3.4.22 \(\int \frac {(d \sec (e+f x))^{5/2}}{(b \tan (e+f x))^{3/2}} \, dx\) [322]

Optimal. Leaf size=171 \[ -\frac {2 d^2 \sqrt {d \sec (e+f x)}}{b f \sqrt {b \tan (e+f x)}}-\frac {d^3 \text {ArcTan}\left (\frac {\sqrt {b \sin (e+f x)}}{\sqrt {b}}\right ) \sqrt {b \tan (e+f x)}}{b^{3/2} f \sqrt {d \sec (e+f x)} \sqrt {b \sin (e+f x)}}+\frac {d^3 \tanh ^{-1}\left (\frac {\sqrt {b \sin (e+f x)}}{\sqrt {b}}\right ) \sqrt {b \tan (e+f x)}}{b^{3/2} f \sqrt {d \sec (e+f x)} \sqrt {b \sin (e+f x)}} \]

[Out]

-2*d^2*(d*sec(f*x+e))^(1/2)/b/f/(b*tan(f*x+e))^(1/2)-d^3*arctan((b*sin(f*x+e))^(1/2)/b^(1/2))*(b*tan(f*x+e))^(
1/2)/b^(3/2)/f/(d*sec(f*x+e))^(1/2)/(b*sin(f*x+e))^(1/2)+d^3*arctanh((b*sin(f*x+e))^(1/2)/b^(1/2))*(b*tan(f*x+
e))^(1/2)/b^(3/2)/f/(d*sec(f*x+e))^(1/2)/(b*sin(f*x+e))^(1/2)

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Rubi [A]
time = 0.12, antiderivative size = 171, 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 = {2688, 2696, 2644, 335, 304, 209, 212} \begin {gather*} -\frac {d^3 \sqrt {b \tan (e+f x)} \text {ArcTan}\left (\frac {\sqrt {b \sin (e+f x)}}{\sqrt {b}}\right )}{b^{3/2} f \sqrt {b \sin (e+f x)} \sqrt {d \sec (e+f x)}}+\frac {d^3 \sqrt {b \tan (e+f x)} \tanh ^{-1}\left (\frac {\sqrt {b \sin (e+f x)}}{\sqrt {b}}\right )}{b^{3/2} f \sqrt {b \sin (e+f x)} \sqrt {d \sec (e+f x)}}-\frac {2 d^2 \sqrt {d \sec (e+f x)}}{b f \sqrt {b \tan (e+f x)}} \end {gather*}

Antiderivative was successfully verified.

[In]

Int[(d*Sec[e + f*x])^(5/2)/(b*Tan[e + f*x])^(3/2),x]

[Out]

(-2*d^2*Sqrt[d*Sec[e + f*x]])/(b*f*Sqrt[b*Tan[e + f*x]]) - (d^3*ArcTan[Sqrt[b*Sin[e + f*x]]/Sqrt[b]]*Sqrt[b*Ta
n[e + f*x]])/(b^(3/2)*f*Sqrt[d*Sec[e + f*x]]*Sqrt[b*Sin[e + f*x]]) + (d^3*ArcTanh[Sqrt[b*Sin[e + f*x]]/Sqrt[b]
]*Sqrt[b*Tan[e + f*x]])/(b^(3/2)*f*Sqrt[d*Sec[e + f*x]]*Sqrt[b*Sin[e + f*x]])

Rule 209

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

Rule 212

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

Rule 304

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 335

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 2644

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 2688

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 2696

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

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Mathematica [A]
time = 1.21, size = 211, normalized size = 1.23 \begin {gather*} -\frac {d^3 \sin (e+f x) \left (-4 \csc ^2(e+f x)+16 \csc ^2(2 (e+f x))-2 \text {ArcTan}\left (\frac {\sqrt {\sec (e+f x)}}{\sqrt [4]{\tan ^2(e+f x)}}\right ) \sec ^{\frac {3}{2}}(e+f x) \sqrt [4]{\tan ^2(e+f x)}+\log \left (1-\frac {\sqrt {\sec (e+f x)}}{\sqrt [4]{\tan ^2(e+f x)}}\right ) \sec ^{\frac {3}{2}}(e+f x) \sqrt [4]{\tan ^2(e+f x)}-\log \left (1+\frac {\sqrt {\sec (e+f x)}}{\sqrt [4]{\tan ^2(e+f x)}}\right ) \sec ^{\frac {3}{2}}(e+f x) \sqrt [4]{\tan ^2(e+f x)}\right )}{2 f \sqrt {d \sec (e+f x)} (b \tan (e+f x))^{3/2}} \end {gather*}

Antiderivative was successfully verified.

[In]

Integrate[(d*Sec[e + f*x])^(5/2)/(b*Tan[e + f*x])^(3/2),x]

[Out]

-1/2*(d^3*Sin[e + f*x]*(-4*Csc[e + f*x]^2 + 16*Csc[2*(e + f*x)]^2 - 2*ArcTan[Sqrt[Sec[e + f*x]]/(Tan[e + f*x]^
2)^(1/4)]*Sec[e + f*x]^(3/2)*(Tan[e + f*x]^2)^(1/4) + Log[1 - Sqrt[Sec[e + f*x]]/(Tan[e + f*x]^2)^(1/4)]*Sec[e
+ f*x]^(3/2)*(Tan[e + f*x]^2)^(1/4) - Log[1 + Sqrt[Sec[e + f*x]]/(Tan[e + f*x]^2)^(1/4)]*Sec[e + f*x]^(3/2)*(
Tan[e + f*x]^2)^(1/4)))/(f*Sqrt[d*Sec[e + f*x]]*(b*Tan[e + f*x])^(3/2))

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Maple [C] Result contains higher order function than in optimal. Order 4 vs. order 3.
time = 0.36, size = 1061, normalized size = 6.20

method	result	size
default	\(\text {Expression too large to display}\)	\(1061\)

Verification of antiderivative is not currently implemented for this CAS.

[In]

int((d*sec(f*x+e))^(5/2)/(b*tan(f*x+e))^(3/2),x,method=_RETURNVERBOSE)

[Out]

-1/2/f*(I*((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)*Ellip
ticPi(((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))*(-I*(cos(f*x+e)-1)/sin(f*x+e))^(1/
2)*cos(f*x+e)-I*((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))*(-I*(cos(f*x+e)-1)/sin(f*x+e
))^(1/2)*cos(f*x+e)+I*((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))*(-I*(cos(f*x+e)-1)/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))*(-I*(cos(f*x+e)-1)/sin(f*x+e)
)^(1/2)*cos(f*x+e)-I*((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))*(-I*(cos(f*x+e)-1)/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)*E
llipticPi(((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))*(-I*(cos(f*x+e)-1)/sin(f*x+e))
^(1/2)*cos(f*x+e)-((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))*(-I*(cos(f*x+e)-1)/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)*Elli
pticPi(((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))*(-I*(cos(f*x+e)-1)/sin(f*x+e))^(1
/2)+2*2^(1/2))*(d/cos(f*x+e))^(5/2)*sin(f*x+e)*cos(f*x+e)/(b*sin(f*x+e)/cos(f*x+e))^(3/2)*2^(1/2)

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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((d*sec(f*x+e))^(5/2)/(b*tan(f*x+e))^(3/2),x, algorithm="maxima")

[Out]

integrate((d*sec(f*x + e))^(5/2)/(b*tan(f*x + e))^(3/2), x)

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Fricas [B] Leaf count of result is larger than twice the leaf count of optimal. 428 vs. \(2 (153) = 306\).
time = 0.53, size = 864, normalized size = 5.05 \begin {gather*} \left [-\frac {2 \, b d^{2} \sqrt {-\frac {d}{b}} \arctan \left (\frac {{\left (\cos \left (f x + e\right )^{3} - 5 \, \cos \left (f x + e\right )^{2} - {\left (\cos \left (f x + e\right )^{2} + 6 \, \cos \left (f x + e\right ) + 4\right )} \sin \left (f x + e\right ) - 2 \, \cos \left (f x + e\right ) + 4\right )} \sqrt {\frac {b \sin \left (f x + e\right )}{\cos \left (f x + e\right )}} \sqrt {-\frac {d}{b}} \sqrt {\frac {d}{\cos \left (f x + e\right )}}}{4 \, {\left (d \cos \left (f x + e\right )^{2} - {\left (d \cos \left (f x + e\right ) + d\right )} \sin \left (f x + e\right ) - d\right )}}\right ) \sin \left (f x + e\right ) - b d^{2} \sqrt {-\frac {d}{b}} \log \left (\frac {d \cos \left (f x + e\right )^{4} - 72 \, d \cos \left (f x + e\right )^{2} - 8 \, {\left (7 \, \cos \left (f x + e\right )^{3} - {\left (\cos \left (f x + e\right )^{3} - 8 \, \cos \left (f x + e\right )\right )} \sin \left (f x + e\right ) - 8 \, \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}{b}} \sqrt {\frac {d}{\cos \left (f x + e\right )}} + 28 \, {\left (d \cos \left (f x + e\right )^{2} - 2 \, d\right )} \sin \left (f x + e\right ) + 72 \, d}{\cos \left (f x + e\right )^{4} - 8 \, \cos \left (f x + e\right )^{2} - 4 \, {\left (\cos \left (f x + e\right )^{2} - 2\right )} \sin \left (f x + e\right ) + 8}\right ) \sin \left (f x + e\right ) + 16 \, d^{2} \sqrt {\frac {b \sin \left (f x + e\right )}{\cos \left (f x + e\right )}} \sqrt {\frac {d}{\cos \left (f x + e\right )}} \cos \left (f x + e\right )}{8 \, b^{2} f \sin \left (f x + e\right )}, -\frac {2 \, b d^{2} \sqrt {\frac {d}{b}} \arctan \left (\frac {{\left (\cos \left (f x + e\right )^{3} - 5 \, \cos \left (f x + e\right )^{2} + {\left (\cos \left (f x + e\right )^{2} + 6 \, \cos \left (f x + e\right ) + 4\right )} \sin \left (f x + e\right ) - 2 \, \cos \left (f x + e\right ) + 4\right )} \sqrt {\frac {b \sin \left (f x + e\right )}{\cos \left (f x + e\right )}} \sqrt {\frac {d}{b}} \sqrt {\frac {d}{\cos \left (f x + e\right )}}}{4 \, {\left (d \cos \left (f x + e\right )^{2} + {\left (d \cos \left (f x + e\right ) + d\right )} \sin \left (f x + e\right ) - d\right )}}\right ) \sin \left (f x + e\right ) - b d^{2} \sqrt {\frac {d}{b}} \log \left (\frac {d \cos \left (f x + e\right )^{4} - 72 \, d \cos \left (f x + e\right )^{2} - 8 \, {\left (7 \, \cos \left (f x + e\right )^{3} + {\left (\cos \left (f x + e\right )^{3} - 8 \, \cos \left (f x + e\right )\right )} \sin \left (f x + e\right ) - 8 \, \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}{b}} \sqrt {\frac {d}{\cos \left (f x + e\right )}} - 28 \, {\left (d \cos \left (f x + e\right )^{2} - 2 \, d\right )} \sin \left (f x + e\right ) + 72 \, d}{\cos \left (f x + e\right )^{4} - 8 \, \cos \left (f x + e\right )^{2} + 4 \, {\left (\cos \left (f x + e\right )^{2} - 2\right )} \sin \left (f x + e\right ) + 8}\right ) \sin \left (f x + e\right ) + 16 \, d^{2} \sqrt {\frac {b \sin \left (f x + e\right )}{\cos \left (f x + e\right )}} \sqrt {\frac {d}{\cos \left (f x + e\right )}} \cos \left (f x + e\right )}{8 \, b^{2} f \sin \left (f x + e\right )}\right ] \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate((d*sec(f*x+e))^(5/2)/(b*tan(f*x+e))^(3/2),x, algorithm="fricas")

[Out]

[-1/8*(2*b*d^2*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))*sin(f*x + e) - b*d^2*sqrt(-d/b)*log((d*cos(f*x + e)^4 - 7
2*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))*sq
rt(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) + 7
2*d)/(cos(f*x + e)^4 - 8*cos(f*x + e)^2 - 4*(cos(f*x + e)^2 - 2)*sin(f*x + e) + 8))*sin(f*x + e) + 16*d^2*sqrt
(b*sin(f*x + e)/cos(f*x + e))*sqrt(d/cos(f*x + e))*cos(f*x + e))/(b^2*f*sin(f*x + e)), -1/8*(2*b*d^2*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))*sin(f*x + e) - b*d^2*sqrt(d/b)*log((d*cos(f*x + e)^4 - 72*d*cos(f*x + e)^2 - 8*(7*c
os(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))*sin(f*x + e) + 16*d^2*sqrt(b*sin(f*x + e)/cos(f*x + e)
)*sqrt(d/cos(f*x + e))*cos(f*x + e))/(b^2*f*sin(f*x + e))]

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Sympy [F(-2)]
time = 0.00, size = 0, normalized size = 0.00 \begin {gather*} \text {Exception raised: SystemError} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate((d*sec(f*x+e))**(5/2)/(b*tan(f*x+e))**(3/2),x)

[Out]

Exception raised: SystemError >> excessive stack use: stack is 4371 deep

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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((d*sec(f*x+e))^(5/2)/(b*tan(f*x+e))^(3/2),x, algorithm="giac")

[Out]

integrate((d*sec(f*x + e))^(5/2)/(b*tan(f*x + e))^(3/2), x)

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Mupad [F]
time = 0.00, size = -1, normalized size = -0.01 \begin {gather*} \int \frac {{\left (\frac {d}{\cos \left (e+f\,x\right )}\right )}^{5/2}}{{\left (b\,\mathrm {tan}\left (e+f\,x\right )\right )}^{3/2}} \,d x \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

int((d/cos(e + f*x))^(5/2)/(b*tan(e + f*x))^(3/2),x)

[Out]

int((d/cos(e + f*x))^(5/2)/(b*tan(e + f*x))^(3/2), x)

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