3.133 \(\int \sec ^{\frac{7}{2}}(c+d x) \sqrt{b \sec (c+d x)} \, dx\)

Optimal. Leaf size=70 \[ \frac{\sin ^3(c+d x) \sec ^{\frac{5}{2}}(c+d x) \sqrt{b \sec (c+d x)}}{3 d}+\frac{\sin (c+d x) \sqrt{\sec (c+d x)} \sqrt{b \sec (c+d x)}}{d} \]

[Out]

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

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Rubi [A]  time = 0.0165725, antiderivative size = 70, 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} \[ \frac{\sin ^3(c+d x) \sec ^{\frac{5}{2}}(c+d x) \sqrt{b \sec (c+d x)}}{3 d}+\frac{\sin (c+d x) \sqrt{\sec (c+d x)} \sqrt{b \sec (c+d x)}}{d} \]

Antiderivative was successfully verified.

[In]

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

[Out]

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

Mathematica [A]  time = 0.0867656, size = 45, normalized size = 0.64 \[ \frac{\left (\frac{1}{3} \tan ^3(c+d x)+\tan (c+d x)\right ) \sqrt{b \sec (c+d x)}}{d \sqrt{\sec (c+d x)}} \]

Antiderivative was successfully verified.

[In]

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

[Out]

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

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Maple [A]  time = 0.129, size = 52, normalized size = 0.7 \begin{align*}{\frac{ \left ( 2\, \left ( \cos \left ( dx+c \right ) \right ) ^{2}+1 \right ) \cos \left ( dx+c \right ) \sin \left ( dx+c \right ) }{3\,d} \left ( \left ( \cos \left ( dx+c \right ) \right ) ^{-1} \right ) ^{{\frac{7}{2}}}\sqrt{{\frac{b}{\cos \left ( dx+c \right ) }}}} \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

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

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Maxima [B]  time = 2.07997, size = 397, normalized size = 5.67 \begin{align*} \frac{4 \,{\left ({\left (3 \, \cos \left (2 \, d x + 2 \, c\right ) + 1\right )} \sin \left (6 \, d x + 6 \, c\right ) + 3 \,{\left (3 \, \cos \left (2 \, d x + 2 \, c\right ) + 1\right )} \sin \left (4 \, d x + 4 \, c\right ) - 3 \, \cos \left (6 \, d x + 6 \, c\right ) \sin \left (2 \, d x + 2 \, c\right ) - 9 \, \cos \left (4 \, d x + 4 \, c\right ) \sin \left (2 \, d x + 2 \, 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(sec(d*x+c)^(7/2)*(b*sec(d*x+c))^(1/2),x, algorithm="maxima")

[Out]

4/3*((3*cos(2*d*x + 2*c) + 1)*sin(6*d*x + 6*c) + 3*(3*cos(2*d*x + 2*c) + 1)*sin(4*d*x + 4*c) - 3*cos(6*d*x + 6
*c)*sin(2*d*x + 2*c) - 9*cos(4*d*x + 4*c)*sin(2*d*x + 2*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 = 1.41191, size = 115, normalized size = 1.64 \begin{align*} \frac{{\left (2 \, \cos \left (d x + c\right )^{2} + 1\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{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate(sec(d*x+c)^(7/2)*(b*sec(d*x+c))^(1/2),x, algorithm="fricas")

[Out]

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

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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(sec(d*x+c)**(7/2)*(b*sec(d*x+c))**(1/2),x)

[Out]

Timed out

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Giac [F]  time = 0., size = 0, normalized size = 0. \begin{align*} \int \sqrt{b \sec \left (d x + c\right )} \sec \left (d x + c\right )^{\frac{7}{2}}\,{d x} \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate(sec(d*x+c)^(7/2)*(b*sec(d*x+c))^(1/2),x, algorithm="giac")

[Out]

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