3.115 \(\int \cos ^4(c+d x) (a+a \sec (c+d x))^{5/2} \, dx\)

Optimal. Leaf size=182 \[ \frac {163 a^{5/2} \tan ^{-1}\left (\frac {\sqrt {a} \tan (c+d x)}{\sqrt {a \sec (c+d x)+a}}\right )}{64 d}+\frac {163 a^3 \sin (c+d x)}{64 d \sqrt {a \sec (c+d x)+a}}+\frac {17 a^3 \sin (c+d x) \cos ^2(c+d x)}{24 d \sqrt {a \sec (c+d x)+a}}+\frac {163 a^3 \sin (c+d x) \cos (c+d x)}{96 d \sqrt {a \sec (c+d x)+a}}+\frac {a^2 \sin (c+d x) \cos ^3(c+d x) \sqrt {a \sec (c+d x)+a}}{4 d} \]

[Out]

163/64*a^(5/2)*arctan(a^(1/2)*tan(d*x+c)/(a+a*sec(d*x+c))^(1/2))/d+163/64*a^3*sin(d*x+c)/d/(a+a*sec(d*x+c))^(1
/2)+163/96*a^3*cos(d*x+c)*sin(d*x+c)/d/(a+a*sec(d*x+c))^(1/2)+17/24*a^3*cos(d*x+c)^2*sin(d*x+c)/d/(a+a*sec(d*x
+c))^(1/2)+1/4*a^2*cos(d*x+c)^3*sin(d*x+c)*(a+a*sec(d*x+c))^(1/2)/d

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Rubi [A]  time = 0.29, antiderivative size = 182, normalized size of antiderivative = 1.00, number of steps used = 6, number of rules used = 5, integrand size = 23, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.217, Rules used = {3813, 4015, 3805, 3774, 203} \[ \frac {163 a^3 \sin (c+d x)}{64 d \sqrt {a \sec (c+d x)+a}}+\frac {163 a^{5/2} \tan ^{-1}\left (\frac {\sqrt {a} \tan (c+d x)}{\sqrt {a \sec (c+d x)+a}}\right )}{64 d}+\frac {17 a^3 \sin (c+d x) \cos ^2(c+d x)}{24 d \sqrt {a \sec (c+d x)+a}}+\frac {a^2 \sin (c+d x) \cos ^3(c+d x) \sqrt {a \sec (c+d x)+a}}{4 d}+\frac {163 a^3 \sin (c+d x) \cos (c+d x)}{96 d \sqrt {a \sec (c+d x)+a}} \]

Antiderivative was successfully verified.

[In]

Int[Cos[c + d*x]^4*(a + a*Sec[c + d*x])^(5/2),x]

[Out]

(163*a^(5/2)*ArcTan[(Sqrt[a]*Tan[c + d*x])/Sqrt[a + a*Sec[c + d*x]]])/(64*d) + (163*a^3*Sin[c + d*x])/(64*d*Sq
rt[a + a*Sec[c + d*x]]) + (163*a^3*Cos[c + d*x]*Sin[c + d*x])/(96*d*Sqrt[a + a*Sec[c + d*x]]) + (17*a^3*Cos[c
+ d*x]^2*Sin[c + d*x])/(24*d*Sqrt[a + a*Sec[c + d*x]]) + (a^2*Cos[c + d*x]^3*Sqrt[a + a*Sec[c + d*x]]*Sin[c +
d*x])/(4*d)

Rule 203

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

Rule 3774

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

Rule 3805

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

Rule 3813

Int[(csc[(e_.) + (f_.)*(x_)]*(d_.))^(n_)*(csc[(e_.) + (f_.)*(x_)]*(b_.) + (a_))^(m_), x_Symbol] :> Simp[(b^2*C
ot[e + f*x]*(a + b*Csc[e + f*x])^(m - 2)*(d*Csc[e + f*x])^n)/(f*n), x] - Dist[a/(d*n), Int[(a + b*Csc[e + f*x]
)^(m - 2)*(d*Csc[e + f*x])^(n + 1)*(b*(m - 2*n - 2) - a*(m + 2*n - 1)*Csc[e + f*x]), x], x] /; FreeQ[{a, b, d,
e, f}, x] && EqQ[a^2 - b^2, 0] && GtQ[m, 1] && (LtQ[n, -1] || (EqQ[m, 3/2] && EqQ[n, -2^(-1)])) && IntegerQ[2
*m]

Rule 4015

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

Rubi steps

\begin {align*} \int \cos ^4(c+d x) (a+a \sec (c+d x))^{5/2} \, dx &=\frac {a^2 \cos ^3(c+d x) \sqrt {a+a \sec (c+d x)} \sin (c+d x)}{4 d}+\frac {1}{4} a \int \cos ^3(c+d x) \sqrt {a+a \sec (c+d x)} \left (\frac {17 a}{2}+\frac {13}{2} a \sec (c+d x)\right ) \, dx\\ &=\frac {17 a^3 \cos ^2(c+d x) \sin (c+d x)}{24 d \sqrt {a+a \sec (c+d x)}}+\frac {a^2 \cos ^3(c+d x) \sqrt {a+a \sec (c+d x)} \sin (c+d x)}{4 d}+\frac {1}{48} \left (163 a^2\right ) \int \cos ^2(c+d x) \sqrt {a+a \sec (c+d x)} \, dx\\ &=\frac {163 a^3 \cos (c+d x) \sin (c+d x)}{96 d \sqrt {a+a \sec (c+d x)}}+\frac {17 a^3 \cos ^2(c+d x) \sin (c+d x)}{24 d \sqrt {a+a \sec (c+d x)}}+\frac {a^2 \cos ^3(c+d x) \sqrt {a+a \sec (c+d x)} \sin (c+d x)}{4 d}+\frac {1}{64} \left (163 a^2\right ) \int \cos (c+d x) \sqrt {a+a \sec (c+d x)} \, dx\\ &=\frac {163 a^3 \sin (c+d x)}{64 d \sqrt {a+a \sec (c+d x)}}+\frac {163 a^3 \cos (c+d x) \sin (c+d x)}{96 d \sqrt {a+a \sec (c+d x)}}+\frac {17 a^3 \cos ^2(c+d x) \sin (c+d x)}{24 d \sqrt {a+a \sec (c+d x)}}+\frac {a^2 \cos ^3(c+d x) \sqrt {a+a \sec (c+d x)} \sin (c+d x)}{4 d}+\frac {1}{128} \left (163 a^2\right ) \int \sqrt {a+a \sec (c+d x)} \, dx\\ &=\frac {163 a^3 \sin (c+d x)}{64 d \sqrt {a+a \sec (c+d x)}}+\frac {163 a^3 \cos (c+d x) \sin (c+d x)}{96 d \sqrt {a+a \sec (c+d x)}}+\frac {17 a^3 \cos ^2(c+d x) \sin (c+d x)}{24 d \sqrt {a+a \sec (c+d x)}}+\frac {a^2 \cos ^3(c+d x) \sqrt {a+a \sec (c+d x)} \sin (c+d x)}{4 d}-\frac {\left (163 a^3\right ) \operatorname {Subst}\left (\int \frac {1}{a+x^2} \, dx,x,-\frac {a \tan (c+d x)}{\sqrt {a+a \sec (c+d x)}}\right )}{64 d}\\ &=\frac {163 a^{5/2} \tan ^{-1}\left (\frac {\sqrt {a} \tan (c+d x)}{\sqrt {a+a \sec (c+d x)}}\right )}{64 d}+\frac {163 a^3 \sin (c+d x)}{64 d \sqrt {a+a \sec (c+d x)}}+\frac {163 a^3 \cos (c+d x) \sin (c+d x)}{96 d \sqrt {a+a \sec (c+d x)}}+\frac {17 a^3 \cos ^2(c+d x) \sin (c+d x)}{24 d \sqrt {a+a \sec (c+d x)}}+\frac {a^2 \cos ^3(c+d x) \sqrt {a+a \sec (c+d x)} \sin (c+d x)}{4 d}\\ \end {align*}

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Mathematica [C]  time = 0.82, size = 161, normalized size = 0.88 \[ \frac {a^2 \sin (c+d x) \sqrt {a (\sec (c+d x)+1)} \left (512 \sqrt {1-\sec (c+d x)} \, _2F_1\left (\frac {1}{2},5;\frac {3}{2};1-\sec (c+d x)\right )+(849 \cos (c+d x)+233 \cos (2 (c+d x))+58 \cos (3 (c+d x))+2 \cos (4 (c+d x))+231) \sqrt {1-\sec (c+d x)}+675 \tanh ^{-1}\left (\sqrt {1-\sec (c+d x)}\right )\right )}{320 d (\cos (c+d x)+1) \sqrt {1-\sec (c+d x)}} \]

Antiderivative was successfully verified.

[In]

Integrate[Cos[c + d*x]^4*(a + a*Sec[c + d*x])^(5/2),x]

[Out]

(a^2*(675*ArcTanh[Sqrt[1 - Sec[c + d*x]]] + (231 + 849*Cos[c + d*x] + 233*Cos[2*(c + d*x)] + 58*Cos[3*(c + d*x
)] + 2*Cos[4*(c + d*x)])*Sqrt[1 - Sec[c + d*x]] + 512*Hypergeometric2F1[1/2, 5, 3/2, 1 - Sec[c + d*x]]*Sqrt[1
- Sec[c + d*x]])*Sqrt[a*(1 + Sec[c + d*x])]*Sin[c + d*x])/(320*d*(1 + Cos[c + d*x])*Sqrt[1 - Sec[c + d*x]])

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fricas [A]  time = 0.53, size = 346, normalized size = 1.90 \[ \left [\frac {489 \, {\left (a^{2} \cos \left (d x + c\right ) + a^{2}\right )} \sqrt {-a} \log \left (\frac {2 \, a \cos \left (d x + c\right )^{2} - 2 \, \sqrt {-a} \sqrt {\frac {a \cos \left (d x + c\right ) + a}{\cos \left (d x + c\right )}} \cos \left (d x + c\right ) \sin \left (d x + c\right ) + a \cos \left (d x + c\right ) - a}{\cos \left (d x + c\right ) + 1}\right ) + 2 \, {\left (48 \, a^{2} \cos \left (d x + c\right )^{4} + 184 \, a^{2} \cos \left (d x + c\right )^{3} + 326 \, a^{2} \cos \left (d x + c\right )^{2} + 489 \, a^{2} \cos \left (d x + c\right )\right )} \sqrt {\frac {a \cos \left (d x + c\right ) + a}{\cos \left (d x + c\right )}} \sin \left (d x + c\right )}{384 \, {\left (d \cos \left (d x + c\right ) + d\right )}}, -\frac {489 \, {\left (a^{2} \cos \left (d x + c\right ) + a^{2}\right )} \sqrt {a} \arctan \left (\frac {\sqrt {\frac {a \cos \left (d x + c\right ) + a}{\cos \left (d x + c\right )}} \cos \left (d x + c\right )}{\sqrt {a} \sin \left (d x + c\right )}\right ) - {\left (48 \, a^{2} \cos \left (d x + c\right )^{4} + 184 \, a^{2} \cos \left (d x + c\right )^{3} + 326 \, a^{2} \cos \left (d x + c\right )^{2} + 489 \, a^{2} \cos \left (d x + c\right )\right )} \sqrt {\frac {a \cos \left (d x + c\right ) + a}{\cos \left (d x + c\right )}} \sin \left (d x + c\right )}{192 \, {\left (d \cos \left (d x + c\right ) + d\right )}}\right ] \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate(cos(d*x+c)^4*(a+a*sec(d*x+c))^(5/2),x, algorithm="fricas")

[Out]

[1/384*(489*(a^2*cos(d*x + c) + a^2)*sqrt(-a)*log((2*a*cos(d*x + c)^2 - 2*sqrt(-a)*sqrt((a*cos(d*x + c) + a)/c
os(d*x + c))*cos(d*x + c)*sin(d*x + c) + a*cos(d*x + c) - a)/(cos(d*x + c) + 1)) + 2*(48*a^2*cos(d*x + c)^4 +
184*a^2*cos(d*x + c)^3 + 326*a^2*cos(d*x + c)^2 + 489*a^2*cos(d*x + c))*sqrt((a*cos(d*x + c) + a)/cos(d*x + c)
)*sin(d*x + c))/(d*cos(d*x + c) + d), -1/192*(489*(a^2*cos(d*x + c) + a^2)*sqrt(a)*arctan(sqrt((a*cos(d*x + c)
+ a)/cos(d*x + c))*cos(d*x + c)/(sqrt(a)*sin(d*x + c))) - (48*a^2*cos(d*x + c)^4 + 184*a^2*cos(d*x + c)^3 + 3
26*a^2*cos(d*x + c)^2 + 489*a^2*cos(d*x + c))*sqrt((a*cos(d*x + c) + a)/cos(d*x + c))*sin(d*x + c))/(d*cos(d*x
+ c) + d)]

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giac [F(-2)]  time = 0.00, size = 0, normalized size = 0.00 \[ \text {Exception raised: TypeError} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate(cos(d*x+c)^4*(a+a*sec(d*x+c))^(5/2),x, algorithm="giac")

[Out]

Exception raised: TypeError >> An error occurred running a Giac command:INPUT:sage2:=int(sage0,x):;OUTPUT:Warn
ing, integration of abs or sign assumes constant sign by intervals (correct if the argument is real):Check [ab
s(cos(d*x+c))]Warning, assuming -2*a+a is positive. Hint: run assume to make assumptions on a variableWarning,
assuming -2*a+a is positive. Hint: run assume to make assumptions on a variableWarning, assuming -2*a+a is po
sitive. Hint: run assume to make assumptions on a variableWarning, assuming -2*a+a is positive. Hint: run assu
me to make assumptions on a variableWarning, assuming -2*a+a is positive. Hint: run assume to make assumptions
on a variableWarning, assuming -2*a+a is positive. Hint: run assume to make assumptions on a variableWarning,
assuming -2*a+a is positive. Hint: run assume to make assumptions on a variableWarning, assuming -2*a+a is po
sitive. Hint: run assume to make assumptions on a variableEvaluation time: 1.43Unable to divide, perhaps due t
o rounding error%%%{%%{[%%%{%%{[-2309237210123256509497344,0]:[1,0,-2]%%},[35]%%%},0]:[1,0,%%%{1,[1]%%%}]%%},[
0,1]%%%} / %%%{%%%{%%{[-14167099448608935641088,0]:[1,0,-2]%%},[32]%%%},[0,0]%%%} Error: Bad Argument Value

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maple [B]  time = 1.14, size = 402, normalized size = 2.21 \[ \frac {\left (489 \left (-\frac {2 \cos \left (d x +c \right )}{1+\cos \left (d x +c \right )}\right )^{\frac {7}{2}} \arctanh \left (\frac {\sqrt {-\frac {2 \cos \left (d x +c \right )}{1+\cos \left (d x +c \right )}}\, \sin \left (d x +c \right ) \sqrt {2}}{2 \cos \left (d x +c \right )}\right ) \left (\cos ^{3}\left (d x +c \right )\right ) \sin \left (d x +c \right ) \sqrt {2}+1467 \left (-\frac {2 \cos \left (d x +c \right )}{1+\cos \left (d x +c \right )}\right )^{\frac {7}{2}} \arctanh \left (\frac {\sqrt {-\frac {2 \cos \left (d x +c \right )}{1+\cos \left (d x +c \right )}}\, \sin \left (d x +c \right ) \sqrt {2}}{2 \cos \left (d x +c \right )}\right ) \left (\cos ^{2}\left (d x +c \right )\right ) \sin \left (d x +c \right ) \sqrt {2}+1467 \left (-\frac {2 \cos \left (d x +c \right )}{1+\cos \left (d x +c \right )}\right )^{\frac {7}{2}} \arctanh \left (\frac {\sqrt {-\frac {2 \cos \left (d x +c \right )}{1+\cos \left (d x +c \right )}}\, \sin \left (d x +c \right ) \sqrt {2}}{2 \cos \left (d x +c \right )}\right ) \cos \left (d x +c \right ) \sin \left (d x +c \right ) \sqrt {2}+489 \left (-\frac {2 \cos \left (d x +c \right )}{1+\cos \left (d x +c \right )}\right )^{\frac {7}{2}} \sqrt {2}\, \arctanh \left (\frac {\sqrt {-\frac {2 \cos \left (d x +c \right )}{1+\cos \left (d x +c \right )}}\, \sin \left (d x +c \right ) \sqrt {2}}{2 \cos \left (d x +c \right )}\right ) \sin \left (d x +c \right )-768 \left (\cos ^{8}\left (d x +c \right )\right )-2176 \left (\cos ^{7}\left (d x +c \right )\right )-2272 \left (\cos ^{6}\left (d x +c \right )\right )-2608 \left (\cos ^{5}\left (d x +c \right )\right )+7824 \left (\cos ^{4}\left (d x +c \right )\right )\right ) \sqrt {\frac {a \left (1+\cos \left (d x +c \right )\right )}{\cos \left (d x +c \right )}}\, a^{2}}{3072 d \sin \left (d x +c \right ) \cos \left (d x +c \right )^{3}} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

int(cos(d*x+c)^4*(a+a*sec(d*x+c))^(5/2),x)

[Out]

1/3072/d*(489*(-2*cos(d*x+c)/(1+cos(d*x+c)))^(7/2)*arctanh(1/2*(-2*cos(d*x+c)/(1+cos(d*x+c)))^(1/2)*sin(d*x+c)
/cos(d*x+c)*2^(1/2))*cos(d*x+c)^3*sin(d*x+c)*2^(1/2)+1467*(-2*cos(d*x+c)/(1+cos(d*x+c)))^(7/2)*arctanh(1/2*(-2
*cos(d*x+c)/(1+cos(d*x+c)))^(1/2)*sin(d*x+c)/cos(d*x+c)*2^(1/2))*cos(d*x+c)^2*sin(d*x+c)*2^(1/2)+1467*(-2*cos(
d*x+c)/(1+cos(d*x+c)))^(7/2)*arctanh(1/2*(-2*cos(d*x+c)/(1+cos(d*x+c)))^(1/2)*sin(d*x+c)/cos(d*x+c)*2^(1/2))*c
os(d*x+c)*sin(d*x+c)*2^(1/2)+489*(-2*cos(d*x+c)/(1+cos(d*x+c)))^(7/2)*2^(1/2)*arctanh(1/2*(-2*cos(d*x+c)/(1+co
s(d*x+c)))^(1/2)*sin(d*x+c)/cos(d*x+c)*2^(1/2))*sin(d*x+c)-768*cos(d*x+c)^8-2176*cos(d*x+c)^7-2272*cos(d*x+c)^
6-2608*cos(d*x+c)^5+7824*cos(d*x+c)^4)*(a*(1+cos(d*x+c))/cos(d*x+c))^(1/2)/sin(d*x+c)/cos(d*x+c)^3*a^2

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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)^4*(a+a*sec(d*x+c))^(5/2),x, algorithm="maxima")

[Out]

Timed out

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mupad [F]  time = 0.00, size = -1, normalized size = -0.01 \[ \int {\cos \left (c+d\,x\right )}^4\,{\left (a+\frac {a}{\cos \left (c+d\,x\right )}\right )}^{5/2} \,d x \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

int(cos(c + d*x)^4*(a + a/cos(c + d*x))^(5/2),x)

[Out]

int(cos(c + d*x)^4*(a + a/cos(c + d*x))^(5/2), 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(cos(d*x+c)**4*(a+a*sec(d*x+c))**(5/2),x)

[Out]

Timed out

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