3.158 \(\int \frac{(b \cos (c+d x))^{3/2}}{\cos ^{\frac{11}{2}}(c+d x)} \, dx\)

Optimal. Leaf size=72 \[ \frac{b \sin ^3(c+d x) \sqrt{b \cos (c+d x)}}{3 d \cos ^{\frac{7}{2}}(c+d x)}+\frac{b \sin (c+d x) \sqrt{b \cos (c+d x)}}{d \cos ^{\frac{3}{2}}(c+d x)} \]

[Out]

(b*Sqrt[b*Cos[c + d*x]]*Sin[c + d*x])/(d*Cos[c + d*x]^(3/2)) + (b*Sqrt[b*Cos[c + d*x]]*Sin[c + d*x]^3)/(3*d*Co
s[c + d*x]^(7/2))

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Rubi [A]  time = 0.0183373, antiderivative size = 72, 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 = {17, 3767} \[ \frac{b \sin ^3(c+d x) \sqrt{b \cos (c+d x)}}{3 d \cos ^{\frac{7}{2}}(c+d x)}+\frac{b \sin (c+d x) \sqrt{b \cos (c+d x)}}{d \cos ^{\frac{3}{2}}(c+d x)} \]

Antiderivative was successfully verified.

[In]

Int[(b*Cos[c + d*x])^(3/2)/Cos[c + d*x]^(11/2),x]

[Out]

(b*Sqrt[b*Cos[c + d*x]]*Sin[c + d*x])/(d*Cos[c + d*x]^(3/2)) + (b*Sqrt[b*Cos[c + d*x]]*Sin[c + d*x]^3)/(3*d*Co
s[c + d*x]^(7/2))

Rule 17

Int[(u_.)*((a_.)*(v_))^(m_)*((b_.)*(v_))^(n_), x_Symbol] :> Dist[(a^(m + 1/2)*b^(n - 1/2)*Sqrt[b*v])/Sqrt[a*v]
, Int[u*v^(m + n), x], x] /; FreeQ[{a, b, m}, x] &&  !IntegerQ[m] && IGtQ[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{(b \cos (c+d x))^{3/2}}{\cos ^{\frac{11}{2}}(c+d x)} \, dx &=\frac{\left (b \sqrt{b \cos (c+d x)}\right ) \int \sec ^4(c+d x) \, dx}{\sqrt{\cos (c+d x)}}\\ &=-\frac{\left (b \sqrt{b \cos (c+d x)}\right ) \operatorname{Subst}\left (\int \left (1+x^2\right ) \, dx,x,-\tan (c+d x)\right )}{d \sqrt{\cos (c+d x)}}\\ &=\frac{b \sqrt{b \cos (c+d x)} \sin (c+d x)}{d \cos ^{\frac{3}{2}}(c+d x)}+\frac{b \sqrt{b \cos (c+d x)} \sin ^3(c+d x)}{3 d \cos ^{\frac{7}{2}}(c+d x)}\\ \end{align*}

Mathematica [A]  time = 0.0604555, size = 45, normalized size = 0.62 \[ \frac{\left (\frac{1}{3} \tan ^3(c+d x)+\tan (c+d x)\right ) (b \cos (c+d x))^{3/2}}{d \cos ^{\frac{3}{2}}(c+d x)} \]

Antiderivative was successfully verified.

[In]

Integrate[(b*Cos[c + d*x])^(3/2)/Cos[c + d*x]^(11/2),x]

[Out]

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

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Maple [A]  time = 0.165, size = 42, normalized size = 0.6 \begin{align*}{\frac{ \left ( 2\, \left ( \cos \left ( dx+c \right ) \right ) ^{2}+1 \right ) \sin \left ( dx+c \right ) }{3\,d} \left ( b\cos \left ( dx+c \right ) \right ) ^{{\frac{3}{2}}} \left ( \cos \left ( dx+c \right ) \right ) ^{-{\frac{9}{2}}}} \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

int((b*cos(d*x+c))^(3/2)/cos(d*x+c)^(11/2),x)

[Out]

1/3/d*(2*cos(d*x+c)^2+1)*(b*cos(d*x+c))^(3/2)*sin(d*x+c)/cos(d*x+c)^(9/2)

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Maxima [B]  time = 1.82876, size = 404, normalized size = 5.61 \begin{align*} -\frac{4 \,{\left (3 \, b \cos \left (6 \, d x + 6 \, c\right ) \sin \left (2 \, d x + 2 \, c\right ) + 9 \, b \cos \left (4 \, d x + 4 \, c\right ) \sin \left (2 \, d x + 2 \, c\right ) -{\left (3 \, b \cos \left (2 \, d x + 2 \, c\right ) + b\right )} \sin \left (6 \, d x + 6 \, c\right ) - 3 \,{\left (3 \, b \cos \left (2 \, d x + 2 \, c\right ) + b\right )} \sin \left (4 \, d x + 4 \, c\right )\right )} \sqrt{b}}{3 \,{\left (2 \,{\left (3 \, \cos \left (4 \, d x + 4 \, c\right ) + 3 \, \cos \left (2 \, d x + 2 \, c\right ) + 1\right )} \cos \left (6 \, d x + 6 \, c\right ) + \cos \left (6 \, d x + 6 \, c\right )^{2} + 6 \,{\left (3 \, \cos \left (2 \, d x + 2 \, c\right ) + 1\right )} \cos \left (4 \, d x + 4 \, c\right ) + 9 \, \cos \left (4 \, d x + 4 \, c\right )^{2} + 9 \, \cos \left (2 \, d x + 2 \, c\right )^{2} + 6 \,{\left (\sin \left (4 \, d x + 4 \, c\right ) + \sin \left (2 \, d x + 2 \, c\right )\right )} \sin \left (6 \, d x + 6 \, c\right ) + \sin \left (6 \, d x + 6 \, c\right )^{2} + 9 \, \sin \left (4 \, d x + 4 \, c\right )^{2} + 18 \, \sin \left (4 \, d x + 4 \, c\right ) \sin \left (2 \, d x + 2 \, c\right ) + 9 \, \sin \left (2 \, d x + 2 \, c\right )^{2} + 6 \, \cos \left (2 \, d x + 2 \, c\right ) + 1\right )} d} \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate((b*cos(d*x+c))^(3/2)/cos(d*x+c)^(11/2),x, algorithm="maxima")

[Out]

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

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Fricas [A]  time = 1.60906, size = 117, normalized size = 1.62 \begin{align*} \frac{{\left (2 \, b \cos \left (d x + c\right )^{2} + b\right )} \sqrt{b \cos \left (d x + c\right )} \sin \left (d x + c\right )}{3 \, d \cos \left (d x + c\right )^{\frac{7}{2}}} \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate((b*cos(d*x+c))^(3/2)/cos(d*x+c)^(11/2),x, algorithm="fricas")

[Out]

1/3*(2*b*cos(d*x + c)^2 + b)*sqrt(b*cos(d*x + c))*sin(d*x + c)/(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((b*cos(d*x+c))**(3/2)/cos(d*x+c)**(11/2),x)

[Out]

Timed out

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Giac [F]  time = 0., size = 0, normalized size = 0. \begin{align*} \int \frac{\left (b \cos \left (d x + c\right )\right )^{\frac{3}{2}}}{\cos \left (d x + c\right )^{\frac{11}{2}}}\,{d x} \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate((b*cos(d*x+c))^(3/2)/cos(d*x+c)^(11/2),x, algorithm="giac")

[Out]

integrate((b*cos(d*x + c))^(3/2)/cos(d*x + c)^(11/2), x)