3.114 \(\int \cos ^3(c+d x) (a+a \sec (c+d x))^{5/2} \, dx\)

Optimal. Leaf size=144 \[ \frac{25 a^3 \sin (c+d x)}{8 d \sqrt{a \sec (c+d x)+a}}+\frac{25 a^{5/2} \tan ^{-1}\left (\frac{\sqrt{a} \tan (c+d x)}{\sqrt{a \sec (c+d x)+a}}\right )}{8 d}+\frac{a^2 \sin (c+d x) \cos ^2(c+d x) \sqrt{a \sec (c+d x)+a}}{3 d}+\frac{13 a^3 \sin (c+d x) \cos (c+d x)}{12 d \sqrt{a \sec (c+d x)+a}} \]

[Out]

(25*a^(5/2)*ArcTan[(Sqrt[a]*Tan[c + d*x])/Sqrt[a + a*Sec[c + d*x]]])/(8*d) + (25*a^3*Sin[c + d*x])/(8*d*Sqrt[a
+ a*Sec[c + d*x]]) + (13*a^3*Cos[c + d*x]*Sin[c + d*x])/(12*d*Sqrt[a + a*Sec[c + d*x]]) + (a^2*Cos[c + d*x]^2
*Sqrt[a + a*Sec[c + d*x]]*Sin[c + d*x])/(3*d)

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Rubi [A]  time = 0.234547, antiderivative size = 144, normalized size of antiderivative = 1., 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 = {3813, 4015, 3805, 3774, 203} \[ \frac{25 a^3 \sin (c+d x)}{8 d \sqrt{a \sec (c+d x)+a}}+\frac{25 a^{5/2} \tan ^{-1}\left (\frac{\sqrt{a} \tan (c+d x)}{\sqrt{a \sec (c+d x)+a}}\right )}{8 d}+\frac{a^2 \sin (c+d x) \cos ^2(c+d x) \sqrt{a \sec (c+d x)+a}}{3 d}+\frac{13 a^3 \sin (c+d x) \cos (c+d x)}{12 d \sqrt{a \sec (c+d x)+a}} \]

Antiderivative was successfully verified.

[In]

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

[Out]

(25*a^(5/2)*ArcTan[(Sqrt[a]*Tan[c + d*x])/Sqrt[a + a*Sec[c + d*x]]])/(8*d) + (25*a^3*Sin[c + d*x])/(8*d*Sqrt[a
+ a*Sec[c + d*x]]) + (13*a^3*Cos[c + d*x]*Sin[c + d*x])/(12*d*Sqrt[a + a*Sec[c + d*x]]) + (a^2*Cos[c + d*x]^2
*Sqrt[a + a*Sec[c + d*x]]*Sin[c + d*x])/(3*d)

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]

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 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 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])

Rubi steps

\begin{align*} \int \cos ^3(c+d x) (a+a \sec (c+d x))^{5/2} \, dx &=\frac{a^2 \cos ^2(c+d x) \sqrt{a+a \sec (c+d x)} \sin (c+d x)}{3 d}+\frac{1}{3} a \int \cos ^2(c+d x) \sqrt{a+a \sec (c+d x)} \left (\frac{13 a}{2}+\frac{9}{2} a \sec (c+d x)\right ) \, dx\\ &=\frac{13 a^3 \cos (c+d x) \sin (c+d x)}{12 d \sqrt{a+a \sec (c+d x)}}+\frac{a^2 \cos ^2(c+d x) \sqrt{a+a \sec (c+d x)} \sin (c+d x)}{3 d}+\frac{1}{8} \left (25 a^2\right ) \int \cos (c+d x) \sqrt{a+a \sec (c+d x)} \, dx\\ &=\frac{25 a^3 \sin (c+d x)}{8 d \sqrt{a+a \sec (c+d x)}}+\frac{13 a^3 \cos (c+d x) \sin (c+d x)}{12 d \sqrt{a+a \sec (c+d x)}}+\frac{a^2 \cos ^2(c+d x) \sqrt{a+a \sec (c+d x)} \sin (c+d x)}{3 d}+\frac{1}{16} \left (25 a^2\right ) \int \sqrt{a+a \sec (c+d x)} \, dx\\ &=\frac{25 a^3 \sin (c+d x)}{8 d \sqrt{a+a \sec (c+d x)}}+\frac{13 a^3 \cos (c+d x) \sin (c+d x)}{12 d \sqrt{a+a \sec (c+d x)}}+\frac{a^2 \cos ^2(c+d x) \sqrt{a+a \sec (c+d x)} \sin (c+d x)}{3 d}-\frac{\left (25 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 )}{8 d}\\ &=\frac{25 a^{5/2} \tan ^{-1}\left (\frac{\sqrt{a} \tan (c+d x)}{\sqrt{a+a \sec (c+d x)}}\right )}{8 d}+\frac{25 a^3 \sin (c+d x)}{8 d \sqrt{a+a \sec (c+d x)}}+\frac{13 a^3 \cos (c+d x) \sin (c+d x)}{12 d \sqrt{a+a \sec (c+d x)}}+\frac{a^2 \cos ^2(c+d x) \sqrt{a+a \sec (c+d x)} \sin (c+d x)}{3 d}\\ \end{align*}

Mathematica [C]  time = 0.771256, size = 151, normalized size = 1.05 \[ \frac{a^2 \sin (c+d x) \sqrt{a (\sec (c+d x)+1)} \left (192 \sqrt{1-\sec (c+d x)} \text{Hypergeometric2F1}\left (\frac{1}{2},4,\frac{3}{2},1-\sec (c+d x)\right )+(159 \cos (c+d x)+31 \cos (2 (c+d x))-2 \cos (3 (c+d x))+31) \sqrt{1-\sec (c+d x)}+165 \tanh ^{-1}\left (\sqrt{1-\sec (c+d x)}\right )\right )}{72 d (\cos (c+d x)+1) \sqrt{1-\sec (c+d x)}} \]

Antiderivative was successfully verified.

[In]

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

[Out]

(a^2*(165*ArcTanh[Sqrt[1 - Sec[c + d*x]]] + (31 + 159*Cos[c + d*x] + 31*Cos[2*(c + d*x)] - 2*Cos[3*(c + d*x)])
*Sqrt[1 - Sec[c + d*x]] + 192*Hypergeometric2F1[1/2, 4, 3/2, 1 - Sec[c + d*x]]*Sqrt[1 - Sec[c + d*x]])*Sqrt[a*
(1 + Sec[c + d*x])]*Sin[c + d*x])/(72*d*(1 + Cos[c + d*x])*Sqrt[1 - Sec[c + d*x]])

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Maple [B]  time = 0.214, size = 313, normalized size = 2.2 \begin{align*} -{\frac{{a}^{2}}{192\,d \left ( \cos \left ( dx+c \right ) \right ) ^{2}\sin \left ( dx+c \right ) } \left ( 75\, \left ( -2\,{\frac{\cos \left ( dx+c \right ) }{\cos \left ( dx+c \right ) +1}} \right ) ^{5/2}\sqrt{2}{\it Artanh} \left ( 1/2\,{\frac{\sqrt{2}\sin \left ( dx+c \right ) }{\cos \left ( dx+c \right ) }\sqrt{-2\,{\frac{\cos \left ( dx+c \right ) }{\cos \left ( dx+c \right ) +1}}}} \right ) \sin \left ( dx+c \right ) \left ( \cos \left ( dx+c \right ) \right ) ^{2}+150\, \left ( -2\,{\frac{\cos \left ( dx+c \right ) }{\cos \left ( dx+c \right ) +1}} \right ) ^{5/2}\sqrt{2}{\it Artanh} \left ( 1/2\,{\frac{\sqrt{2}\sin \left ( dx+c \right ) }{\cos \left ( dx+c \right ) }\sqrt{-2\,{\frac{\cos \left ( dx+c \right ) }{\cos \left ( dx+c \right ) +1}}}} \right ) \sin \left ( dx+c \right ) \cos \left ( dx+c \right ) +75\,\sqrt{2}{\it Artanh} \left ( 1/2\,{\frac{\sqrt{2}\sin \left ( dx+c \right ) }{\cos \left ( dx+c \right ) }\sqrt{-2\,{\frac{\cos \left ( dx+c \right ) }{\cos \left ( dx+c \right ) +1}}}} \right ) \left ( -2\,{\frac{\cos \left ( dx+c \right ) }{\cos \left ( dx+c \right ) +1}} \right ) ^{5/2}\sin \left ( dx+c \right ) +64\, \left ( \cos \left ( dx+c \right ) \right ) ^{6}+208\, \left ( \cos \left ( dx+c \right ) \right ) ^{5}+328\, \left ( \cos \left ( dx+c \right ) \right ) ^{4}-600\, \left ( \cos \left ( dx+c \right ) \right ) ^{3} \right ) \sqrt{{\frac{a \left ( \cos \left ( dx+c \right ) +1 \right ) }{\cos \left ( dx+c \right ) }}}} \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

-1/192/d*a^2*(75*(-2*cos(d*x+c)/(cos(d*x+c)+1))^(5/2)*2^(1/2)*arctanh(1/2*2^(1/2)*(-2*cos(d*x+c)/(cos(d*x+c)+1
))^(1/2)*sin(d*x+c)/cos(d*x+c))*sin(d*x+c)*cos(d*x+c)^2+150*(-2*cos(d*x+c)/(cos(d*x+c)+1))^(5/2)*2^(1/2)*arcta
nh(1/2*2^(1/2)*(-2*cos(d*x+c)/(cos(d*x+c)+1))^(1/2)*sin(d*x+c)/cos(d*x+c))*sin(d*x+c)*cos(d*x+c)+75*2^(1/2)*ar
ctanh(1/2*2^(1/2)*(-2*cos(d*x+c)/(cos(d*x+c)+1))^(1/2)*sin(d*x+c)/cos(d*x+c))*(-2*cos(d*x+c)/(cos(d*x+c)+1))^(
5/2)*sin(d*x+c)+64*cos(d*x+c)^6+208*cos(d*x+c)^5+328*cos(d*x+c)^4-600*cos(d*x+c)^3)*(a*(cos(d*x+c)+1)/cos(d*x+
c))^(1/2)/cos(d*x+c)^2/sin(d*x+c)

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Maxima [F(-1)]  time = 0., size = 0, normalized size = 0. \begin{align*} \text{Timed out} \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

Timed out

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Fricas [A]  time = 2.01304, size = 830, normalized size = 5.76 \begin{align*} \left [\frac{75 \,{\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 (8 \, a^{2} \cos \left (d x + c\right )^{3} + 34 \, a^{2} \cos \left (d x + c\right )^{2} + 75 \, 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 )}{48 \,{\left (d \cos \left (d x + c\right ) + d\right )}}, -\frac{75 \,{\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 (8 \, a^{2} \cos \left (d x + c\right )^{3} + 34 \, a^{2} \cos \left (d x + c\right )^{2} + 75 \, 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 )}{24 \,{\left (d \cos \left (d x + c\right ) + d\right )}}\right ] \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

[1/48*(75*(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)/cos
(d*x + c))*cos(d*x + c)*sin(d*x + c) + a*cos(d*x + c) - a)/(cos(d*x + c) + 1)) + 2*(8*a^2*cos(d*x + c)^3 + 34*
a^2*cos(d*x + c)^2 + 75*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), -1/24*(75*(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))) - (8*a^2*cos(d*x + c)^3 + 34*a^2*cos(d*x + c)^2 + 75*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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Sympy [F(-1)]  time = 0., size = 0, normalized size = 0. \begin{align*} \text{Timed out} \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

Timed out

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Giac [B]  time = 8.19531, size = 707, normalized size = 4.91 \begin{align*} \text{result too large to display} \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

-1/48*(75*sqrt(-a)*a^2*log(abs((sqrt(-a)*tan(1/2*d*x + 1/2*c) - sqrt(-a*tan(1/2*d*x + 1/2*c)^2 + a))^2 - a*(2*
sqrt(2) + 3)))*sgn(cos(d*x + c)) - 75*sqrt(-a)*a^2*log(abs((sqrt(-a)*tan(1/2*d*x + 1/2*c) - sqrt(-a*tan(1/2*d*
x + 1/2*c)^2 + a))^2 + a*(2*sqrt(2) - 3)))*sgn(cos(d*x + c)) + 4*sqrt(2)*(75*(sqrt(-a)*tan(1/2*d*x + 1/2*c) -
sqrt(-a*tan(1/2*d*x + 1/2*c)^2 + a))^10*sqrt(-a)*a^3*sgn(cos(d*x + c)) - 1125*(sqrt(-a)*tan(1/2*d*x + 1/2*c) -
sqrt(-a*tan(1/2*d*x + 1/2*c)^2 + a))^8*sqrt(-a)*a^4*sgn(cos(d*x + c)) + 6174*(sqrt(-a)*tan(1/2*d*x + 1/2*c) -
sqrt(-a*tan(1/2*d*x + 1/2*c)^2 + a))^6*sqrt(-a)*a^5*sgn(cos(d*x + c)) - 4314*(sqrt(-a)*tan(1/2*d*x + 1/2*c) -
sqrt(-a*tan(1/2*d*x + 1/2*c)^2 + a))^4*sqrt(-a)*a^6*sgn(cos(d*x + c)) + 807*(sqrt(-a)*tan(1/2*d*x + 1/2*c) -
sqrt(-a*tan(1/2*d*x + 1/2*c)^2 + a))^2*sqrt(-a)*a^7*sgn(cos(d*x + c)) - 49*sqrt(-a)*a^8*sgn(cos(d*x + c)))/((s
qrt(-a)*tan(1/2*d*x + 1/2*c) - sqrt(-a*tan(1/2*d*x + 1/2*c)^2 + a))^4 - 6*(sqrt(-a)*tan(1/2*d*x + 1/2*c) - sqr
t(-a*tan(1/2*d*x + 1/2*c)^2 + a))^2*a + a^2)^3)/d