3.2.5 \(\int \cos ^2(c+d x) (a+a \sec (c+d x))^{3/2} \, dx\) [105]

Optimal. Leaf size=106 \[ \frac {7 a^{3/2} \text {ArcTan}\left (\frac {\sqrt {a} \tan (c+d x)}{\sqrt {a+a \sec (c+d x)}}\right )}{4 d}+\frac {7 a^2 \sin (c+d x)}{4 d \sqrt {a+a \sec (c+d x)}}+\frac {a^2 \cos (c+d x) \sin (c+d x)}{2 d \sqrt {a+a \sec (c+d x)}} \]

[Out]

7/4*a^(3/2)*arctan(a^(1/2)*tan(d*x+c)/(a+a*sec(d*x+c))^(1/2))/d+7/4*a^2*sin(d*x+c)/d/(a+a*sec(d*x+c))^(1/2)+1/
2*a^2*cos(d*x+c)*sin(d*x+c)/d/(a+a*sec(d*x+c))^(1/2)

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Rubi [A]
time = 0.09, antiderivative size = 106, normalized size of antiderivative = 1.00, number of steps used = 5, number of rules used = 5, integrand size = 23, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.217, Rules used = {3898, 21, 3890, 3859, 209} \begin {gather*} \frac {7 a^{3/2} \text {ArcTan}\left (\frac {\sqrt {a} \tan (c+d x)}{\sqrt {a \sec (c+d x)+a}}\right )}{4 d}+\frac {7 a^2 \sin (c+d x)}{4 d \sqrt {a \sec (c+d x)+a}}+\frac {a^2 \sin (c+d x) \cos (c+d x)}{2 d \sqrt {a \sec (c+d x)+a}} \end {gather*}

Antiderivative was successfully verified.

[In]

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

[Out]

(7*a^(3/2)*ArcTan[(Sqrt[a]*Tan[c + d*x])/Sqrt[a + a*Sec[c + d*x]]])/(4*d) + (7*a^2*Sin[c + d*x])/(4*d*Sqrt[a +
a*Sec[c + d*x]]) + (a^2*Cos[c + d*x]*Sin[c + d*x])/(2*d*Sqrt[a + a*Sec[c + d*x]])

Rule 21

Int[(u_.)*((a_) + (b_.)*(v_))^(m_.)*((c_) + (d_.)*(v_))^(n_.), x_Symbol] :> Dist[(b/d)^m, Int[u*(c + d*v)^(m +
n), x], x] /; FreeQ[{a, b, c, d, n}, x] && EqQ[b*c - a*d, 0] && IntegerQ[m] && ( !IntegerQ[n] || SimplerQ[c +
d*x, a + b*x])

Rule 209

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

Rule 3859

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 3890

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 3898

Int[(csc[(e_.) + (f_.)*(x_)]*(d_.))^(n_)*(csc[(e_.) + (f_.)*(x_)]*(b_.) + (a_))^(m_), x_Symbol] :> Simp[b^2*Co
t[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]

Rubi steps

\begin {align*} \int \cos ^2(c+d x) (a+a \sec (c+d x))^{3/2} \, dx &=\frac {a^2 \cos (c+d x) \sin (c+d x)}{2 d \sqrt {a+a \sec (c+d x)}}+\frac {1}{2} a \int \frac {\cos (c+d x) \left (\frac {7 a}{2}+\frac {7}{2} a \sec (c+d x)\right )}{\sqrt {a+a \sec (c+d x)}} \, dx\\ &=\frac {a^2 \cos (c+d x) \sin (c+d x)}{2 d \sqrt {a+a \sec (c+d x)}}+\frac {1}{4} (7 a) \int \cos (c+d x) \sqrt {a+a \sec (c+d x)} \, dx\\ &=\frac {7 a^2 \sin (c+d x)}{4 d \sqrt {a+a \sec (c+d x)}}+\frac {a^2 \cos (c+d x) \sin (c+d x)}{2 d \sqrt {a+a \sec (c+d x)}}+\frac {1}{8} (7 a) \int \sqrt {a+a \sec (c+d x)} \, dx\\ &=\frac {7 a^2 \sin (c+d x)}{4 d \sqrt {a+a \sec (c+d x)}}+\frac {a^2 \cos (c+d x) \sin (c+d x)}{2 d \sqrt {a+a \sec (c+d x)}}-\frac {\left (7 a^2\right ) \text {Subst}\left (\int \frac {1}{a+x^2} \, dx,x,-\frac {a \tan (c+d x)}{\sqrt {a+a \sec (c+d x)}}\right )}{4 d}\\ &=\frac {7 a^{3/2} \tan ^{-1}\left (\frac {\sqrt {a} \tan (c+d x)}{\sqrt {a+a \sec (c+d x)}}\right )}{4 d}+\frac {7 a^2 \sin (c+d x)}{4 d \sqrt {a+a \sec (c+d x)}}+\frac {a^2 \cos (c+d x) \sin (c+d x)}{2 d \sqrt {a+a \sec (c+d x)}}\\ \end {align*}

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Mathematica [A]
time = 0.38, size = 108, normalized size = 1.02 \begin {gather*} \frac {a \cos (c+d x) \sqrt {a (1+\sec (c+d x))} \left (\sqrt {1-\sec (c+d x)} (7 \sin (c+d x)+\sin (2 (c+d x)))+7 \tanh ^{-1}\left (\sqrt {1-\sec (c+d x)}\right ) \tan (c+d x)\right )}{4 d (1+\cos (c+d x)) \sqrt {1-\sec (c+d x)}} \end {gather*}

Antiderivative was successfully verified.

[In]

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

[Out]

(a*Cos[c + d*x]*Sqrt[a*(1 + Sec[c + d*x])]*(Sqrt[1 - Sec[c + d*x]]*(7*Sin[c + d*x] + Sin[2*(c + d*x)]) + 7*Arc
Tanh[Sqrt[1 - Sec[c + d*x]]]*Tan[c + d*x]))/(4*d*(1 + Cos[c + d*x])*Sqrt[1 - Sec[c + d*x]])

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Maple [B] Leaf count of result is larger than twice the leaf count of optimal. \(221\) vs. \(2(90)=180\).
time = 0.13, size = 222, normalized size = 2.09

method	result	size
default	\(\frac {\left (7 \sin \left (d x +c \right ) \cos \left (d x +c \right ) \sqrt {2}\, \arctanh \left (\frac {\sqrt {-\frac {2 \cos \left (d x +c \right )}{1+\cos \left (d x +c \right )}}\, \sin \left (d x +c \right ) \sqrt {2}}{2 \cos \left (d x +c \right )}\right ) \left (-\frac {2 \cos \left (d x +c \right )}{1+\cos \left (d x +c \right )}\right )^{\frac {3}{2}}+7 \sqrt {2}\, \arctanh \left (\frac {\sqrt {-\frac {2 \cos \left (d x +c \right )}{1+\cos \left (d x +c \right )}}\, \sin \left (d x +c \right ) \sqrt {2}}{2 \cos \left (d x +c \right )}\right ) \left (-\frac {2 \cos \left (d x +c \right )}{1+\cos \left (d x +c \right )}\right )^{\frac {3}{2}} \sin \left (d x +c \right )-8 \left (\cos ^{4}\left (d x +c \right )\right )-20 \left (\cos ^{3}\left (d x +c \right )\right )+28 \left (\cos ^{2}\left (d x +c \right )\right )\right ) \sqrt {\frac {a \left (1+\cos \left (d x +c \right )\right )}{\cos \left (d x +c \right )}}\, a}{16 d \cos \left (d x +c \right ) \sin \left (d x +c \right )}\)	\(222\)

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

1/16/d*(7*sin(d*x+c)*cos(d*x+c)*2^(1/2)*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))*(-2*cos(d*x+c)/(1+cos(d*x+c)))^(3/2)+7*2^(1/2)*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))*(-2*cos(d*x+c)/(1+cos(d*x+c)))^(3/2)*sin(d*x+c)-8*cos(d*x+c)^4-20*cos(d*x+c)^3+28*c
os(d*x+c)^2)*(a*(1+cos(d*x+c))/cos(d*x+c))^(1/2)/cos(d*x+c)/sin(d*x+c)*a

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Maxima [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)^2*(a+a*sec(d*x+c))^(3/2),x, algorithm="maxima")

[Out]

Timed out

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Fricas [A]
time = 2.75, size = 278, normalized size = 2.62 \begin {gather*} \left [\frac {7 \, {\left (a \cos \left (d x + c\right ) + a\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 (2 \, a \cos \left (d x + c\right )^{2} + 7 \, a \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 )}{8 \, {\left (d \cos \left (d x + c\right ) + d\right )}}, -\frac {7 \, {\left (a \cos \left (d x + c\right ) + a\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 (2 \, a \cos \left (d x + c\right )^{2} + 7 \, a \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 )}{4 \, {\left (d \cos \left (d x + c\right ) + d\right )}}\right ] \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

[1/8*(7*(a*cos(d*x + c) + a)*sqrt(-a)*log((2*a*cos(d*x + c)^2 - 2*sqrt(-a)*sqrt((a*cos(d*x + c) + a)/cos(d*x +
c))*cos(d*x + c)*sin(d*x + c) + a*cos(d*x + c) - a)/(cos(d*x + c) + 1)) + 2*(2*a*cos(d*x + c)^2 + 7*a*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), -1/4*(7*(a*cos(d*x + c) + a
)*sqrt(a)*arctan(sqrt((a*cos(d*x + c) + a)/cos(d*x + c))*cos(d*x + c)/(sqrt(a)*sin(d*x + c))) - (2*a*cos(d*x +
c)^2 + 7*a*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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Sympy [F(-2)]
time = 0.00, size = 0, normalized size = 0.00 \begin {gather*} \text {Exception raised: SystemError} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

Exception raised: SystemError >> excessive stack use: stack is 3004 deep

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

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

Exception raised: TypeError >> An error occurred running a Giac command:INPUT:sage2:=int(sage0,sageVARx):;OUTP
UT:Warning, integration of abs or sign assumes constant sign by intervals (correct if the argument is real):Ch
eck [abs(co

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Mupad [F]
time = 0.00, size = -1, normalized size = -0.01 \begin {gather*} \int {\cos \left (c+d\,x\right )}^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)^2*(a + a/cos(c + d*x))^(3/2),x)

[Out]

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

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