3.328 \(\int \frac {(d \sec (e+f x))^{7/2}}{(b \tan (e+f x))^{5/2}} \, dx\)
Optimal. Leaf size=172 \[ \frac {d^3 \sqrt {b \sin (e+f x)} \sqrt {d \sec (e+f x)} \tan ^{-1}\left (\frac {\sqrt {b \sin (e+f x)}}{\sqrt {b}}\right )}{b^{5/2} f \sqrt {b \tan (e+f x)}}+\frac {d^3 \sqrt {b \sin (e+f x)} \sqrt {d \sec (e+f x)} \tanh ^{-1}\left (\frac {\sqrt {b \sin (e+f x)}}{\sqrt {b}}\right )}{b^{5/2} f \sqrt {b \tan (e+f x)}}-\frac {2 d^2 (d \sec (e+f x))^{3/2}}{3 b f (b \tan (e+f x))^{3/2}} \]
[Out]
d^3*arctan((b*sin(f*x+e))^(1/2)/b^(1/2))*(d*sec(f*x+e))^(1/2)*(b*sin(f*x+e))^(1/2)/b^(5/2)/f/(b*tan(f*x+e))^(1
/2)+d^3*arctanh((b*sin(f*x+e))^(1/2)/b^(1/2))*(d*sec(f*x+e))^(1/2)*(b*sin(f*x+e))^(1/2)/b^(5/2)/f/(b*tan(f*x+e
))^(1/2)-2/3*d^2*(d*sec(f*x+e))^(3/2)/b/f/(b*tan(f*x+e))^(3/2)
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Rubi [A] time = 0.18, antiderivative size = 172, normalized size of antiderivative = 1.00,
number of steps used = 7, number of rules used = 7, integrand size = 25, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.280, Rules used =
{2608, 2616, 2564, 329, 212, 206, 203} \[ \frac {d^3 \sqrt {b \sin (e+f x)} \sqrt {d \sec (e+f x)} \tan ^{-1}\left (\frac {\sqrt {b \sin (e+f x)}}{\sqrt {b}}\right )}{b^{5/2} f \sqrt {b \tan (e+f x)}}+\frac {d^3 \sqrt {b \sin (e+f x)} \sqrt {d \sec (e+f x)} \tanh ^{-1}\left (\frac {\sqrt {b \sin (e+f x)}}{\sqrt {b}}\right )}{b^{5/2} f \sqrt {b \tan (e+f x)}}-\frac {2 d^2 (d \sec (e+f x))^{3/2}}{3 b f (b \tan (e+f x))^{3/2}} \]
Antiderivative was successfully verified.
[In]
Int[(d*Sec[e + f*x])^(7/2)/(b*Tan[e + f*x])^(5/2),x]
[Out]
(-2*d^2*(d*Sec[e + f*x])^(3/2))/(3*b*f*(b*Tan[e + f*x])^(3/2)) + (d^3*ArcTan[Sqrt[b*Sin[e + f*x]]/Sqrt[b]]*Sqr
t[d*Sec[e + f*x]]*Sqrt[b*Sin[e + f*x]])/(b^(5/2)*f*Sqrt[b*Tan[e + f*x]]) + (d^3*ArcTanh[Sqrt[b*Sin[e + f*x]]/S
qrt[b]]*Sqrt[d*Sec[e + f*x]]*Sqrt[b*Sin[e + f*x]])/(b^(5/2)*f*Sqrt[b*Tan[e + f*x]])
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])
Rule 212
Int[((a_) + (b_.)*(x_)^4)^(-1), x_Symbol] :> With[{r = Numerator[Rt[-(a/b), 2]], s = Denominator[Rt[-(a/b), 2]
]}, Dist[r/(2*a), Int[1/(r - s*x^2), x], x] + Dist[r/(2*a), Int[1/(r + s*x^2), x], x]] /; FreeQ[{a, b}, x] &&
!GtQ[a/b, 0]
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 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 2608
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*(n + 1)), x] - Dist[(a^2*(m - 2))/(b^2*(n + 1)), Int[(a*Sec[e
+ f*x])^(m - 2)*(b*Tan[e + f*x])^(n + 2), x], x] /; FreeQ[{a, b, e, f}, x] && LtQ[n, -1] && (GtQ[m, 1] || (Eq
Q[m, 1] && EqQ[n, -3/2])) && 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]
Rubi steps
\begin {align*} \int \frac {(d \sec (e+f x))^{7/2}}{(b \tan (e+f x))^{5/2}} \, dx &=-\frac {2 d^2 (d \sec (e+f x))^{3/2}}{3 b f (b \tan (e+f x))^{3/2}}+\frac {d^2 \int \frac {(d \sec (e+f x))^{3/2}}{\sqrt {b \tan (e+f x)}} \, dx}{b^2}\\ &=-\frac {2 d^2 (d \sec (e+f x))^{3/2}}{3 b f (b \tan (e+f x))^{3/2}}+\frac {\left (d^3 \sqrt {d \sec (e+f x)} \sqrt {b \sin (e+f x)}\right ) \int \frac {\sec (e+f x)}{\sqrt {b \sin (e+f x)}} \, dx}{b^2 \sqrt {b \tan (e+f x)}}\\ &=-\frac {2 d^2 (d \sec (e+f x))^{3/2}}{3 b f (b \tan (e+f x))^{3/2}}+\frac {\left (d^3 \sqrt {d \sec (e+f x)} \sqrt {b \sin (e+f x)}\right ) \operatorname {Subst}\left (\int \frac {1}{\sqrt {x} \left (1-\frac {x^2}{b^2}\right )} \, dx,x,b \sin (e+f x)\right )}{b^3 f \sqrt {b \tan (e+f x)}}\\ &=-\frac {2 d^2 (d \sec (e+f x))^{3/2}}{3 b f (b \tan (e+f x))^{3/2}}+\frac {\left (2 d^3 \sqrt {d \sec (e+f x)} \sqrt {b \sin (e+f x)}\right ) \operatorname {Subst}\left (\int \frac {1}{1-\frac {x^4}{b^2}} \, dx,x,\sqrt {b \sin (e+f x)}\right )}{b^3 f \sqrt {b \tan (e+f x)}}\\ &=-\frac {2 d^2 (d \sec (e+f x))^{3/2}}{3 b f (b \tan (e+f x))^{3/2}}+\frac {\left (d^3 \sqrt {d \sec (e+f x)} \sqrt {b \sin (e+f x)}\right ) \operatorname {Subst}\left (\int \frac {1}{b-x^2} \, dx,x,\sqrt {b \sin (e+f x)}\right )}{b^2 f \sqrt {b \tan (e+f x)}}+\frac {\left (d^3 \sqrt {d \sec (e+f x)} \sqrt {b \sin (e+f x)}\right ) \operatorname {Subst}\left (\int \frac {1}{b+x^2} \, dx,x,\sqrt {b \sin (e+f x)}\right )}{b^2 f \sqrt {b \tan (e+f x)}}\\ &=-\frac {2 d^2 (d \sec (e+f x))^{3/2}}{3 b f (b \tan (e+f x))^{3/2}}+\frac {d^3 \tan ^{-1}\left (\frac {\sqrt {b \sin (e+f x)}}{\sqrt {b}}\right ) \sqrt {d \sec (e+f x)} \sqrt {b \sin (e+f x)}}{b^{5/2} f \sqrt {b \tan (e+f x)}}+\frac {d^3 \tanh ^{-1}\left (\frac {\sqrt {b \sin (e+f x)}}{\sqrt {b}}\right ) \sqrt {d \sec (e+f x)} \sqrt {b \sin (e+f x)}}{b^{5/2} f \sqrt {b \tan (e+f x)}}\\ \end {align*}
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Mathematica [A] time = 1.21, size = 144, normalized size = 0.84 \[ -\frac {d^4 \sqrt {b \tan (e+f x)} \left (2 \sqrt [4]{\tan ^2(e+f x)} \csc ^2(e+f x)+3 \sqrt {\sec (e+f x)} \tan ^{-1}\left (\frac {\sqrt {\sec (e+f x)}}{\sqrt [4]{\tan ^2(e+f x)}}\right )-3 \sqrt {\sec (e+f x)} \tanh ^{-1}\left (\frac {\sqrt {\sec (e+f x)}}{\sqrt [4]{\tan ^2(e+f x)}}\right )\right )}{3 b^3 f \sqrt [4]{\tan ^2(e+f x)} \sqrt {d \sec (e+f x)}} \]
Antiderivative was successfully verified.
[In]
Integrate[(d*Sec[e + f*x])^(7/2)/(b*Tan[e + f*x])^(5/2),x]
[Out]
-1/3*(d^4*Sqrt[b*Tan[e + f*x]]*(3*ArcTan[Sqrt[Sec[e + f*x]]/(Tan[e + f*x]^2)^(1/4)]*Sqrt[Sec[e + f*x]] - 3*Arc
Tanh[Sqrt[Sec[e + f*x]]/(Tan[e + f*x]^2)^(1/4)]*Sqrt[Sec[e + f*x]] + 2*Csc[e + f*x]^2*(Tan[e + f*x]^2)^(1/4)))
/(b^3*f*Sqrt[d*Sec[e + f*x]]*(Tan[e + f*x]^2)^(1/4))
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fricas [B] time = 0.99, size = 850, normalized size = 4.94 \[ \text {result too large to display} \]
Verification of antiderivative is not currently implemented for this CAS.
[In]
integrate((d*sec(f*x+e))^(7/2)/(b*tan(f*x+e))^(5/2),x, algorithm="fricas")
[Out]
[1/24*(16*d^3*sqrt(b*sin(f*x + e)/cos(f*x + e))*sqrt(d/cos(f*x + e))*cos(f*x + e) - 6*(b*d^3*cos(f*x + e)^2 -
b*d^3)*sqrt(-d/b)*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*sin(f*x + e)/cos(f*x + e))*sqrt(-d/b)*sqrt(d/cos(f*x + e))/(d*cos(f*x + e)
^2 - (d*cos(f*x + e) + d)*sin(f*x + e) - d)) + 3*(b*d^3*cos(f*x + e)^2 - b*d^3)*sqrt(-d/b)*log((d*cos(f*x + e)
^4 - 72*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*sin(f*x + e)/cos(f*x + e))*sqrt(-d/b)*sqrt(d/cos(f*x + e)) + 28*(d*cos(f*x + e)^2 - 2*d)*sin(f*x +
e) + 72*d)/(cos(f*x + e)^4 - 8*cos(f*x + e)^2 - 4*(cos(f*x + e)^2 - 2)*sin(f*x + e) + 8)))/(b^3*f*cos(f*x + e)
^2 - b^3*f), 1/24*(16*d^3*sqrt(b*sin(f*x + e)/cos(f*x + e))*sqrt(d/cos(f*x + e))*cos(f*x + e) + 6*(b*d^3*cos(f
*x + e)^2 - b*d^3)*sqrt(d/b)*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*sin(f*x + e)/cos(f*x + e))*sqrt(d/b)*sqrt(d/cos(f*x + e))/(d*co
s(f*x + e)^2 + (d*cos(f*x + e) + d)*sin(f*x + e) - d)) + 3*(b*d^3*cos(f*x + e)^2 - b*d^3)*sqrt(d/b)*log((d*cos
(f*x + e)^4 - 72*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*c
os(f*x + e))*sqrt(b*sin(f*x + e)/cos(f*x + e))*sqrt(d/b)*sqrt(d/cos(f*x + e)) - 28*(d*cos(f*x + e)^2 - 2*d)*si
n(f*x + e) + 72*d)/(cos(f*x + e)^4 - 8*cos(f*x + e)^2 + 4*(cos(f*x + e)^2 - 2)*sin(f*x + e) + 8)))/(b^3*f*cos(
f*x + e)^2 - b^3*f)]
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giac [F] time = 0.00, size = 0, normalized size = 0.00 \[ \int \frac {\left (d \sec \left (f x + e\right )\right )^{\frac {7}{2}}}{\left (b \tan \left (f x + e\right )\right )^{\frac {5}{2}}}\,{d x} \]
Verification of antiderivative is not currently implemented for this CAS.
[In]
integrate((d*sec(f*x+e))^(7/2)/(b*tan(f*x+e))^(5/2),x, algorithm="giac")
[Out]
integrate((d*sec(f*x + e))^(7/2)/(b*tan(f*x + e))^(5/2), x)
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maple [C] time = 0.66, size = 1367, normalized size = 7.95 \[ \text {result too large to display} \]
Verification of antiderivative is not currently implemented for this CAS.
[In]
int((d*sec(f*x+e))^(7/2)/(b*tan(f*x+e))^(5/2),x)
[Out]
1/6/f*(3*I*cos(f*x+e)*sin(f*x+e)*(-I*(-1+cos(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)-I-sin(f*x+e))/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)*sin(f*x+e)*(-I*(-1+cos(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)-I-sin(f*x+e))/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))-6*I*cos(f*x+e)*sin(f*x+e)*(-(I*cos(f*x+e)-I-sin(f*x+e))/sin(f*x
+e))^(1/2)*EllipticF(((I*cos(f*x+e)-I+sin(f*x+e))/sin(f*x+e))^(1/2),1/2*2^(1/2))*((I*cos(f*x+e)-I+sin(f*x+e))/
sin(f*x+e))^(1/2)*(-I*(-1+cos(f*x+e))/sin(f*x+e))^(1/2)+3*I*sin(f*x+e)*(-I*(-1+cos(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)-I-sin(f*x+e))/sin(f*x+e))^(1/2)*EllipticPi(((I*c
os(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*sin(f*x+e)*(-I*(-1+cos(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)-I-sin(f*x+e))/sin(f*x+e))^(1/2)*Ellipti
cPi(((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))-6*I*sin(f*x+e)*(-I*(-1+cos(f*x+e))/s
in(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)-I-sin(f*x+e))/sin(f*x+e))^(1/2
)*EllipticF(((I*cos(f*x+e)-I+sin(f*x+e))/sin(f*x+e))^(1/2),1/2*2^(1/2))-3*cos(f*x+e)*sin(f*x+e)*(-I*(-1+cos(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)-I-sin(f*x+e))/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)*sin(f*x
+e)*(-I*(-1+cos(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)-I-sin
(f*x+e))/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*
sin(f*x+e)*(-I*(-1+cos(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
)-I-sin(f*x+e))/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*sin(f*x+e)*(-I*(-1+cos(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*co
s(f*x+e)-I-sin(f*x+e))/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))-2*2^(1/2))*(d/cos(f*x+e))^(7/2)*sin(f*x+e)*cos(f*x+e)/(b*sin(f*x+e)/cos(f*x+e))^(5/2)*2^(1/2)
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maxima [F] time = 0.00, size = 0, normalized size = 0.00 \[ \int \frac {\left (d \sec \left (f x + e\right )\right )^{\frac {7}{2}}}{\left (b \tan \left (f x + e\right )\right )^{\frac {5}{2}}}\,{d x} \]
Verification of antiderivative is not currently implemented for this CAS.
[In]
integrate((d*sec(f*x+e))^(7/2)/(b*tan(f*x+e))^(5/2),x, algorithm="maxima")
[Out]
integrate((d*sec(f*x + e))^(7/2)/(b*tan(f*x + e))^(5/2), x)
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mupad [F] time = 0.00, size = -1, normalized size = -0.01 \[ \int \frac {{\left (\frac {d}{\cos \left (e+f\,x\right )}\right )}^{7/2}}{{\left (b\,\mathrm {tan}\left (e+f\,x\right )\right )}^{5/2}} \,d x \]
Verification of antiderivative is not currently implemented for this CAS.
[In]
int((d/cos(e + f*x))^(7/2)/(b*tan(e + f*x))^(5/2),x)
[Out]
int((d/cos(e + f*x))^(7/2)/(b*tan(e + f*x))^(5/2), x)
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sympy [F(-1)] time = 0.00, size = 0, normalized size = 0.00 \[ \text {Timed out} \]
Verification of antiderivative is not currently implemented for this CAS.
[In]
integrate((d*sec(f*x+e))**(7/2)/(b*tan(f*x+e))**(5/2),x)
[Out]
Timed out
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