3.2.43 \(\int \sec ^{\frac {5}{2}}(c+d x) (b \sec (c+d x))^{3/2} \, dx\) [143]

Optimal. Leaf size=72 \[ \frac {b \sqrt {\sec (c+d x)} \sqrt {b \sec (c+d x)} \sin (c+d x)}{d}+\frac {b \sec ^{\frac {5}{2}}(c+d x) \sqrt {b \sec (c+d x)} \sin ^3(c+d x)}{3 d} \]

[Out]

1/3*b*sec(d*x+c)^(5/2)*sin(d*x+c)^3*(b*sec(d*x+c))^(1/2)/d+b*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 = 0.01, antiderivative size = 72, 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, 3852} \begin {gather*} \frac {b \sin ^3(c+d x) \sec ^{\frac {5}{2}}(c+d x) \sqrt {b \sec (c+d x)}}{3 d}+\frac {b \sin (c+d x) \sqrt {\sec (c+d x)} \sqrt {b \sec (c+d x)}}{d} \end {gather*}

Antiderivative was successfully verified.

[In]

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

[Out]

(b*Sqrt[Sec[c + d*x]]*Sqrt[b*Sec[c + d*x]]*Sin[c + d*x])/d + (b*Sec[c + d*x]^(5/2)*Sqrt[b*Sec[c + d*x]]*Sin[c
+ d*x]^3)/(3*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 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*} \int \sec ^{\frac {5}{2}}(c+d x) (b \sec (c+d x))^{3/2} \, dx &=\frac {\left (b \sqrt {b \sec (c+d x)}\right ) \int \sec ^4(c+d x) \, dx}{\sqrt {\sec (c+d x)}}\\ &=-\frac {\left (b \sqrt {b \sec (c+d x)}\right ) \text {Subst}\left (\int \left (1+x^2\right ) \, dx,x,-\tan (c+d x)\right )}{d \sqrt {\sec (c+d x)}}\\ &=\frac {b \sqrt {\sec (c+d x)} \sqrt {b \sec (c+d x)} \sin (c+d x)}{d}+\frac {b \sec ^{\frac {5}{2}}(c+d x) \sqrt {b \sec (c+d x)} \sin ^3(c+d x)}{3 d}\\ \end {align*}

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Mathematica [A]
time = 0.10, size = 45, normalized size = 0.62 \begin {gather*} \frac {(b \sec (c+d x))^{3/2} \left (\tan (c+d x)+\frac {1}{3} \tan ^3(c+d x)\right )}{d \sec ^{\frac {3}{2}}(c+d x)} \end {gather*}

Antiderivative was successfully verified.

[In]

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

[Out]

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

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Maple [A]
time = 34.25, size = 52, normalized size = 0.72

method	result	size
default	\(\frac {\left (2 \left (\cos ^{2}\left (d x +c \right )\right )+1\right ) \cos \left (d x +c \right ) \left (\frac {1}{\cos \left (d x +c \right )}\right )^{\frac {5}{2}} \left (\frac {b}{\cos \left (d x +c \right )}\right )^{\frac {3}{2}} \sin \left (d x +c \right )}{3 d}\)	\(52\)
risch	\(\frac {4 i b \sqrt {\frac {{\mathrm e}^{i \left (d x +c \right )}}{{\mathrm e}^{2 i \left (d x +c \right )}+1}}\, \sqrt {\frac {b \,{\mathrm e}^{i \left (d x +c \right )}}{{\mathrm e}^{2 i \left (d x +c \right )}+1}}\, \left (4 \cos \left (d x +c \right )+2 i \sin \left (d x +c \right )\right )}{3 \left ({\mathrm e}^{2 i \left (d x +c \right )}+1\right )^{2} d}\)	\(90\)

Verification of antiderivative is not currently implemented for this CAS.

[In]

int(sec(d*x+c)^(5/2)*(b*sec(d*x+c))^(3/2),x,method=_RETURNVERBOSE)

[Out]

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

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Maxima [B] Leaf count of result is larger than twice the leaf count of optimal. 299 vs. \(2 (62) = 124\).
time = 0.66, size = 299, normalized size = 4.15 \begin {gather*} -\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 {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate(sec(d*x+c)^(5/2)*(b*sec(d*x+c))^(3/2),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 = 2.85, size = 44, normalized size = 0.61 \begin {gather*} \frac {{\left (2 \, b \cos \left (d x + c\right )^{2} + b\right )} \sqrt {\frac {b}{\cos \left (d x + c\right )}} \sin \left (d x + c\right )}{3 \, d \cos \left (d x + c\right )^{\frac {5}{2}}} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

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Sympy [F(-2)]
time = 0.00, size = 0, normalized size = 0.00 \begin {gather*} \text {Exception raised: SystemError} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate(sec(d*x+c)**(5/2)*(b*sec(d*x+c))**(3/2),x)

[Out]

Exception raised: SystemError >> excessive stack use: stack is 8569 deep

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Giac [F]
time = 0.00, size = 0, normalized size = 0.00 \begin {gather*} \text {could not integrate} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

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

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Mupad [B]
time = 1.36, size = 127, normalized size = 1.76 \begin {gather*} \frac {2\,b\,\cos \left (c+d\,x\right )\,\sqrt {\frac {b}{\cos \left (c+d\,x\right )}}\,\sqrt {\frac {1}{\cos \left (c+d\,x\right )}}\,\left (4\,\sin \left (c+d\,x\right )+5\,\sin \left (3\,c+3\,d\,x\right )+\sin \left (5\,c+5\,d\,x\right )+\cos \left (c+d\,x\right )\,10{}\mathrm {i}+\cos \left (3\,c+3\,d\,x\right )\,5{}\mathrm {i}+\cos \left (5\,c+5\,d\,x\right )\,1{}\mathrm {i}\right )}{3\,d\,\left (10\,\cos \left (c+d\,x\right )+5\,\cos \left (3\,c+3\,d\,x\right )+\cos \left (5\,c+5\,d\,x\right )\right )} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

(2*b*cos(c + d*x)*(b/cos(c + d*x))^(1/2)*(1/cos(c + d*x))^(1/2)*(cos(c + d*x)*10i + 4*sin(c + d*x) + cos(3*c +
3*d*x)*5i + cos(5*c + 5*d*x)*1i + 5*sin(3*c + 3*d*x) + sin(5*c + 5*d*x)))/(3*d*(10*cos(c + d*x) + 5*cos(3*c +
3*d*x) + cos(5*c + 5*d*x)))

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