3.2.13 \(\int \frac {(b \cos (c+d x))^{5/2} (A+C \cos ^2(c+d x))}{\cos ^{\frac {11}{2}}(c+d x)} \, dx\) [113]

Optimal. Leaf size=84 \[ \frac {b^2 (A+2 C) \tanh ^{-1}(\sin (c+d x)) \sqrt {b \cos (c+d x)}}{2 d \sqrt {\cos (c+d x)}}+\frac {A b^2 \sqrt {b \cos (c+d x)} \sin (c+d x)}{2 d \cos ^{\frac {5}{2}}(c+d x)} \]

[Out]

1/2*A*b^2*sin(d*x+c)*(b*cos(d*x+c))^(1/2)/d/cos(d*x+c)^(5/2)+1/2*b^2*(A+2*C)*arctanh(sin(d*x+c))*(b*cos(d*x+c)
)^(1/2)/d/cos(d*x+c)^(1/2)

________________________________________________________________________________________

Rubi [A]
time = 0.03, antiderivative size = 84, normalized size of antiderivative = 1.00, number of steps used = 3, number of rules used = 3, integrand size = 35, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.086, Rules used = {17, 3091, 3855} \begin {gather*} \frac {b^2 (A+2 C) \sqrt {b \cos (c+d x)} \tanh ^{-1}(\sin (c+d x))}{2 d \sqrt {\cos (c+d x)}}+\frac {A b^2 \sin (c+d x) \sqrt {b \cos (c+d x)}}{2 d \cos ^{\frac {5}{2}}(c+d x)} \end {gather*}

Antiderivative was successfully verified.

[In]

Int[((b*Cos[c + d*x])^(5/2)*(A + C*Cos[c + d*x]^2))/Cos[c + d*x]^(11/2),x]

[Out]

(b^2*(A + 2*C)*ArcTanh[Sin[c + d*x]]*Sqrt[b*Cos[c + d*x]])/(2*d*Sqrt[Cos[c + d*x]]) + (A*b^2*Sqrt[b*Cos[c + d*
x]]*Sin[c + d*x])/(2*d*Cos[c + d*x]^(5/2))

Rule 17

Int[(u_.)*((a_.)*(v_))^(m_)*((b_.)*(v_))^(n_), x_Symbol] :> Dist[a^(m + 1/2)*b^(n - 1/2)*(Sqrt[b*v]/Sqrt[a*v])
, Int[u*v^(m + n), x], x] /; FreeQ[{a, b, m}, x] &&  !IntegerQ[m] && IGtQ[n + 1/2, 0] && IntegerQ[m + n]

Rule 3091

Int[((b_.)*sin[(e_.) + (f_.)*(x_)])^(m_)*((A_) + (C_.)*sin[(e_.) + (f_.)*(x_)]^2), x_Symbol] :> Simp[A*Cos[e +
f*x]*((b*Sin[e + f*x])^(m + 1)/(b*f*(m + 1))), x] + Dist[(A*(m + 2) + C*(m + 1))/(b^2*(m + 1)), Int[(b*Sin[e
+ f*x])^(m + 2), x], x] /; FreeQ[{b, e, f, A, C}, x] && LtQ[m, -1]

Rule 3855

Int[csc[(c_.) + (d_.)*(x_)], x_Symbol] :> Simp[-ArcTanh[Cos[c + d*x]]/d, x] /; FreeQ[{c, d}, x]

Rubi steps

\begin {align*} \int \frac {(b \cos (c+d x))^{5/2} \left (A+C \cos ^2(c+d x)\right )}{\cos ^{\frac {11}{2}}(c+d x)} \, dx &=\frac {\left (b^2 \sqrt {b \cos (c+d x)}\right ) \int \left (A+C \cos ^2(c+d x)\right ) \sec ^3(c+d x) \, dx}{\sqrt {\cos (c+d x)}}\\ &=\frac {A b^2 \sqrt {b \cos (c+d x)} \sin (c+d x)}{2 d \cos ^{\frac {5}{2}}(c+d x)}+\frac {\left (b^2 (A+2 C) \sqrt {b \cos (c+d x)}\right ) \int \sec (c+d x) \, dx}{2 \sqrt {\cos (c+d x)}}\\ &=\frac {b^2 (A+2 C) \tanh ^{-1}(\sin (c+d x)) \sqrt {b \cos (c+d x)}}{2 d \sqrt {\cos (c+d x)}}+\frac {A b^2 \sqrt {b \cos (c+d x)} \sin (c+d x)}{2 d \cos ^{\frac {5}{2}}(c+d x)}\\ \end {align*}

________________________________________________________________________________________

Mathematica [A]
time = 0.11, size = 59, normalized size = 0.70 \begin {gather*} \frac {(b \cos (c+d x))^{5/2} \left ((A+2 C) \tanh ^{-1}(\sin (c+d x)) \cos ^2(c+d x)+A \sin (c+d x)\right )}{2 d \cos ^{\frac {9}{2}}(c+d x)} \end {gather*}

Antiderivative was successfully verified.

[In]

Integrate[((b*Cos[c + d*x])^(5/2)*(A + C*Cos[c + d*x]^2))/Cos[c + d*x]^(11/2),x]

[Out]

((b*Cos[c + d*x])^(5/2)*((A + 2*C)*ArcTanh[Sin[c + d*x]]*Cos[c + d*x]^2 + A*Sin[c + d*x]))/(2*d*Cos[c + d*x]^(
9/2))

________________________________________________________________________________________

Maple [A]
time = 0.29, size = 134, normalized size = 1.60

method	result	size
default	\(\frac {\left (-A \left (\cos ^{2}\left (d x +c \right )\right ) \ln \left (-\frac {-1+\cos \left (d x +c \right )+\sin \left (d x +c \right )}{\sin \left (d x +c \right )}\right )+A \left (\cos ^{2}\left (d x +c \right )\right ) \ln \left (\frac {1-\cos \left (d x +c \right )+\sin \left (d x +c \right )}{\sin \left (d x +c \right )}\right )-4 C \left (\cos ^{2}\left (d x +c \right )\right ) \arctanh \left (\frac {-1+\cos \left (d x +c \right )}{\sin \left (d x +c \right )}\right )+A \sin \left (d x +c \right )\right ) \left (b \cos \left (d x +c \right )\right )^{\frac {5}{2}}}{2 d \cos \left (d x +c \right )^{\frac {9}{2}}}\)	\(134\)
risch	\(-\frac {i b^{2} \sqrt {b \cos \left (d x +c \right )}\, A \left ({\mathrm e}^{3 i \left (d x +c \right )}-{\mathrm e}^{i \left (d x +c \right )}\right )}{\sqrt {\cos \left (d x +c \right )}\, d \left ({\mathrm e}^{2 i \left (d x +c \right )}+1\right )^{2}}-\frac {b^{2} \sqrt {b \cos \left (d x +c \right )}\, \left (A +2 C \right ) \ln \left ({\mathrm e}^{i \left (d x +c \right )}-i\right )}{2 \sqrt {\cos \left (d x +c \right )}\, d}+\frac {b^{2} \sqrt {b \cos \left (d x +c \right )}\, \left (A +2 C \right ) \ln \left ({\mathrm e}^{i \left (d x +c \right )}+i\right )}{2 \sqrt {\cos \left (d x +c \right )}\, d}\)	\(152\)

Verification of antiderivative is not currently implemented for this CAS.

[In]

int((b*cos(d*x+c))^(5/2)*(A+C*cos(d*x+c)^2)/cos(d*x+c)^(11/2),x,method=_RETURNVERBOSE)

[Out]

1/2/d*(-A*cos(d*x+c)^2*ln(-(-1+cos(d*x+c)+sin(d*x+c))/sin(d*x+c))+A*cos(d*x+c)^2*ln((1-cos(d*x+c)+sin(d*x+c))/
sin(d*x+c))-4*C*cos(d*x+c)^2*arctanh((-1+cos(d*x+c))/sin(d*x+c))+A*sin(d*x+c))*(b*cos(d*x+c))^(5/2)/cos(d*x+c)
^(9/2)

________________________________________________________________________________________

Maxima [B] Leaf count of result is larger than twice the leaf count of optimal. 821 vs. \(2 (72) = 144\).
time = 0.65, size = 821, normalized size = 9.77 \begin {gather*} \frac {2 \, {\left (b^{2} \log \left (\cos \left (d x + c\right )^{2} + \sin \left (d x + c\right )^{2} + 2 \, \sin \left (d x + c\right ) + 1\right ) - b^{2} \log \left (\cos \left (d x + c\right )^{2} + \sin \left (d x + c\right )^{2} - 2 \, \sin \left (d x + c\right ) + 1\right )\right )} C \sqrt {b} - \frac {{\left (4 \, {\left (b^{2} \sin \left (4 \, d x + 4 \, c\right ) + 2 \, b^{2} \sin \left (2 \, d x + 2 \, c\right )\right )} \cos \left (\frac {3}{2} \, \arctan \left (\sin \left (2 \, d x + 2 \, c\right ), \cos \left (2 \, d x + 2 \, c\right )\right )\right ) - 4 \, {\left (b^{2} \sin \left (4 \, d x + 4 \, c\right ) + 2 \, b^{2} \sin \left (2 \, d x + 2 \, c\right )\right )} \cos \left (\frac {1}{2} \, \arctan \left (\sin \left (2 \, d x + 2 \, c\right ), \cos \left (2 \, d x + 2 \, c\right )\right )\right ) - {\left (b^{2} \cos \left (4 \, d x + 4 \, c\right )^{2} + 4 \, b^{2} \cos \left (2 \, d x + 2 \, c\right )^{2} + b^{2} \sin \left (4 \, d x + 4 \, c\right )^{2} + 4 \, b^{2} \sin \left (4 \, d x + 4 \, c\right ) \sin \left (2 \, d x + 2 \, c\right ) + 4 \, b^{2} \sin \left (2 \, d x + 2 \, c\right )^{2} + 4 \, b^{2} \cos \left (2 \, d x + 2 \, c\right ) + b^{2} + 2 \, {\left (2 \, b^{2} \cos \left (2 \, d x + 2 \, c\right ) + b^{2}\right )} \cos \left (4 \, d x + 4 \, c\right )\right )} \log \left (\cos \left (\frac {1}{2} \, \arctan \left (\sin \left (2 \, d x + 2 \, c\right ), \cos \left (2 \, d x + 2 \, c\right )\right )\right )^{2} + \sin \left (\frac {1}{2} \, \arctan \left (\sin \left (2 \, d x + 2 \, c\right ), \cos \left (2 \, d x + 2 \, c\right )\right )\right )^{2} + 2 \, \sin \left (\frac {1}{2} \, \arctan \left (\sin \left (2 \, d x + 2 \, c\right ), \cos \left (2 \, d x + 2 \, c\right )\right )\right ) + 1\right ) + {\left (b^{2} \cos \left (4 \, d x + 4 \, c\right )^{2} + 4 \, b^{2} \cos \left (2 \, d x + 2 \, c\right )^{2} + b^{2} \sin \left (4 \, d x + 4 \, c\right )^{2} + 4 \, b^{2} \sin \left (4 \, d x + 4 \, c\right ) \sin \left (2 \, d x + 2 \, c\right ) + 4 \, b^{2} \sin \left (2 \, d x + 2 \, c\right )^{2} + 4 \, b^{2} \cos \left (2 \, d x + 2 \, c\right ) + b^{2} + 2 \, {\left (2 \, b^{2} \cos \left (2 \, d x + 2 \, c\right ) + b^{2}\right )} \cos \left (4 \, d x + 4 \, c\right )\right )} \log \left (\cos \left (\frac {1}{2} \, \arctan \left (\sin \left (2 \, d x + 2 \, c\right ), \cos \left (2 \, d x + 2 \, c\right )\right )\right )^{2} + \sin \left (\frac {1}{2} \, \arctan \left (\sin \left (2 \, d x + 2 \, c\right ), \cos \left (2 \, d x + 2 \, c\right )\right )\right )^{2} - 2 \, \sin \left (\frac {1}{2} \, \arctan \left (\sin \left (2 \, d x + 2 \, c\right ), \cos \left (2 \, d x + 2 \, c\right )\right )\right ) + 1\right ) - 4 \, {\left (b^{2} \cos \left (4 \, d x + 4 \, c\right ) + 2 \, b^{2} \cos \left (2 \, d x + 2 \, c\right ) + b^{2}\right )} \sin \left (\frac {3}{2} \, \arctan \left (\sin \left (2 \, d x + 2 \, c\right ), \cos \left (2 \, d x + 2 \, c\right )\right )\right ) + 4 \, {\left (b^{2} \cos \left (4 \, d x + 4 \, c\right ) + 2 \, b^{2} \cos \left (2 \, d x + 2 \, c\right ) + b^{2}\right )} \sin \left (\frac {1}{2} \, \arctan \left (\sin \left (2 \, d x + 2 \, c\right ), \cos \left (2 \, d x + 2 \, c\right )\right )\right )\right )} A \sqrt {b}}{2 \, {\left (2 \, \cos \left (2 \, d x + 2 \, c\right ) + 1\right )} \cos \left (4 \, d x + 4 \, c\right ) + \cos \left (4 \, d x + 4 \, c\right )^{2} + 4 \, \cos \left (2 \, d x + 2 \, c\right )^{2} + \sin \left (4 \, d x + 4 \, c\right )^{2} + 4 \, \sin \left (4 \, d x + 4 \, c\right ) \sin \left (2 \, d x + 2 \, c\right ) + 4 \, \sin \left (2 \, d x + 2 \, c\right )^{2} + 4 \, \cos \left (2 \, d x + 2 \, c\right ) + 1}}{4 \, d} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate((b*cos(d*x+c))^(5/2)*(A+C*cos(d*x+c)^2)/cos(d*x+c)^(11/2),x, algorithm="maxima")

[Out]

1/4*(2*(b^2*log(cos(d*x + c)^2 + sin(d*x + c)^2 + 2*sin(d*x + c) + 1) - b^2*log(cos(d*x + c)^2 + sin(d*x + c)^
2 - 2*sin(d*x + c) + 1))*C*sqrt(b) - (4*(b^2*sin(4*d*x + 4*c) + 2*b^2*sin(2*d*x + 2*c))*cos(3/2*arctan2(sin(2*
d*x + 2*c), cos(2*d*x + 2*c))) - 4*(b^2*sin(4*d*x + 4*c) + 2*b^2*sin(2*d*x + 2*c))*cos(1/2*arctan2(sin(2*d*x +
2*c), cos(2*d*x + 2*c))) - (b^2*cos(4*d*x + 4*c)^2 + 4*b^2*cos(2*d*x + 2*c)^2 + b^2*sin(4*d*x + 4*c)^2 + 4*b^
2*sin(4*d*x + 4*c)*sin(2*d*x + 2*c) + 4*b^2*sin(2*d*x + 2*c)^2 + 4*b^2*cos(2*d*x + 2*c) + b^2 + 2*(2*b^2*cos(2
*d*x + 2*c) + b^2)*cos(4*d*x + 4*c))*log(cos(1/2*arctan2(sin(2*d*x + 2*c), cos(2*d*x + 2*c)))^2 + sin(1/2*arct
an2(sin(2*d*x + 2*c), cos(2*d*x + 2*c)))^2 + 2*sin(1/2*arctan2(sin(2*d*x + 2*c), cos(2*d*x + 2*c))) + 1) + (b^
2*cos(4*d*x + 4*c)^2 + 4*b^2*cos(2*d*x + 2*c)^2 + b^2*sin(4*d*x + 4*c)^2 + 4*b^2*sin(4*d*x + 4*c)*sin(2*d*x +
2*c) + 4*b^2*sin(2*d*x + 2*c)^2 + 4*b^2*cos(2*d*x + 2*c) + b^2 + 2*(2*b^2*cos(2*d*x + 2*c) + b^2)*cos(4*d*x +
4*c))*log(cos(1/2*arctan2(sin(2*d*x + 2*c), cos(2*d*x + 2*c)))^2 + sin(1/2*arctan2(sin(2*d*x + 2*c), cos(2*d*x
+ 2*c)))^2 - 2*sin(1/2*arctan2(sin(2*d*x + 2*c), cos(2*d*x + 2*c))) + 1) - 4*(b^2*cos(4*d*x + 4*c) + 2*b^2*co
s(2*d*x + 2*c) + b^2)*sin(3/2*arctan2(sin(2*d*x + 2*c), cos(2*d*x + 2*c))) + 4*(b^2*cos(4*d*x + 4*c) + 2*b^2*c
os(2*d*x + 2*c) + b^2)*sin(1/2*arctan2(sin(2*d*x + 2*c), cos(2*d*x + 2*c))))*A*sqrt(b)/(2*(2*cos(2*d*x + 2*c)
+ 1)*cos(4*d*x + 4*c) + cos(4*d*x + 4*c)^2 + 4*cos(2*d*x + 2*c)^2 + sin(4*d*x + 4*c)^2 + 4*sin(4*d*x + 4*c)*si
n(2*d*x + 2*c) + 4*sin(2*d*x + 2*c)^2 + 4*cos(2*d*x + 2*c) + 1))/d

________________________________________________________________________________________

Fricas [A]
time = 0.42, size = 222, normalized size = 2.64 \begin {gather*} \left [\frac {{\left (A + 2 \, C\right )} b^{\frac {5}{2}} \cos \left (d x + c\right )^{3} \log \left (-\frac {b \cos \left (d x + c\right )^{3} - 2 \, \sqrt {b \cos \left (d x + c\right )} \sqrt {b} \sqrt {\cos \left (d x + c\right )} \sin \left (d x + c\right ) - 2 \, b \cos \left (d x + c\right )}{\cos \left (d x + c\right )^{3}}\right ) + 2 \, \sqrt {b \cos \left (d x + c\right )} A b^{2} \sqrt {\cos \left (d x + c\right )} \sin \left (d x + c\right )}{4 \, d \cos \left (d x + c\right )^{3}}, -\frac {{\left (A + 2 \, C\right )} \sqrt {-b} b^{2} \arctan \left (\frac {\sqrt {b \cos \left (d x + c\right )} \sqrt {-b} \sin \left (d x + c\right )}{b \sqrt {\cos \left (d x + c\right )}}\right ) \cos \left (d x + c\right )^{3} - \sqrt {b \cos \left (d x + c\right )} A b^{2} \sqrt {\cos \left (d x + c\right )} \sin \left (d x + c\right )}{2 \, d \cos \left (d x + c\right )^{3}}\right ] \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate((b*cos(d*x+c))^(5/2)*(A+C*cos(d*x+c)^2)/cos(d*x+c)^(11/2),x, algorithm="fricas")

[Out]

[1/4*((A + 2*C)*b^(5/2)*cos(d*x + c)^3*log(-(b*cos(d*x + c)^3 - 2*sqrt(b*cos(d*x + c))*sqrt(b)*sqrt(cos(d*x +
c))*sin(d*x + c) - 2*b*cos(d*x + c))/cos(d*x + c)^3) + 2*sqrt(b*cos(d*x + c))*A*b^2*sqrt(cos(d*x + c))*sin(d*x
+ c))/(d*cos(d*x + c)^3), -1/2*((A + 2*C)*sqrt(-b)*b^2*arctan(sqrt(b*cos(d*x + c))*sqrt(-b)*sin(d*x + c)/(b*s
qrt(cos(d*x + c))))*cos(d*x + c)^3 - sqrt(b*cos(d*x + c))*A*b^2*sqrt(cos(d*x + c))*sin(d*x + c))/(d*cos(d*x +
c)^3)]

________________________________________________________________________________________

Sympy [F(-1)] Timed out
time = 0.00, size = 0, normalized size = 0.00 \begin {gather*} \text {Timed out} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate((b*cos(d*x+c))**(5/2)*(A+C*cos(d*x+c)**2)/cos(d*x+c)**(11/2),x)

[Out]

Timed out

________________________________________________________________________________________

Giac [F]
time = 0.00, size = 0, normalized size = 0.00 \begin {gather*} \text {could not integrate} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate((b*cos(d*x+c))^(5/2)*(A+C*cos(d*x+c)^2)/cos(d*x+c)^(11/2),x, algorithm="giac")

[Out]

integrate((C*cos(d*x + c)^2 + A)*(b*cos(d*x + c))^(5/2)/cos(d*x + c)^(11/2), x)

________________________________________________________________________________________

Mupad [F]
time = 0.00, size = -1, normalized size = -0.01 \begin {gather*} \int \frac {\left (C\,{\cos \left (c+d\,x\right )}^2+A\right )\,{\left (b\,\cos \left (c+d\,x\right )\right )}^{5/2}}{{\cos \left (c+d\,x\right )}^{11/2}} \,d x \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

int(((A + C*cos(c + d*x)^2)*(b*cos(c + d*x))^(5/2))/cos(c + d*x)^(11/2),x)

[Out]

int(((A + C*cos(c + d*x)^2)*(b*cos(c + d*x))^(5/2))/cos(c + d*x)^(11/2), x)

________________________________________________________________________________________