3.152 \(\int \sec ^{\frac {7}{2}}(c+d x) (b \sec (c+d x))^{5/2} \, dx\)

Optimal. Leaf size=116 \[ \frac {b^2 \sin ^5(c+d x) \sec ^{\frac {9}{2}}(c+d x) \sqrt {b \sec (c+d x)}}{5 d}+\frac {2 b^2 \sin ^3(c+d x) \sec ^{\frac {5}{2}}(c+d x) \sqrt {b \sec (c+d x)}}{3 d}+\frac {b^2 \sin (c+d x) \sqrt {\sec (c+d x)} \sqrt {b \sec (c+d x)}}{d} \]

[Out]

2/3*b^2*sec(d*x+c)^(5/2)*sin(d*x+c)^3*(b*sec(d*x+c))^(1/2)/d+1/5*b^2*sec(d*x+c)^(9/2)*sin(d*x+c)^5*(b*sec(d*x+
c))^(1/2)/d+b^2*sin(d*x+c)*sec(d*x+c)^(1/2)*(b*sec(d*x+c))^(1/2)/d

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Rubi [A]  time = 0.02, antiderivative size = 116, normalized size of antiderivative = 1.00, 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} \[ \frac {b^2 \sin ^5(c+d x) \sec ^{\frac {9}{2}}(c+d x) \sqrt {b \sec (c+d x)}}{5 d}+\frac {2 b^2 \sin ^3(c+d x) \sec ^{\frac {5}{2}}(c+d x) \sqrt {b \sec (c+d x)}}{3 d}+\frac {b^2 \sin (c+d x) \sqrt {\sec (c+d x)} \sqrt {b \sec (c+d x)}}{d} \]

Antiderivative was successfully verified.

[In]

Int[Sec[c + d*x]^(7/2)*(b*Sec[c + d*x])^(5/2),x]

[Out]

(b^2*Sqrt[Sec[c + d*x]]*Sqrt[b*Sec[c + d*x]]*Sin[c + d*x])/d + (2*b^2*Sec[c + d*x]^(5/2)*Sqrt[b*Sec[c + d*x]]*
Sin[c + d*x]^3)/(3*d) + (b^2*Sec[c + d*x]^(9/2)*Sqrt[b*Sec[c + d*x]]*Sin[c + d*x]^5)/(5*d)

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 \sec ^{\frac {7}{2}}(c+d x) (b \sec (c+d x))^{5/2} \, dx &=\frac {\left (b^2 \sqrt {b \sec (c+d x)}\right ) \int \sec ^6(c+d x) \, dx}{\sqrt {\sec (c+d x)}}\\ &=-\frac {\left (b^2 \sqrt {b \sec (c+d x)}\right ) \operatorname {Subst}\left (\int \left (1+2 x^2+x^4\right ) \, dx,x,-\tan (c+d x)\right )}{d \sqrt {\sec (c+d x)}}\\ &=\frac {b^2 \sqrt {\sec (c+d x)} \sqrt {b \sec (c+d x)} \sin (c+d x)}{d}+\frac {2 b^2 \sec ^{\frac {5}{2}}(c+d x) \sqrt {b \sec (c+d x)} \sin ^3(c+d x)}{3 d}+\frac {b^2 \sec ^{\frac {9}{2}}(c+d x) \sqrt {b \sec (c+d x)} \sin ^5(c+d x)}{5 d}\\ \end {align*}

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Mathematica [A]  time = 0.23, size = 57, normalized size = 0.49 \[ \frac {\left (\frac {1}{5} \tan ^5(c+d x)+\frac {2}{3} \tan ^3(c+d x)+\tan (c+d x)\right ) (b \sec (c+d x))^{5/2}}{d \sec ^{\frac {5}{2}}(c+d x)} \]

Antiderivative was successfully verified.

[In]

Integrate[Sec[c + d*x]^(7/2)*(b*Sec[c + d*x])^(5/2),x]

[Out]

((b*Sec[c + d*x])^(5/2)*(Tan[c + d*x] + (2*Tan[c + d*x]^3)/3 + Tan[c + d*x]^5/5))/(d*Sec[c + d*x]^(5/2))

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fricas [A]  time = 0.55, size = 63, normalized size = 0.54 \[ \frac {{\left (8 \, b^{2} \cos \left (d x + c\right )^{4} + 4 \, b^{2} \cos \left (d x + c\right )^{2} + 3 \, b^{2}\right )} \sqrt {\frac {b}{\cos \left (d x + c\right )}} \sin \left (d x + c\right )}{15 \, d \cos \left (d x + c\right )^{\frac {9}{2}}} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate(sec(d*x+c)^(7/2)*(b*sec(d*x+c))^(5/2),x, algorithm="fricas")

[Out]

1/15*(8*b^2*cos(d*x + c)^4 + 4*b^2*cos(d*x + c)^2 + 3*b^2)*sqrt(b/cos(d*x + c))*sin(d*x + c)/(d*cos(d*x + c)^(
9/2))

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giac [F]  time = 0.00, size = 0, normalized size = 0.00 \[ \int \left (b \sec \left (d x + c\right )\right )^{\frac {5}{2}} \sec \left (d x + c\right )^{\frac {7}{2}}\,{d x} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate(sec(d*x+c)^(7/2)*(b*sec(d*x+c))^(5/2),x, algorithm="giac")

[Out]

integrate((b*sec(d*x + c))^(5/2)*sec(d*x + c)^(7/2), x)

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maple [A]  time = 0.94, size = 62, normalized size = 0.53 \[ \frac {\left (8 \left (\cos ^{4}\left (d x +c \right )\right )+4 \left (\cos ^{2}\left (d x +c \right )\right )+3\right ) \sin \left (d x +c \right ) \cos \left (d x +c \right ) \left (\frac {b}{\cos \left (d x +c \right )}\right )^{\frac {5}{2}} \left (\frac {1}{\cos \left (d x +c \right )}\right )^{\frac {7}{2}}}{15 d} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

int(sec(d*x+c)^(7/2)*(b*sec(d*x+c))^(5/2),x)

[Out]

1/15/d*(8*cos(d*x+c)^4+4*cos(d*x+c)^2+3)*sin(d*x+c)*cos(d*x+c)*(b/cos(d*x+c))^(5/2)*(1/cos(d*x+c))^(7/2)

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maxima [B]  time = 1.18, size = 705, normalized size = 6.08 \[ -\frac {16 \, {\left (5 \, {\left (2 \, b^{2} \sin \left (4 \, d x + 4 \, c\right ) + b^{2} \sin \left (2 \, d x + 2 \, c\right )\right )} \cos \left (10 \, d x + 10 \, c\right ) + 25 \, {\left (2 \, b^{2} \sin \left (4 \, d x + 4 \, c\right ) + b^{2} \sin \left (2 \, d x + 2 \, c\right )\right )} \cos \left (8 \, d x + 8 \, c\right ) + 50 \, {\left (2 \, b^{2} \sin \left (4 \, d x + 4 \, c\right ) + b^{2} \sin \left (2 \, d x + 2 \, c\right )\right )} \cos \left (6 \, d x + 6 \, c\right ) - {\left (10 \, b^{2} \cos \left (4 \, d x + 4 \, c\right ) + 5 \, b^{2} \cos \left (2 \, d x + 2 \, c\right ) + b^{2}\right )} \sin \left (10 \, d x + 10 \, c\right ) - 5 \, {\left (10 \, b^{2} \cos \left (4 \, d x + 4 \, c\right ) + 5 \, b^{2} \cos \left (2 \, d x + 2 \, c\right ) + b^{2}\right )} \sin \left (8 \, d x + 8 \, c\right ) - 10 \, {\left (10 \, b^{2} \cos \left (4 \, d x + 4 \, c\right ) + 5 \, b^{2} \cos \left (2 \, d x + 2 \, c\right ) + b^{2}\right )} \sin \left (6 \, d x + 6 \, c\right )\right )} \sqrt {b}}{15 \, {\left (2 \, {\left (5 \, \cos \left (8 \, d x + 8 \, c\right ) + 10 \, \cos \left (6 \, d x + 6 \, c\right ) + 10 \, \cos \left (4 \, d x + 4 \, c\right ) + 5 \, \cos \left (2 \, d x + 2 \, c\right ) + 1\right )} \cos \left (10 \, d x + 10 \, c\right ) + \cos \left (10 \, d x + 10 \, c\right )^{2} + 10 \, {\left (10 \, \cos \left (6 \, d x + 6 \, c\right ) + 10 \, \cos \left (4 \, d x + 4 \, c\right ) + 5 \, \cos \left (2 \, d x + 2 \, c\right ) + 1\right )} \cos \left (8 \, d x + 8 \, c\right ) + 25 \, \cos \left (8 \, d x + 8 \, c\right )^{2} + 20 \, {\left (10 \, \cos \left (4 \, d x + 4 \, c\right ) + 5 \, \cos \left (2 \, d x + 2 \, c\right ) + 1\right )} \cos \left (6 \, d x + 6 \, c\right ) + 100 \, \cos \left (6 \, d x + 6 \, c\right )^{2} + 20 \, {\left (5 \, \cos \left (2 \, d x + 2 \, c\right ) + 1\right )} \cos \left (4 \, d x + 4 \, c\right ) + 100 \, \cos \left (4 \, d x + 4 \, c\right )^{2} + 25 \, \cos \left (2 \, d x + 2 \, c\right )^{2} + 10 \, {\left (\sin \left (8 \, d x + 8 \, c\right ) + 2 \, \sin \left (6 \, d x + 6 \, c\right ) + 2 \, \sin \left (4 \, d x + 4 \, c\right ) + \sin \left (2 \, d x + 2 \, c\right )\right )} \sin \left (10 \, d x + 10 \, c\right ) + \sin \left (10 \, d x + 10 \, c\right )^{2} + 50 \, {\left (2 \, \sin \left (6 \, d x + 6 \, c\right ) + 2 \, \sin \left (4 \, d x + 4 \, c\right ) + \sin \left (2 \, d x + 2 \, c\right )\right )} \sin \left (8 \, d x + 8 \, c\right ) + 25 \, \sin \left (8 \, d x + 8 \, c\right )^{2} + 100 \, {\left (2 \, \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 ) + 100 \, \sin \left (6 \, d x + 6 \, c\right )^{2} + 100 \, \sin \left (4 \, d x + 4 \, c\right )^{2} + 100 \, \sin \left (4 \, d x + 4 \, c\right ) \sin \left (2 \, d x + 2 \, c\right ) + 25 \, \sin \left (2 \, d x + 2 \, c\right )^{2} + 10 \, \cos \left (2 \, d x + 2 \, c\right ) + 1\right )} d} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate(sec(d*x+c)^(7/2)*(b*sec(d*x+c))^(5/2),x, algorithm="maxima")

[Out]

-16/15*(5*(2*b^2*sin(4*d*x + 4*c) + b^2*sin(2*d*x + 2*c))*cos(10*d*x + 10*c) + 25*(2*b^2*sin(4*d*x + 4*c) + b^
2*sin(2*d*x + 2*c))*cos(8*d*x + 8*c) + 50*(2*b^2*sin(4*d*x + 4*c) + b^2*sin(2*d*x + 2*c))*cos(6*d*x + 6*c) - (
10*b^2*cos(4*d*x + 4*c) + 5*b^2*cos(2*d*x + 2*c) + b^2)*sin(10*d*x + 10*c) - 5*(10*b^2*cos(4*d*x + 4*c) + 5*b^
2*cos(2*d*x + 2*c) + b^2)*sin(8*d*x + 8*c) - 10*(10*b^2*cos(4*d*x + 4*c) + 5*b^2*cos(2*d*x + 2*c) + b^2)*sin(6
*d*x + 6*c))*sqrt(b)/((2*(5*cos(8*d*x + 8*c) + 10*cos(6*d*x + 6*c) + 10*cos(4*d*x + 4*c) + 5*cos(2*d*x + 2*c)
+ 1)*cos(10*d*x + 10*c) + cos(10*d*x + 10*c)^2 + 10*(10*cos(6*d*x + 6*c) + 10*cos(4*d*x + 4*c) + 5*cos(2*d*x +
2*c) + 1)*cos(8*d*x + 8*c) + 25*cos(8*d*x + 8*c)^2 + 20*(10*cos(4*d*x + 4*c) + 5*cos(2*d*x + 2*c) + 1)*cos(6*
d*x + 6*c) + 100*cos(6*d*x + 6*c)^2 + 20*(5*cos(2*d*x + 2*c) + 1)*cos(4*d*x + 4*c) + 100*cos(4*d*x + 4*c)^2 +
25*cos(2*d*x + 2*c)^2 + 10*(sin(8*d*x + 8*c) + 2*sin(6*d*x + 6*c) + 2*sin(4*d*x + 4*c) + sin(2*d*x + 2*c))*sin
(10*d*x + 10*c) + sin(10*d*x + 10*c)^2 + 50*(2*sin(6*d*x + 6*c) + 2*sin(4*d*x + 4*c) + sin(2*d*x + 2*c))*sin(8
*d*x + 8*c) + 25*sin(8*d*x + 8*c)^2 + 100*(2*sin(4*d*x + 4*c) + sin(2*d*x + 2*c))*sin(6*d*x + 6*c) + 100*sin(6
*d*x + 6*c)^2 + 100*sin(4*d*x + 4*c)^2 + 100*sin(4*d*x + 4*c)*sin(2*d*x + 2*c) + 25*sin(2*d*x + 2*c)^2 + 10*co
s(2*d*x + 2*c) + 1)*d)

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mupad [B]  time = 4.70, size = 205, normalized size = 1.77 \[ -\frac {\sqrt {-\frac {b}{2\,{\sin \left (\frac {c}{2}+\frac {d\,x}{2}\right )}^2-1}}\,\left (\frac {b^2\,\sqrt {-\frac {1}{2\,{\sin \left (\frac {c}{2}+\frac {d\,x}{2}\right )}^2-1}}\,8{}\mathrm {i}}{15\,d}+\frac {b^2\,\sqrt {-\frac {1}{2\,{\sin \left (\frac {c}{2}+\frac {d\,x}{2}\right )}^2-1}}\,\left (-2\,{\sin \left (2\,c+2\,d\,x\right )}^2+\sin \left (4\,c+4\,d\,x\right )\,1{}\mathrm {i}+1\right )\,16{}\mathrm {i}}{3\,d}+\frac {b^2\,\sqrt {-\frac {1}{2\,{\sin \left (\frac {c}{2}+\frac {d\,x}{2}\right )}^2-1}}\,\left (-2\,{\sin \left (c+d\,x\right )}^2+\sin \left (2\,c+2\,d\,x\right )\,1{}\mathrm {i}+1\right )\,8{}\mathrm {i}}{3\,d}\right )\,\left (2\,{\sin \left (\frac {5\,c}{2}+\frac {5\,d\,x}{2}\right )}^2+\sin \left (5\,c+5\,d\,x\right )\,1{}\mathrm {i}-1\right )}{16\,{\left ({\sin \left (c+d\,x\right )}^2-1\right )}^2} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

int((b/cos(c + d*x))^(5/2)*(1/cos(c + d*x))^(7/2),x)

[Out]

-((-b/(2*sin(c/2 + (d*x)/2)^2 - 1))^(1/2)*((b^2*(-1/(2*sin(c/2 + (d*x)/2)^2 - 1))^(1/2)*8i)/(15*d) + (b^2*(-1/
(2*sin(c/2 + (d*x)/2)^2 - 1))^(1/2)*(sin(4*c + 4*d*x)*1i - 2*sin(2*c + 2*d*x)^2 + 1)*16i)/(3*d) + (b^2*(-1/(2*
sin(c/2 + (d*x)/2)^2 - 1))^(1/2)*(sin(2*c + 2*d*x)*1i - 2*sin(c + d*x)^2 + 1)*8i)/(3*d))*(sin(5*c + 5*d*x)*1i
+ 2*sin((5*c)/2 + (5*d*x)/2)^2 - 1))/(16*(sin(c + d*x)^2 - 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(sec(d*x+c)**(7/2)*(b*sec(d*x+c))**(5/2),x)

[Out]

Timed out

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