3.357 \(\int (a+a \cos (c+d x))^{5/2} \sec ^{\frac{3}{2}}(c+d x) \, dx\)

Optimal. Leaf size=134 \[ \frac{5 a^{5/2} \sqrt{\cos (c+d x)} \sqrt{\sec (c+d x)} \sin ^{-1}\left (\frac{\sqrt{a} \sin (c+d x)}{\sqrt{a \cos (c+d x)+a}}\right )}{d}-\frac{a^3 \sin (c+d x)}{d \sqrt{\sec (c+d x)} \sqrt{a \cos (c+d x)+a}}+\frac{2 a^2 \sin (c+d x) \sqrt{\sec (c+d x)} \sqrt{a \cos (c+d x)+a}}{d} \]

[Out]

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

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Rubi [A]  time = 0.282289, antiderivative size = 134, normalized size of antiderivative = 1., number of steps used = 5, number of rules used = 5, integrand size = 25, \(\frac{\text{number of rules}}{\text{integrand size}}\) = 0.2, Rules used = {4222, 2762, 2981, 2774, 216} \[ \frac{5 a^{5/2} \sqrt{\cos (c+d x)} \sqrt{\sec (c+d x)} \sin ^{-1}\left (\frac{\sqrt{a} \sin (c+d x)}{\sqrt{a \cos (c+d x)+a}}\right )}{d}-\frac{a^3 \sin (c+d x)}{d \sqrt{\sec (c+d x)} \sqrt{a \cos (c+d x)+a}}+\frac{2 a^2 \sin (c+d x) \sqrt{\sec (c+d x)} \sqrt{a \cos (c+d x)+a}}{d} \]

Antiderivative was successfully verified.

[In]

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

[Out]

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

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 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 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]

Rule 2774

Int[Sqrt[(a_) + (b_.)*sin[(e_.) + (f_.)*(x_)]]/Sqrt[(d_.)*sin[(e_.) + (f_.)*(x_)]], x_Symbol] :> Dist[-2/f, Su
bst[Int[1/Sqrt[1 - x^2/a], x], x, (b*Cos[e + f*x])/Sqrt[a + b*Sin[e + f*x]]], x] /; FreeQ[{a, b, d, e, f}, x]
&& EqQ[a^2 - b^2, 0] && EqQ[d, a/b]

Rule 216

Int[1/Sqrt[(a_) + (b_.)*(x_)^2], x_Symbol] :> Simp[ArcSin[(Rt[-b, 2]*x)/Sqrt[a]]/Rt[-b, 2], x] /; FreeQ[{a, b}
, x] && GtQ[a, 0] && NegQ[b]

Rubi steps

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

Mathematica [C]  time = 3.07871, size = 202, normalized size = 1.51 \[ \frac{\sqrt{\cos (c+d x)} \tan \left (\frac{1}{2} (c+d x)\right ) \sec ^4\left (\frac{1}{2} (c+d x)\right ) \sqrt{\sec (c+d x)} (a (\cos (c+d x)+1))^{5/2} \left (6 \sin ^4(c+d x) \csc ^2\left (\frac{1}{2} (c+d x)\right ) \, _3F_2\left (\frac{3}{2},2,\frac{5}{2};1,\frac{9}{2};2 \sin ^2\left (\frac{1}{2} (c+d x)\right )\right )+24 \sin ^2(c+d x) (\cos (c+d x)+3) \, _2F_1\left (\frac{3}{2},\frac{5}{2};\frac{9}{2};2 \sin ^2\left (\frac{1}{2} (c+d x)\right )\right )+7 (28 \cos (c+d x)+3 \cos (2 (c+d x))+89) \, _2F_1\left (\frac{1}{2},\frac{3}{2};\frac{7}{2};2 \sin ^2\left (\frac{1}{2} (c+d x)\right )\right )\right )}{420 d} \]

Warning: Unable to verify antiderivative.

[In]

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

[Out]

(Sqrt[Cos[c + d*x]]*(a*(1 + Cos[c + d*x]))^(5/2)*Sec[(c + d*x)/2]^4*Sqrt[Sec[c + d*x]]*(7*(89 + 28*Cos[c + d*x
] + 3*Cos[2*(c + d*x)])*Hypergeometric2F1[1/2, 3/2, 7/2, 2*Sin[(c + d*x)/2]^2] + 24*(3 + Cos[c + d*x])*Hyperge
ometric2F1[3/2, 5/2, 9/2, 2*Sin[(c + d*x)/2]^2]*Sin[c + d*x]^2 + 6*Csc[(c + d*x)/2]^2*HypergeometricPFQ[{3/2,
2, 5/2}, {1, 9/2}, 2*Sin[(c + d*x)/2]^2]*Sin[c + d*x]^4)*Tan[(c + d*x)/2])/(420*d)

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Maple [A]  time = 0.414, size = 186, normalized size = 1.4 \begin{align*}{\frac{{a}^{2}\cos \left ( dx+c \right ) }{d \left ( 1+\cos \left ( dx+c \right ) \right ) } \left ( 5\,\cos \left ( dx+c \right ) \arctan \left ({\frac{\sin \left ( dx+c \right ) }{\cos \left ( dx+c \right ) }\sqrt{{\frac{\cos \left ( dx+c \right ) }{1+\cos \left ( dx+c \right ) }}}} \right ) \sqrt{{\frac{\cos \left ( dx+c \right ) }{1+\cos \left ( dx+c \right ) }}}+\cos \left ( dx+c \right ) \sin \left ( dx+c \right ) +5\,\arctan \left ({\frac{\sin \left ( dx+c \right ) }{\cos \left ( dx+c \right ) }\sqrt{{\frac{\cos \left ( dx+c \right ) }{1+\cos \left ( dx+c \right ) }}}} \right ) \sqrt{{\frac{\cos \left ( dx+c \right ) }{1+\cos \left ( dx+c \right ) }}}+2\,\sin \left ( dx+c \right ) \right ) \left ( \left ( \cos \left ( dx+c \right ) \right ) ^{-1} \right ) ^{{\frac{3}{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((a+cos(d*x+c)*a)^(5/2)*sec(d*x+c)^(3/2),x)

[Out]

1/d*a^2*(5*cos(d*x+c)*arctan(sin(d*x+c)*(cos(d*x+c)/(1+cos(d*x+c)))^(1/2)/cos(d*x+c))*(cos(d*x+c)/(1+cos(d*x+c
)))^(1/2)+cos(d*x+c)*sin(d*x+c)+5*arctan(sin(d*x+c)*(cos(d*x+c)/(1+cos(d*x+c)))^(1/2)/cos(d*x+c))*(cos(d*x+c)/
(1+cos(d*x+c)))^(1/2)+2*sin(d*x+c))*cos(d*x+c)*(1/cos(d*x+c))^(3/2)*(a*(1+cos(d*x+c)))^(1/2)/(1+cos(d*x+c))

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Maxima [B]  time = 2.05394, size = 1314, normalized size = 9.81 \begin{align*} \text{result too large to display} \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

1/4*(2*(a^2*cos(1/2*arctan2(sin(2*d*x + 2*c), cos(2*d*x + 2*c) + 1))*sin(d*x + c) - (a^2*cos(d*x + c) - a^2)*s
in(1/2*arctan2(sin(2*d*x + 2*c), cos(2*d*x + 2*c) + 1)))*sqrt(cos(2*d*x + 2*c)^2 + sin(2*d*x + 2*c)^2 + 2*cos(
2*d*x + 2*c) + 1)*sqrt(a) + 5*(a^2*arctan2(-(cos(2*d*x + 2*c)^2 + sin(2*d*x + 2*c)^2 + 2*cos(2*d*x + 2*c) + 1)
^(1/4)*(cos(1/2*arctan2(sin(2*d*x + 2*c), cos(2*d*x + 2*c) + 1))*sin(d*x + c) - cos(d*x + c)*sin(1/2*arctan2(s
in(2*d*x + 2*c), cos(2*d*x + 2*c) + 1))), (cos(2*d*x + 2*c)^2 + sin(2*d*x + 2*c)^2 + 2*cos(2*d*x + 2*c) + 1)^(
1/4)*(cos(d*x + c)*cos(1/2*arctan2(sin(2*d*x + 2*c), cos(2*d*x + 2*c) + 1)) + sin(d*x + c)*sin(1/2*arctan2(sin
(2*d*x + 2*c), cos(2*d*x + 2*c) + 1))) + 1) - a^2*arctan2(-(cos(2*d*x + 2*c)^2 + sin(2*d*x + 2*c)^2 + 2*cos(2*
d*x + 2*c) + 1)^(1/4)*(cos(1/2*arctan2(sin(2*d*x + 2*c), cos(2*d*x + 2*c) + 1))*sin(d*x + c) - cos(d*x + c)*si
n(1/2*arctan2(sin(2*d*x + 2*c), cos(2*d*x + 2*c) + 1))), (cos(2*d*x + 2*c)^2 + sin(2*d*x + 2*c)^2 + 2*cos(2*d*
x + 2*c) + 1)^(1/4)*(cos(d*x + c)*cos(1/2*arctan2(sin(2*d*x + 2*c), cos(2*d*x + 2*c) + 1)) + sin(d*x + c)*sin(
1/2*arctan2(sin(2*d*x + 2*c), cos(2*d*x + 2*c) + 1))) - 1) - a^2*arctan2((cos(2*d*x + 2*c)^2 + sin(2*d*x + 2*c
)^2 + 2*cos(2*d*x + 2*c) + 1)^(1/4)*sin(1/2*arctan2(sin(2*d*x + 2*c), cos(2*d*x + 2*c) + 1)), (cos(2*d*x + 2*c
)^2 + sin(2*d*x + 2*c)^2 + 2*cos(2*d*x + 2*c) + 1)^(1/4)*cos(1/2*arctan2(sin(2*d*x + 2*c), cos(2*d*x + 2*c) +
1)) + 1) + a^2*arctan2((cos(2*d*x + 2*c)^2 + sin(2*d*x + 2*c)^2 + 2*cos(2*d*x + 2*c) + 1)^(1/4)*sin(1/2*arctan
2(sin(2*d*x + 2*c), cos(2*d*x + 2*c) + 1)), (cos(2*d*x + 2*c)^2 + sin(2*d*x + 2*c)^2 + 2*cos(2*d*x + 2*c) + 1)
^(1/4)*cos(1/2*arctan2(sin(2*d*x + 2*c), cos(2*d*x + 2*c) + 1)) - 1))*(cos(2*d*x + 2*c)^2 + sin(2*d*x + 2*c)^2
+ 2*cos(2*d*x + 2*c) + 1)^(1/4)*sqrt(a) + 8*(a^2*cos(1/2*arctan2(sin(2*d*x + 2*c), cos(2*d*x + 2*c) + 1))*sin
(d*x + c) - (a^2*cos(d*x + c) - a^2)*sin(1/2*arctan2(sin(2*d*x + 2*c), cos(2*d*x + 2*c) + 1)))*sqrt(a))/((cos(
2*d*x + 2*c)^2 + sin(2*d*x + 2*c)^2 + 2*cos(2*d*x + 2*c) + 1)^(1/4)*d)

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Fricas [A]  time = 1.75789, size = 298, normalized size = 2.22 \begin{align*} -\frac{5 \,{\left (a^{2} \cos \left (d x + c\right ) + a^{2}\right )} \sqrt{a} \arctan \left (\frac{\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{{\left (a^{2} \cos \left (d x + c\right ) + 2 \, a^{2}\right )} \sqrt{a \cos \left (d x + c\right ) + a} \sin \left (d x + c\right )}{\sqrt{\cos \left (d x + c\right )}}}{d \cos \left (d x + c\right ) + d} \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

-(5*(a^2*cos(d*x + c) + a^2)*sqrt(a)*arctan(sqrt(a*cos(d*x + c) + a)*sqrt(cos(d*x + c))/(sqrt(a)*sin(d*x + c))
) - (a^2*cos(d*x + c) + 2*a^2)*sqrt(a*cos(d*x + c) + a)*sin(d*x + c)/sqrt(cos(d*x + c)))/(d*cos(d*x + c) + d)

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

[Out]

Timed out

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

[Out]

Timed out