3.1276 \(\int \frac {(c+d \tan (e+f x))^{3/2}}{(a+b \tan (e+f x))^{7/2}} \, dx\)

Optimal. Leaf size=391 \[ -\frac {4 \left (-2 a^2 d+5 a b c+3 b^2 d\right ) \sqrt {c+d \tan (e+f x)}}{15 f \left (a^2+b^2\right )^2 (a+b \tan (e+f x))^{3/2}}-\frac {2 (b c-a d) \sqrt {c+d \tan (e+f x)}}{5 f \left (a^2+b^2\right ) (a+b \tan (e+f x))^{5/2}}+\frac {2 \left (-8 a^4 d^2+50 a^3 b c d-a^2 b^2 \left (45 c^2-49 d^2\right )-70 a b^3 c d+3 b^4 \left (5 c^2-d^2\right )\right ) \sqrt {c+d \tan (e+f x)}}{15 f \left (a^2+b^2\right )^3 (b c-a d) \sqrt {a+b \tan (e+f x)}}-\frac {i (c-i d)^{3/2} \tanh ^{-1}\left (\frac {\sqrt {c-i d} \sqrt {a+b \tan (e+f x)}}{\sqrt {a-i b} \sqrt {c+d \tan (e+f x)}}\right )}{f (a-i b)^{7/2}}+\frac {i (c+i d)^{3/2} \tanh ^{-1}\left (\frac {\sqrt {c+i d} \sqrt {a+b \tan (e+f x)}}{\sqrt {a+i b} \sqrt {c+d \tan (e+f x)}}\right )}{f (a+i b)^{7/2}} \]

[Out]

-I*(c-I*d)^(3/2)*arctanh((c-I*d)^(1/2)*(a+b*tan(f*x+e))^(1/2)/(a-I*b)^(1/2)/(c+d*tan(f*x+e))^(1/2))/(a-I*b)^(7
/2)/f+I*(c+I*d)^(3/2)*arctanh((c+I*d)^(1/2)*(a+b*tan(f*x+e))^(1/2)/(a+I*b)^(1/2)/(c+d*tan(f*x+e))^(1/2))/(a+I*
b)^(7/2)/f+2/15*(50*a^3*b*c*d-70*a*b^3*c*d-8*a^4*d^2-a^2*b^2*(45*c^2-49*d^2)+3*b^4*(5*c^2-d^2))*(c+d*tan(f*x+e
))^(1/2)/(a^2+b^2)^3/(-a*d+b*c)/f/(a+b*tan(f*x+e))^(1/2)-2/5*(-a*d+b*c)*(c+d*tan(f*x+e))^(1/2)/(a^2+b^2)/f/(a+
b*tan(f*x+e))^(5/2)-4/15*(-2*a^2*d+5*a*b*c+3*b^2*d)*(c+d*tan(f*x+e))^(1/2)/(a^2+b^2)^2/f/(a+b*tan(f*x+e))^(3/2
)

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Rubi [A]  time = 2.10, antiderivative size = 391, normalized size of antiderivative = 1.00, number of steps used = 10, number of rules used = 6, integrand size = 29, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.207, Rules used = {3567, 3649, 3616, 3615, 93, 208} \[ \frac {2 \left (-a^2 b^2 \left (45 c^2-49 d^2\right )+50 a^3 b c d-8 a^4 d^2-70 a b^3 c d+3 b^4 \left (5 c^2-d^2\right )\right ) \sqrt {c+d \tan (e+f x)}}{15 f \left (a^2+b^2\right )^3 (b c-a d) \sqrt {a+b \tan (e+f x)}}-\frac {4 \left (-2 a^2 d+5 a b c+3 b^2 d\right ) \sqrt {c+d \tan (e+f x)}}{15 f \left (a^2+b^2\right )^2 (a+b \tan (e+f x))^{3/2}}-\frac {2 (b c-a d) \sqrt {c+d \tan (e+f x)}}{5 f \left (a^2+b^2\right ) (a+b \tan (e+f x))^{5/2}}-\frac {i (c-i d)^{3/2} \tanh ^{-1}\left (\frac {\sqrt {c-i d} \sqrt {a+b \tan (e+f x)}}{\sqrt {a-i b} \sqrt {c+d \tan (e+f x)}}\right )}{f (a-i b)^{7/2}}+\frac {i (c+i d)^{3/2} \tanh ^{-1}\left (\frac {\sqrt {c+i d} \sqrt {a+b \tan (e+f x)}}{\sqrt {a+i b} \sqrt {c+d \tan (e+f x)}}\right )}{f (a+i b)^{7/2}} \]

Antiderivative was successfully verified.

[In]

Int[(c + d*Tan[e + f*x])^(3/2)/(a + b*Tan[e + f*x])^(7/2),x]

[Out]

((-I)*(c - I*d)^(3/2)*ArcTanh[(Sqrt[c - I*d]*Sqrt[a + b*Tan[e + f*x]])/(Sqrt[a - I*b]*Sqrt[c + d*Tan[e + f*x]]
)])/((a - I*b)^(7/2)*f) + (I*(c + I*d)^(3/2)*ArcTanh[(Sqrt[c + I*d]*Sqrt[a + b*Tan[e + f*x]])/(Sqrt[a + I*b]*S
qrt[c + d*Tan[e + f*x]])])/((a + I*b)^(7/2)*f) - (2*(b*c - a*d)*Sqrt[c + d*Tan[e + f*x]])/(5*(a^2 + b^2)*f*(a
+ b*Tan[e + f*x])^(5/2)) - (4*(5*a*b*c - 2*a^2*d + 3*b^2*d)*Sqrt[c + d*Tan[e + f*x]])/(15*(a^2 + b^2)^2*f*(a +
b*Tan[e + f*x])^(3/2)) + (2*(50*a^3*b*c*d - 70*a*b^3*c*d - 8*a^4*d^2 - a^2*b^2*(45*c^2 - 49*d^2) + 3*b^4*(5*c
^2 - d^2))*Sqrt[c + d*Tan[e + f*x]])/(15*(a^2 + b^2)^3*(b*c - a*d)*f*Sqrt[a + b*Tan[e + f*x]])

Rule 93

Int[(((a_.) + (b_.)*(x_))^(m_)*((c_.) + (d_.)*(x_))^(n_))/((e_.) + (f_.)*(x_)), x_Symbol] :> With[{q = Denomin
ator[m]}, Dist[q, Subst[Int[x^(q*(m + 1) - 1)/(b*e - a*f - (d*e - c*f)*x^q), x], x, (a + b*x)^(1/q)/(c + d*x)^
(1/q)], x]] /; FreeQ[{a, b, c, d, e, f}, x] && EqQ[m + n + 1, 0] && RationalQ[n] && LtQ[-1, m, 0] && SimplerQ[
a + b*x, c + d*x]

Rule 208

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

Rule 3567

Int[((a_.) + (b_.)*tan[(e_.) + (f_.)*(x_)])^(m_)*((c_.) + (d_.)*tan[(e_.) + (f_.)*(x_)])^(n_), x_Symbol] :> Si
mp[((b*c - a*d)*(a + b*Tan[e + f*x])^(m + 1)*(c + d*Tan[e + f*x])^(n - 1))/(f*(m + 1)*(a^2 + b^2)), x] + Dist[
1/((m + 1)*(a^2 + b^2)), Int[(a + b*Tan[e + f*x])^(m + 1)*(c + d*Tan[e + f*x])^(n - 2)*Simp[a*c^2*(m + 1) + a*
d^2*(n - 1) + b*c*d*(m - n + 2) - (b*c^2 - 2*a*c*d - b*d^2)*(m + 1)*Tan[e + f*x] - d*(b*c - a*d)*(m + n)*Tan[e
+ f*x]^2, x], x], x] /; FreeQ[{a, b, c, d, e, f}, x] && NeQ[b*c - a*d, 0] && NeQ[a^2 + b^2, 0] && NeQ[c^2 + d
^2, 0] && LtQ[m, -1] && LtQ[1, n, 2] && IntegerQ[2*m]

Rule 3615

Int[((a_.) + (b_.)*tan[(e_.) + (f_.)*(x_)])^(m_)*((A_.) + (B_.)*tan[(e_.) + (f_.)*(x_)])*((c_.) + (d_.)*tan[(e
_.) + (f_.)*(x_)])^(n_), x_Symbol] :> Dist[A^2/f, Subst[Int[((a + b*x)^m*(c + d*x)^n)/(A - B*x), x], x, Tan[e
+ f*x]], x] /; FreeQ[{a, b, c, d, e, f, A, B, m, n}, x] && NeQ[b*c - a*d, 0] && NeQ[a^2 + b^2, 0] && EqQ[A^2 +
B^2, 0]

Rule 3616

Int[((a_.) + (b_.)*tan[(e_.) + (f_.)*(x_)])^(m_)*((A_.) + (B_.)*tan[(e_.) + (f_.)*(x_)])*((c_.) + (d_.)*tan[(e
_.) + (f_.)*(x_)])^(n_), x_Symbol] :> Dist[(A + I*B)/2, Int[(a + b*Tan[e + f*x])^m*(c + d*Tan[e + f*x])^n*(1 -
I*Tan[e + f*x]), x], x] + Dist[(A - I*B)/2, Int[(a + b*Tan[e + f*x])^m*(c + d*Tan[e + f*x])^n*(1 + I*Tan[e +
f*x]), x], x] /; FreeQ[{a, b, c, d, e, f, A, B, m, n}, x] && NeQ[b*c - a*d, 0] && NeQ[a^2 + b^2, 0] && NeQ[A^2
+ B^2, 0]

Rule 3649

Int[((a_.) + (b_.)*tan[(e_.) + (f_.)*(x_)])^(m_)*((c_.) + (d_.)*tan[(e_.) + (f_.)*(x_)])^(n_)*((A_.) + (B_.)*t
an[(e_.) + (f_.)*(x_)] + (C_.)*tan[(e_.) + (f_.)*(x_)]^2), x_Symbol] :> Simp[((A*b^2 - a*(b*B - a*C))*(a + b*T
an[e + f*x])^(m + 1)*(c + d*Tan[e + f*x])^(n + 1))/(f*(m + 1)*(b*c - a*d)*(a^2 + b^2)), x] + Dist[1/((m + 1)*(
b*c - a*d)*(a^2 + b^2)), Int[(a + b*Tan[e + f*x])^(m + 1)*(c + d*Tan[e + f*x])^n*Simp[A*(a*(b*c - a*d)*(m + 1)
- b^2*d*(m + n + 2)) + (b*B - a*C)*(b*c*(m + 1) + a*d*(n + 1)) - (m + 1)*(b*c - a*d)*(A*b - a*B - b*C)*Tan[e
+ f*x] - d*(A*b^2 - a*(b*B - a*C))*(m + n + 2)*Tan[e + f*x]^2, x], x], x] /; FreeQ[{a, b, c, d, e, f, A, B, C,
n}, x] && NeQ[b*c - a*d, 0] && NeQ[a^2 + b^2, 0] && NeQ[c^2 + d^2, 0] && LtQ[m, -1] &&  !(ILtQ[n, -1] && ( !I
ntegerQ[m] || (EqQ[c, 0] && NeQ[a, 0])))

Rubi steps

\begin {align*} \int \frac {(c+d \tan (e+f x))^{3/2}}{(a+b \tan (e+f x))^{7/2}} \, dx &=-\frac {2 (b c-a d) \sqrt {c+d \tan (e+f x)}}{5 \left (a^2+b^2\right ) f (a+b \tan (e+f x))^{5/2}}-\frac {2 \int \frac {\frac {1}{2} \left (-6 b c d-a \left (5 c^2-d^2\right )\right )-\frac {5}{2} \left (2 a c d-b \left (c^2-d^2\right )\right ) \tan (e+f x)+2 d (b c-a d) \tan ^2(e+f x)}{(a+b \tan (e+f x))^{5/2} \sqrt {c+d \tan (e+f x)}} \, dx}{5 \left (a^2+b^2\right )}\\ &=-\frac {2 (b c-a d) \sqrt {c+d \tan (e+f x)}}{5 \left (a^2+b^2\right ) f (a+b \tan (e+f x))^{5/2}}-\frac {4 \left (5 a b c-2 a^2 d+3 b^2 d\right ) \sqrt {c+d \tan (e+f x)}}{15 \left (a^2+b^2\right )^2 f (a+b \tan (e+f x))^{3/2}}+\frac {4 \int \frac {\frac {1}{4} (b c-a d) \left (15 a^2 c^2-15 b^2 c^2+40 a b c d-7 a^2 d^2+3 b^2 d^2\right )-\frac {15}{2} (b c-a d)^2 (a c+b d) \tan (e+f x)-d (b c-a d) \left (5 a b c-2 a^2 d+3 b^2 d\right ) \tan ^2(e+f x)}{(a+b \tan (e+f x))^{3/2} \sqrt {c+d \tan (e+f x)}} \, dx}{15 \left (a^2+b^2\right )^2 (b c-a d)}\\ &=-\frac {2 (b c-a d) \sqrt {c+d \tan (e+f x)}}{5 \left (a^2+b^2\right ) f (a+b \tan (e+f x))^{5/2}}-\frac {4 \left (5 a b c-2 a^2 d+3 b^2 d\right ) \sqrt {c+d \tan (e+f x)}}{15 \left (a^2+b^2\right )^2 f (a+b \tan (e+f x))^{3/2}}+\frac {2 \left (50 a^3 b c d-70 a b^3 c d-8 a^4 d^2-a^2 b^2 \left (45 c^2-49 d^2\right )+3 b^4 \left (5 c^2-d^2\right )\right ) \sqrt {c+d \tan (e+f x)}}{15 \left (a^2+b^2\right )^3 (b c-a d) f \sqrt {a+b \tan (e+f x)}}-\frac {8 \int \frac {-\frac {15}{8} (b c-a d)^2 \left (6 a^2 b c d-2 b^3 c d+a^3 \left (c^2-d^2\right )-3 a b^2 \left (c^2-d^2\right )\right )+\frac {15}{8} (b c-a d)^2 \left (3 a^2 b c^2-b^3 c^2-2 a^3 c d+6 a b^2 c d-3 a^2 b d^2+b^3 d^2\right ) \tan (e+f x)}{\sqrt {a+b \tan (e+f x)} \sqrt {c+d \tan (e+f x)}} \, dx}{15 \left (a^2+b^2\right )^3 (b c-a d)^2}\\ &=-\frac {2 (b c-a d) \sqrt {c+d \tan (e+f x)}}{5 \left (a^2+b^2\right ) f (a+b \tan (e+f x))^{5/2}}-\frac {4 \left (5 a b c-2 a^2 d+3 b^2 d\right ) \sqrt {c+d \tan (e+f x)}}{15 \left (a^2+b^2\right )^2 f (a+b \tan (e+f x))^{3/2}}+\frac {2 \left (50 a^3 b c d-70 a b^3 c d-8 a^4 d^2-a^2 b^2 \left (45 c^2-49 d^2\right )+3 b^4 \left (5 c^2-d^2\right )\right ) \sqrt {c+d \tan (e+f x)}}{15 \left (a^2+b^2\right )^3 (b c-a d) f \sqrt {a+b \tan (e+f x)}}+\frac {(c-i d)^2 \int \frac {1+i \tan (e+f x)}{\sqrt {a+b \tan (e+f x)} \sqrt {c+d \tan (e+f x)}} \, dx}{2 (a-i b)^3}+\frac {(c+i d)^2 \int \frac {1-i \tan (e+f x)}{\sqrt {a+b \tan (e+f x)} \sqrt {c+d \tan (e+f x)}} \, dx}{2 (a+i b)^3}\\ &=-\frac {2 (b c-a d) \sqrt {c+d \tan (e+f x)}}{5 \left (a^2+b^2\right ) f (a+b \tan (e+f x))^{5/2}}-\frac {4 \left (5 a b c-2 a^2 d+3 b^2 d\right ) \sqrt {c+d \tan (e+f x)}}{15 \left (a^2+b^2\right )^2 f (a+b \tan (e+f x))^{3/2}}+\frac {2 \left (50 a^3 b c d-70 a b^3 c d-8 a^4 d^2-a^2 b^2 \left (45 c^2-49 d^2\right )+3 b^4 \left (5 c^2-d^2\right )\right ) \sqrt {c+d \tan (e+f x)}}{15 \left (a^2+b^2\right )^3 (b c-a d) f \sqrt {a+b \tan (e+f x)}}+\frac {(c-i d)^2 \operatorname {Subst}\left (\int \frac {1}{(1-i x) \sqrt {a+b x} \sqrt {c+d x}} \, dx,x,\tan (e+f x)\right )}{2 (a-i b)^3 f}+\frac {(c+i d)^2 \operatorname {Subst}\left (\int \frac {1}{(1+i x) \sqrt {a+b x} \sqrt {c+d x}} \, dx,x,\tan (e+f x)\right )}{2 (a+i b)^3 f}\\ &=-\frac {2 (b c-a d) \sqrt {c+d \tan (e+f x)}}{5 \left (a^2+b^2\right ) f (a+b \tan (e+f x))^{5/2}}-\frac {4 \left (5 a b c-2 a^2 d+3 b^2 d\right ) \sqrt {c+d \tan (e+f x)}}{15 \left (a^2+b^2\right )^2 f (a+b \tan (e+f x))^{3/2}}+\frac {2 \left (50 a^3 b c d-70 a b^3 c d-8 a^4 d^2-a^2 b^2 \left (45 c^2-49 d^2\right )+3 b^4 \left (5 c^2-d^2\right )\right ) \sqrt {c+d \tan (e+f x)}}{15 \left (a^2+b^2\right )^3 (b c-a d) f \sqrt {a+b \tan (e+f x)}}+\frac {(c-i d)^2 \operatorname {Subst}\left (\int \frac {1}{i a+b-(i c+d) x^2} \, dx,x,\frac {\sqrt {a+b \tan (e+f x)}}{\sqrt {c+d \tan (e+f x)}}\right )}{(a-i b)^3 f}+\frac {(c+i d)^2 \operatorname {Subst}\left (\int \frac {1}{-i a+b-(-i c+d) x^2} \, dx,x,\frac {\sqrt {a+b \tan (e+f x)}}{\sqrt {c+d \tan (e+f x)}}\right )}{(a+i b)^3 f}\\ &=-\frac {i (c-i d)^{3/2} \tanh ^{-1}\left (\frac {\sqrt {c-i d} \sqrt {a+b \tan (e+f x)}}{\sqrt {a-i b} \sqrt {c+d \tan (e+f x)}}\right )}{(a-i b)^{7/2} f}+\frac {i (c+i d)^{3/2} \tanh ^{-1}\left (\frac {\sqrt {c+i d} \sqrt {a+b \tan (e+f x)}}{\sqrt {a+i b} \sqrt {c+d \tan (e+f x)}}\right )}{(a+i b)^{7/2} f}-\frac {2 (b c-a d) \sqrt {c+d \tan (e+f x)}}{5 \left (a^2+b^2\right ) f (a+b \tan (e+f x))^{5/2}}-\frac {4 \left (5 a b c-2 a^2 d+3 b^2 d\right ) \sqrt {c+d \tan (e+f x)}}{15 \left (a^2+b^2\right )^2 f (a+b \tan (e+f x))^{3/2}}+\frac {2 \left (50 a^3 b c d-70 a b^3 c d-8 a^4 d^2-a^2 b^2 \left (45 c^2-49 d^2\right )+3 b^4 \left (5 c^2-d^2\right )\right ) \sqrt {c+d \tan (e+f x)}}{15 \left (a^2+b^2\right )^3 (b c-a d) f \sqrt {a+b \tan (e+f x)}}\\ \end {align*}

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Mathematica [A]  time = 6.27, size = 498, normalized size = 1.27 \[ \frac {b (c+d \tan (e+f x))^{5/2}}{5 f (-b+i a) (b c-a d) (a+b \tan (e+f x))^{5/2}}-\frac {b (c+d \tan (e+f x))^{5/2}}{5 f (b+i a) (b c-a d) (a+b \tan (e+f x))^{5/2}}-\frac {\frac {(c+d \tan (e+f x))^{3/2}}{(a-i b) (a+b \tan (e+f x))^{3/2}}+\frac {3 (c-i d) \left (\frac {\sqrt {c+d \tan (e+f x)}}{(a-i b) \sqrt {a+b \tan (e+f x)}}+\frac {\sqrt {-c+i d} \tanh ^{-1}\left (\frac {\sqrt {-c+i d} \sqrt {a+b \tan (e+f x)}}{\sqrt {-a+i b} \sqrt {c+d \tan (e+f x)}}\right )}{(-a+i b)^{3/2}}\right )}{a-i b}}{3 f (b+i a)}+\frac {\frac {(c+d \tan (e+f x))^{3/2}}{(a+i b) (a+b \tan (e+f x))^{3/2}}-\frac {3 (c+i d) \left (\frac {\sqrt {c+i d} \tanh ^{-1}\left (\frac {\sqrt {c+i d} \sqrt {a+b \tan (e+f x)}}{\sqrt {a+i b} \sqrt {c+d \tan (e+f x)}}\right )}{(a+i b)^{3/2}}-\frac {\sqrt {c+d \tan (e+f x)}}{(a+i b) \sqrt {a+b \tan (e+f x)}}\right )}{a+i b}}{3 f (-b+i a)} \]

Antiderivative was successfully verified.

[In]

Integrate[(c + d*Tan[e + f*x])^(3/2)/(a + b*Tan[e + f*x])^(7/2),x]

[Out]

(b*(c + d*Tan[e + f*x])^(5/2))/(5*(I*a - b)*(b*c - a*d)*f*(a + b*Tan[e + f*x])^(5/2)) - (b*(c + d*Tan[e + f*x]
)^(5/2))/(5*(I*a + b)*(b*c - a*d)*f*(a + b*Tan[e + f*x])^(5/2)) - ((c + d*Tan[e + f*x])^(3/2)/((a - I*b)*(a +
b*Tan[e + f*x])^(3/2)) + (3*(c - I*d)*((Sqrt[-c + I*d]*ArcTanh[(Sqrt[-c + I*d]*Sqrt[a + b*Tan[e + f*x]])/(Sqrt
[-a + I*b]*Sqrt[c + d*Tan[e + f*x]])])/(-a + I*b)^(3/2) + Sqrt[c + d*Tan[e + f*x]]/((a - I*b)*Sqrt[a + b*Tan[e
+ f*x]])))/(a - I*b))/(3*(I*a + b)*f) + ((c + d*Tan[e + f*x])^(3/2)/((a + I*b)*(a + b*Tan[e + f*x])^(3/2)) -
(3*(c + I*d)*((Sqrt[c + I*d]*ArcTanh[(Sqrt[c + I*d]*Sqrt[a + b*Tan[e + f*x]])/(Sqrt[a + I*b]*Sqrt[c + d*Tan[e
+ f*x]])])/(a + I*b)^(3/2) - Sqrt[c + d*Tan[e + f*x]]/((a + I*b)*Sqrt[a + b*Tan[e + f*x]])))/(a + I*b))/(3*(I*
a - b)*f)

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fricas [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((c+d*tan(f*x+e))^(3/2)/(a+b*tan(f*x+e))^(7/2),x, algorithm="fricas")

[Out]

Timed out

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giac [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((c+d*tan(f*x+e))^(3/2)/(a+b*tan(f*x+e))^(7/2),x, algorithm="giac")

[Out]

Timed out

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maple [F(-1)]  time = 180.00, size = 0, normalized size = 0.00 \[ \int \frac {\left (c +d \tan \left (f x +e \right )\right )^{\frac {3}{2}}}{\left (a +b \tan \left (f x +e \right )\right )^{\frac {7}{2}}}\, dx \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

int((c+d*tan(f*x+e))^(3/2)/(a+b*tan(f*x+e))^(7/2),x)

[Out]

int((c+d*tan(f*x+e))^(3/2)/(a+b*tan(f*x+e))^(7/2),x)

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maxima [F(-2)]  time = 0.00, size = 0, normalized size = 0.00 \[ \text {Exception raised: ValueError} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate((c+d*tan(f*x+e))^(3/2)/(a+b*tan(f*x+e))^(7/2),x, algorithm="maxima")

[Out]

Exception raised: ValueError >> Computation failed since Maxima requested additional constraints; using the 'a
ssume' command before evaluation *may* help (example of legal syntax is 'assume(((2*b*d+2*a*c)^2>0)', see `ass
ume?` for more details)Is ((2*b*d+2*a*c)^2    -4*((a*c-b*d)^2       -((-a*d)-b*c)        *(a*d+b*c)))    ^2 po
sitive or zero?

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mupad [F(-1)]  time = 0.00, size = -1, normalized size = -0.00 \[ \text {Hanged} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

int((c + d*tan(e + f*x))^(3/2)/(a + b*tan(e + f*x))^(7/2),x)

[Out]

\text{Hanged}

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sympy [F]  time = 0.00, size = 0, normalized size = 0.00 \[ \int \frac {\left (c + d \tan {\left (e + f x \right )}\right )^{\frac {3}{2}}}{\left (a + b \tan {\left (e + f x \right )}\right )^{\frac {7}{2}}}\, dx \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate((c+d*tan(f*x+e))**(3/2)/(a+b*tan(f*x+e))**(7/2),x)

[Out]

Integral((c + d*tan(e + f*x))**(3/2)/(a + b*tan(e + f*x))**(7/2), x)

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