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

Optimal. Leaf size=81 \[ \frac {2 a^2 \sin (c+d x) \sec ^{\frac {3}{2}}(c+d x)}{3 d \sqrt {a \cos (c+d x)+a}}+\frac {10 a^2 \sin (c+d x) \sqrt {\sec (c+d x)}}{3 d \sqrt {a \cos (c+d x)+a}} \]

[Out]

2/3*a^2*sec(d*x+c)^(3/2)*sin(d*x+c)/d/(a+a*cos(d*x+c))^(1/2)+10/3*a^2*sin(d*x+c)*sec(d*x+c)^(1/2)/d/(a+a*cos(d
*x+c))^(1/2)

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Rubi [A]  time = 0.17, antiderivative size = 81, normalized size of antiderivative = 1.00, number of steps used = 4, number of rules used = 4, integrand size = 25, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.160, Rules used = {4222, 2762, 21, 2771} \[ \frac {2 a^2 \sin (c+d x) \sec ^{\frac {3}{2}}(c+d x)}{3 d \sqrt {a \cos (c+d x)+a}}+\frac {10 a^2 \sin (c+d x) \sqrt {\sec (c+d x)}}{3 d \sqrt {a \cos (c+d x)+a}} \]

Antiderivative was successfully verified.

[In]

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

[Out]

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

Rule 21

Int[(u_.)*((a_) + (b_.)*(v_))^(m_.)*((c_) + (d_.)*(v_))^(n_.), x_Symbol] :> Dist[(b/d)^m, Int[u*(c + d*v)^(m +
n), x], x] /; FreeQ[{a, b, c, d, n}, x] && EqQ[b*c - a*d, 0] && IntegerQ[m] && ( !IntegerQ[n] || SimplerQ[c +
d*x, a + b*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 2771

Int[Sqrt[(a_) + (b_.)*sin[(e_.) + (f_.)*(x_)]]/((c_.) + (d_.)*sin[(e_.) + (f_.)*(x_)])^(3/2), x_Symbol] :> Sim
p[(-2*b^2*Cos[e + f*x])/(f*(b*c + a*d)*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 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 (a+a \cos (c+d x))^{3/2} \sec ^{\frac {5}{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))^{3/2}}{\cos ^{\frac {5}{2}}(c+d x)} \, dx\\ &=\frac {2 a^2 \sec ^{\frac {3}{2}}(c+d x) \sin (c+d x)}{3 d \sqrt {a+a \cos (c+d x)}}-\frac {1}{3} \left (2 a \sqrt {\cos (c+d x)} \sqrt {\sec (c+d x)}\right ) \int \frac {-\frac {5 a}{2}-\frac {5}{2} a \cos (c+d x)}{\cos ^{\frac {3}{2}}(c+d x) \sqrt {a+a \cos (c+d x)}} \, dx\\ &=\frac {2 a^2 \sec ^{\frac {3}{2}}(c+d x) \sin (c+d x)}{3 d \sqrt {a+a \cos (c+d x)}}+\frac {1}{3} \left (5 a \sqrt {\cos (c+d x)} \sqrt {\sec (c+d x)}\right ) \int \frac {\sqrt {a+a \cos (c+d x)}}{\cos ^{\frac {3}{2}}(c+d x)} \, dx\\ &=\frac {10 a^2 \sqrt {\sec (c+d x)} \sin (c+d x)}{3 d \sqrt {a+a \cos (c+d x)}}+\frac {2 a^2 \sec ^{\frac {3}{2}}(c+d x) \sin (c+d x)}{3 d \sqrt {a+a \cos (c+d x)}}\\ \end {align*}

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Mathematica [A]  time = 0.14, size = 52, normalized size = 0.64 \[ \frac {2 a (5 \cos (c+d x)+1) \tan \left (\frac {1}{2} (c+d x)\right ) \sec ^{\frac {3}{2}}(c+d x) \sqrt {a (\cos (c+d x)+1)}}{3 d} \]

Antiderivative was successfully verified.

[In]

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

[Out]

(2*a*Sqrt[a*(1 + Cos[c + d*x])]*(1 + 5*Cos[c + d*x])*Sec[c + d*x]^(3/2)*Tan[(c + d*x)/2])/(3*d)

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fricas [A]  time = 0.90, size = 60, normalized size = 0.74 \[ \frac {2 \, {\left (5 \, a \cos \left (d x + c\right ) + a\right )} \sqrt {a \cos \left (d x + c\right ) + a} \sin \left (d x + c\right )}{3 \, {\left (d \cos \left (d x + c\right )^{2} + d \cos \left (d x + c\right )\right )} \sqrt {\cos \left (d x + c\right )}} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

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

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

[Out]

Timed out

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maple [A]  time = 0.18, size = 63, normalized size = 0.78 \[ -\frac {2 \left (5 \left (\cos ^{2}\left (d x +c \right )\right )-4 \cos \left (d x +c \right )-1\right ) \cos \left (d x +c \right ) \sqrt {a \left (1+\cos \left (d x +c \right )\right )}\, \left (\frac {1}{\cos \left (d x +c \right )}\right )^{\frac {5}{2}} a}{3 d \sin \left (d x +c \right )} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

-2/3/d*(5*cos(d*x+c)^2-4*cos(d*x+c)-1)*cos(d*x+c)*(a*(1+cos(d*x+c)))^(1/2)*(1/cos(d*x+c))^(5/2)/sin(d*x+c)*a

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maxima [A]  time = 1.01, size = 125, normalized size = 1.54 \[ \frac {4 \, {\left (\frac {3 \, \sqrt {2} a^{\frac {3}{2}} \sin \left (d x + c\right )}{\cos \left (d x + c\right ) + 1} - \frac {5 \, \sqrt {2} a^{\frac {3}{2}} \sin \left (d x + c\right )^{3}}{{\left (\cos \left (d x + c\right ) + 1\right )}^{3}} + \frac {2 \, \sqrt {2} a^{\frac {3}{2}} \sin \left (d x + c\right )^{5}}{{\left (\cos \left (d x + c\right ) + 1\right )}^{5}}\right )}}{3 \, d {\left (\frac {\sin \left (d x + c\right )}{\cos \left (d x + c\right ) + 1} + 1\right )}^{\frac {5}{2}} {\left (-\frac {\sin \left (d x + c\right )}{\cos \left (d x + c\right ) + 1} + 1\right )}^{\frac {5}{2}}} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

4/3*(3*sqrt(2)*a^(3/2)*sin(d*x + c)/(cos(d*x + c) + 1) - 5*sqrt(2)*a^(3/2)*sin(d*x + c)^3/(cos(d*x + c) + 1)^3
+ 2*sqrt(2)*a^(3/2)*sin(d*x + c)^5/(cos(d*x + c) + 1)^5)/(d*(sin(d*x + c)/(cos(d*x + c) + 1) + 1)^(5/2)*(-sin
(d*x + c)/(cos(d*x + c) + 1) + 1)^(5/2))

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mupad [B]  time = 0.79, size = 91, normalized size = 1.12 \[ \frac {2\,a\,\sqrt {a\,\left (\cos \left (c+d\,x\right )+1\right )}\,\sqrt {\frac {1}{\cos \left (c+d\,x\right )}}\,\left (5\,\sin \left (c+d\,x\right )+2\,\sin \left (2\,c+2\,d\,x\right )+5\,\sin \left (3\,c+3\,d\,x\right )\right )}{3\,d\,\left (3\,\cos \left (c+d\,x\right )+2\,\cos \left (2\,c+2\,d\,x\right )+\cos \left (3\,c+3\,d\,x\right )+2\right )} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

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

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

[Out]

Timed out

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