3.97 \(\int \frac {\sqrt {b \cos (c+d x)} (A+C \cos ^2(c+d x))}{\cos ^{\frac {11}{2}}(c+d x)} \, dx\)
Optimal. Leaf size=122 \[ \frac {(3 A+4 C) \sin (c+d x) \sqrt {b \cos (c+d x)}}{8 d \cos ^{\frac {5}{2}}(c+d x)}+\frac {(3 A+4 C) \sqrt {b \cos (c+d x)} \tanh ^{-1}(\sin (c+d x))}{8 d \sqrt {\cos (c+d x)}}+\frac {A \sin (c+d x) \sqrt {b \cos (c+d x)}}{4 d \cos ^{\frac {9}{2}}(c+d x)} \]
[Out]
1/4*A*sin(d*x+c)*(b*cos(d*x+c))^(1/2)/d/cos(d*x+c)^(9/2)+1/8*(3*A+4*C)*sin(d*x+c)*(b*cos(d*x+c))^(1/2)/d/cos(d
*x+c)^(5/2)+1/8*(3*A+4*C)*arctanh(sin(d*x+c))*(b*cos(d*x+c))^(1/2)/d/cos(d*x+c)^(1/2)
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Rubi [A] time = 0.06, antiderivative size = 122, normalized size of antiderivative = 1.00,
number of steps used = 4, number of rules used = 4, integrand size = 35, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.114, Rules used =
{17, 3012, 3768, 3770} \[ \frac {(3 A+4 C) \sin (c+d x) \sqrt {b \cos (c+d x)}}{8 d \cos ^{\frac {5}{2}}(c+d x)}+\frac {(3 A+4 C) \sqrt {b \cos (c+d x)} \tanh ^{-1}(\sin (c+d x))}{8 d \sqrt {\cos (c+d x)}}+\frac {A \sin (c+d x) \sqrt {b \cos (c+d x)}}{4 d \cos ^{\frac {9}{2}}(c+d x)} \]
Antiderivative was successfully verified.
[In]
Int[(Sqrt[b*Cos[c + d*x]]*(A + C*Cos[c + d*x]^2))/Cos[c + d*x]^(11/2),x]
[Out]
((3*A + 4*C)*ArcTanh[Sin[c + d*x]]*Sqrt[b*Cos[c + d*x]])/(8*d*Sqrt[Cos[c + d*x]]) + (A*Sqrt[b*Cos[c + d*x]]*Si
n[c + d*x])/(4*d*Cos[c + d*x]^(9/2)) + ((3*A + 4*C)*Sqrt[b*Cos[c + d*x]]*Sin[c + d*x])/(8*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 3012
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 3768
Int[(csc[(c_.) + (d_.)*(x_)]*(b_.))^(n_), x_Symbol] :> -Simp[(b*Cos[c + d*x]*(b*Csc[c + d*x])^(n - 1))/(d*(n -
1)), x] + Dist[(b^2*(n - 2))/(n - 1), Int[(b*Csc[c + d*x])^(n - 2), x], x] /; FreeQ[{b, c, d}, x] && GtQ[n, 1
] && IntegerQ[2*n]
Rule 3770
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 {\sqrt {b \cos (c+d x)} \left (A+C \cos ^2(c+d x)\right )}{\cos ^{\frac {11}{2}}(c+d x)} \, dx &=\frac {\sqrt {b \cos (c+d x)} \int \left (A+C \cos ^2(c+d x)\right ) \sec ^5(c+d x) \, dx}{\sqrt {\cos (c+d x)}}\\ &=\frac {A \sqrt {b \cos (c+d x)} \sin (c+d x)}{4 d \cos ^{\frac {9}{2}}(c+d x)}+\frac {\left ((3 A+4 C) \sqrt {b \cos (c+d x)}\right ) \int \sec ^3(c+d x) \, dx}{4 \sqrt {\cos (c+d x)}}\\ &=\frac {A \sqrt {b \cos (c+d x)} \sin (c+d x)}{4 d \cos ^{\frac {9}{2}}(c+d x)}+\frac {(3 A+4 C) \sqrt {b \cos (c+d x)} \sin (c+d x)}{8 d \cos ^{\frac {5}{2}}(c+d x)}+\frac {\left ((3 A+4 C) \sqrt {b \cos (c+d x)}\right ) \int \sec (c+d x) \, dx}{8 \sqrt {\cos (c+d x)}}\\ &=\frac {(3 A+4 C) \tanh ^{-1}(\sin (c+d x)) \sqrt {b \cos (c+d x)}}{8 d \sqrt {\cos (c+d x)}}+\frac {A \sqrt {b \cos (c+d x)} \sin (c+d x)}{4 d \cos ^{\frac {9}{2}}(c+d x)}+\frac {(3 A+4 C) \sqrt {b \cos (c+d x)} \sin (c+d x)}{8 d \cos ^{\frac {5}{2}}(c+d x)}\\ \end {align*}
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Mathematica [A] time = 0.26, size = 80, normalized size = 0.66 \[ \frac {\sqrt {b \cos (c+d x)} \left (\sin (c+d x) \left ((3 A+4 C) \cos ^2(c+d x)+2 A\right )+(3 A+4 C) \cos ^4(c+d x) \tanh ^{-1}(\sin (c+d x))\right )}{8 d \cos ^{\frac {9}{2}}(c+d x)} \]
Antiderivative was successfully verified.
[In]
Integrate[(Sqrt[b*Cos[c + d*x]]*(A + C*Cos[c + d*x]^2))/Cos[c + d*x]^(11/2),x]
[Out]
(Sqrt[b*Cos[c + d*x]]*((3*A + 4*C)*ArcTanh[Sin[c + d*x]]*Cos[c + d*x]^4 + (2*A + (3*A + 4*C)*Cos[c + d*x]^2)*S
in[c + d*x]))/(8*d*Cos[c + d*x]^(9/2))
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fricas [A] time = 0.46, size = 255, normalized size = 2.09 \[ \left [\frac {{\left (3 \, A + 4 \, C\right )} \sqrt {b} \cos \left (d x + c\right )^{5} \log \left (-\frac {b \cos \left (d x + c\right )^{3} - 2 \, \sqrt {b \cos \left (d x + c\right )} \sqrt {b} \sqrt {\cos \left (d x + c\right )} \sin \left (d x + c\right ) - 2 \, b \cos \left (d x + c\right )}{\cos \left (d x + c\right )^{3}}\right ) + 2 \, {\left ({\left (3 \, A + 4 \, C\right )} \cos \left (d x + c\right )^{2} + 2 \, A\right )} \sqrt {b \cos \left (d x + c\right )} \sqrt {\cos \left (d x + c\right )} \sin \left (d x + c\right )}{16 \, d \cos \left (d x + c\right )^{5}}, -\frac {{\left (3 \, A + 4 \, C\right )} \sqrt {-b} \arctan \left (\frac {\sqrt {b \cos \left (d x + c\right )} \sqrt {-b} \sin \left (d x + c\right )}{b \sqrt {\cos \left (d x + c\right )}}\right ) \cos \left (d x + c\right )^{5} - {\left ({\left (3 \, A + 4 \, C\right )} \cos \left (d x + c\right )^{2} + 2 \, A\right )} \sqrt {b \cos \left (d x + c\right )} \sqrt {\cos \left (d x + c\right )} \sin \left (d x + c\right )}{8 \, d \cos \left (d x + c\right )^{5}}\right ] \]
Verification of antiderivative is not currently implemented for this CAS.
[In]
integrate((A+C*cos(d*x+c)^2)*(b*cos(d*x+c))^(1/2)/cos(d*x+c)^(11/2),x, algorithm="fricas")
[Out]
[1/16*((3*A + 4*C)*sqrt(b)*cos(d*x + c)^5*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*((3*A + 4*C)*cos(d*x + c)^2 + 2*A)*sqrt(b*cos(d*x
+ c))*sqrt(cos(d*x + c))*sin(d*x + c))/(d*cos(d*x + c)^5), -1/8*((3*A + 4*C)*sqrt(-b)*arctan(sqrt(b*cos(d*x +
c))*sqrt(-b)*sin(d*x + c)/(b*sqrt(cos(d*x + c))))*cos(d*x + c)^5 - ((3*A + 4*C)*cos(d*x + c)^2 + 2*A)*sqrt(b*c
os(d*x + c))*sqrt(cos(d*x + c))*sin(d*x + c))/(d*cos(d*x + c)^5)]
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giac [F] time = 0.00, size = 0, normalized size = 0.00 \[ \int \frac {{\left (C \cos \left (d x + c\right )^{2} + A\right )} \sqrt {b \cos \left (d x + c\right )}}{\cos \left (d x + c\right )^{\frac {11}{2}}}\,{d x} \]
Verification of antiderivative is not currently implemented for this CAS.
[In]
integrate((A+C*cos(d*x+c)^2)*(b*cos(d*x+c))^(1/2)/cos(d*x+c)^(11/2),x, algorithm="giac")
[Out]
integrate((C*cos(d*x + c)^2 + A)*sqrt(b*cos(d*x + c))/cos(d*x + c)^(11/2), x)
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maple [B] time = 0.24, size = 214, normalized size = 1.75 \[ -\frac {\left (3 A \left (\cos ^{4}\left (d x +c \right )\right ) \ln \left (-\frac {-1+\cos \left (d x +c \right )+\sin \left (d x +c \right )}{\sin \left (d x +c \right )}\right )-3 A \left (\cos ^{4}\left (d x +c \right )\right ) \ln \left (\frac {1-\cos \left (d x +c \right )+\sin \left (d x +c \right )}{\sin \left (d x +c \right )}\right )+4 C \left (\cos ^{4}\left (d x +c \right )\right ) \ln \left (-\frac {-1+\cos \left (d x +c \right )+\sin \left (d x +c \right )}{\sin \left (d x +c \right )}\right )-4 C \left (\cos ^{4}\left (d x +c \right )\right ) \ln \left (\frac {1-\cos \left (d x +c \right )+\sin \left (d x +c \right )}{\sin \left (d x +c \right )}\right )-3 A \left (\cos ^{2}\left (d x +c \right )\right ) \sin \left (d x +c \right )-4 C \sin \left (d x +c \right ) \left (\cos ^{2}\left (d x +c \right )\right )-2 A \sin \left (d x +c \right )\right ) \sqrt {b \cos \left (d x +c \right )}}{8 d \cos \left (d x +c \right )^{\frac {9}{2}}} \]
Verification of antiderivative is not currently implemented for this CAS.
[In]
int((A+C*cos(d*x+c)^2)*(b*cos(d*x+c))^(1/2)/cos(d*x+c)^(11/2),x)
[Out]
-1/8/d*(3*A*cos(d*x+c)^4*ln(-(-1+cos(d*x+c)+sin(d*x+c))/sin(d*x+c))-3*A*cos(d*x+c)^4*ln((1-cos(d*x+c)+sin(d*x+
c))/sin(d*x+c))+4*C*cos(d*x+c)^4*ln(-(-1+cos(d*x+c)+sin(d*x+c))/sin(d*x+c))-4*C*cos(d*x+c)^4*ln((1-cos(d*x+c)+
sin(d*x+c))/sin(d*x+c))-3*A*cos(d*x+c)^2*sin(d*x+c)-4*C*sin(d*x+c)*cos(d*x+c)^2-2*A*sin(d*x+c))*(b*cos(d*x+c))
^(1/2)/cos(d*x+c)^(9/2)
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maxima [B] time = 1.07, size = 2318, normalized size = 19.00 \[ \text {result too large to display} \]
Verification of antiderivative is not currently implemented for this CAS.
[In]
integrate((A+C*cos(d*x+c)^2)*(b*cos(d*x+c))^(1/2)/cos(d*x+c)^(11/2),x, algorithm="maxima")
[Out]
-1/16*((12*(sin(8*d*x + 8*c) + 4*sin(6*d*x + 6*c) + 6*sin(4*d*x + 4*c) + 4*sin(2*d*x + 2*c))*cos(7/2*arctan2(s
in(2*d*x + 2*c), cos(2*d*x + 2*c))) + 44*(sin(8*d*x + 8*c) + 4*sin(6*d*x + 6*c) + 6*sin(4*d*x + 4*c) + 4*sin(2
*d*x + 2*c))*cos(5/2*arctan2(sin(2*d*x + 2*c), cos(2*d*x + 2*c))) - 44*(sin(8*d*x + 8*c) + 4*sin(6*d*x + 6*c)
+ 6*sin(4*d*x + 4*c) + 4*sin(2*d*x + 2*c))*cos(3/2*arctan2(sin(2*d*x + 2*c), cos(2*d*x + 2*c))) - 12*(sin(8*d*
x + 8*c) + 4*sin(6*d*x + 6*c) + 6*sin(4*d*x + 4*c) + 4*sin(2*d*x + 2*c))*cos(1/2*arctan2(sin(2*d*x + 2*c), cos
(2*d*x + 2*c))) - 3*(2*(4*cos(6*d*x + 6*c) + 6*cos(4*d*x + 4*c) + 4*cos(2*d*x + 2*c) + 1)*cos(8*d*x + 8*c) + c
os(8*d*x + 8*c)^2 + 8*(6*cos(4*d*x + 4*c) + 4*cos(2*d*x + 2*c) + 1)*cos(6*d*x + 6*c) + 16*cos(6*d*x + 6*c)^2 +
12*(4*cos(2*d*x + 2*c) + 1)*cos(4*d*x + 4*c) + 36*cos(4*d*x + 4*c)^2 + 16*cos(2*d*x + 2*c)^2 + 4*(2*sin(6*d*x
+ 6*c) + 3*sin(4*d*x + 4*c) + 2*sin(2*d*x + 2*c))*sin(8*d*x + 8*c) + sin(8*d*x + 8*c)^2 + 16*(3*sin(4*d*x + 4
*c) + 2*sin(2*d*x + 2*c))*sin(6*d*x + 6*c) + 16*sin(6*d*x + 6*c)^2 + 36*sin(4*d*x + 4*c)^2 + 48*sin(4*d*x + 4*
c)*sin(2*d*x + 2*c) + 16*sin(2*d*x + 2*c)^2 + 8*cos(2*d*x + 2*c) + 1)*log(cos(1/2*arctan2(sin(2*d*x + 2*c), co
s(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) + 3*(2*(4*cos(6*d*x + 6*c) + 6*cos(4*d*x + 4*c) + 4*cos(2*d*x + 2*c) + 1)*cos(8*d*
x + 8*c) + cos(8*d*x + 8*c)^2 + 8*(6*cos(4*d*x + 4*c) + 4*cos(2*d*x + 2*c) + 1)*cos(6*d*x + 6*c) + 16*cos(6*d*
x + 6*c)^2 + 12*(4*cos(2*d*x + 2*c) + 1)*cos(4*d*x + 4*c) + 36*cos(4*d*x + 4*c)^2 + 16*cos(2*d*x + 2*c)^2 + 4*
(2*sin(6*d*x + 6*c) + 3*sin(4*d*x + 4*c) + 2*sin(2*d*x + 2*c))*sin(8*d*x + 8*c) + sin(8*d*x + 8*c)^2 + 16*(3*s
in(4*d*x + 4*c) + 2*sin(2*d*x + 2*c))*sin(6*d*x + 6*c) + 16*sin(6*d*x + 6*c)^2 + 36*sin(4*d*x + 4*c)^2 + 48*si
n(4*d*x + 4*c)*sin(2*d*x + 2*c) + 16*sin(2*d*x + 2*c)^2 + 8*cos(2*d*x + 2*c) + 1)*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(si
n(2*d*x + 2*c), cos(2*d*x + 2*c))) + 1) - 12*(cos(8*d*x + 8*c) + 4*cos(6*d*x + 6*c) + 6*cos(4*d*x + 4*c) + 4*c
os(2*d*x + 2*c) + 1)*sin(7/2*arctan2(sin(2*d*x + 2*c), cos(2*d*x + 2*c))) - 44*(cos(8*d*x + 8*c) + 4*cos(6*d*x
+ 6*c) + 6*cos(4*d*x + 4*c) + 4*cos(2*d*x + 2*c) + 1)*sin(5/2*arctan2(sin(2*d*x + 2*c), cos(2*d*x + 2*c))) +
44*(cos(8*d*x + 8*c) + 4*cos(6*d*x + 6*c) + 6*cos(4*d*x + 4*c) + 4*cos(2*d*x + 2*c) + 1)*sin(3/2*arctan2(sin(2
*d*x + 2*c), cos(2*d*x + 2*c))) + 12*(cos(8*d*x + 8*c) + 4*cos(6*d*x + 6*c) + 6*cos(4*d*x + 4*c) + 4*cos(2*d*x
+ 2*c) + 1)*sin(1/2*arctan2(sin(2*d*x + 2*c), cos(2*d*x + 2*c))))*A*sqrt(b)/(2*(4*cos(6*d*x + 6*c) + 6*cos(4*
d*x + 4*c) + 4*cos(2*d*x + 2*c) + 1)*cos(8*d*x + 8*c) + cos(8*d*x + 8*c)^2 + 8*(6*cos(4*d*x + 4*c) + 4*cos(2*d
*x + 2*c) + 1)*cos(6*d*x + 6*c) + 16*cos(6*d*x + 6*c)^2 + 12*(4*cos(2*d*x + 2*c) + 1)*cos(4*d*x + 4*c) + 36*co
s(4*d*x + 4*c)^2 + 16*cos(2*d*x + 2*c)^2 + 4*(2*sin(6*d*x + 6*c) + 3*sin(4*d*x + 4*c) + 2*sin(2*d*x + 2*c))*si
n(8*d*x + 8*c) + sin(8*d*x + 8*c)^2 + 16*(3*sin(4*d*x + 4*c) + 2*sin(2*d*x + 2*c))*sin(6*d*x + 6*c) + 16*sin(6
*d*x + 6*c)^2 + 36*sin(4*d*x + 4*c)^2 + 48*sin(4*d*x + 4*c)*sin(2*d*x + 2*c) + 16*sin(2*d*x + 2*c)^2 + 8*cos(2
*d*x + 2*c) + 1) + 4*(4*(sin(4*d*x + 4*c) + 2*sin(2*d*x + 2*c))*cos(3/2*arctan2(sin(2*d*x + 2*c), cos(2*d*x +
2*c))) - 4*(sin(4*d*x + 4*c) + 2*sin(2*d*x + 2*c))*cos(1/2*arctan2(sin(2*d*x + 2*c), cos(2*d*x + 2*c))) - (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)*sin(2*d*x + 2*c) + 4*sin(2*d*x + 2*c)^2 + 4*cos(2*d*x + 2*c) + 1)*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(s
in(2*d*x + 2*c), cos(2*d*x + 2*c))) + 1) + (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)*sin(2*d*x + 2*c) + 4*sin(2*d*x + 2*c)^2 + 4*co
s(2*d*x + 2*c) + 1)*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*(cos(4*d*x + 4*c)
+ 2*cos(2*d*x + 2*c) + 1)*sin(3/2*arctan2(sin(2*d*x + 2*c), cos(2*d*x + 2*c))) + 4*(cos(4*d*x + 4*c) + 2*cos(2
*d*x + 2*c) + 1)*sin(1/2*arctan2(sin(2*d*x + 2*c), cos(2*d*x + 2*c))))*C*sqrt(b)/(2*(2*cos(2*d*x + 2*c) + 1)*c
os(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)*sin(2*d*
x + 2*c) + 4*sin(2*d*x + 2*c)^2 + 4*cos(2*d*x + 2*c) + 1))/d
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mupad [F] time = 0.00, size = -1, normalized size = -0.01 \[ \int \frac {\left (C\,{\cos \left (c+d\,x\right )}^2+A\right )\,\sqrt {b\,\cos \left (c+d\,x\right )}}{{\cos \left (c+d\,x\right )}^{11/2}} \,d x \]
Verification of antiderivative is not currently implemented for this CAS.
[In]
int(((A + C*cos(c + d*x)^2)*(b*cos(c + d*x))^(1/2))/cos(c + d*x)^(11/2),x)
[Out]
int(((A + C*cos(c + d*x)^2)*(b*cos(c + d*x))^(1/2))/cos(c + d*x)^(11/2), x)
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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+C*cos(d*x+c)**2)*(b*cos(d*x+c))**(1/2)/cos(d*x+c)**(11/2),x)
[Out]
Timed out
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