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

Optimal. Leaf size=197 \[ -\frac {75 \sqrt {\cos (c+d x)} \sqrt {\sec (c+d x)} \tanh ^{-1}\left (\frac {\sqrt {a} \sin (c+d x) \sqrt {\sec (c+d x)}}{\sqrt {2} \sqrt {a \sec (c+d x)+a}}\right )}{16 \sqrt {2} a^{5/2} d}+\frac {49 \sin (c+d x)}{16 a^2 d \sqrt {\cos (c+d x)} \sqrt {a \sec (c+d x)+a}}-\frac {13 \sin (c+d x)}{16 a d \sqrt {\cos (c+d x)} (a \sec (c+d x)+a)^{3/2}}-\frac {\sin (c+d x)}{4 d \sqrt {\cos (c+d x)} (a \sec (c+d x)+a)^{5/2}} \]

[Out]

-1/4*sin(d*x+c)/d/(a+a*sec(d*x+c))^(5/2)/cos(d*x+c)^(1/2)-13/16*sin(d*x+c)/a/d/(a+a*sec(d*x+c))^(3/2)/cos(d*x+
c)^(1/2)-75/32*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^(5/2)/d*2^(1/2)+49/16*sin(d*x+c)/a^2/d/cos(d*x+c)^(1/2)/(a+a*sec(d*x+c))^(1/2)

________________________________________________________________________________________

Rubi [A]  time = 0.46, antiderivative size = 197, normalized size of antiderivative = 1.00, number of steps used = 6, number of rules used = 6, integrand size = 25, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.240, Rules used = {4264, 3817, 4020, 4013, 3808, 206} \[ \frac {49 \sin (c+d x)}{16 a^2 d \sqrt {\cos (c+d x)} \sqrt {a \sec (c+d x)+a}}-\frac {75 \sqrt {\cos (c+d x)} \sqrt {\sec (c+d x)} \tanh ^{-1}\left (\frac {\sqrt {a} \sin (c+d x) \sqrt {\sec (c+d x)}}{\sqrt {2} \sqrt {a \sec (c+d x)+a}}\right )}{16 \sqrt {2} a^{5/2} d}-\frac {13 \sin (c+d x)}{16 a d \sqrt {\cos (c+d x)} (a \sec (c+d x)+a)^{3/2}}-\frac {\sin (c+d x)}{4 d \sqrt {\cos (c+d x)} (a \sec (c+d x)+a)^{5/2}} \]

Antiderivative was successfully verified.

[In]

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

[Out]

(-75*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]])/(16*Sqrt[2]*a^(5/2)*d) - Sin[c + d*x]/(4*d*Sqrt[Cos[c + d*x]]*(a + a*Sec[c + d*x])^(5/2))
- (13*Sin[c + d*x])/(16*a*d*Sqrt[Cos[c + d*x]]*(a + a*Sec[c + d*x])^(3/2)) + (49*Sin[c + d*x])/(16*a^2*d*Sqrt[
Cos[c + d*x]]*Sqrt[a + a*Sec[c + d*x]])

Rule 206

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

Rule 3808

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 3817

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*Csc
[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 4013

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 4020

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

Rule 4264

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 {\sqrt {\cos (c+d x)}}{(a+a \sec (c+d x))^{5/2}} \, dx &=\left (\sqrt {\cos (c+d x)} \sqrt {\sec (c+d x)}\right ) \int \frac {1}{\sqrt {\sec (c+d x)} (a+a \sec (c+d x))^{5/2}} \, dx\\ &=-\frac {\sin (c+d x)}{4 d \sqrt {\cos (c+d x)} (a+a \sec (c+d x))^{5/2}}-\frac {\left (\sqrt {\cos (c+d x)} \sqrt {\sec (c+d x)}\right ) \int \frac {-\frac {9 a}{2}+2 a \sec (c+d x)}{\sqrt {\sec (c+d x)} (a+a \sec (c+d x))^{3/2}} \, dx}{4 a^2}\\ &=-\frac {\sin (c+d x)}{4 d \sqrt {\cos (c+d x)} (a+a \sec (c+d x))^{5/2}}-\frac {13 \sin (c+d x)}{16 a d \sqrt {\cos (c+d x)} (a+a \sec (c+d x))^{3/2}}-\frac {\left (\sqrt {\cos (c+d x)} \sqrt {\sec (c+d x)}\right ) \int \frac {-\frac {49 a^2}{4}+\frac {13}{2} a^2 \sec (c+d x)}{\sqrt {\sec (c+d x)} \sqrt {a+a \sec (c+d x)}} \, dx}{8 a^4}\\ &=-\frac {\sin (c+d x)}{4 d \sqrt {\cos (c+d x)} (a+a \sec (c+d x))^{5/2}}-\frac {13 \sin (c+d x)}{16 a d \sqrt {\cos (c+d x)} (a+a \sec (c+d x))^{3/2}}+\frac {49 \sin (c+d x)}{16 a^2 d \sqrt {\cos (c+d x)} \sqrt {a+a \sec (c+d x)}}-\frac {\left (75 \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}{32 a^2}\\ &=-\frac {\sin (c+d x)}{4 d \sqrt {\cos (c+d x)} (a+a \sec (c+d x))^{5/2}}-\frac {13 \sin (c+d x)}{16 a d \sqrt {\cos (c+d x)} (a+a \sec (c+d x))^{3/2}}+\frac {49 \sin (c+d x)}{16 a^2 d \sqrt {\cos (c+d x)} \sqrt {a+a \sec (c+d x)}}+\frac {\left (75 \sqrt {\cos (c+d x)} \sqrt {\sec (c+d x)}\right ) \operatorname {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 )}{16 a^2 d}\\ &=-\frac {75 \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)}}{16 \sqrt {2} a^{5/2} d}-\frac {\sin (c+d x)}{4 d \sqrt {\cos (c+d x)} (a+a \sec (c+d x))^{5/2}}-\frac {13 \sin (c+d x)}{16 a d \sqrt {\cos (c+d x)} (a+a \sec (c+d x))^{3/2}}+\frac {49 \sin (c+d x)}{16 a^2 d \sqrt {\cos (c+d x)} \sqrt {a+a \sec (c+d x)}}\\ \end {align*}

________________________________________________________________________________________

Mathematica [A]  time = 0.96, size = 141, normalized size = 0.72 \[ \frac {\sqrt {1-\sec (c+d x)} (32 \sin (c+d x)+\tan (c+d x) (49 \sec (c+d x)+85))+150 \sqrt {2} \sin (c+d x) \cos ^4\left (\frac {1}{2} (c+d x)\right ) \sec ^{\frac {5}{2}}(c+d x) \tan ^{-1}\left (\frac {\sqrt {2} \sqrt {\sec (c+d x)}}{\sqrt {1-\sec (c+d x)}}\right )}{16 d \sqrt {\cos (c+d x)-1} (a (\sec (c+d x)+1))^{5/2}} \]

Antiderivative was successfully verified.

[In]

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

[Out]

(150*Sqrt[2]*ArcTan[(Sqrt[2]*Sqrt[Sec[c + d*x]])/Sqrt[1 - Sec[c + d*x]]]*Cos[(c + d*x)/2]^4*Sec[c + d*x]^(5/2)
*Sin[c + d*x] + Sqrt[1 - Sec[c + d*x]]*(32*Sin[c + d*x] + (85 + 49*Sec[c + d*x])*Tan[c + d*x]))/(16*d*Sqrt[-1
+ Cos[c + d*x]]*(a*(1 + Sec[c + d*x]))^(5/2))

________________________________________________________________________________________

fricas [A]  time = 0.46, size = 428, normalized size = 2.17 \[ \left [\frac {75 \, \sqrt {2} {\left (\cos \left (d x + c\right )^{3} + 3 \, \cos \left (d x + c\right )^{2} + 3 \, \cos \left (d x + c\right ) + 1\right )} \sqrt {a} \log \left (-\frac {a \cos \left (d x + c\right )^{2} + 2 \, \sqrt {2} \sqrt {a} \sqrt {\frac {a \cos \left (d x + c\right ) + a}{\cos \left (d x + c\right )}} \sqrt {\cos \left (d x + c\right )} \sin \left (d x + c\right ) - 2 \, a \cos \left (d x + c\right ) - 3 \, a}{\cos \left (d x + c\right )^{2} + 2 \, \cos \left (d x + c\right ) + 1}\right ) + 4 \, {\left (32 \, \cos \left (d x + c\right )^{2} + 85 \, \cos \left (d x + c\right ) + 49\right )} \sqrt {\frac {a \cos \left (d x + c\right ) + a}{\cos \left (d x + c\right )}} \sqrt {\cos \left (d x + c\right )} \sin \left (d x + c\right )}{64 \, {\left (a^{3} d \cos \left (d x + c\right )^{3} + 3 \, a^{3} d \cos \left (d x + c\right )^{2} + 3 \, a^{3} d \cos \left (d x + c\right ) + a^{3} d\right )}}, \frac {75 \, \sqrt {2} {\left (\cos \left (d x + c\right )^{3} + 3 \, \cos \left (d x + c\right )^{2} + 3 \, \cos \left (d x + c\right ) + 1\right )} \sqrt {-a} \arctan \left (\frac {\sqrt {2} \sqrt {-a} \sqrt {\frac {a \cos \left (d x + c\right ) + a}{\cos \left (d x + c\right )}} \sqrt {\cos \left (d x + c\right )}}{a \sin \left (d x + c\right )}\right ) + 2 \, {\left (32 \, \cos \left (d x + c\right )^{2} + 85 \, \cos \left (d x + c\right ) + 49\right )} \sqrt {\frac {a \cos \left (d x + c\right ) + a}{\cos \left (d x + c\right )}} \sqrt {\cos \left (d x + c\right )} \sin \left (d x + c\right )}{32 \, {\left (a^{3} d \cos \left (d x + c\right )^{3} + 3 \, a^{3} d \cos \left (d x + c\right )^{2} + 3 \, a^{3} d \cos \left (d x + c\right ) + a^{3} d\right )}}\right ] \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate(cos(d*x+c)^(1/2)/(a+a*sec(d*x+c))^(5/2),x, algorithm="fricas")

[Out]

[1/64*(75*sqrt(2)*(cos(d*x + c)^3 + 3*cos(d*x + c)^2 + 3*cos(d*x + c) + 1)*sqrt(a)*log(-(a*cos(d*x + c)^2 + 2*
sqrt(2)*sqrt(a)*sqrt((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*(32*cos(d*x + c)^2 + 85*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^3*d*cos(d*x + c)^3 + 3*a^3*d*cos(d*x + c)^2 + 3*a^3*d*
cos(d*x + c) + a^3*d), 1/32*(75*sqrt(2)*(cos(d*x + c)^3 + 3*cos(d*x + c)^2 + 3*cos(d*x + c) + 1)*sqrt(-a)*arct
an(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*(32*cos(d
*x + c)^2 + 85*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^
3*d*cos(d*x + c)^3 + 3*a^3*d*cos(d*x + c)^2 + 3*a^3*d*cos(d*x + c) + a^3*d)]

________________________________________________________________________________________

giac [F]  time = 0.00, size = 0, normalized size = 0.00 \[ \int \frac {\sqrt {\cos \left (d x + c\right )}}{{\left (a \sec \left (d x + c\right ) + a\right )}^{\frac {5}{2}}}\,{d x} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate(cos(d*x+c)^(1/2)/(a+a*sec(d*x+c))^(5/2),x, algorithm="giac")

[Out]

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

________________________________________________________________________________________

maple [A]  time = 1.16, size = 234, normalized size = 1.19 \[ -\frac {\left (\sqrt {\cos }\left (d x +c \right )\right ) \sqrt {\frac {a \left (1+\cos \left (d x +c \right )\right )}{\cos \left (d x +c \right )}}\, \left (-1+\cos \left (d x +c \right )\right )^{2} \left (-75 \arctan \left (\frac {\sin \left (d x +c \right ) \sqrt {-\frac {2}{1+\cos \left (d x +c \right )}}}{2}\right ) \sqrt {-\frac {2}{1+\cos \left (d x +c \right )}}\, \left (\cos ^{2}\left (d x +c \right )\right ) \sin \left (d x +c \right )-150 \cos \left (d x +c \right ) \sin \left (d x +c \right ) \arctan \left (\frac {\sin \left (d x +c \right ) \sqrt {-\frac {2}{1+\cos \left (d x +c \right )}}}{2}\right ) \sqrt {-\frac {2}{1+\cos \left (d x +c \right )}}+64 \left (\cos ^{3}\left (d x +c \right )\right )-75 \arctan \left (\frac {\sin \left (d x +c \right ) \sqrt {-\frac {2}{1+\cos \left (d x +c \right )}}}{2}\right ) \sqrt {-\frac {2}{1+\cos \left (d x +c \right )}}\, \sin \left (d x +c \right )+106 \left (\cos ^{2}\left (d x +c \right )\right )-72 \cos \left (d x +c \right )-98\right )}{32 d \sin \left (d x +c \right )^{5} a^{3}} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

-1/32/d*cos(d*x+c)^(1/2)*(a*(1+cos(d*x+c))/cos(d*x+c))^(1/2)*(-1+cos(d*x+c))^2*(-75*arctan(1/2*sin(d*x+c)*(-2/
(1+cos(d*x+c)))^(1/2))*(-2/(1+cos(d*x+c)))^(1/2)*cos(d*x+c)^2*sin(d*x+c)-150*cos(d*x+c)*sin(d*x+c)*arctan(1/2*
sin(d*x+c)*(-2/(1+cos(d*x+c)))^(1/2))*(-2/(1+cos(d*x+c)))^(1/2)+64*cos(d*x+c)^3-75*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)+106*cos(d*x+c)^2-72*cos(d*x+c)-98)/sin(d*x+c)^5/a^3

________________________________________________________________________________________

maxima [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(cos(d*x+c)^(1/2)/(a+a*sec(d*x+c))^(5/2),x, algorithm="maxima")

[Out]

Timed out

________________________________________________________________________________________

mupad [F]  time = 0.00, size = -1, normalized size = -0.01 \[ \int \frac {\sqrt {\cos \left (c+d\,x\right )}}{{\left (a+\frac {a}{\cos \left (c+d\,x\right )}\right )}^{5/2}} \,d x \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

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

________________________________________________________________________________________

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(cos(d*x+c)**(1/2)/(a+a*sec(d*x+c))**(5/2),x)

[Out]

Timed out

________________________________________________________________________________________