3.367 \(\int \frac{\sec ^{\frac{7}{2}}(c+d x)}{\sqrt{a+a \cos (c+d x)}} \, dx\)

Optimal. Leaf size=189 \[ \frac{2 \sin (c+d x) \sec ^{\frac{5}{2}}(c+d x)}{5 d \sqrt{a \cos (c+d x)+a}}-\frac{2 \sin (c+d x) \sec ^{\frac{3}{2}}(c+d x)}{15 d \sqrt{a \cos (c+d x)+a}}+\frac{26 \sin (c+d x) \sqrt{\sec (c+d x)}}{15 d \sqrt{a \cos (c+d x)+a}}-\frac{\sqrt{2} \sqrt{\cos (c+d x)} \sqrt{\sec (c+d x)} \tan ^{-1}\left (\frac{\sqrt{a} \sin (c+d x)}{\sqrt{2} \sqrt{\cos (c+d x)} \sqrt{a \cos (c+d x)+a}}\right )}{\sqrt{a} d} \]

[Out]

-((Sqrt[2]*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]]*Sqrt[Sec[c + d*x]])/(Sqrt[a]*d)) + (26*Sqrt[Sec[c + d*x]]*Sin[c + d*x])/(15*d*Sqrt[a + a*Cos[c + d*x]]) -
(2*Sec[c + d*x]^(3/2)*Sin[c + d*x])/(15*d*Sqrt[a + a*Cos[c + d*x]]) + (2*Sec[c + d*x]^(5/2)*Sin[c + d*x])/(5*
d*Sqrt[a + a*Cos[c + d*x]])

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Rubi [A]  time = 0.4211, antiderivative size = 189, normalized size of antiderivative = 1., number of steps used = 7, number of rules used = 6, integrand size = 25, \(\frac{\text{number of rules}}{\text{integrand size}}\) = 0.24, Rules used = {4222, 2779, 2984, 12, 2782, 205} \[ \frac{2 \sin (c+d x) \sec ^{\frac{5}{2}}(c+d x)}{5 d \sqrt{a \cos (c+d x)+a}}-\frac{2 \sin (c+d x) \sec ^{\frac{3}{2}}(c+d x)}{15 d \sqrt{a \cos (c+d x)+a}}+\frac{26 \sin (c+d x) \sqrt{\sec (c+d x)}}{15 d \sqrt{a \cos (c+d x)+a}}-\frac{\sqrt{2} \sqrt{\cos (c+d x)} \sqrt{\sec (c+d x)} \tan ^{-1}\left (\frac{\sqrt{a} \sin (c+d x)}{\sqrt{2} \sqrt{\cos (c+d x)} \sqrt{a \cos (c+d x)+a}}\right )}{\sqrt{a} d} \]

Antiderivative was successfully verified.

[In]

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

[Out]

-((Sqrt[2]*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]]*Sqrt[Sec[c + d*x]])/(Sqrt[a]*d)) + (26*Sqrt[Sec[c + d*x]]*Sin[c + d*x])/(15*d*Sqrt[a + a*Cos[c + d*x]]) -
(2*Sec[c + d*x]^(3/2)*Sin[c + d*x])/(15*d*Sqrt[a + a*Cos[c + d*x]]) + (2*Sec[c + d*x]^(5/2)*Sin[c + d*x])/(5*
d*Sqrt[a + a*Cos[c + d*x]])

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]

Rule 2779

Int[((c_.) + (d_.)*sin[(e_.) + (f_.)*(x_)])^(n_)/Sqrt[(a_) + (b_.)*sin[(e_.) + (f_.)*(x_)]], x_Symbol] :> -Sim
p[(d*Cos[e + f*x]*(c + d*Sin[e + f*x])^(n + 1))/(f*(n + 1)*(c^2 - d^2)*Sqrt[a + b*Sin[e + f*x]]), x] - Dist[1/
(2*b*(n + 1)*(c^2 - d^2)), Int[((c + d*Sin[e + f*x])^(n + 1)*Simp[a*d - 2*b*c*(n + 1) + b*d*(2*n + 3)*Sin[e +
f*x], x])/Sqrt[a + b*Sin[e + f*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[n, -1] && IntegerQ[2*n]

Rule 2984

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])

Rule 12

Int[(a_)*(u_), x_Symbol] :> Dist[a, Int[u, x], x] /; FreeQ[a, x] &&  !MatchQ[u, (b_)*(v_) /; FreeQ[b, x]]

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 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]

Rubi steps

\begin{align*} \int \frac{\sec ^{\frac{7}{2}}(c+d x)}{\sqrt{a+a \cos (c+d x)}} \, dx &=\left (\sqrt{\cos (c+d x)} \sqrt{\sec (c+d x)}\right ) \int \frac{1}{\cos ^{\frac{7}{2}}(c+d x) \sqrt{a+a \cos (c+d x)}} \, dx\\ &=\frac{2 \sec ^{\frac{5}{2}}(c+d x) \sin (c+d x)}{5 d \sqrt{a+a \cos (c+d x)}}-\frac{\left (\sqrt{\cos (c+d x)} \sqrt{\sec (c+d x)}\right ) \int \frac{a-4 a \cos (c+d x)}{\cos ^{\frac{5}{2}}(c+d x) \sqrt{a+a \cos (c+d x)}} \, dx}{5 a}\\ &=-\frac{2 \sec ^{\frac{3}{2}}(c+d x) \sin (c+d x)}{15 d \sqrt{a+a \cos (c+d x)}}+\frac{2 \sec ^{\frac{5}{2}}(c+d x) \sin (c+d x)}{5 d \sqrt{a+a \cos (c+d x)}}-\frac{\left (2 \sqrt{\cos (c+d x)} \sqrt{\sec (c+d x)}\right ) \int \frac{-\frac{13 a^2}{2}+a^2 \cos (c+d x)}{\cos ^{\frac{3}{2}}(c+d x) \sqrt{a+a \cos (c+d x)}} \, dx}{15 a^2}\\ &=\frac{26 \sqrt{\sec (c+d x)} \sin (c+d x)}{15 d \sqrt{a+a \cos (c+d x)}}-\frac{2 \sec ^{\frac{3}{2}}(c+d x) \sin (c+d x)}{15 d \sqrt{a+a \cos (c+d x)}}+\frac{2 \sec ^{\frac{5}{2}}(c+d x) \sin (c+d x)}{5 d \sqrt{a+a \cos (c+d x)}}-\frac{\left (4 \sqrt{\cos (c+d x)} \sqrt{\sec (c+d x)}\right ) \int \frac{15 a^3}{4 \sqrt{\cos (c+d x)} \sqrt{a+a \cos (c+d x)}} \, dx}{15 a^3}\\ &=\frac{26 \sqrt{\sec (c+d x)} \sin (c+d x)}{15 d \sqrt{a+a \cos (c+d x)}}-\frac{2 \sec ^{\frac{3}{2}}(c+d x) \sin (c+d x)}{15 d \sqrt{a+a \cos (c+d x)}}+\frac{2 \sec ^{\frac{5}{2}}(c+d x) \sin (c+d x)}{5 d \sqrt{a+a \cos (c+d x)}}-\left (\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\\ &=\frac{26 \sqrt{\sec (c+d x)} \sin (c+d x)}{15 d \sqrt{a+a \cos (c+d x)}}-\frac{2 \sec ^{\frac{3}{2}}(c+d x) \sin (c+d x)}{15 d \sqrt{a+a \cos (c+d x)}}+\frac{2 \sec ^{\frac{5}{2}}(c+d x) \sin (c+d x)}{5 d \sqrt{a+a \cos (c+d x)}}+\frac{\left (2 a \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 )}{d}\\ &=-\frac{\sqrt{2} \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)}}{\sqrt{a} d}+\frac{26 \sqrt{\sec (c+d x)} \sin (c+d x)}{15 d \sqrt{a+a \cos (c+d x)}}-\frac{2 \sec ^{\frac{3}{2}}(c+d x) \sin (c+d x)}{15 d \sqrt{a+a \cos (c+d x)}}+\frac{2 \sec ^{\frac{5}{2}}(c+d x) \sin (c+d x)}{5 d \sqrt{a+a \cos (c+d x)}}\\ \end{align*}

Mathematica [C]  time = 7.76042, size = 1542, normalized size = 8.16 \[ \text{result too large to display} \]

Warning: Unable to verify antiderivative.

[In]

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

[Out]

(-2*Cot[c/2 + (d*x)/2]*Csc[c/2 + (d*x)/2]^6*((1 - 2*Sin[c/2 + (d*x)/2]^2)^(-1))^(7/2)*(4725*Sin[c/2 + (d*x)/2]
^2 - 48825*Sin[c/2 + (d*x)/2]^4 + 210105*Sin[c/2 + (d*x)/2]^6 - 486630*Sin[c/2 + (d*x)/2]^8 + 655812*Sin[c/2 +
(d*x)/2]^10 - 710*Hypergeometric2F1[2, 9/2, 11/2, Sin[c/2 + (d*x)/2]^2/(-1 + 2*Sin[c/2 + (d*x)/2]^2)]*Sin[c/2
+ (d*x)/2]^10 - 40*Cos[(c + d*x)/2]^6*HypergeometricPFQ[{2, 2, 2, 9/2}, {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]^10 - 518760*Sin[c/2 + (d*x)/2]^12 + 1770*Hypergeometric2F1[2,
9/2, 11/2, Sin[c/2 + (d*x)/2]^2/(-1 + 2*Sin[c/2 + (d*x)/2]^2)]*Sin[c/2 + (d*x)/2]^12 + 226656*Sin[c/2 + (d*x)/
2]^14 - 1500*Hypergeometric2F1[2, 9/2, 11/2, Sin[c/2 + (d*x)/2]^2/(-1 + 2*Sin[c/2 + (d*x)/2]^2)]*Sin[c/2 + (d*
x)/2]^14 - 42048*Sin[c/2 + (d*x)/2]^16 + 440*Hypergeometric2F1[2, 9/2, 11/2, Sin[c/2 + (d*x)/2]^2/(-1 + 2*Sin[
c/2 + (d*x)/2]^2)]*Sin[c/2 + (d*x)/2]^16 + 4725*ArcTanh[Sqrt[Sin[c/2 + (d*x)/2]^2/(-1 + 2*Sin[c/2 + (d*x)/2]^2
)]]*Sqrt[Sin[c/2 + (d*x)/2]^2/(-1 + 2*Sin[c/2 + (d*x)/2]^2)] - 56700*ArcTanh[Sqrt[Sin[c/2 + (d*x)/2]^2/(-1 + 2
*Sin[c/2 + (d*x)/2]^2)]]*Sin[c/2 + (d*x)/2]^2*Sqrt[Sin[c/2 + (d*x)/2]^2/(-1 + 2*Sin[c/2 + (d*x)/2]^2)] + 29106
0*ArcTanh[Sqrt[Sin[c/2 + (d*x)/2]^2/(-1 + 2*Sin[c/2 + (d*x)/2]^2)]]*Sin[c/2 + (d*x)/2]^4*Sqrt[Sin[c/2 + (d*x)/
2]^2/(-1 + 2*Sin[c/2 + (d*x)/2]^2)] - 833760*ArcTanh[Sqrt[Sin[c/2 + (d*x)/2]^2/(-1 + 2*Sin[c/2 + (d*x)/2]^2)]]
*Sin[c/2 + (d*x)/2]^6*Sqrt[Sin[c/2 + (d*x)/2]^2/(-1 + 2*Sin[c/2 + (d*x)/2]^2)] + 1458000*ArcTanh[Sqrt[Sin[c/2
+ (d*x)/2]^2/(-1 + 2*Sin[c/2 + (d*x)/2]^2)]]*Sin[c/2 + (d*x)/2]^8*Sqrt[Sin[c/2 + (d*x)/2]^2/(-1 + 2*Sin[c/2 +
(d*x)/2]^2)] - 1598400*ArcTanh[Sqrt[Sin[c/2 + (d*x)/2]^2/(-1 + 2*Sin[c/2 + (d*x)/2]^2)]]*Sin[c/2 + (d*x)/2]^10
*Sqrt[Sin[c/2 + (d*x)/2]^2/(-1 + 2*Sin[c/2 + (d*x)/2]^2)] + 1080000*ArcTanh[Sqrt[Sin[c/2 + (d*x)/2]^2/(-1 + 2*
Sin[c/2 + (d*x)/2]^2)]]*Sin[c/2 + (d*x)/2]^12*Sqrt[Sin[c/2 + (d*x)/2]^2/(-1 + 2*Sin[c/2 + (d*x)/2]^2)] - 41472
0*ArcTanh[Sqrt[Sin[c/2 + (d*x)/2]^2/(-1 + 2*Sin[c/2 + (d*x)/2]^2)]]*Sin[c/2 + (d*x)/2]^14*Sqrt[Sin[c/2 + (d*x)
/2]^2/(-1 + 2*Sin[c/2 + (d*x)/2]^2)] + 69120*ArcTanh[Sqrt[Sin[c/2 + (d*x)/2]^2/(-1 + 2*Sin[c/2 + (d*x)/2]^2)]]
*Sin[c/2 + (d*x)/2]^16*Sqrt[Sin[c/2 + (d*x)/2]^2/(-1 + 2*Sin[c/2 + (d*x)/2]^2)] + 60*Cos[(c + d*x)/2]^4*Hyperg
eometricPFQ[{2, 2, 9/2}, {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]^10*
(-5 + 4*Sin[c/2 + (d*x)/2]^2)))/(675*d*Sqrt[a*(1 + Cos[c + d*x])]*(-1 + 2*Sin[c/2 + (d*x)/2]^2))

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Maple [A]  time = 0.436, size = 294, normalized size = 1.6 \begin{align*}{\frac{\sqrt{2}\cos \left ( dx+c \right ) \left ( \sin \left ( dx+c \right ) \right ) ^{4}}{15\,da \left ( -1+\cos \left ( dx+c \right ) \right ) ^{2} \left ( 1+\cos \left ( dx+c \right ) \right ) ^{3}} \left ( 15\,\arcsin \left ({\frac{-1+\cos \left ( dx+c \right ) }{\sin \left ( dx+c \right ) }} \right ) \left ( \cos \left ( dx+c \right ) \right ) ^{3} \left ({\frac{\cos \left ( dx+c \right ) }{1+\cos \left ( dx+c \right ) }} \right ) ^{5/2}+45\,\arcsin \left ({\frac{-1+\cos \left ( dx+c \right ) }{\sin \left ( dx+c \right ) }} \right ) \left ( \cos \left ( dx+c \right ) \right ) ^{2} \left ({\frac{\cos \left ( dx+c \right ) }{1+\cos \left ( dx+c \right ) }} \right ) ^{5/2}+45\,\arcsin \left ({\frac{-1+\cos \left ( dx+c \right ) }{\sin \left ( dx+c \right ) }} \right ) \cos \left ( dx+c \right ) \left ({\frac{\cos \left ( dx+c \right ) }{1+\cos \left ( dx+c \right ) }} \right ) ^{5/2}+15\,\arcsin \left ({\frac{-1+\cos \left ( dx+c \right ) }{\sin \left ( dx+c \right ) }} \right ) \left ({\frac{\cos \left ( dx+c \right ) }{1+\cos \left ( dx+c \right ) }} \right ) ^{5/2}+13\,\sqrt{2} \left ( \cos \left ( dx+c \right ) \right ) ^{2}\sin \left ( dx+c \right ) -\cos \left ( dx+c \right ) \sin \left ( dx+c \right ) \sqrt{2}+3\,\sqrt{2}\sin \left ( dx+c \right ) \right ) \left ( \left ( \cos \left ( dx+c \right ) \right ) ^{-1} \right ) ^{{\frac{7}{2}}}\sqrt{a \left ( 1+\cos \left ( dx+c \right ) \right ) }} \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

1/15/d*2^(1/2)/a*(15*arcsin((-1+cos(d*x+c))/sin(d*x+c))*cos(d*x+c)^3*(cos(d*x+c)/(1+cos(d*x+c)))^(5/2)+45*arcs
in((-1+cos(d*x+c))/sin(d*x+c))*cos(d*x+c)^2*(cos(d*x+c)/(1+cos(d*x+c)))^(5/2)+45*arcsin((-1+cos(d*x+c))/sin(d*
x+c))*cos(d*x+c)*(cos(d*x+c)/(1+cos(d*x+c)))^(5/2)+15*arcsin((-1+cos(d*x+c))/sin(d*x+c))*(cos(d*x+c)/(1+cos(d*
x+c)))^(5/2)+13*2^(1/2)*cos(d*x+c)^2*sin(d*x+c)-cos(d*x+c)*sin(d*x+c)*2^(1/2)+3*2^(1/2)*sin(d*x+c))*cos(d*x+c)
*sin(d*x+c)^4*(1/cos(d*x+c))^(7/2)*(a*(1+cos(d*x+c)))^(1/2)/(-1+cos(d*x+c))^2/(1+cos(d*x+c))^3

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Maxima [F(-2)]  time = 0., size = 0, normalized size = 0. \begin{align*} \text{Exception raised: ValueError} \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

Exception raised: ValueError

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Fricas [A]  time = 1.88342, size = 392, normalized size = 2.07 \begin{align*} \frac{\frac{15 \, \sqrt{2}{\left (a \cos \left (d x + c\right )^{3} + a \cos \left (d x + c\right )^{2}\right )} \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 )}{\sqrt{a}} + \frac{2 \, \sqrt{a \cos \left (d x + c\right ) + a}{\left (13 \, \cos \left (d x + c\right )^{2} - \cos \left (d x + c\right ) + 3\right )} \sin \left (d x + c\right )}{\sqrt{\cos \left (d x + c\right )}}}{15 \,{\left (a d \cos \left (d x + c\right )^{3} + a d \cos \left (d x + c\right )^{2}\right )}} \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

1/15*(15*sqrt(2)*(a*cos(d*x + c)^3 + a*cos(d*x + c)^2)*arctan(sqrt(2)*sqrt(a*cos(d*x + c) + a)*sqrt(cos(d*x +
c))/(sqrt(a)*sin(d*x + c)))/sqrt(a) + 2*sqrt(a*cos(d*x + c) + a)*(13*cos(d*x + c)^2 - cos(d*x + c) + 3)*sin(d*
x + c)/sqrt(cos(d*x + c)))/(a*d*cos(d*x + c)^3 + a*d*cos(d*x + c)^2)

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

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

Timed out

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Giac [F]  time = 0., size = 0, normalized size = 0. \begin{align*} \int \frac{\sec \left (d x + c\right )^{\frac{7}{2}}}{\sqrt{a \cos \left (d x + c\right ) + a}}\,{d x} \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

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