3.307 \(\int (d \sec (e+f x))^{5/2} (b \tan (e+f x))^{5/2} \, dx\)

Optimal. Leaf size=208 \[ \frac{3 b^{5/2} d^3 \sqrt{b \tan (e+f x)} \tan ^{-1}\left (\frac{\sqrt{b \sin (e+f x)}}{\sqrt{b}}\right )}{32 f \sqrt{b \sin (e+f x)} \sqrt{d \sec (e+f x)}}-\frac{3 b^{5/2} d^3 \sqrt{b \tan (e+f x)} \tanh ^{-1}\left (\frac{\sqrt{b \sin (e+f x)}}{\sqrt{b}}\right )}{32 f \sqrt{b \sin (e+f x)} \sqrt{d \sec (e+f x)}}-\frac{3 b d^2 (b \tan (e+f x))^{3/2} \sqrt{d \sec (e+f x)}}{16 f}+\frac{b (b \tan (e+f x))^{3/2} (d \sec (e+f x))^{5/2}}{4 f} \]

[Out]

(3*b^(5/2)*d^3*ArcTan[Sqrt[b*Sin[e + f*x]]/Sqrt[b]]*Sqrt[b*Tan[e + f*x]])/(32*f*Sqrt[d*Sec[e + f*x]]*Sqrt[b*Si
n[e + f*x]]) - (3*b^(5/2)*d^3*ArcTanh[Sqrt[b*Sin[e + f*x]]/Sqrt[b]]*Sqrt[b*Tan[e + f*x]])/(32*f*Sqrt[d*Sec[e +
f*x]]*Sqrt[b*Sin[e + f*x]]) - (3*b*d^2*Sqrt[d*Sec[e + f*x]]*(b*Tan[e + f*x])^(3/2))/(16*f) + (b*(d*Sec[e + f*
x])^(5/2)*(b*Tan[e + f*x])^(3/2))/(4*f)

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Rubi [A]  time = 0.228034, antiderivative size = 208, normalized size of antiderivative = 1., number of steps used = 8, number of rules used = 8, integrand size = 25, \(\frac{\text{number of rules}}{\text{integrand size}}\) = 0.32, Rules used = {2611, 2613, 2616, 2564, 329, 298, 203, 206} \[ \frac{3 b^{5/2} d^3 \sqrt{b \tan (e+f x)} \tan ^{-1}\left (\frac{\sqrt{b \sin (e+f x)}}{\sqrt{b}}\right )}{32 f \sqrt{b \sin (e+f x)} \sqrt{d \sec (e+f x)}}-\frac{3 b^{5/2} d^3 \sqrt{b \tan (e+f x)} \tanh ^{-1}\left (\frac{\sqrt{b \sin (e+f x)}}{\sqrt{b}}\right )}{32 f \sqrt{b \sin (e+f x)} \sqrt{d \sec (e+f x)}}-\frac{3 b d^2 (b \tan (e+f x))^{3/2} \sqrt{d \sec (e+f x)}}{16 f}+\frac{b (b \tan (e+f x))^{3/2} (d \sec (e+f x))^{5/2}}{4 f} \]

Antiderivative was successfully verified.

[In]

Int[(d*Sec[e + f*x])^(5/2)*(b*Tan[e + f*x])^(5/2),x]

[Out]

(3*b^(5/2)*d^3*ArcTan[Sqrt[b*Sin[e + f*x]]/Sqrt[b]]*Sqrt[b*Tan[e + f*x]])/(32*f*Sqrt[d*Sec[e + f*x]]*Sqrt[b*Si
n[e + f*x]]) - (3*b^(5/2)*d^3*ArcTanh[Sqrt[b*Sin[e + f*x]]/Sqrt[b]]*Sqrt[b*Tan[e + f*x]])/(32*f*Sqrt[d*Sec[e +
f*x]]*Sqrt[b*Sin[e + f*x]]) - (3*b*d^2*Sqrt[d*Sec[e + f*x]]*(b*Tan[e + f*x])^(3/2))/(16*f) + (b*(d*Sec[e + f*
x])^(5/2)*(b*Tan[e + f*x])^(3/2))/(4*f)

Rule 2611

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 + n - 1)), x] - Dist[(b^2*(n - 1))/(m + n - 1), Int[(a*Sec[e + f*x])
^m*(b*Tan[e + f*x])^(n - 2), x], x] /; FreeQ[{a, b, e, f, m}, x] && GtQ[n, 1] && NeQ[m + n - 1, 0] && Integers
Q[2*m, 2*n]

Rule 2613

Int[((a_.)*sec[(e_.) + (f_.)*(x_)])^(m_.)*((b_.)*tan[(e_.) + (f_.)*(x_)])^(n_), x_Symbol] :> Simp[(a^2*(a*Sec[
e + f*x])^(m - 2)*(b*Tan[e + f*x])^(n + 1))/(b*f*(m + n - 1)), x] + Dist[(a^2*(m - 2))/(m + n - 1), Int[(a*Sec
[e + f*x])^(m - 2)*(b*Tan[e + f*x])^n, x], x] /; FreeQ[{a, b, e, f, n}, x] && (GtQ[m, 1] || (EqQ[m, 1] && EqQ[
n, 1/2])) && NeQ[m + n - 1, 0] && IntegersQ[2*m, 2*n]

Rule 2616

Int[((a_.)*sec[(e_.) + (f_.)*(x_)])^(m_)*((b_.)*tan[(e_.) + (f_.)*(x_)])^(n_), x_Symbol] :> Dist[(a^(m + n)*(b
*Tan[e + f*x])^n)/((a*Sec[e + f*x])^n*(b*Sin[e + f*x])^n), Int[(b*Sin[e + f*x])^n/Cos[e + f*x]^(m + n), x], x]
/; FreeQ[{a, b, e, f, m, n}, x] && IntegerQ[n + 1/2] && IntegerQ[m + 1/2]

Rule 2564

Int[cos[(e_.) + (f_.)*(x_)]^(n_.)*((a_.)*sin[(e_.) + (f_.)*(x_)])^(m_.), x_Symbol] :> Dist[1/(a*f), Subst[Int[
x^m*(1 - x^2/a^2)^((n - 1)/2), x], x, a*Sin[e + f*x]], x] /; FreeQ[{a, e, f, m}, x] && IntegerQ[(n - 1)/2] &&
!(IntegerQ[(m - 1)/2] && LtQ[0, m, n])

Rule 329

Int[((c_.)*(x_))^(m_)*((a_) + (b_.)*(x_)^(n_))^(p_), x_Symbol] :> With[{k = Denominator[m]}, Dist[k/c, Subst[I
nt[x^(k*(m + 1) - 1)*(a + (b*x^(k*n))/c^n)^p, x], x, (c*x)^(1/k)], x]] /; FreeQ[{a, b, c, p}, x] && IGtQ[n, 0]
&& FractionQ[m] && IntBinomialQ[a, b, c, n, m, p, x]

Rule 298

Int[(x_)^2/((a_) + (b_.)*(x_)^4), x_Symbol] :> With[{r = Numerator[Rt[-(a/b), 2]], s = Denominator[Rt[-(a/b),
2]]}, Dist[s/(2*b), Int[1/(r + s*x^2), x], x] - Dist[s/(2*b), Int[1/(r - s*x^2), x], x]] /; FreeQ[{a, b}, x] &
&  !GtQ[a/b, 0]

Rule 203

Int[((a_) + (b_.)*(x_)^2)^(-1), x_Symbol] :> Simp[(1*ArcTan[(Rt[b, 2]*x)/Rt[a, 2]])/(Rt[a, 2]*Rt[b, 2]), x] /;
FreeQ[{a, b}, x] && PosQ[a/b] && (GtQ[a, 0] || GtQ[b, 0])

Rule 206

Int[((a_) + (b_.)*(x_)^2)^(-1), x_Symbol] :> Simp[(1*ArcTanh[(Rt[-b, 2]*x)/Rt[a, 2]])/(Rt[a, 2]*Rt[-b, 2]), x]
/; FreeQ[{a, b}, x] && NegQ[a/b] && (GtQ[a, 0] || LtQ[b, 0])

Rubi steps

\begin{align*} \int (d \sec (e+f x))^{5/2} (b \tan (e+f x))^{5/2} \, dx &=\frac{b (d \sec (e+f x))^{5/2} (b \tan (e+f x))^{3/2}}{4 f}-\frac{1}{8} \left (3 b^2\right ) \int (d \sec (e+f x))^{5/2} \sqrt{b \tan (e+f x)} \, dx\\ &=-\frac{3 b d^2 \sqrt{d \sec (e+f x)} (b \tan (e+f x))^{3/2}}{16 f}+\frac{b (d \sec (e+f x))^{5/2} (b \tan (e+f x))^{3/2}}{4 f}-\frac{1}{32} \left (3 b^2 d^2\right ) \int \sqrt{d \sec (e+f x)} \sqrt{b \tan (e+f x)} \, dx\\ &=-\frac{3 b d^2 \sqrt{d \sec (e+f x)} (b \tan (e+f x))^{3/2}}{16 f}+\frac{b (d \sec (e+f x))^{5/2} (b \tan (e+f x))^{3/2}}{4 f}-\frac{\left (3 b^2 d^3 \sqrt{b \tan (e+f x)}\right ) \int \sec (e+f x) \sqrt{b \sin (e+f x)} \, dx}{32 \sqrt{d \sec (e+f x)} \sqrt{b \sin (e+f x)}}\\ &=-\frac{3 b d^2 \sqrt{d \sec (e+f x)} (b \tan (e+f x))^{3/2}}{16 f}+\frac{b (d \sec (e+f x))^{5/2} (b \tan (e+f x))^{3/2}}{4 f}-\frac{\left (3 b d^3 \sqrt{b \tan (e+f x)}\right ) \operatorname{Subst}\left (\int \frac{\sqrt{x}}{1-\frac{x^2}{b^2}} \, dx,x,b \sin (e+f x)\right )}{32 f \sqrt{d \sec (e+f x)} \sqrt{b \sin (e+f x)}}\\ &=-\frac{3 b d^2 \sqrt{d \sec (e+f x)} (b \tan (e+f x))^{3/2}}{16 f}+\frac{b (d \sec (e+f x))^{5/2} (b \tan (e+f x))^{3/2}}{4 f}-\frac{\left (3 b d^3 \sqrt{b \tan (e+f x)}\right ) \operatorname{Subst}\left (\int \frac{x^2}{1-\frac{x^4}{b^2}} \, dx,x,\sqrt{b \sin (e+f x)}\right )}{16 f \sqrt{d \sec (e+f x)} \sqrt{b \sin (e+f x)}}\\ &=-\frac{3 b d^2 \sqrt{d \sec (e+f x)} (b \tan (e+f x))^{3/2}}{16 f}+\frac{b (d \sec (e+f x))^{5/2} (b \tan (e+f x))^{3/2}}{4 f}-\frac{\left (3 b^3 d^3 \sqrt{b \tan (e+f x)}\right ) \operatorname{Subst}\left (\int \frac{1}{b-x^2} \, dx,x,\sqrt{b \sin (e+f x)}\right )}{32 f \sqrt{d \sec (e+f x)} \sqrt{b \sin (e+f x)}}+\frac{\left (3 b^3 d^3 \sqrt{b \tan (e+f x)}\right ) \operatorname{Subst}\left (\int \frac{1}{b+x^2} \, dx,x,\sqrt{b \sin (e+f x)}\right )}{32 f \sqrt{d \sec (e+f x)} \sqrt{b \sin (e+f x)}}\\ &=\frac{3 b^{5/2} d^3 \tan ^{-1}\left (\frac{\sqrt{b \sin (e+f x)}}{\sqrt{b}}\right ) \sqrt{b \tan (e+f x)}}{32 f \sqrt{d \sec (e+f x)} \sqrt{b \sin (e+f x)}}-\frac{3 b^{5/2} d^3 \tanh ^{-1}\left (\frac{\sqrt{b \sin (e+f x)}}{\sqrt{b}}\right ) \sqrt{b \tan (e+f x)}}{32 f \sqrt{d \sec (e+f x)} \sqrt{b \sin (e+f x)}}-\frac{3 b d^2 \sqrt{d \sec (e+f x)} (b \tan (e+f x))^{3/2}}{16 f}+\frac{b (d \sec (e+f x))^{5/2} (b \tan (e+f x))^{3/2}}{4 f}\\ \end{align*}

Mathematica [A]  time = 3.14155, size = 189, normalized size = 0.91 \[ \frac{b^3 (d \sec (e+f x))^{5/2} \left (16 \sec ^{\frac{9}{2}}(e+f x)-28 \sec ^{\frac{5}{2}}(e+f x)+12 \sqrt{\sec (e+f x)}-6 \sqrt [4]{\tan ^2(e+f x)} \tan ^{-1}\left (\frac{\sqrt{\sec (e+f x)}}{\sqrt [4]{\tan ^2(e+f x)}}\right )+3 \sqrt [4]{\tan ^2(e+f x)} \left (\log \left (1-\frac{\sqrt{\sec (e+f x)}}{\sqrt [4]{\tan ^2(e+f x)}}\right )-\log \left (\frac{\sqrt{\sec (e+f x)}}{\sqrt [4]{\tan ^2(e+f x)}}+1\right )\right )\right )}{64 f \sec ^{\frac{5}{2}}(e+f x) \sqrt{b \tan (e+f x)}} \]

Antiderivative was successfully verified.

[In]

Integrate[(d*Sec[e + f*x])^(5/2)*(b*Tan[e + f*x])^(5/2),x]

[Out]

(b^3*(d*Sec[e + f*x])^(5/2)*(12*Sqrt[Sec[e + f*x]] - 28*Sec[e + f*x]^(5/2) + 16*Sec[e + f*x]^(9/2) - 6*ArcTan[
Sqrt[Sec[e + f*x]]/(Tan[e + f*x]^2)^(1/4)]*(Tan[e + f*x]^2)^(1/4) + 3*(Log[1 - Sqrt[Sec[e + f*x]]/(Tan[e + f*x
]^2)^(1/4)] - Log[1 + Sqrt[Sec[e + f*x]]/(Tan[e + f*x]^2)^(1/4)])*(Tan[e + f*x]^2)^(1/4)))/(64*f*Sec[e + f*x]^
(5/2)*Sqrt[b*Tan[e + f*x]])

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Maple [C]  time = 0.229, size = 628, normalized size = 3. \begin{align*} \text{result too large to display} \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

int((d*sec(f*x+e))^(5/2)*(b*tan(f*x+e))^(5/2),x)

[Out]

1/64/f*2^(1/2)*(3*I*cos(f*x+e)^4*((I*cos(f*x+e)-I+sin(f*x+e))/sin(f*x+e))^(1/2)*(-(I*cos(f*x+e)-I-sin(f*x+e))/
sin(f*x+e))^(1/2)*(-I*(cos(f*x+e)-1)/sin(f*x+e))^(1/2)*EllipticPi(((I*cos(f*x+e)-I+sin(f*x+e))/sin(f*x+e))^(1/
2),1/2+1/2*I,1/2*2^(1/2))-3*I*cos(f*x+e)^4*((I*cos(f*x+e)-I+sin(f*x+e))/sin(f*x+e))^(1/2)*(-(I*cos(f*x+e)-I-si
n(f*x+e))/sin(f*x+e))^(1/2)*(-I*(cos(f*x+e)-1)/sin(f*x+e))^(1/2)*EllipticPi(((I*cos(f*x+e)-I+sin(f*x+e))/sin(f
*x+e))^(1/2),1/2-1/2*I,1/2*2^(1/2))+3*cos(f*x+e)^4*((I*cos(f*x+e)-I+sin(f*x+e))/sin(f*x+e))^(1/2)*(-(I*cos(f*x
+e)-I-sin(f*x+e))/sin(f*x+e))^(1/2)*(-I*(cos(f*x+e)-1)/sin(f*x+e))^(1/2)*EllipticPi(((I*cos(f*x+e)-I+sin(f*x+e
))/sin(f*x+e))^(1/2),1/2+1/2*I,1/2*2^(1/2))+3*cos(f*x+e)^4*((I*cos(f*x+e)-I+sin(f*x+e))/sin(f*x+e))^(1/2)*(-(I
*cos(f*x+e)-I-sin(f*x+e))/sin(f*x+e))^(1/2)*(-I*(cos(f*x+e)-1)/sin(f*x+e))^(1/2)*EllipticPi(((I*cos(f*x+e)-I+s
in(f*x+e))/sin(f*x+e))^(1/2),1/2-1/2*I,1/2*2^(1/2))-6*cos(f*x+e)^3*2^(1/2)+6*cos(f*x+e)^2*2^(1/2)+8*cos(f*x+e)
*2^(1/2)-8*2^(1/2))*cos(f*x+e)*(d/cos(f*x+e))^(5/2)*(b*sin(f*x+e)/cos(f*x+e))^(5/2)/(cos(f*x+e)-1)/sin(f*x+e)

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Maxima [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((d*sec(f*x+e))^(5/2)*(b*tan(f*x+e))^(5/2),x, algorithm="maxima")

[Out]

Timed out

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Fricas [B]  time = 3.56756, size = 2182, normalized size = 10.49 \begin{align*} \text{result too large to display} \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate((d*sec(f*x+e))^(5/2)*(b*tan(f*x+e))^(5/2),x, algorithm="fricas")

[Out]

[1/256*(6*sqrt(-b*d)*b^2*d^2*arctan(1/4*(cos(f*x + e)^3 - 5*cos(f*x + e)^2 - (cos(f*x + e)^2 + 6*cos(f*x + e)
+ 4)*sin(f*x + e) - 2*cos(f*x + e) + 4)*sqrt(-b*d)*sqrt(b*sin(f*x + e)/cos(f*x + e))*sqrt(d/cos(f*x + e))/(b*d
*cos(f*x + e)^2 - b*d - (b*d*cos(f*x + e) + b*d)*sin(f*x + e)))*cos(f*x + e)^3 + 3*sqrt(-b*d)*b^2*d^2*cos(f*x
+ e)^3*log((b*d*cos(f*x + e)^4 - 72*b*d*cos(f*x + e)^2 + 8*(7*cos(f*x + e)^3 - (cos(f*x + e)^3 - 8*cos(f*x + e
))*sin(f*x + e) - 8*cos(f*x + e))*sqrt(-b*d)*sqrt(b*sin(f*x + e)/cos(f*x + e))*sqrt(d/cos(f*x + e)) + 72*b*d +
28*(b*d*cos(f*x + e)^2 - 2*b*d)*sin(f*x + e))/(cos(f*x + e)^4 - 8*cos(f*x + e)^2 - 4*(cos(f*x + e)^2 - 2)*sin
(f*x + e) + 8)) - 16*(3*b^2*d^2*cos(f*x + e)^2 - 4*b^2*d^2)*sqrt(b*sin(f*x + e)/cos(f*x + e))*sqrt(d/cos(f*x +
e))*sin(f*x + e))/(f*cos(f*x + e)^3), 1/256*(6*sqrt(b*d)*b^2*d^2*arctan(1/4*(cos(f*x + e)^3 - 5*cos(f*x + e)^
2 + (cos(f*x + e)^2 + 6*cos(f*x + e) + 4)*sin(f*x + e) - 2*cos(f*x + e) + 4)*sqrt(b*d)*sqrt(b*sin(f*x + e)/cos
(f*x + e))*sqrt(d/cos(f*x + e))/(b*d*cos(f*x + e)^2 - b*d + (b*d*cos(f*x + e) + b*d)*sin(f*x + e)))*cos(f*x +
e)^3 + 3*sqrt(b*d)*b^2*d^2*cos(f*x + e)^3*log((b*d*cos(f*x + e)^4 - 72*b*d*cos(f*x + e)^2 + 8*(7*cos(f*x + e)^
3 + (cos(f*x + e)^3 - 8*cos(f*x + e))*sin(f*x + e) - 8*cos(f*x + e))*sqrt(b*d)*sqrt(b*sin(f*x + e)/cos(f*x + e
))*sqrt(d/cos(f*x + e)) + 72*b*d - 28*(b*d*cos(f*x + e)^2 - 2*b*d)*sin(f*x + e))/(cos(f*x + e)^4 - 8*cos(f*x +
e)^2 + 4*(cos(f*x + e)^2 - 2)*sin(f*x + e) + 8)) - 16*(3*b^2*d^2*cos(f*x + e)^2 - 4*b^2*d^2)*sqrt(b*sin(f*x +
e)/cos(f*x + e))*sqrt(d/cos(f*x + e))*sin(f*x + e))/(f*cos(f*x + e)^3)]

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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((d*sec(f*x+e))**(5/2)*(b*tan(f*x+e))**(5/2),x)

[Out]

Timed out

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Giac [F]  time = 0., size = 0, normalized size = 0. \begin{align*} \int \left (d \sec \left (f x + e\right )\right )^{\frac{5}{2}} \left (b \tan \left (f x + e\right )\right )^{\frac{5}{2}}\,{d x} \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate((d*sec(f*x+e))^(5/2)*(b*tan(f*x+e))^(5/2),x, algorithm="giac")

[Out]

integrate((d*sec(f*x + e))^(5/2)*(b*tan(f*x + e))^(5/2), x)