3.4.6 \(\int \frac {(b \tan (e+f x))^{3/2}}{(d \sec (e+f x))^{9/2}} \, dx\) [306]

Optimal. Leaf size=103 \[ -\frac {2 b \sqrt {b \tan (e+f x)}}{9 f (d \sec (e+f x))^{9/2}}+\frac {2 b \sqrt {b \tan (e+f x)}}{45 d^2 f (d \sec (e+f x))^{5/2}}+\frac {8 b \sqrt {b \tan (e+f x)}}{45 d^4 f \sqrt {d \sec (e+f x)}} \]

[Out]

-2/9*b*(b*tan(f*x+e))^(1/2)/f/(d*sec(f*x+e))^(9/2)+2/45*b*(b*tan(f*x+e))^(1/2)/d^2/f/(d*sec(f*x+e))^(5/2)+8/45
*b*(b*tan(f*x+e))^(1/2)/d^4/f/(d*sec(f*x+e))^(1/2)

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Rubi [A]
time = 0.11, antiderivative size = 103, normalized size of antiderivative = 1.00, number of steps used = 3, number of rules used = 3, integrand size = 25, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.120, Rules used = {2690, 2692, 2685} \begin {gather*} \frac {8 b \sqrt {b \tan (e+f x)}}{45 d^4 f \sqrt {d \sec (e+f x)}}+\frac {2 b \sqrt {b \tan (e+f x)}}{45 d^2 f (d \sec (e+f x))^{5/2}}-\frac {2 b \sqrt {b \tan (e+f x)}}{9 f (d \sec (e+f x))^{9/2}} \end {gather*}

Antiderivative was successfully verified.

[In]

Int[(b*Tan[e + f*x])^(3/2)/(d*Sec[e + f*x])^(9/2),x]

[Out]

(-2*b*Sqrt[b*Tan[e + f*x]])/(9*f*(d*Sec[e + f*x])^(9/2)) + (2*b*Sqrt[b*Tan[e + f*x]])/(45*d^2*f*(d*Sec[e + f*x
])^(5/2)) + (8*b*Sqrt[b*Tan[e + f*x]])/(45*d^4*f*Sqrt[d*Sec[e + f*x]])

Rule 2685

Int[((a_.)*sec[(e_.) + (f_.)*(x_)])^(m_.)*((b_.)*tan[(e_.) + (f_.)*(x_)])^(n_.), x_Symbol] :> Simp[(-(a*Sec[e
+ f*x])^m)*((b*Tan[e + f*x])^(n + 1)/(b*f*m)), x] /; FreeQ[{a, b, e, f, m, n}, x] && EqQ[m + n + 1, 0]

Rule 2690

Int[((a_.)*sec[(e_.) + (f_.)*(x_)])^(m_)*((b_.)*tan[(e_.) + (f_.)*(x_)])^(n_), x_Symbol] :> Simp[b*(a*Sec[e +
f*x])^m*((b*Tan[e + f*x])^(n - 1)/(f*m)), x] - Dist[b^2*((n - 1)/(a^2*m)), Int[(a*Sec[e + f*x])^(m + 2)*(b*Tan
[e + f*x])^(n - 2), x], x] /; FreeQ[{a, b, e, f}, x] && GtQ[n, 1] && (LtQ[m, -1] || (EqQ[m, -1] && EqQ[n, 3/2]
)) && IntegersQ[2*m, 2*n]

Rule 2692

Int[((a_.)*sec[(e_.) + (f_.)*(x_)])^(m_)*((b_.)*tan[(e_.) + (f_.)*(x_)])^(n_), x_Symbol] :> Simp[(-(a*Sec[e +
f*x])^m)*((b*Tan[e + f*x])^(n + 1)/(b*f*m)), x] + Dist[(m + n + 1)/(a^2*m), Int[(a*Sec[e + f*x])^(m + 2)*(b*Ta
n[e + f*x])^n, x], x] /; FreeQ[{a, b, e, f, n}, x] && (LtQ[m, -1] || (EqQ[m, -1] && EqQ[n, -2^(-1)])) && Integ
ersQ[2*m, 2*n]

Rubi steps

\begin {align*} \int \frac {(b \tan (e+f x))^{3/2}}{(d \sec (e+f x))^{9/2}} \, dx &=-\frac {2 b \sqrt {b \tan (e+f x)}}{9 f (d \sec (e+f x))^{9/2}}+\frac {b^2 \int \frac {1}{(d \sec (e+f x))^{5/2} \sqrt {b \tan (e+f x)}} \, dx}{9 d^2}\\ &=-\frac {2 b \sqrt {b \tan (e+f x)}}{9 f (d \sec (e+f x))^{9/2}}+\frac {2 b \sqrt {b \tan (e+f x)}}{45 d^2 f (d \sec (e+f x))^{5/2}}+\frac {\left (4 b^2\right ) \int \frac {1}{\sqrt {d \sec (e+f x)} \sqrt {b \tan (e+f x)}} \, dx}{45 d^4}\\ &=-\frac {2 b \sqrt {b \tan (e+f x)}}{9 f (d \sec (e+f x))^{9/2}}+\frac {2 b \sqrt {b \tan (e+f x)}}{45 d^2 f (d \sec (e+f x))^{5/2}}+\frac {8 b \sqrt {b \tan (e+f x)}}{45 d^4 f \sqrt {d \sec (e+f x)}}\\ \end {align*}

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Mathematica [A]
time = 3.50, size = 158, normalized size = 1.53 \begin {gather*} -\frac {b \sqrt {\sec (e+f x)} \left (16 \sqrt {\frac {1}{1+\cos (e+f x)}} \sqrt {\sec (e+f x)}+\sqrt {\frac {1}{1+\cos (e+f x)}} (21 \cos (3 (e+f x))+5 \cos (5 (e+f x))) \sec ^{\frac {3}{2}}(e+f x)-21 \sec ^2\left (\frac {1}{2} (e+f x)\right ) \sqrt {1+\sec (e+f x)}\right ) \sqrt {b \tan (e+f x)}}{360 d^3 f \sqrt {\frac {1}{1+\cos (e+f x)}} (d \sec (e+f x))^{3/2}} \end {gather*}

Antiderivative was successfully verified.

[In]

Integrate[(b*Tan[e + f*x])^(3/2)/(d*Sec[e + f*x])^(9/2),x]

[Out]

-1/360*(b*Sqrt[Sec[e + f*x]]*(16*Sqrt[(1 + Cos[e + f*x])^(-1)]*Sqrt[Sec[e + f*x]] + Sqrt[(1 + Cos[e + f*x])^(-
1)]*(21*Cos[3*(e + f*x)] + 5*Cos[5*(e + f*x)])*Sec[e + f*x]^(3/2) - 21*Sec[(e + f*x)/2]^2*Sqrt[1 + Sec[e + f*x
]])*Sqrt[b*Tan[e + f*x]])/(d^3*f*Sqrt[(1 + Cos[e + f*x])^(-1)]*(d*Sec[e + f*x])^(3/2))

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Maple [A]
time = 0.36, size = 62, normalized size = 0.60

method	result	size
default	\(\frac {2 \left (5 \left (\cos ^{2}\left (f x +e \right )\right )+4\right ) \sin \left (f x +e \right ) \left (\frac {b \sin \left (f x +e \right )}{\cos \left (f x +e \right )}\right )^{\frac {3}{2}}}{45 f \left (\frac {d}{\cos \left (f x +e \right )}\right )^{\frac {9}{2}} \cos \left (f x +e \right )^{3}}\)	\(62\)

Verification of antiderivative is not currently implemented for this CAS.

[In]

int((b*tan(f*x+e))^(3/2)/(d*sec(f*x+e))^(9/2),x,method=_RETURNVERBOSE)

[Out]

2/45/f*(5*cos(f*x+e)^2+4)*sin(f*x+e)*(b*sin(f*x+e)/cos(f*x+e))^(3/2)/(d/cos(f*x+e))^(9/2)/cos(f*x+e)^3

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Maxima [F]
time = 0.00, size = 0, normalized size = 0.00 \begin {gather*} \text {Failed to integrate} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate((b*tan(f*x+e))^(3/2)/(d*sec(f*x+e))^(9/2),x, algorithm="maxima")

[Out]

integrate((b*tan(f*x + e))^(3/2)/(d*sec(f*x + e))^(9/2), x)

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Fricas [A]
time = 0.42, size = 76, normalized size = 0.74 \begin {gather*} -\frac {2 \, {\left (5 \, b \cos \left (f x + e\right )^{5} - b \cos \left (f x + e\right )^{3} - 4 \, b \cos \left (f x + e\right )\right )} \sqrt {\frac {b \sin \left (f x + e\right )}{\cos \left (f x + e\right )}} \sqrt {\frac {d}{\cos \left (f x + e\right )}}}{45 \, d^{5} f} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate((b*tan(f*x+e))^(3/2)/(d*sec(f*x+e))^(9/2),x, algorithm="fricas")

[Out]

-2/45*(5*b*cos(f*x + e)^5 - b*cos(f*x + e)^3 - 4*b*cos(f*x + e))*sqrt(b*sin(f*x + e)/cos(f*x + e))*sqrt(d/cos(
f*x + e))/(d^5*f)

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Sympy [F(-1)] Timed out
time = 0.00, size = 0, normalized size = 0.00 \begin {gather*} \text {Timed out} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate((b*tan(f*x+e))**(3/2)/(d*sec(f*x+e))**(9/2),x)

[Out]

Timed out

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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((b*tan(f*x+e))^(3/2)/(d*sec(f*x+e))^(9/2),x, algorithm="giac")

[Out]

integrate((b*tan(f*x + e))^(3/2)/(d*sec(f*x + e))^(9/2), x)

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Mupad [B]
time = 3.52, size = 78, normalized size = 0.76 \begin {gather*} -\frac {b\,\sqrt {\frac {d}{\cos \left (e+f\,x\right )}}\,\sqrt {\frac {b\,\sin \left (2\,e+2\,f\,x\right )}{\cos \left (2\,e+2\,f\,x\right )+1}}\,\left (21\,\cos \left (3\,e+3\,f\,x\right )-26\,\cos \left (e+f\,x\right )+5\,\cos \left (5\,e+5\,f\,x\right )\right )}{360\,d^5\,f} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

int((b*tan(e + f*x))^(3/2)/(d/cos(e + f*x))^(9/2),x)

[Out]

-(b*(d/cos(e + f*x))^(1/2)*((b*sin(2*e + 2*f*x))/(cos(2*e + 2*f*x) + 1))^(1/2)*(21*cos(3*e + 3*f*x) - 26*cos(e
+ f*x) + 5*cos(5*e + 5*f*x)))/(360*d^5*f)

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