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

Optimal. Leaf size=101 \[ \frac {3 b x \sqrt {b \sec (c+d x)}}{8 \sqrt {\sec (c+d x)}}+\frac {3 b \sin (c+d x) \sqrt {b \sec (c+d x)}}{8 d \sec ^{\frac {3}{2}}(c+d x)}+\frac {b \sin (c+d x) \sqrt {b \sec (c+d x)}}{4 d \sec ^{\frac {7}{2}}(c+d x)} \]

[Out]

1/4*b*sin(d*x+c)*(b*sec(d*x+c))^(1/2)/d/sec(d*x+c)^(7/2)+3/8*b*sin(d*x+c)*(b*sec(d*x+c))^(1/2)/d/sec(d*x+c)^(3
/2)+3/8*b*x*(b*sec(d*x+c))^(1/2)/sec(d*x+c)^(1/2)

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Rubi [A]  time = 0.03, antiderivative size = 101, normalized size of antiderivative = 1.00, number of steps used = 4, number of rules used = 3, integrand size = 23, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.130, Rules used = {17, 2635, 8} \[ \frac {3 b x \sqrt {b \sec (c+d x)}}{8 \sqrt {\sec (c+d x)}}+\frac {3 b \sin (c+d x) \sqrt {b \sec (c+d x)}}{8 d \sec ^{\frac {3}{2}}(c+d x)}+\frac {b \sin (c+d x) \sqrt {b \sec (c+d x)}}{4 d \sec ^{\frac {7}{2}}(c+d x)} \]

Antiderivative was successfully verified.

[In]

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

[Out]

(3*b*x*Sqrt[b*Sec[c + d*x]])/(8*Sqrt[Sec[c + d*x]]) + (b*Sqrt[b*Sec[c + d*x]]*Sin[c + d*x])/(4*d*Sec[c + d*x]^
(7/2)) + (3*b*Sqrt[b*Sec[c + d*x]]*Sin[c + d*x])/(8*d*Sec[c + d*x]^(3/2))

Rule 8

Int[a_, x_Symbol] :> Simp[a*x, x] /; FreeQ[a, x]

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 2635

Int[((b_.)*sin[(c_.) + (d_.)*(x_)])^(n_), x_Symbol] :> -Simp[(b*Cos[c + d*x]*(b*Sin[c + d*x])^(n - 1))/(d*n),
x] + Dist[(b^2*(n - 1))/n, Int[(b*Sin[c + d*x])^(n - 2), x], x] /; FreeQ[{b, c, d}, x] && GtQ[n, 1] && Integer
Q[2*n]

Rubi steps

\begin {align*} \int \frac {(b \sec (c+d x))^{3/2}}{\sec ^{\frac {11}{2}}(c+d x)} \, dx &=\frac {\left (b \sqrt {b \sec (c+d x)}\right ) \int \cos ^4(c+d x) \, dx}{\sqrt {\sec (c+d x)}}\\ &=\frac {b \sqrt {b \sec (c+d x)} \sin (c+d x)}{4 d \sec ^{\frac {7}{2}}(c+d x)}+\frac {\left (3 b \sqrt {b \sec (c+d x)}\right ) \int \cos ^2(c+d x) \, dx}{4 \sqrt {\sec (c+d x)}}\\ &=\frac {b \sqrt {b \sec (c+d x)} \sin (c+d x)}{4 d \sec ^{\frac {7}{2}}(c+d x)}+\frac {3 b \sqrt {b \sec (c+d x)} \sin (c+d x)}{8 d \sec ^{\frac {3}{2}}(c+d x)}+\frac {\left (3 b \sqrt {b \sec (c+d x)}\right ) \int 1 \, dx}{8 \sqrt {\sec (c+d x)}}\\ &=\frac {3 b x \sqrt {b \sec (c+d x)}}{8 \sqrt {\sec (c+d x)}}+\frac {b \sqrt {b \sec (c+d x)} \sin (c+d x)}{4 d \sec ^{\frac {7}{2}}(c+d x)}+\frac {3 b \sqrt {b \sec (c+d x)} \sin (c+d x)}{8 d \sec ^{\frac {3}{2}}(c+d x)}\\ \end {align*}

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Mathematica [A]  time = 0.16, size = 55, normalized size = 0.54 \[ \frac {(12 (c+d x)+8 \sin (2 (c+d x))+\sin (4 (c+d x))) (b \sec (c+d x))^{3/2}}{32 d \sec ^{\frac {3}{2}}(c+d x)} \]

Antiderivative was successfully verified.

[In]

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

[Out]

((b*Sec[c + d*x])^(3/2)*(12*(c + d*x) + 8*Sin[2*(c + d*x)] + Sin[4*(c + d*x)]))/(32*d*Sec[c + d*x]^(3/2))

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fricas [A]  time = 0.73, size = 207, normalized size = 2.05 \[ \left [\frac {3 \, \sqrt {-b} b \log \left (-2 \, \sqrt {-b} \sqrt {\frac {b}{\cos \left (d x + c\right )}} \cos \left (d x + c\right )^{\frac {3}{2}} \sin \left (d x + c\right ) + 2 \, b \cos \left (d x + c\right )^{2} - b\right ) + \frac {2 \, {\left (2 \, b \cos \left (d x + c\right )^{4} + 3 \, b \cos \left (d x + c\right )^{2}\right )} \sqrt {\frac {b}{\cos \left (d x + c\right )}} \sin \left (d x + c\right )}{\sqrt {\cos \left (d x + c\right )}}}{16 \, d}, \frac {3 \, b^{\frac {3}{2}} \arctan \left (\frac {\sqrt {\frac {b}{\cos \left (d x + c\right )}} \sin \left (d x + c\right )}{\sqrt {b} \sqrt {\cos \left (d x + c\right )}}\right ) + \frac {{\left (2 \, b \cos \left (d x + c\right )^{4} + 3 \, b \cos \left (d x + c\right )^{2}\right )} \sqrt {\frac {b}{\cos \left (d x + c\right )}} \sin \left (d x + c\right )}{\sqrt {\cos \left (d x + c\right )}}}{8 \, d}\right ] \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

[1/16*(3*sqrt(-b)*b*log(-2*sqrt(-b)*sqrt(b/cos(d*x + c))*cos(d*x + c)^(3/2)*sin(d*x + c) + 2*b*cos(d*x + c)^2
- b) + 2*(2*b*cos(d*x + c)^4 + 3*b*cos(d*x + c)^2)*sqrt(b/cos(d*x + c))*sin(d*x + c)/sqrt(cos(d*x + c)))/d, 1/
8*(3*b^(3/2)*arctan(sqrt(b/cos(d*x + c))*sin(d*x + c)/(sqrt(b)*sqrt(cos(d*x + c)))) + (2*b*cos(d*x + c)^4 + 3*
b*cos(d*x + c)^2)*sqrt(b/cos(d*x + c))*sin(d*x + c)/sqrt(cos(d*x + c)))/d]

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giac [F]  time = 0.00, size = 0, normalized size = 0.00 \[ \int \frac {\left (b \sec \left (d x + c\right )\right )^{\frac {3}{2}}}{\sec \left (d x + c\right )^{\frac {11}{2}}}\,{d x} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

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

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maple [A]  time = 1.19, size = 74, normalized size = 0.73 \[ \frac {\left (2 \left (\cos ^{3}\left (d x +c \right )\right ) \sin \left (d x +c \right )+3 \cos \left (d x +c \right ) \sin \left (d x +c \right )+3 d x +3 c \right ) \left (\frac {b}{\cos \left (d x +c \right )}\right )^{\frac {3}{2}}}{8 d \left (\frac {1}{\cos \left (d x +c \right )}\right )^{\frac {11}{2}} \cos \left (d x +c \right )^{4}} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

1/8/d*(2*cos(d*x+c)^3*sin(d*x+c)+3*cos(d*x+c)*sin(d*x+c)+3*d*x+3*c)*(b/cos(d*x+c))^(3/2)/(1/cos(d*x+c))^(11/2)
/cos(d*x+c)^4

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maxima [A]  time = 1.02, size = 53, normalized size = 0.52 \[ \frac {{\left (12 \, {\left (d x + c\right )} b + b \sin \left (4 \, d x + 4 \, c\right ) + 8 \, b \sin \left (\frac {1}{2} \, \arctan \left (\sin \left (4 \, d x + 4 \, c\right ), \cos \left (4 \, d x + 4 \, c\right )\right )\right )\right )} \sqrt {b}}{32 \, d} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

1/32*(12*(d*x + c)*b + b*sin(4*d*x + 4*c) + 8*b*sin(1/2*arctan2(sin(4*d*x + 4*c), cos(4*d*x + 4*c))))*sqrt(b)/
d

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mupad [B]  time = 0.59, size = 53, normalized size = 0.52 \[ \frac {b\,\sqrt {\frac {b}{\cos \left (c+d\,x\right )}}\,\left (8\,\sin \left (2\,c+2\,d\,x\right )+\sin \left (4\,c+4\,d\,x\right )+12\,d\,x\right )}{32\,d\,\sqrt {\frac {1}{\cos \left (c+d\,x\right )}}} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

(b*(b/cos(c + d*x))^(1/2)*(8*sin(2*c + 2*d*x) + sin(4*c + 4*d*x) + 12*d*x))/(32*d*(1/cos(c + d*x))^(1/2))

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sympy [F(-1)]  time = 0.00, size = 0, normalized size = 0.00 \[ \text {Timed out} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

Timed out

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