3.3.100 \(\int (d \sec (e+f x))^{3/2} (b \tan (e+f x))^{3/2} \, dx\) [300]

Optimal. Leaf size=169 \[ -\frac {b^{3/2} d \text {ArcTan}\left (\frac {\sqrt {b \sin (e+f x)}}{\sqrt {b}}\right ) \sqrt {d \sec (e+f x)} \sqrt {b \sin (e+f x)}}{4 f \sqrt {b \tan (e+f x)}}-\frac {b^{3/2} d \tanh ^{-1}\left (\frac {\sqrt {b \sin (e+f x)}}{\sqrt {b}}\right ) \sqrt {d \sec (e+f x)} \sqrt {b \sin (e+f x)}}{4 f \sqrt {b \tan (e+f x)}}+\frac {b (d \sec (e+f x))^{3/2} \sqrt {b \tan (e+f x)}}{2 f} \]

[Out]

-1/4*b^(3/2)*d*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)/f/(b*tan(f*x+e))
^(1/2)-1/4*b^(3/2)*d*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)/f/(b*tan(
f*x+e))^(1/2)+1/2*b*(d*sec(f*x+e))^(3/2)*(b*tan(f*x+e))^(1/2)/f

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Rubi [A]
time = 0.12, antiderivative size = 169, 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 = {2691, 2696, 2644, 335, 218, 212, 209} \begin {gather*} -\frac {b^{3/2} d \sqrt {b \sin (e+f x)} \sqrt {d \sec (e+f x)} \text {ArcTan}\left (\frac {\sqrt {b \sin (e+f x)}}{\sqrt {b}}\right )}{4 f \sqrt {b \tan (e+f x)}}-\frac {b^{3/2} d \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 )}{4 f \sqrt {b \tan (e+f x)}}+\frac {b \sqrt {b \tan (e+f x)} (d \sec (e+f x))^{3/2}}{2 f} \end {gather*}

Antiderivative was successfully verified.

[In]

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

[Out]

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

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 218

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] &&  !Gt
Q[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 2691

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

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Mathematica [A]
time = 6.81, size = 129, normalized size = 0.76 \begin {gather*} \frac {b (d \sec (e+f x))^{3/2} \sqrt {b \tan (e+f x)} \left (\text {ArcTan}\left (\frac {\sqrt {\sec (e+f x)}}{\sqrt [4]{\tan ^2(e+f x)}}\right )-\tanh ^{-1}\left (\frac {\sqrt {\sec (e+f x)}}{\sqrt [4]{\tan ^2(e+f x)}}\right )+2 \sec ^{\frac {3}{2}}(e+f x) \sqrt [4]{\tan ^2(e+f x)}\right )}{4 f \sec ^{\frac {3}{2}}(e+f x) \sqrt [4]{\tan ^2(e+f x)}} \end {gather*}

Antiderivative was successfully verified.

[In]

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

[Out]

(b*(d*Sec[e + f*x])^(3/2)*Sqrt[b*Tan[e + f*x]]*(ArcTan[Sqrt[Sec[e + f*x]]/(Tan[e + f*x]^2)^(1/4)] - ArcTanh[Sq
rt[Sec[e + f*x]]/(Tan[e + f*x]^2)^(1/4)] + 2*Sec[e + f*x]^(3/2)*(Tan[e + f*x]^2)^(1/4)))/(4*f*Sec[e + f*x]^(3/
2)*(Tan[e + f*x]^2)^(1/4))

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Maple [C] Result contains higher order function than in optimal. Order 4 vs. order 3.
time = 0.36, size = 759, normalized size = 4.49

method	result	size
default	\(\frac {\left (-2 i \left (\cos ^{2}\left (f x +e \right )\right ) \sin \left (f x +e \right ) \sqrt {\frac {i \cos \left (f x +e \right )-i+\sin \left (f x +e \right )}{\sin \left (f x +e \right )}}\, \sqrt {-\frac {i \cos \left (f x +e \right )-i-\sin \left (f x +e \right )}{\sin \left (f x +e \right )}}\, \sqrt {-\frac {i \left (\cos \left (f x +e \right )-1\right )}{\sin \left (f x +e \right )}}\, \EllipticF \left (\sqrt {\frac {i \cos \left (f x +e \right )-i+\sin \left (f x +e \right )}{\sin \left (f x +e \right )}}, \frac {\sqrt {2}}{2}\right )+i \left (\cos ^{2}\left (f x +e \right )\right ) \sin \left (f x +e \right ) \sqrt {\frac {i \cos \left (f x +e \right )-i+\sin \left (f x +e \right )}{\sin \left (f x +e \right )}}\, \sqrt {-\frac {i \cos \left (f x +e \right )-i-\sin \left (f x +e \right )}{\sin \left (f x +e \right )}}\, \sqrt {-\frac {i \left (\cos \left (f x +e \right )-1\right )}{\sin \left (f x +e \right )}}\, \EllipticPi \left (\sqrt {\frac {i \cos \left (f x +e \right )-i+\sin \left (f x +e \right )}{\sin \left (f x +e \right )}}, \frac {1}{2}-\frac {i}{2}, \frac {\sqrt {2}}{2}\right )+i \left (\cos ^{2}\left (f x +e \right )\right ) \sin \left (f x +e \right ) \sqrt {\frac {i \cos \left (f x +e \right )-i+\sin \left (f x +e \right )}{\sin \left (f x +e \right )}}\, \sqrt {-\frac {i \cos \left (f x +e \right )-i-\sin \left (f x +e \right )}{\sin \left (f x +e \right )}}\, \sqrt {-\frac {i \left (\cos \left (f x +e \right )-1\right )}{\sin \left (f x +e \right )}}\, \EllipticPi \left (\sqrt {\frac {i \cos \left (f x +e \right )-i+\sin \left (f x +e \right )}{\sin \left (f x +e \right )}}, \frac {1}{2}+\frac {i}{2}, \frac {\sqrt {2}}{2}\right )-\left (\cos ^{2}\left (f x +e \right )\right ) \sin \left (f x +e \right ) \sqrt {\frac {i \cos \left (f x +e \right )-i+\sin \left (f x +e \right )}{\sin \left (f x +e \right )}}\, \sqrt {-\frac {i \cos \left (f x +e \right )-i-\sin \left (f x +e \right )}{\sin \left (f x +e \right )}}\, \sqrt {-\frac {i \left (\cos \left (f x +e \right )-1\right )}{\sin \left (f x +e \right )}}\, \EllipticPi \left (\sqrt {\frac {i \cos \left (f x +e \right )-i+\sin \left (f x +e \right )}{\sin \left (f x +e \right )}}, \frac {1}{2}-\frac {i}{2}, \frac {\sqrt {2}}{2}\right )+\left (\cos ^{2}\left (f x +e \right )\right ) \sin \left (f x +e \right ) \sqrt {\frac {i \cos \left (f x +e \right )-i+\sin \left (f x +e \right )}{\sin \left (f x +e \right )}}\, \sqrt {-\frac {i \cos \left (f x +e \right )-i-\sin \left (f x +e \right )}{\sin \left (f x +e \right )}}\, \sqrt {-\frac {i \left (\cos \left (f x +e \right )-1\right )}{\sin \left (f x +e \right )}}\, \EllipticPi \left (\sqrt {\frac {i \cos \left (f x +e \right )-i+\sin \left (f x +e \right )}{\sin \left (f x +e \right )}}, \frac {1}{2}+\frac {i}{2}, \frac {\sqrt {2}}{2}\right )+2 \cos \left (f x +e \right ) \sqrt {2}-2 \sqrt {2}\right ) \cos \left (f x +e \right ) \left (\frac {d}{\cos \left (f x +e \right )}\right )^{\frac {3}{2}} \left (\frac {b \sin \left (f x +e \right )}{\cos \left (f x +e \right )}\right )^{\frac {3}{2}} \sqrt {2}}{8 f \left (\cos \left (f x +e \right )-1\right ) \sin \left (f x +e \right )}\)	\(759\)

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

1/8/f*(-2*I*cos(f*x+e)^2*sin(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)*(-I*(cos(f*x+e)-1)/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)^2*sin(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-s
in(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))+I*cos(f*x+e)^2*sin(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)*(-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))-cos(f*x+e)^2*sin(f*x+e)*((I*cos(f*x+e)-I+sin(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)-1)/sin(f*x+e))^(1/2)*Elliptic
Pi(((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))+cos(f*x+e)^2*sin(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)*(-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))+2*cos(f*x+e)*2^(1/2)
-2*2^(1/2))*cos(f*x+e)*(d/cos(f*x+e))^(3/2)*(b*sin(f*x+e)/cos(f*x+e))^(3/2)/(cos(f*x+e)-1)/sin(f*x+e)*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))^(3/2)*(b*tan(f*x+e))^(3/2),x, algorithm="maxima")

[Out]

integrate((d*sec(f*x + e))^(3/2)*(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. 415 vs. \(2 (145) = 290\).
time = 0.54, size = 837, normalized size = 4.95 \begin {gather*} \left [\frac {2 \, \sqrt {-b d} b d \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 {-b d} \sqrt {\frac {b \sin \left (f x + e\right )}{\cos \left (f x + e\right )}} \sqrt {\frac {d}{\cos \left (f x + e\right )}}}{4 \, {\left (b d \cos \left (f x + e\right )^{2} - b d - {\left (b d \cos \left (f x + e\right ) + b d\right )} \sin \left (f x + e\right )\right )}}\right ) \cos \left (f x + e\right ) + \sqrt {-b d} b d \cos \left (f x + e\right ) \log \left (\frac {b d \cos \left (f x + e\right )^{4} - 72 \, b 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 {-b d} \sqrt {\frac {b \sin \left (f x + e\right )}{\cos \left (f x + e\right )}} \sqrt {\frac {d}{\cos \left (f x + e\right )}} + 72 \, b d + 28 \, {\left (b d \cos \left (f x + e\right )^{2} - 2 \, b d\right )} \sin \left (f x + e\right )}{\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 ) + 16 \, b d \sqrt {\frac {b \sin \left (f x + e\right )}{\cos \left (f x + e\right )}} \sqrt {\frac {d}{\cos \left (f x + e\right )}}}{32 \, f \cos \left (f x + e\right )}, -\frac {2 \, \sqrt {b d} b d \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 {b d} \sqrt {\frac {b \sin \left (f x + e\right )}{\cos \left (f x + e\right )}} \sqrt {\frac {d}{\cos \left (f x + e\right )}}}{4 \, {\left (b d \cos \left (f x + e\right )^{2} - b d + {\left (b d \cos \left (f x + e\right ) + b d\right )} \sin \left (f x + e\right )\right )}}\right ) \cos \left (f x + e\right ) - \sqrt {b d} b d \cos \left (f x + e\right ) \log \left (\frac {b d \cos \left (f x + e\right )^{4} - 72 \, b 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 {b d} \sqrt {\frac {b \sin \left (f x + e\right )}{\cos \left (f x + e\right )}} \sqrt {\frac {d}{\cos \left (f x + e\right )}} + 72 \, b d - 28 \, {\left (b d \cos \left (f x + e\right )^{2} - 2 \, b d\right )} \sin \left (f x + e\right )}{\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 ) - 16 \, b d \sqrt {\frac {b \sin \left (f x + e\right )}{\cos \left (f x + e\right )}} \sqrt {\frac {d}{\cos \left (f x + e\right )}}}{32 \, f \cos \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))^(3/2)*(b*tan(f*x+e))^(3/2),x, algorithm="fricas")

[Out]

[1/32*(2*sqrt(-b*d)*b*d*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) + sqrt(-b*d)*b*d*cos(f*x + e)*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*b*d*sqrt(b*sin(f*x + e)/cos(f*x + e))*sqrt(d/cos(f*x + e)))/(f*cos(f*x + e)), -1/32*(2*sqrt(b*d)*b*d*arc
tan(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) - sqrt(b*d)*b*d*cos(f*x + e)*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*b*d*sqrt(b*sin(f*x + e)/
cos(f*x + e))*sqrt(d/cos(f*x + e)))/(f*cos(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))**(3/2)*(b*tan(f*x+e))**(3/2),x)

[Out]

Exception raised: SystemError >> excessive stack use: stack is 3005 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))^(3/2)*(b*tan(f*x+e))^(3/2),x, algorithm="giac")

[Out]

integrate((d*sec(f*x + e))^(3/2)*(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 {\left (b\,\mathrm {tan}\left (e+f\,x\right )\right )}^{3/2}\,{\left (\frac {d}{\cos \left (e+f\,x\right )}\right )}^{3/2} \,d x \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))^(3/2),x)

[Out]

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

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