3.5.27 \(\int \frac {\cos ^{\frac {5}{2}}(c+d x)}{(a+a \sec (c+d x))^{3/2}} \, dx\) [427]

Optimal. Leaf size=237 \[ -\frac {15 \tanh ^{-1}\left (\frac {\sqrt {a} \sqrt {\sec (c+d x)} \sin (c+d x)}{\sqrt {2} \sqrt {a+a \sec (c+d x)}}\right ) \sqrt {\cos (c+d x)} \sqrt {\sec (c+d x)}}{2 \sqrt {2} a^{3/2} d}-\frac {\cos ^{\frac {3}{2}}(c+d x) \sin (c+d x)}{2 d (a+a \sec (c+d x))^{3/2}}+\frac {49 \sin (c+d x)}{10 a d \sqrt {\cos (c+d x)} \sqrt {a+a \sec (c+d x)}}-\frac {13 \sqrt {\cos (c+d x)} \sin (c+d x)}{10 a d \sqrt {a+a \sec (c+d x)}}+\frac {9 \cos ^{\frac {3}{2}}(c+d x) \sin (c+d x)}{10 a d \sqrt {a+a \sec (c+d x)}} \]

[Out]

-1/2*cos(d*x+c)^(3/2)*sin(d*x+c)/d/(a+a*sec(d*x+c))^(3/2)-15/4*arctanh(1/2*sin(d*x+c)*a^(1/2)*sec(d*x+c)^(1/2)
*2^(1/2)/(a+a*sec(d*x+c))^(1/2))*cos(d*x+c)^(1/2)*sec(d*x+c)^(1/2)/a^(3/2)/d*2^(1/2)+9/10*cos(d*x+c)^(3/2)*sin
(d*x+c)/a/d/(a+a*sec(d*x+c))^(1/2)+49/10*sin(d*x+c)/a/d/cos(d*x+c)^(1/2)/(a+a*sec(d*x+c))^(1/2)-13/10*sin(d*x+
c)*cos(d*x+c)^(1/2)/a/d/(a+a*sec(d*x+c))^(1/2)

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Rubi [A]
time = 0.41, antiderivative size = 237, normalized size of antiderivative = 1.00, number of steps used = 7, number of rules used = 6, integrand size = 25, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.240, Rules used = {4349, 3902, 4107, 4098, 3893, 212} \begin {gather*} -\frac {15 \sqrt {\cos (c+d x)} \sqrt {\sec (c+d x)} \tanh ^{-1}\left (\frac {\sqrt {a} \sin (c+d x) \sqrt {\sec (c+d x)}}{\sqrt {2} \sqrt {a \sec (c+d x)+a}}\right )}{2 \sqrt {2} a^{3/2} d}+\frac {9 \sin (c+d x) \cos ^{\frac {3}{2}}(c+d x)}{10 a d \sqrt {a \sec (c+d x)+a}}-\frac {\sin (c+d x) \cos ^{\frac {3}{2}}(c+d x)}{2 d (a \sec (c+d x)+a)^{3/2}}-\frac {13 \sin (c+d x) \sqrt {\cos (c+d x)}}{10 a d \sqrt {a \sec (c+d x)+a}}+\frac {49 \sin (c+d x)}{10 a d \sqrt {\cos (c+d x)} \sqrt {a \sec (c+d x)+a}} \end {gather*}

Antiderivative was successfully verified.

[In]

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

[Out]

(-15*ArcTanh[(Sqrt[a]*Sqrt[Sec[c + d*x]]*Sin[c + d*x])/(Sqrt[2]*Sqrt[a + a*Sec[c + d*x]])]*Sqrt[Cos[c + d*x]]*
Sqrt[Sec[c + d*x]])/(2*Sqrt[2]*a^(3/2)*d) - (Cos[c + d*x]^(3/2)*Sin[c + d*x])/(2*d*(a + a*Sec[c + d*x])^(3/2))
+ (49*Sin[c + d*x])/(10*a*d*Sqrt[Cos[c + d*x]]*Sqrt[a + a*Sec[c + d*x]]) - (13*Sqrt[Cos[c + d*x]]*Sin[c + d*x
])/(10*a*d*Sqrt[a + a*Sec[c + d*x]]) + (9*Cos[c + d*x]^(3/2)*Sin[c + d*x])/(10*a*d*Sqrt[a + a*Sec[c + d*x]])

Rule 212

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

Rule 3893

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

Rule 3902

Int[(csc[(e_.) + (f_.)*(x_)]*(d_.))^(n_)*(csc[(e_.) + (f_.)*(x_)]*(b_.) + (a_))^(m_), x_Symbol] :> Simp[(-Cot[
e + f*x])*(a + b*Csc[e + f*x])^m*((d*Csc[e + f*x])^n/(f*(2*m + 1))), x] + Dist[1/(a^2*(2*m + 1)), Int[(a + b*C
sc[e + f*x])^(m + 1)*(d*Csc[e + f*x])^n*(a*(2*m + n + 1) - b*(m + n + 1)*Csc[e + f*x]), x], x] /; FreeQ[{a, b,
d, e, f, n}, x] && EqQ[a^2 - b^2, 0] && LtQ[m, -1] && (IntegersQ[2*m, 2*n] || IntegerQ[m])

Rule 4098

Int[(csc[(e_.) + (f_.)*(x_)]*(d_.))^(n_)*(csc[(e_.) + (f_.)*(x_)]*(b_.) + (a_))^(m_)*(csc[(e_.) + (f_.)*(x_)]*
(B_.) + (A_)), x_Symbol] :> Simp[A*Cot[e + f*x]*(a + b*Csc[e + f*x])^m*((d*Csc[e + f*x])^n/(f*n)), x] - Dist[(
a*A*m - b*B*n)/(b*d*n), Int[(a + b*Csc[e + f*x])^m*(d*Csc[e + f*x])^(n + 1), x], x] /; FreeQ[{a, b, d, e, f, A
, B, m, n}, x] && NeQ[A*b - a*B, 0] && EqQ[a^2 - b^2, 0] && EqQ[m + n + 1, 0] &&  !LeQ[m, -1]

Rule 4107

Int[(csc[(e_.) + (f_.)*(x_)]*(d_.))^(n_)*(csc[(e_.) + (f_.)*(x_)]*(b_.) + (a_))^(m_)*(csc[(e_.) + (f_.)*(x_)]*
(B_.) + (A_)), x_Symbol] :> Simp[A*Cot[e + f*x]*(a + b*Csc[e + f*x])^m*((d*Csc[e + f*x])^n/(f*n)), x] - Dist[1
/(b*d*n), Int[(a + b*Csc[e + f*x])^m*(d*Csc[e + f*x])^(n + 1)*Simp[a*A*m - b*B*n - A*b*(m + n + 1)*Csc[e + f*x
], x], x], x] /; FreeQ[{a, b, d, e, f, A, B, m}, x] && NeQ[A*b - a*B, 0] && EqQ[a^2 - b^2, 0] && LtQ[n, 0]

Rule 4349

Int[(u_)*((c_.)*sin[(a_.) + (b_.)*(x_)])^(m_.), x_Symbol] :> Dist[(c*Csc[a + b*x])^m*(c*Sin[a + b*x])^m, Int[A
ctivateTrig[u]/(c*Csc[a + b*x])^m, x], x] /; FreeQ[{a, b, c, m}, x] &&  !IntegerQ[m] && KnownSecantIntegrandQ[
u, x]

Rubi steps

\begin {align*} \int \frac {\cos ^{\frac {5}{2}}(c+d x)}{(a+a \sec (c+d x))^{3/2}} \, dx &=\left (\sqrt {\cos (c+d x)} \sqrt {\sec (c+d x)}\right ) \int \frac {1}{\sec ^{\frac {5}{2}}(c+d x) (a+a \sec (c+d x))^{3/2}} \, dx\\ &=-\frac {\cos ^{\frac {3}{2}}(c+d x) \sin (c+d x)}{2 d (a+a \sec (c+d x))^{3/2}}-\frac {\left (\sqrt {\cos (c+d x)} \sqrt {\sec (c+d x)}\right ) \int \frac {-\frac {9 a}{2}+3 a \sec (c+d x)}{\sec ^{\frac {5}{2}}(c+d x) \sqrt {a+a \sec (c+d x)}} \, dx}{2 a^2}\\ &=-\frac {\cos ^{\frac {3}{2}}(c+d x) \sin (c+d x)}{2 d (a+a \sec (c+d x))^{3/2}}+\frac {9 \cos ^{\frac {3}{2}}(c+d x) \sin (c+d x)}{10 a d \sqrt {a+a \sec (c+d x)}}-\frac {\left (\sqrt {\cos (c+d x)} \sqrt {\sec (c+d x)}\right ) \int \frac {\frac {39 a^2}{4}-9 a^2 \sec (c+d x)}{\sec ^{\frac {3}{2}}(c+d x) \sqrt {a+a \sec (c+d x)}} \, dx}{5 a^3}\\ &=-\frac {\cos ^{\frac {3}{2}}(c+d x) \sin (c+d x)}{2 d (a+a \sec (c+d x))^{3/2}}-\frac {13 \sqrt {\cos (c+d x)} \sin (c+d x)}{10 a d \sqrt {a+a \sec (c+d x)}}+\frac {9 \cos ^{\frac {3}{2}}(c+d x) \sin (c+d x)}{10 a d \sqrt {a+a \sec (c+d x)}}-\frac {\left (2 \sqrt {\cos (c+d x)} \sqrt {\sec (c+d x)}\right ) \int \frac {-\frac {147 a^3}{8}+\frac {39}{4} a^3 \sec (c+d x)}{\sqrt {\sec (c+d x)} \sqrt {a+a \sec (c+d x)}} \, dx}{15 a^4}\\ &=-\frac {\cos ^{\frac {3}{2}}(c+d x) \sin (c+d x)}{2 d (a+a \sec (c+d x))^{3/2}}+\frac {49 \sin (c+d x)}{10 a d \sqrt {\cos (c+d x)} \sqrt {a+a \sec (c+d x)}}-\frac {13 \sqrt {\cos (c+d x)} \sin (c+d x)}{10 a d \sqrt {a+a \sec (c+d x)}}+\frac {9 \cos ^{\frac {3}{2}}(c+d x) \sin (c+d x)}{10 a d \sqrt {a+a \sec (c+d x)}}-\frac {\left (15 \sqrt {\cos (c+d x)} \sqrt {\sec (c+d x)}\right ) \int \frac {\sqrt {\sec (c+d x)}}{\sqrt {a+a \sec (c+d x)}} \, dx}{4 a}\\ &=-\frac {\cos ^{\frac {3}{2}}(c+d x) \sin (c+d x)}{2 d (a+a \sec (c+d x))^{3/2}}+\frac {49 \sin (c+d x)}{10 a d \sqrt {\cos (c+d x)} \sqrt {a+a \sec (c+d x)}}-\frac {13 \sqrt {\cos (c+d x)} \sin (c+d x)}{10 a d \sqrt {a+a \sec (c+d x)}}+\frac {9 \cos ^{\frac {3}{2}}(c+d x) \sin (c+d x)}{10 a d \sqrt {a+a \sec (c+d x)}}+\frac {\left (15 \sqrt {\cos (c+d x)} \sqrt {\sec (c+d x)}\right ) \text {Subst}\left (\int \frac {1}{2 a-x^2} \, dx,x,-\frac {a \sqrt {\sec (c+d x)} \sin (c+d x)}{\sqrt {a+a \sec (c+d x)}}\right )}{2 a d}\\ &=-\frac {15 \tanh ^{-1}\left (\frac {\sqrt {a} \sqrt {\sec (c+d x)} \sin (c+d x)}{\sqrt {2} \sqrt {a+a \sec (c+d x)}}\right ) \sqrt {\cos (c+d x)} \sqrt {\sec (c+d x)}}{2 \sqrt {2} a^{3/2} d}-\frac {\cos ^{\frac {3}{2}}(c+d x) \sin (c+d x)}{2 d (a+a \sec (c+d x))^{3/2}}+\frac {49 \sin (c+d x)}{10 a d \sqrt {\cos (c+d x)} \sqrt {a+a \sec (c+d x)}}-\frac {13 \sqrt {\cos (c+d x)} \sin (c+d x)}{10 a d \sqrt {a+a \sec (c+d x)}}+\frac {9 \cos ^{\frac {3}{2}}(c+d x) \sin (c+d x)}{10 a d \sqrt {a+a \sec (c+d x)}}\\ \end {align*}

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Mathematica [A]
time = 0.97, size = 152, normalized size = 0.64 \begin {gather*} \frac {75 \sqrt {2} \text {ArcTan}\left (\frac {\sqrt {2} \sqrt {\sec (c+d x)}}{\sqrt {1-\sec (c+d x)}}\right ) \cos ^2\left (\frac {1}{2} (c+d x)\right ) \sec ^{\frac {3}{2}}(c+d x) \sin (c+d x)+\sqrt {1-\sec (c+d x)} \left (4 \left (9+\cos ^2(c+d x)\right ) \sin (c+d x)-2 \sin (2 (c+d x))+49 \tan (c+d x)\right )}{10 d \sqrt {-1+\cos (c+d x)} (a (1+\sec (c+d x)))^{3/2}} \end {gather*}

Warning: Unable to verify antiderivative.

[In]

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

[Out]

(75*Sqrt[2]*ArcTan[(Sqrt[2]*Sqrt[Sec[c + d*x]])/Sqrt[1 - Sec[c + d*x]]]*Cos[(c + d*x)/2]^2*Sec[c + d*x]^(3/2)*
Sin[c + d*x] + Sqrt[1 - Sec[c + d*x]]*(4*(9 + Cos[c + d*x]^2)*Sin[c + d*x] - 2*Sin[2*(c + d*x)] + 49*Tan[c + d
*x]))/(10*d*Sqrt[-1 + Cos[c + d*x]]*(a*(1 + Sec[c + d*x]))^(3/2))

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Maple [A]
time = 0.14, size = 193, normalized size = 0.81

method	result	size
default	\(-\frac {\left (75 \left (\cos ^{2}\left (d x +c \right )\right ) \sin \left (d x +c \right ) \arctan \left (\frac {\sin \left (d x +c \right ) \sqrt {-\frac {2}{1+\cos \left (d x +c \right )}}}{2}\right ) \sqrt {-\frac {2}{1+\cos \left (d x +c \right )}}-8 \left (\cos ^{5}\left (d x +c \right )\right )+24 \left (\cos ^{4}\left (d x +c \right )\right )-75 \arctan \left (\frac {\sin \left (d x +c \right ) \sqrt {-\frac {2}{1+\cos \left (d x +c \right )}}}{2}\right ) \sqrt {-\frac {2}{1+\cos \left (d x +c \right )}}\, \sin \left (d x +c \right )-96 \left (\cos ^{3}\left (d x +c \right )\right )+54 \left (\cos ^{2}\left (d x +c \right )\right )+124 \cos \left (d x +c \right )-98\right ) \sqrt {\frac {a \left (1+\cos \left (d x +c \right )\right )}{\cos \left (d x +c \right )}}\, \left (\sqrt {\cos }\left (d x +c \right )\right )}{20 d \sin \left (d x +c \right )^{3} a^{2}}\)	\(193\)

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

-1/20/d*(75*sin(d*x+c)*cos(d*x+c)^2*arctan(1/2*sin(d*x+c)*(-2/(1+cos(d*x+c)))^(1/2))*(-2/(1+cos(d*x+c)))^(1/2)
-8*cos(d*x+c)^5+24*cos(d*x+c)^4-75*arctan(1/2*sin(d*x+c)*(-2/(1+cos(d*x+c)))^(1/2))*(-2/(1+cos(d*x+c)))^(1/2)*
sin(d*x+c)-96*cos(d*x+c)^3+54*cos(d*x+c)^2+124*cos(d*x+c)-98)*(a*(1+cos(d*x+c))/cos(d*x+c))^(1/2)*cos(d*x+c)^(
1/2)/sin(d*x+c)^3/a^2

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Maxima [F(-2)]
time = 0.00, size = 0, normalized size = 0.00 \begin {gather*} \text {Exception raised: RuntimeError} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

Exception raised: RuntimeError >> ECL says: THROW: The catch RAT-ERR is undefined.

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Fricas [A]
time = 4.09, size = 400, normalized size = 1.69 \begin {gather*} \left [\frac {75 \, \sqrt {2} {\left (\cos \left (d x + c\right )^{2} + 2 \, \cos \left (d x + c\right ) + 1\right )} \sqrt {a} \log \left (-\frac {a \cos \left (d x + c\right )^{2} + 2 \, \sqrt {2} \sqrt {a} \sqrt {\frac {a \cos \left (d x + c\right ) + a}{\cos \left (d x + c\right )}} \sqrt {\cos \left (d x + c\right )} \sin \left (d x + c\right ) - 2 \, a \cos \left (d x + c\right ) - 3 \, a}{\cos \left (d x + c\right )^{2} + 2 \, \cos \left (d x + c\right ) + 1}\right ) + 4 \, {\left (4 \, \cos \left (d x + c\right )^{3} - 4 \, \cos \left (d x + c\right )^{2} + 36 \, \cos \left (d x + c\right ) + 49\right )} \sqrt {\frac {a \cos \left (d x + c\right ) + a}{\cos \left (d x + c\right )}} \sqrt {\cos \left (d x + c\right )} \sin \left (d x + c\right )}{40 \, {\left (a^{2} d \cos \left (d x + c\right )^{2} + 2 \, a^{2} d \cos \left (d x + c\right ) + a^{2} d\right )}}, \frac {75 \, \sqrt {2} {\left (\cos \left (d x + c\right )^{2} + 2 \, \cos \left (d x + c\right ) + 1\right )} \sqrt {-a} \arctan \left (\frac {\sqrt {2} \sqrt {-a} \sqrt {\frac {a \cos \left (d x + c\right ) + a}{\cos \left (d x + c\right )}} \sqrt {\cos \left (d x + c\right )}}{a \sin \left (d x + c\right )}\right ) + 2 \, {\left (4 \, \cos \left (d x + c\right )^{3} - 4 \, \cos \left (d x + c\right )^{2} + 36 \, \cos \left (d x + c\right ) + 49\right )} \sqrt {\frac {a \cos \left (d x + c\right ) + a}{\cos \left (d x + c\right )}} \sqrt {\cos \left (d x + c\right )} \sin \left (d x + c\right )}{20 \, {\left (a^{2} d \cos \left (d x + c\right )^{2} + 2 \, a^{2} d \cos \left (d x + c\right ) + a^{2} d\right )}}\right ] \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

[1/40*(75*sqrt(2)*(cos(d*x + c)^2 + 2*cos(d*x + c) + 1)*sqrt(a)*log(-(a*cos(d*x + c)^2 + 2*sqrt(2)*sqrt(a)*sqr
t((a*cos(d*x + c) + a)/cos(d*x + c))*sqrt(cos(d*x + c))*sin(d*x + c) - 2*a*cos(d*x + c) - 3*a)/(cos(d*x + c)^2
+ 2*cos(d*x + c) + 1)) + 4*(4*cos(d*x + c)^3 - 4*cos(d*x + c)^2 + 36*cos(d*x + c) + 49)*sqrt((a*cos(d*x + c)
+ a)/cos(d*x + c))*sqrt(cos(d*x + c))*sin(d*x + c))/(a^2*d*cos(d*x + c)^2 + 2*a^2*d*cos(d*x + c) + a^2*d), 1/2
0*(75*sqrt(2)*(cos(d*x + c)^2 + 2*cos(d*x + c) + 1)*sqrt(-a)*arctan(sqrt(2)*sqrt(-a)*sqrt((a*cos(d*x + c) + a)
/cos(d*x + c))*sqrt(cos(d*x + c))/(a*sin(d*x + c))) + 2*(4*cos(d*x + c)^3 - 4*cos(d*x + c)^2 + 36*cos(d*x + c)
+ 49)*sqrt((a*cos(d*x + c) + a)/cos(d*x + c))*sqrt(cos(d*x + c))*sin(d*x + c))/(a^2*d*cos(d*x + c)^2 + 2*a^2*
d*cos(d*x + c) + a^2*d)]

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

[Out]

Timed out

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

[Out]

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

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Mupad [F]
time = 0.00, size = -1, normalized size = -0.00 \begin {gather*} \int \frac {{\cos \left (c+d\,x\right )}^{5/2}}{{\left (a+\frac {a}{\cos \left (c+d\,x\right )}\right )}^{3/2}} \,d x \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

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

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