3.5.7 \(\int \cos ^{\frac {3}{2}}(c+d x) (a+a \sec (c+d x))^{3/2} \, dx\) [407]

Optimal. Leaf size=79 \[ \frac {8 a^2 \sin (c+d x)}{3 d \sqrt {\cos (c+d x)} \sqrt {a+a \sec (c+d x)}}+\frac {2 a \sqrt {\cos (c+d x)} \sqrt {a+a \sec (c+d x)} \sin (c+d x)}{3 d} \]

[Out]

8/3*a^2*sin(d*x+c)/d/cos(d*x+c)^(1/2)/(a+a*sec(d*x+c))^(1/2)+2/3*a*sin(d*x+c)*cos(d*x+c)^(1/2)*(a+a*sec(d*x+c)
)^(1/2)/d

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Rubi [A]
time = 0.12, antiderivative size = 79, normalized size of antiderivative = 1.00, number of steps used = 3, number of rules used = 3, integrand size = 25, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.120, Rules used = {4349, 3894, 3889} \begin {gather*} \frac {8 a^2 \sin (c+d x)}{3 d \sqrt {\cos (c+d x)} \sqrt {a \sec (c+d x)+a}}+\frac {2 a \sin (c+d x) \sqrt {\cos (c+d x)} \sqrt {a \sec (c+d x)+a}}{3 d} \end {gather*}

Antiderivative was successfully verified.

[In]

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

[Out]

(8*a^2*Sin[c + d*x])/(3*d*Sqrt[Cos[c + d*x]]*Sqrt[a + a*Sec[c + d*x]]) + (2*a*Sqrt[Cos[c + d*x]]*Sqrt[a + a*Se
c[c + d*x]]*Sin[c + d*x])/(3*d)

Rule 3889

Int[Sqrt[csc[(e_.) + (f_.)*(x_)]*(b_.) + (a_)]/Sqrt[csc[(e_.) + (f_.)*(x_)]*(d_.)], x_Symbol] :> Simp[-2*a*(Co
t[e + f*x]/(f*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 3894

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

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

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Mathematica [A]
time = 0.18, size = 50, normalized size = 0.63 \begin {gather*} \frac {2 a \sqrt {\cos (c+d x)} (5+\cos (c+d x)) \sqrt {a (1+\sec (c+d x))} \tan \left (\frac {1}{2} (c+d x)\right )}{3 d} \end {gather*}

Antiderivative was successfully verified.

[In]

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

[Out]

(2*a*Sqrt[Cos[c + d*x]]*(5 + Cos[c + d*x])*Sqrt[a*(1 + Sec[c + d*x])]*Tan[(c + d*x)/2])/(3*d)

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Maple [A]
time = 0.12, size = 61, normalized size = 0.77

method	result	size
default	\(-\frac {2 \left (\cos ^{2}\left (d x +c \right )+4 \cos \left (d x +c \right )-5\right ) \left (\sqrt {\cos }\left (d x +c \right )\right ) \sqrt {\frac {a \left (1+\cos \left (d x +c \right )\right )}{\cos \left (d x +c \right )}}\, a}{3 d \sin \left (d x +c \right )}\)	\(61\)

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

-2/3/d*(cos(d*x+c)^2+4*cos(d*x+c)-5)*cos(d*x+c)^(1/2)*(a*(1+cos(d*x+c))/cos(d*x+c))^(1/2)/sin(d*x+c)*a

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Maxima [A]
time = 0.55, size = 38, normalized size = 0.48 \begin {gather*} \frac {{\left (\sqrt {2} a \sin \left (\frac {3}{2} \, d x + \frac {3}{2} \, c\right ) + 9 \, \sqrt {2} a \sin \left (\frac {1}{2} \, d x + \frac {1}{2} \, c\right )\right )} \sqrt {a}}{3 \, d} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

1/3*(sqrt(2)*a*sin(3/2*d*x + 3/2*c) + 9*sqrt(2)*a*sin(1/2*d*x + 1/2*c))*sqrt(a)/d

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Fricas [A]
time = 3.50, size = 61, normalized size = 0.77 \begin {gather*} \frac {2 \, {\left (a \cos \left (d x + c\right ) + 5 \, a\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 )}{3 \, {\left (d \cos \left (d x + c\right ) + d\right )}} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

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

[Out]

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

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

[Out]

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

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