\(\int (a+a \cos (c+d x))^{3/2} \sec ^{\frac {7}{2}}(c+d x) \, dx\) [347]

Optimal result

Integrand size = 25, antiderivative size = 121 \[ \int (a+a \cos (c+d x))^{3/2} \sec ^{\frac {7}{2}}(c+d x) \, dx=\frac {12 a^2 \sqrt {\sec (c+d x)} \sin (c+d x)}{5 d \sqrt {a+a \cos (c+d x)}}+\frac {6 a^2 \sec ^{\frac {3}{2}}(c+d x) \sin (c+d x)}{5 d \sqrt {a+a \cos (c+d x)}}+\frac {2 a^2 \sec ^{\frac {5}{2}}(c+d x) \sin (c+d x)}{5 d \sqrt {a+a \cos (c+d x)}} \]

[Out]

6/5*a^2*sec(d*x+c)^(3/2)*sin(d*x+c)/d/(a+a*cos(d*x+c))^(1/2)+2/5*a^2*sec(d*x+c)^(5/2)*sin(d*x+c)/d/(a+a*cos(d*
x+c))^(1/2)+12/5*a^2*sin(d*x+c)*sec(d*x+c)^(1/2)/d/(a+a*cos(d*x+c))^(1/2)

Rubi [A] (verified)

Time = 0.30 (sec) , antiderivative size = 121, normalized size of antiderivative = 1.00, number of steps used = 5, number of rules used = 5, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.200, Rules used = {4307, 2841, 21, 2851, 2850} \[ \int (a+a \cos (c+d x))^{3/2} \sec ^{\frac {7}{2}}(c+d x) \, dx=\frac {2 a^2 \sin (c+d x) \sec ^{\frac {5}{2}}(c+d x)}{5 d \sqrt {a \cos (c+d x)+a}}+\frac {6 a^2 \sin (c+d x) \sec ^{\frac {3}{2}}(c+d x)}{5 d \sqrt {a \cos (c+d x)+a}}+\frac {12 a^2 \sin (c+d x) \sqrt {\sec (c+d x)}}{5 d \sqrt {a \cos (c+d x)+a}} \]

[In]

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

[Out]

(12*a^2*Sqrt[Sec[c + d*x]]*Sin[c + d*x])/(5*d*Sqrt[a + a*Cos[c + d*x]]) + (6*a^2*Sec[c + d*x]^(3/2)*Sin[c + d*
x])/(5*d*Sqrt[a + a*Cos[c + d*x]]) + (2*a^2*Sec[c + d*x]^(5/2)*Sin[c + d*x])/(5*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 2841

Int[((a_) + (b_.)*sin[(e_.) + (f_.)*(x_)])^(m_)*((c_.) + (d_.)*sin[(e_.) + (f_.)*(x_)])^(n_), x_Symbol] :> Sim
p[(-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 2850

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 2851

Int[Sqrt[(a_) + (b_.)*sin[(e_.) + (f_.)*(x_)]]*((c_.) + (d_.)*sin[(e_.) + (f_.)*(x_)])^(n_), x_Symbol] :> Simp
[(b*c - a*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[(2*n + 3)*((b*c - a*d)/(2*b*(n + 1)*(c^2 - d^2))), Int[Sqrt[a + b*Sin[e + f*x]]*(c + d*Sin[e + f*x])^(n
+ 1), 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] && NeQ[2*n + 3, 0] && IntegerQ[2*n]

Rule 4307

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*} \text {integral}& = \left (\sqrt {\cos (c+d x)} \sqrt {\sec (c+d x)}\right ) \int \frac {(a+a \cos (c+d x))^{3/2}}{\cos ^{\frac {7}{2}}(c+d x)} \, dx \\ & = \frac {2 a^2 \sec ^{\frac {5}{2}}(c+d x) \sin (c+d x)}{5 d \sqrt {a+a \cos (c+d x)}}-\frac {1}{5} \left (2 a \sqrt {\cos (c+d x)} \sqrt {\sec (c+d x)}\right ) \int \frac {-\frac {9 a}{2}-\frac {9}{2} a \cos (c+d x)}{\cos ^{\frac {5}{2}}(c+d x) \sqrt {a+a \cos (c+d x)}} \, dx \\ & = \frac {2 a^2 \sec ^{\frac {5}{2}}(c+d x) \sin (c+d x)}{5 d \sqrt {a+a \cos (c+d x)}}+\frac {1}{5} \left (9 a \sqrt {\cos (c+d x)} \sqrt {\sec (c+d x)}\right ) \int \frac {\sqrt {a+a \cos (c+d x)}}{\cos ^{\frac {5}{2}}(c+d x)} \, dx \\ & = \frac {6 a^2 \sec ^{\frac {3}{2}}(c+d x) \sin (c+d x)}{5 d \sqrt {a+a \cos (c+d x)}}+\frac {2 a^2 \sec ^{\frac {5}{2}}(c+d x) \sin (c+d x)}{5 d \sqrt {a+a \cos (c+d x)}}+\frac {1}{5} \left (6 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 {12 a^2 \sqrt {\sec (c+d x)} \sin (c+d x)}{5 d \sqrt {a+a \cos (c+d x)}}+\frac {6 a^2 \sec ^{\frac {3}{2}}(c+d x) \sin (c+d x)}{5 d \sqrt {a+a \cos (c+d x)}}+\frac {2 a^2 \sec ^{\frac {5}{2}}(c+d x) \sin (c+d x)}{5 d \sqrt {a+a \cos (c+d x)}} \\ \end{align*}

Mathematica [A] (verified)

Time = 0.12 (sec) , antiderivative size = 62, normalized size of antiderivative = 0.51 \[ \int (a+a \cos (c+d x))^{3/2} \sec ^{\frac {7}{2}}(c+d x) \, dx=\frac {2 a \sqrt {a (1+\cos (c+d x))} (4+3 \cos (c+d x)+3 \cos (2 (c+d x))) \sec ^{\frac {5}{2}}(c+d x) \tan \left (\frac {1}{2} (c+d x)\right )}{5 d} \]

[In]

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

[Out]

(2*a*Sqrt[a*(1 + Cos[c + d*x])]*(4 + 3*Cos[c + d*x] + 3*Cos[2*(c + d*x)])*Sec[c + d*x]^(5/2)*Tan[(c + d*x)/2])
/(5*d)

Maple [A] (verified)

Time = 6.28 (sec) , antiderivative size = 63, normalized size of antiderivative = 0.52

method	result	size
default	\(-\frac {2 \cot \left (d x +c \right ) \sqrt {a \left (1+\cos \left (d x +c \right )\right )}\, \left (\sec ^{\frac {7}{2}}\left (d x +c \right )\right ) \left (6 \left (\cos ^{3}\left (d x +c \right )\right )-3 \left (\cos ^{2}\left (d x +c \right )\right )-2 \cos \left (d x +c \right )-1\right ) a}{5 d}\)	\(63\)

[In]

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

[Out]

-2/5/d*cot(d*x+c)*(a*(1+cos(d*x+c)))^(1/2)*sec(d*x+c)^(7/2)*(6*cos(d*x+c)^3-3*cos(d*x+c)^2-2*cos(d*x+c)-1)*a

Fricas [A] (verification not implemented)

none

Time = 0.28 (sec) , antiderivative size = 73, normalized size of antiderivative = 0.60 \[ \int (a+a \cos (c+d x))^{3/2} \sec ^{\frac {7}{2}}(c+d x) \, dx=\frac {2 \, {\left (6 \, a \cos \left (d x + c\right )^{2} + 3 \, a \cos \left (d x + c\right ) + a\right )} \sqrt {a \cos \left (d x + c\right ) + a} \sin \left (d x + c\right )}{5 \, {\left (d \cos \left (d x + c\right )^{3} + d \cos \left (d x + c\right )^{2}\right )} \sqrt {\cos \left (d x + c\right )}} \]

[In]

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

[Out]

2/5*(6*a*cos(d*x + c)^2 + 3*a*cos(d*x + c) + a)*sqrt(a*cos(d*x + c) + a)*sin(d*x + c)/((d*cos(d*x + c)^3 + d*c
os(d*x + c)^2)*sqrt(cos(d*x + c)))

Sympy [F(-1)]

Timed out. \[ \int (a+a \cos (c+d x))^{3/2} \sec ^{\frac {7}{2}}(c+d x) \, dx=\text {Timed out} \]

[In]

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

[Out]

Timed out

Maxima [B] (verification not implemented)

Leaf count of result is larger than twice the leaf count of optimal. 217 vs. \(2 (103) = 206\).

Time = 0.32 (sec) , antiderivative size = 217, normalized size of antiderivative = 1.79 \[ \int (a+a \cos (c+d x))^{3/2} \sec ^{\frac {7}{2}}(c+d x) \, dx=\frac {4 \, {\left (\frac {5 \, \sqrt {2} a^{\frac {3}{2}} \sin \left (d x + c\right )}{\cos \left (d x + c\right ) + 1} - \frac {10 \, \sqrt {2} a^{\frac {3}{2}} \sin \left (d x + c\right )^{3}}{{\left (\cos \left (d x + c\right ) + 1\right )}^{3}} + \frac {7 \, \sqrt {2} a^{\frac {3}{2}} \sin \left (d x + c\right )^{5}}{{\left (\cos \left (d x + c\right ) + 1\right )}^{5}} - \frac {2 \, \sqrt {2} a^{\frac {3}{2}} \sin \left (d x + c\right )^{7}}{{\left (\cos \left (d x + c\right ) + 1\right )}^{7}}\right )} {\left (\frac {\sin \left (d x + c\right )^{2}}{{\left (\cos \left (d x + c\right ) + 1\right )}^{2}} + 1\right )}^{2}}{5 \, d {\left (\frac {\sin \left (d x + c\right )}{\cos \left (d x + c\right ) + 1} + 1\right )}^{\frac {7}{2}} {\left (-\frac {\sin \left (d x + c\right )}{\cos \left (d x + c\right ) + 1} + 1\right )}^{\frac {7}{2}} {\left (\frac {2 \, \sin \left (d x + c\right )^{2}}{{\left (\cos \left (d x + c\right ) + 1\right )}^{2}} + \frac {\sin \left (d x + c\right )^{4}}{{\left (\cos \left (d x + c\right ) + 1\right )}^{4}} + 1\right )}} \]

[In]

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

[Out]

4/5*(5*sqrt(2)*a^(3/2)*sin(d*x + c)/(cos(d*x + c) + 1) - 10*sqrt(2)*a^(3/2)*sin(d*x + c)^3/(cos(d*x + c) + 1)^
3 + 7*sqrt(2)*a^(3/2)*sin(d*x + c)^5/(cos(d*x + c) + 1)^5 - 2*sqrt(2)*a^(3/2)*sin(d*x + c)^7/(cos(d*x + c) + 1
)^7)*(sin(d*x + c)^2/(cos(d*x + c) + 1)^2 + 1)^2/(d*(sin(d*x + c)/(cos(d*x + c) + 1) + 1)^(7/2)*(-sin(d*x + c)
/(cos(d*x + c) + 1) + 1)^(7/2)*(2*sin(d*x + c)^2/(cos(d*x + c) + 1)^2 + sin(d*x + c)^4/(cos(d*x + c) + 1)^4 +
1))

Giac [F(-1)]

Timed out. \[ \int (a+a \cos (c+d x))^{3/2} \sec ^{\frac {7}{2}}(c+d x) \, dx=\text {Timed out} \]

[In]

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

[Out]

Timed out

Mupad [B] (verification not implemented)

Time = 1.70 (sec) , antiderivative size = 135, normalized size of antiderivative = 1.12 \[ \int (a+a \cos (c+d x))^{3/2} \sec ^{\frac {7}{2}}(c+d x) \, dx=\frac {4\,a\,\sqrt {a\,\left (\cos \left (c+d\,x\right )+1\right )}\,\sqrt {\frac {1}{\cos \left (c+d\,x\right )}}\,\left (8\,\sin \left (c+d\,x\right )+6\,\sin \left (2\,c+2\,d\,x\right )+11\,\sin \left (3\,c+3\,d\,x\right )+3\,\sin \left (4\,c+4\,d\,x\right )+3\,\sin \left (5\,c+5\,d\,x\right )\right )}{5\,d\,\left (10\,\cos \left (c+d\,x\right )+8\,\cos \left (2\,c+2\,d\,x\right )+5\,\cos \left (3\,c+3\,d\,x\right )+2\,\cos \left (4\,c+4\,d\,x\right )+\cos \left (5\,c+5\,d\,x\right )+6\right )} \]

[In]

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

[Out]

(4*a*(a*(cos(c + d*x) + 1))^(1/2)*(1/cos(c + d*x))^(1/2)*(8*sin(c + d*x) + 6*sin(2*c + 2*d*x) + 11*sin(3*c + 3
*d*x) + 3*sin(4*c + 4*d*x) + 3*sin(5*c + 5*d*x)))/(5*d*(10*cos(c + d*x) + 8*cos(2*c + 2*d*x) + 5*cos(3*c + 3*d
*x) + 2*cos(4*c + 4*d*x) + cos(5*c + 5*d*x) + 6))