3.190 \(\int \frac{1}{\cos ^{\frac{5}{2}}(c+d x) (b \cos (c+d x))^{3/2}} \, dx\)

Optimal. Leaf size=76 \[ \frac{\sin ^3(c+d x)}{3 b d \cos ^{\frac{5}{2}}(c+d x) \sqrt{b \cos (c+d x)}}+\frac{\sin (c+d x)}{b d \sqrt{\cos (c+d x)} \sqrt{b \cos (c+d x)}} \]

[Out]

Sin[c + d*x]/(b*d*Sqrt[Cos[c + d*x]]*Sqrt[b*Cos[c + d*x]]) + Sin[c + d*x]^3/(3*b*d*Cos[c + d*x]^(5/2)*Sqrt[b*C
os[c + d*x]])

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Rubi [A]  time = 0.01873, antiderivative size = 76, normalized size of antiderivative = 1., number of steps used = 3, number of rules used = 2, integrand size = 23, \(\frac{\text{number of rules}}{\text{integrand size}}\) = 0.087, Rules used = {18, 3767} \[ \frac{\sin ^3(c+d x)}{3 b d \cos ^{\frac{5}{2}}(c+d x) \sqrt{b \cos (c+d x)}}+\frac{\sin (c+d x)}{b d \sqrt{\cos (c+d x)} \sqrt{b \cos (c+d x)}} \]

Antiderivative was successfully verified.

[In]

Int[1/(Cos[c + d*x]^(5/2)*(b*Cos[c + d*x])^(3/2)),x]

[Out]

Sin[c + d*x]/(b*d*Sqrt[Cos[c + d*x]]*Sqrt[b*Cos[c + d*x]]) + Sin[c + d*x]^3/(3*b*d*Cos[c + d*x]^(5/2)*Sqrt[b*C
os[c + d*x]])

Rule 18

Int[(u_.)*((a_.)*(v_))^(m_)*((b_.)*(v_))^(n_), x_Symbol] :> Dist[(a^(m - 1/2)*b^(n + 1/2)*Sqrt[a*v])/Sqrt[b*v]
, Int[u*v^(m + n), x], x] /; FreeQ[{a, b, m}, x] &&  !IntegerQ[m] && ILtQ[n - 1/2, 0] && IntegerQ[m + n]

Rule 3767

Int[csc[(c_.) + (d_.)*(x_)]^(n_), x_Symbol] :> -Dist[d^(-1), Subst[Int[ExpandIntegrand[(1 + x^2)^(n/2 - 1), x]
, x], x, Cot[c + d*x]], x] /; FreeQ[{c, d}, x] && IGtQ[n/2, 0]

Rubi steps

\begin{align*} \int \frac{1}{\cos ^{\frac{5}{2}}(c+d x) (b \cos (c+d x))^{3/2}} \, dx &=\frac{\sqrt{\cos (c+d x)} \int \sec ^4(c+d x) \, dx}{b \sqrt{b \cos (c+d x)}}\\ &=-\frac{\sqrt{\cos (c+d x)} \operatorname{Subst}\left (\int \left (1+x^2\right ) \, dx,x,-\tan (c+d x)\right )}{b d \sqrt{b \cos (c+d x)}}\\ &=\frac{\sin (c+d x)}{b d \sqrt{\cos (c+d x)} \sqrt{b \cos (c+d x)}}+\frac{\sin ^3(c+d x)}{3 b d \cos ^{\frac{5}{2}}(c+d x) \sqrt{b \cos (c+d x)}}\\ \end{align*}

Mathematica [A]  time = 0.0256597, size = 45, normalized size = 0.59 \[ \frac{\cos ^{\frac{3}{2}}(c+d x) \left (\frac{1}{3} \tan ^3(c+d x)+\tan (c+d x)\right )}{d (b \cos (c+d x))^{3/2}} \]

Antiderivative was successfully verified.

[In]

Integrate[1/(Cos[c + d*x]^(5/2)*(b*Cos[c + d*x])^(3/2)),x]

[Out]

(Cos[c + d*x]^(3/2)*(Tan[c + d*x] + Tan[c + d*x]^3/3))/(d*(b*Cos[c + d*x])^(3/2))

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Maple [A]  time = 0.158, size = 42, normalized size = 0.6 \begin{align*}{\frac{\sin \left ( dx+c \right ) \left ( 2\, \left ( \cos \left ( dx+c \right ) \right ) ^{2}+1 \right ) }{3\,d} \left ( \cos \left ( dx+c \right ) \right ) ^{-{\frac{3}{2}}} \left ( b\cos \left ( dx+c \right ) \right ) ^{-{\frac{3}{2}}}} \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

int(1/cos(d*x+c)^(5/2)/(b*cos(d*x+c))^(3/2),x)

[Out]

1/3/d*sin(d*x+c)*(2*cos(d*x+c)^2+1)/(b*cos(d*x+c))^(3/2)/cos(d*x+c)^(3/2)

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Maxima [B]  time = 1.80497, size = 420, normalized size = 5.53 \begin{align*} \frac{4 \,{\left ({\left (3 \, \cos \left (2 \, d x + 2 \, c\right ) + 1\right )} \sin \left (6 \, d x + 6 \, c\right ) + 3 \,{\left (3 \, \cos \left (2 \, d x + 2 \, c\right ) + 1\right )} \sin \left (4 \, d x + 4 \, c\right ) - 3 \, \cos \left (6 \, d x + 6 \, c\right ) \sin \left (2 \, d x + 2 \, c\right ) - 9 \, \cos \left (4 \, d x + 4 \, c\right ) \sin \left (2 \, d x + 2 \, c\right )\right )}}{3 \,{\left (b \cos \left (6 \, d x + 6 \, c\right )^{2} + 9 \, b \cos \left (4 \, d x + 4 \, c\right )^{2} + 9 \, b \cos \left (2 \, d x + 2 \, c\right )^{2} + b \sin \left (6 \, d x + 6 \, c\right )^{2} + 9 \, b \sin \left (4 \, d x + 4 \, c\right )^{2} + 18 \, b \sin \left (4 \, d x + 4 \, c\right ) \sin \left (2 \, d x + 2 \, c\right ) + 9 \, b \sin \left (2 \, d x + 2 \, c\right )^{2} + 2 \,{\left (3 \, b \cos \left (4 \, d x + 4 \, c\right ) + 3 \, b \cos \left (2 \, d x + 2 \, c\right ) + b\right )} \cos \left (6 \, d x + 6 \, c\right ) + 6 \,{\left (3 \, b \cos \left (2 \, d x + 2 \, c\right ) + b\right )} \cos \left (4 \, d x + 4 \, c\right ) + 6 \, b \cos \left (2 \, d x + 2 \, c\right ) + 6 \,{\left (b \sin \left (4 \, d x + 4 \, c\right ) + b \sin \left (2 \, d x + 2 \, c\right )\right )} \sin \left (6 \, d x + 6 \, c\right ) + b\right )} \sqrt{b} d} \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate(1/cos(d*x+c)^(5/2)/(b*cos(d*x+c))^(3/2),x, algorithm="maxima")

[Out]

4/3*((3*cos(2*d*x + 2*c) + 1)*sin(6*d*x + 6*c) + 3*(3*cos(2*d*x + 2*c) + 1)*sin(4*d*x + 4*c) - 3*cos(6*d*x + 6
*c)*sin(2*d*x + 2*c) - 9*cos(4*d*x + 4*c)*sin(2*d*x + 2*c))/((b*cos(6*d*x + 6*c)^2 + 9*b*cos(4*d*x + 4*c)^2 +
9*b*cos(2*d*x + 2*c)^2 + b*sin(6*d*x + 6*c)^2 + 9*b*sin(4*d*x + 4*c)^2 + 18*b*sin(4*d*x + 4*c)*sin(2*d*x + 2*c
) + 9*b*sin(2*d*x + 2*c)^2 + 2*(3*b*cos(4*d*x + 4*c) + 3*b*cos(2*d*x + 2*c) + b)*cos(6*d*x + 6*c) + 6*(3*b*cos
(2*d*x + 2*c) + b)*cos(4*d*x + 4*c) + 6*b*cos(2*d*x + 2*c) + 6*(b*sin(4*d*x + 4*c) + b*sin(2*d*x + 2*c))*sin(6
*d*x + 6*c) + b)*sqrt(b)*d)

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Fricas [A]  time = 1.9584, size = 120, normalized size = 1.58 \begin{align*} \frac{\sqrt{b \cos \left (d x + c\right )}{\left (2 \, \cos \left (d x + c\right )^{2} + 1\right )} \sin \left (d x + c\right )}{3 \, b^{2} d \cos \left (d x + c\right )^{\frac{7}{2}}} \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate(1/cos(d*x+c)^(5/2)/(b*cos(d*x+c))^(3/2),x, algorithm="fricas")

[Out]

1/3*sqrt(b*cos(d*x + c))*(2*cos(d*x + c)^2 + 1)*sin(d*x + c)/(b^2*d*cos(d*x + c)^(7/2))

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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(1/cos(d*x+c)**(5/2)/(b*cos(d*x+c))**(3/2),x)

[Out]

Timed out

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Giac [F]  time = 0., size = 0, normalized size = 0. \begin{align*} \int \frac{1}{\left (b \cos \left (d x + c\right )\right )^{\frac{3}{2}} \cos \left (d x + c\right )^{\frac{5}{2}}}\,{d x} \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate(1/cos(d*x+c)^(5/2)/(b*cos(d*x+c))^(3/2),x, algorithm="giac")

[Out]

integrate(1/((b*cos(d*x + c))^(3/2)*cos(d*x + c)^(5/2)), x)