3.3.38 \(\int \sec ^2(a+b x) (d \tan (a+b x))^{3/2} \, dx\) [238]

Optimal. Leaf size=22 \[ \frac {2 (d \tan (a+b x))^{5/2}}{5 b d} \]

[Out]

2/5*(d*tan(b*x+a))^(5/2)/b/d

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Rubi [A]
time = 0.03, antiderivative size = 22, normalized size of antiderivative = 1.00, number of steps used = 2, number of rules used = 2, integrand size = 21, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.095, Rules used = {2687, 32} \begin {gather*} \frac {2 (d \tan (a+b x))^{5/2}}{5 b d} \end {gather*}

Antiderivative was successfully verified.

[In]

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

[Out]

(2*(d*Tan[a + b*x])^(5/2))/(5*b*d)

Rule 32

Int[((a_.) + (b_.)*(x_))^(m_), x_Symbol] :> Simp[(a + b*x)^(m + 1)/(b*(m + 1)), x] /; FreeQ[{a, b, m}, x] && N
eQ[m, -1]

Rule 2687

Int[sec[(e_.) + (f_.)*(x_)]^(m_)*((b_.)*tan[(e_.) + (f_.)*(x_)])^(n_.), x_Symbol] :> Dist[1/f, Subst[Int[(b*x)
^n*(1 + x^2)^(m/2 - 1), x], x, Tan[e + f*x]], x] /; FreeQ[{b, e, f, n}, x] && IntegerQ[m/2] &&  !(IntegerQ[(n
- 1)/2] && LtQ[0, n, m - 1])

Rubi steps

\begin {align*} \int \sec ^2(a+b x) (d \tan (a+b x))^{3/2} \, dx &=\frac {\text {Subst}\left (\int (d x)^{3/2} \, dx,x,\tan (a+b x)\right )}{b}\\ &=\frac {2 (d \tan (a+b x))^{5/2}}{5 b d}\\ \end {align*}

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Mathematica [A]
time = 0.06, size = 22, normalized size = 1.00 \begin {gather*} \frac {2 (d \tan (a+b x))^{5/2}}{5 b d} \end {gather*}

Antiderivative was successfully verified.

[In]

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

[Out]

(2*(d*Tan[a + b*x])^(5/2))/(5*b*d)

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Maple [A]
time = 0.12, size = 19, normalized size = 0.86

method	result	size
derivativedivides	\(\frac {2 \left (d \tan \left (b x +a \right )\right )^{\frac {5}{2}}}{5 b d}\)	\(19\)
default	\(\frac {2 \left (d \tan \left (b x +a \right )\right )^{\frac {5}{2}}}{5 b d}\)	\(19\)

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

2/5*(d*tan(b*x+a))^(5/2)/b/d

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Maxima [A]
time = 0.28, size = 18, normalized size = 0.82 \begin {gather*} \frac {2 \, \left (d \tan \left (b x + a\right )\right )^{\frac {5}{2}}}{5 \, b d} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

2/5*(d*tan(b*x + a))^(5/2)/(b*d)

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Fricas [B] Leaf count of result is larger than twice the leaf count of optimal. 45 vs. \(2 (18) = 36\).
time = 0.38, size = 45, normalized size = 2.05 \begin {gather*} -\frac {2 \, {\left (d \cos \left (b x + a\right )^{2} - d\right )} \sqrt {\frac {d \sin \left (b x + a\right )}{\cos \left (b x + a\right )}}}{5 \, b \cos \left (b x + a\right )^{2}} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

-2/5*(d*cos(b*x + a)^2 - d)*sqrt(d*sin(b*x + a)/cos(b*x + a))/(b*cos(b*x + a)^2)

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Sympy [F]
time = 0.00, size = 0, normalized size = 0.00 \begin {gather*} \int \left (d \tan {\left (a + b x \right )}\right )^{\frac {3}{2}} \sec ^{2}{\left (a + b x \right )}\, dx \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate(sec(b*x+a)**2*(d*tan(b*x+a))**(3/2),x)

[Out]

Integral((d*tan(a + b*x))**(3/2)*sec(a + b*x)**2, x)

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Giac [A]
time = 0.48, size = 24, normalized size = 1.09 \begin {gather*} \frac {2 \, \sqrt {d \tan \left (b x + a\right )} d \tan \left (b x + a\right )^{2}}{5 \, b} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

2/5*sqrt(d*tan(b*x + a))*d*tan(b*x + a)^2/b

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Mupad [B]
time = 3.57, size = 100, normalized size = 4.55 \begin {gather*} \frac {2\,d\,\sqrt {\frac {d\,\sin \left (2\,a+2\,b\,x\right )}{\cos \left (2\,a+2\,b\,x\right )+1}}\,\left (\cos \left (2\,a+2\,b\,x\right )-2\,\cos \left (4\,a+4\,b\,x\right )-\cos \left (6\,a+6\,b\,x\right )+2\right )}{5\,b\,\left (15\,\cos \left (2\,a+2\,b\,x\right )+6\,\cos \left (4\,a+4\,b\,x\right )+\cos \left (6\,a+6\,b\,x\right )+10\right )} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

int((d*tan(a + b*x))^(3/2)/cos(a + b*x)^2,x)

[Out]

(2*d*((d*sin(2*a + 2*b*x))/(cos(2*a + 2*b*x) + 1))^(1/2)*(cos(2*a + 2*b*x) - 2*cos(4*a + 4*b*x) - cos(6*a + 6*
b*x) + 2))/(5*b*(15*cos(2*a + 2*b*x) + 6*cos(4*a + 4*b*x) + cos(6*a + 6*b*x) + 10))

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