (c+d x)}{\sqrt {\cos (c+d x)} (a+a \cos (c+d x))^{5/2}} \, dx}{6 a^2}\\ &=-\frac {\sin (c+d x)}{6 d (a+a \cos (c+d x))^{7/2} \sqrt {\sec (c+d x)}}+\frac {3 \sin (c+d x)}{16 a d (a+a \cos (c+d x))^{5/2} \sqrt {\sec (c+d x)}}-\frac {\left (\sqrt {\cos (c+d x)} \sqrt {\sec (c+d x)}\right ) \int \frac {-\frac {a^2}{4}-\frac {9}{2} a^2 \cos (c+d x)}{\sqrt {\cos (c+d x)} (a+a \cos (c+d x))^{3/2}} \, dx}{24 a^4}\\ &=-\frac {\sin (c+d x)}{6 d (a+a \cos (c+d x))^{7/2} \sqrt {\sec (c+d x)}}+\frac {3 \sin (c+d x)}{16 a d (a+a \cos (c+d x))^{5/2} \sqrt {\sec (c+d x)}}+\frac {17 \sin (c+d x)}{192 a^2 d (a+a \cos (c+d x))^{3/2} \sqrt {\sec (c+d x)}}-\frac {\left (\sqrt {\cos (c+d x)} \sqrt {\sec (c+d x)}\right ) \int -\frac {21 a^3}{8 \sqrt {\cos (c+d x)} \sqrt {a+a \cos (c+d x)}} \, dx}{48 a^6}\\ &=-\frac {\sin (c+d x)}{6 d (a+a \cos (c+d x))^{7/2} \sqrt {\sec (c+d x)}}+\frac {3 \sin (c+d x)}{16 a d (a+a \cos (c+d x))^{5/2} \sqrt {\sec (c+d x)}}+\frac {17 \sin (c+d x)}{192 a^2 d (a+a \cos (c+d x))^{3/2} \sqrt {\sec (c+d x)}}+\frac {\left (7 \sqrt {\cos (c+d x)} \sqrt {\sec (c+d x)}\right ) \int \frac {1}{\sqrt {\cos (c+d x)} \sqrt {a+a \cos (c+d x)}} \, dx}{128 a^3}\\ &=-\frac {\sin (c+d x)}{6 d (a+a \cos (c+d x))^{7/2} \sqrt {\sec (c+d x)}}+\frac {3 \sin (c+d x)}{16 a d (a+a \cos (c+d x))^{5/2} \sqrt {\sec (c+d x)}}+\frac {17 \sin (c+d x)}{192 a^2 d (a+a \cos (c+d x))^{3/2} \sqrt {\sec (c+d x)}}-\frac {\left (7 \sqrt {\cos (c+d x)} \sqrt {\sec (c+d x)}\right ) \operatorname {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 )}{64 a^2 d}\\ &=\frac {7 \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 {\cos (c+d x)} \sqrt {\sec (c+d x)}}{64 \sqrt {2} a^{7/2} d}-\frac {\sin (c+d x)}{6 d (a+a \cos (c+d x))^{7/2} \sqrt {\sec (c+d x)}}+\frac {3 \sin (c+d x)}{16 a d (a+a \cos (c+d x))^{5/2} \sqrt {\sec (c+d x)}}+\frac {17 \sin (c+d x)}{192 a^2 d (a+a \cos (c+d x))^{3/2} \sqrt {\sec (c+d x)}}\\ \end {align*}

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Mathematica [A]  time = 3.69, size = 153, normalized size = 0.78 \[ \frac {\tan \left (\frac {1}{2} (c+d x)\right ) \sec ^4\left (\frac {1}{2} (c+d x)\right ) \left (2 (140 \cos (c+d x)+17 \cos (2 (c+d x))+59) \sqrt {2-2 \sec (c+d x)}+672 \cos ^6\left (\frac {1}{2} (c+d x)\right ) \sec (c+d x) \tanh ^{-1}\left (\sqrt {\sin ^2\left (\frac {1}{2} (c+d x)\right ) (-\sec (c+d x))}\right )\right )}{3072 \sqrt {2} a^3 d \sqrt {-((\sec (c+d x)-1) \sec (c+d x))} \sqrt {a (\cos (c+d x)+1)}} \]

Warning: Unable to verify antiderivative.

[In]

Integrate[1/((a + a*Cos[c + d*x])^(7/2)*Sec[c + d*x]^(3/2)),x]

[Out]

(Sec[(c + d*x)/2]^4*(2*(59 + 140*Cos[c + d*x] + 17*Cos[2*(c + d*x)])*Sqrt[2 - 2*Sec[c + d*x]] + 672*ArcTanh[Sq
rt[-(Sec[c + d*x]*Sin[(c + d*x)/2]^2)]]*Cos[(c + d*x)/2]^6*Sec[c + d*x])*Tan[(c + d*x)/2])/(3072*Sqrt[2]*a^3*d
*Sqrt[a*(1 + Cos[c + d*x])]*Sqrt[-((-1 + Sec[c + d*x])*Sec[c + d*x])])

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fricas [A]  time = 1.90, size = 203, normalized size = 1.03 \[ -\frac {21 \, \sqrt {2} {\left (\cos \left (d x + c\right )^{4} + 4 \, \cos \left (d x + c\right )^{3} + 6 \, \cos \left (d x + c\right )^{2} + 4 \, \cos \left (d x + c\right ) + 1\right )} \sqrt {a} \arctan \left (\frac {\sqrt {2} \sqrt {a \cos \left (d x + c\right ) + a} \sqrt {\cos \left (d x + c\right )}}{\sqrt {a} \sin \left (d x + c\right )}\right ) - \frac {2 \, {\left (17 \, \cos \left (d x + c\right )^{3} + 70 \, \cos \left (d x + c\right )^{2} + 21 \, \cos \left (d x + c\right )\right )} \sqrt {a \cos \left (d x + c\right ) + a} \sin \left (d x + c\right )}{\sqrt {\cos \left (d x + c\right )}}}{384 \, {\left (a^{4} d \cos \left (d x + c\right )^{4} + 4 \, a^{4} d \cos \left (d x + c\right )^{3} + 6 \, a^{4} d \cos \left (d x + c\right )^{2} + 4 \, a^{4} d \cos \left (d x + c\right ) + a^{4} d\right )}} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate(1/(a+a*cos(d*x+c))^(7/2)/sec(d*x+c)^(3/2),x, algorithm="fricas")

[Out]

-1/384*(21*sqrt(2)*(cos(d*x + c)^4 + 4*cos(d*x + c)^3 + 6*cos(d*x + c)^2 + 4*cos(d*x + c) + 1)*sqrt(a)*arctan(
sqrt(2)*sqrt(a*cos(d*x + c) + a)*sqrt(cos(d*x + c))/(sqrt(a)*sin(d*x + c))) - 2*(17*cos(d*x + c)^3 + 70*cos(d*
x + c)^2 + 21*cos(d*x + c))*sqrt(a*cos(d*x + c) + a)*sin(d*x + c)/sqrt(cos(d*x + c)))/(a^4*d*cos(d*x + c)^4 +
4*a^4*d*cos(d*x + c)^3 + 6*a^4*d*cos(d*x + c)^2 + 4*a^4*d*cos(d*x + c) + a^4*d)

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giac [F]  time = 0.00, size = 0, normalized size = 0.00 \[ \int \frac {1}{{\left (a \cos \left (d x + c\right ) + a\right )}^{\frac {7}{2}} \sec \left (d x + c\right )^{\frac {3}{2}}}\,{d x} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate(1/(a+a*cos(d*x+c))^(7/2)/sec(d*x+c)^(3/2),x, algorithm="giac")

[Out]

integrate(1/((a*cos(d*x + c) + a)^(7/2)*sec(d*x + c)^(3/2)), x)

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maple [A]  time = 0.22, size = 288, normalized size = 1.46 \[ \frac {\sqrt {a \left (1+\cos \left (d x +c \right )\right )}\, \left (-1+\cos \left (d x +c \right )\right )^{5} \cos \left (d x +c \right ) \left (17 \left (\cos ^{3}\left (d x +c \right )\right ) \sqrt {2}\, \sqrt {\frac {\cos \left (d x +c \right )}{1+\cos \left (d x +c \right )}}+53 \left (\cos ^{2}\left (d x +c \right )\right ) \sqrt {2}\, \sqrt {\frac {\cos \left (d x +c \right )}{1+\cos \left (d x +c \right )}}+21 \arcsin \left (\frac {-1+\cos \left (d x +c \right )}{\sin \left (d x +c \right )}\right ) \left (\cos ^{2}\left (d x +c \right )\right ) \sin \left (d x +c \right )-49 \cos \left (d x +c \right ) \sqrt {2}\, \sqrt {\frac {\cos \left (d x +c \right )}{1+\cos \left (d x +c \right )}}+42 \cos \left (d x +c \right ) \sin \left (d x +c \right ) \arcsin \left (\frac {-1+\cos \left (d x +c \right )}{\sin \left (d x +c \right )}\right )-21 \sqrt {2}\, \sqrt {\frac {\cos \left (d x +c \right )}{1+\cos \left (d x +c \right )}}+21 \arcsin \left (\frac {-1+\cos \left (d x +c \right )}{\sin \left (d x +c \right )}\right ) \sin \left (d x +c \right )\right ) \sqrt {2}}{384 d \left (\frac {1}{\cos \left (d x +c \right )}\right )^{\frac {3}{2}} \left (\frac {\cos \left (d x +c \right )}{1+\cos \left (d x +c \right )}\right )^{\frac {5}{2}} \sin \left (d x +c \right )^{11} a^{4}} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

int(1/(a+a*cos(d*x+c))^(7/2)/sec(d*x+c)^(3/2),x)

[Out]

1/384/d*(a*(1+cos(d*x+c)))^(1/2)*(-1+cos(d*x+c))^5*cos(d*x+c)*(17*cos(d*x+c)^3*2^(1/2)*(cos(d*x+c)/(1+cos(d*x+
c)))^(1/2)+53*cos(d*x+c)^2*2^(1/2)*(cos(d*x+c)/(1+cos(d*x+c)))^(1/2)+21*arcsin((-1+cos(d*x+c))/sin(d*x+c))*cos
(d*x+c)^2*sin(d*x+c)-49*cos(d*x+c)*2^(1/2)*(cos(d*x+c)/(1+cos(d*x+c)))^(1/2)+42*cos(d*x+c)*sin(d*x+c)*arcsin((
-1+cos(d*x+c))/sin(d*x+c))-21*2^(1/2)*(cos(d*x+c)/(1+cos(d*x+c)))^(1/2)+21*arcsin((-1+cos(d*x+c))/sin(d*x+c))*
sin(d*x+c))/(1/cos(d*x+c))^(3/2)/(cos(d*x+c)/(1+cos(d*x+c)))^(5/2)/sin(d*x+c)^11*2^(1/2)/a^4

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maxima [F]  time = 0.00, size = 0, normalized size = 0.00 \[ \int \frac {1}{{\left (a \cos \left (d x + c\right ) + a\right )}^{\frac {7}{2}} \sec \left (d x + c\right )^{\frac {3}{2}}}\,{d x} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate(1/(a+a*cos(d*x+c))^(7/2)/sec(d*x+c)^(3/2),x, algorithm="maxima")

[Out]

integrate(1/((a*cos(d*x + c) + a)^(7/2)*sec(d*x + c)^(3/2)), x)

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mupad [F]  time = 0.00, size = -1, normalized size = -0.01 \[ \int \frac {1}{{\left (\frac {1}{\cos \left (c+d\,x\right )}\right )}^{3/2}\,{\left (a+a\,\cos \left (c+d\,x\right )\right )}^{7/2}} \,d x \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

int(1/((1/cos(c + d*x))^(3/2)*(a + a*cos(c + d*x))^(7/2)),x)

[Out]

int(1/((1/cos(c + d*x))^(3/2)*(a + a*cos(c + d*x))^(7/2)), 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(1/(a+a*cos(d*x+c))**(7/2)/sec(d*x+c)**(3/2),x)

[Out]

Timed out

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