\(\int \frac {(b \cos (c+d x))^{3/2}}{\cos ^{\frac {11}{2}}(c+d x)} \, dx\) [158]

Optimal result

Integrand size = 23, antiderivative size = 72 \[ \int \frac {(b \cos (c+d x))^{3/2}}{\cos ^{\frac {11}{2}}(c+d x)} \, dx=\frac {b \sqrt {b \cos (c+d x)} \sin (c+d x)}{d \cos ^{\frac {3}{2}}(c+d x)}+\frac {b \sqrt {b \cos (c+d x)} \sin ^3(c+d x)}{3 d \cos ^{\frac {7}{2}}(c+d x)} \]

[Out]

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

Rubi [A] (verified)

Time = 0.02 (sec) , antiderivative size = 72, normalized size of antiderivative = 1.00, number of steps used = 3, number of rules used = 2, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.087, Rules used = {17, 3852} \[ \int \frac {(b \cos (c+d x))^{3/2}}{\cos ^{\frac {11}{2}}(c+d x)} \, dx=\frac {b \sin ^3(c+d x) \sqrt {b \cos (c+d x)}}{3 d \cos ^{\frac {7}{2}}(c+d x)}+\frac {b \sin (c+d x) \sqrt {b \cos (c+d x)}}{d \cos ^{\frac {3}{2}}(c+d x)} \]

[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 3852

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*} \text {integral}& = \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 ) \text {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] (verified)

Time = 0.07 (sec) , antiderivative size = 45, normalized size of antiderivative = 0.62 \[ \int \frac {(b \cos (c+d x))^{3/2}}{\cos ^{\frac {11}{2}}(c+d x)} \, dx=\frac {(b \cos (c+d x))^{3/2} \left (\tan (c+d x)+\frac {1}{3} \tan ^3(c+d x)\right )}{d \cos ^{\frac {3}{2}}(c+d x)} \]

[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))

Maple [A] (verified)

Time = 2.77 (sec) , antiderivative size = 43, normalized size of antiderivative = 0.60

method	result	size
default	\(\frac {b \left (2 \left (\cos ^{2}\left (d x +c \right )\right )+1\right ) \sqrt {\cos \left (d x +c \right ) b}\, \sin \left (d x +c \right )}{3 d \cos \left (d x +c \right )^{\frac {7}{2}}}\)	\(43\)
risch	\(\frac {4 i b \sqrt {\cos \left (d x +c \right ) b}\, \left (3 \,{\mathrm e}^{2 i \left (d x +c \right )}+1\right )}{3 \sqrt {\cos \left (d x +c \right )}\, d \left ({\mathrm e}^{2 i \left (d x +c \right )}+1\right )^{3}}\)	\(52\)

[In]

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

[Out]

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

Fricas [A] (verification not implemented)

none

Time = 0.28 (sec) , antiderivative size = 42, normalized size of antiderivative = 0.58 \[ \int \frac {(b \cos (c+d x))^{3/2}}{\cos ^{\frac {11}{2}}(c+d x)} \, dx=\frac {{\left (2 \, b \cos \left (d x + c\right )^{2} + b\right )} \sqrt {b \cos \left (d x + c\right )} \sin \left (d x + c\right )}{3 \, d \cos \left (d x + c\right )^{\frac {7}{2}}} \]

[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))

Sympy [F(-1)]

Timed out. \[ \int \frac {(b \cos (c+d x))^{3/2}}{\cos ^{\frac {11}{2}}(c+d x)} \, dx=\text {Timed out} \]

[In]

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

[Out]

Timed out

Maxima [B] (verification not implemented)

Leaf count of result is larger than twice the leaf count of optimal. 299 vs. \(2 (62) = 124\).

Time = 0.41 (sec) , antiderivative size = 299, normalized size of antiderivative = 4.15 \[ \int \frac {(b \cos (c+d x))^{3/2}}{\cos ^{\frac {11}{2}}(c+d x)} \, dx=-\frac {4 \, {\left (3 \, b \cos \left (6 \, d x + 6 \, c\right ) \sin \left (2 \, d x + 2 \, c\right ) + 9 \, b \cos \left (4 \, d x + 4 \, c\right ) \sin \left (2 \, d x + 2 \, c\right ) - {\left (3 \, b \cos \left (2 \, d x + 2 \, c\right ) + b\right )} \sin \left (6 \, d x + 6 \, c\right ) - 3 \, {\left (3 \, b \cos \left (2 \, d x + 2 \, c\right ) + b\right )} \sin \left (4 \, d x + 4 \, c\right )\right )} \sqrt {b}}{3 \, {\left (2 \, {\left (3 \, \cos \left (4 \, d x + 4 \, c\right ) + 3 \, \cos \left (2 \, d x + 2 \, c\right ) + 1\right )} \cos \left (6 \, d x + 6 \, c\right ) + \cos \left (6 \, d x + 6 \, c\right )^{2} + 6 \, {\left (3 \, \cos \left (2 \, d x + 2 \, c\right ) + 1\right )} \cos \left (4 \, d x + 4 \, c\right ) + 9 \, \cos \left (4 \, d x + 4 \, c\right )^{2} + 9 \, \cos \left (2 \, d x + 2 \, c\right )^{2} + 6 \, {\left (\sin \left (4 \, d x + 4 \, c\right ) + \sin \left (2 \, d x + 2 \, c\right )\right )} \sin \left (6 \, d x + 6 \, c\right ) + \sin \left (6 \, d x + 6 \, c\right )^{2} + 9 \, \sin \left (4 \, d x + 4 \, c\right )^{2} + 18 \, \sin \left (4 \, d x + 4 \, c\right ) \sin \left (2 \, d x + 2 \, c\right ) + 9 \, \sin \left (2 \, d x + 2 \, c\right )^{2} + 6 \, \cos \left (2 \, d x + 2 \, c\right ) + 1\right )} d} \]

[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)

Giac [F]

\[ \int \frac {(b \cos (c+d x))^{3/2}}{\cos ^{\frac {11}{2}}(c+d x)} \, dx=\int { \frac {\left (b \cos \left (d x + c\right )\right )^{\frac {3}{2}}}{\cos \left (d x + c\right )^{\frac {11}{2}}} \,d x } \]

[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)

Mupad [B] (verification not implemented)

Time = 15.35 (sec) , antiderivative size = 129, normalized size of antiderivative = 1.79 \[ \int \frac {(b \cos (c+d x))^{3/2}}{\cos ^{\frac {11}{2}}(c+d x)} \, dx=\frac {2\,b\,\sqrt {b\,\cos \left (c+d\,x\right )}\,\left (\cos \left (2\,c+2\,d\,x\right )\,15{}\mathrm {i}+\cos \left (4\,c+4\,d\,x\right )\,6{}\mathrm {i}+\cos \left (6\,c+6\,d\,x\right )\,1{}\mathrm {i}+9\,\sin \left (2\,c+2\,d\,x\right )+6\,\sin \left (4\,c+4\,d\,x\right )+\sin \left (6\,c+6\,d\,x\right )+10{}\mathrm {i}\right )}{3\,d\,\sqrt {\cos \left (c+d\,x\right )}\,\left (15\,\cos \left (2\,c+2\,d\,x\right )+6\,\cos \left (4\,c+4\,d\,x\right )+\cos \left (6\,c+6\,d\,x\right )+10\right )} \]

[In]

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

[Out]

(2*b*(b*cos(c + d*x))^(1/2)*(cos(2*c + 2*d*x)*15i + cos(4*c + 4*d*x)*6i + cos(6*c + 6*d*x)*1i + 9*sin(2*c + 2*
d*x) + 6*sin(4*c + 4*d*x) + sin(6*c + 6*d*x) + 10i))/(3*d*cos(c + d*x)^(1/2)*(15*cos(2*c + 2*d*x) + 6*cos(4*c
+ 4*d*x) + cos(6*c + 6*d*x) + 10))