\(\int \frac {\sqrt {a+b \tan (c+d x)}}{\tan ^{\frac {5}{2}}(c+d x)} \, dx\) [613]

Optimal result

Integrand size = 25, antiderivative size = 181 \[ \int \frac {\sqrt {a+b \tan (c+d x)}}{\tan ^{\frac {5}{2}}(c+d x)} \, dx=\frac {i \sqrt {i a-b} \arctan \left (\frac {\sqrt {i a-b} \sqrt {\tan (c+d x)}}{\sqrt {a+b \tan (c+d x)}}\right )}{d}+\frac {i \sqrt {i a+b} \text {arctanh}\left (\frac {\sqrt {i a+b} \sqrt {\tan (c+d x)}}{\sqrt {a+b \tan (c+d x)}}\right )}{d}-\frac {2 \sqrt {a+b \tan (c+d x)}}{3 d \tan ^{\frac {3}{2}}(c+d x)}-\frac {2 b \sqrt {a+b \tan (c+d x)}}{3 a d \sqrt {\tan (c+d x)}} \]

[Out]

I*arctan((I*a-b)^(1/2)*tan(d*x+c)^(1/2)/(a+b*tan(d*x+c))^(1/2))*(I*a-b)^(1/2)/d+I*arctanh((I*a+b)^(1/2)*tan(d*
x+c)^(1/2)/(a+b*tan(d*x+c))^(1/2))*(I*a+b)^(1/2)/d-2/3*b*(a+b*tan(d*x+c))^(1/2)/a/d/tan(d*x+c)^(1/2)-2/3*(a+b*
tan(d*x+c))^(1/2)/d/tan(d*x+c)^(3/2)

Rubi [A] (verified)

Time = 0.45 (sec) , antiderivative size = 181, normalized size of antiderivative = 1.00, number of steps used = 10, number of rules used = 8, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.320, Rules used = {3649, 3730, 21, 3656, 924, 95, 211, 214} \[ \int \frac {\sqrt {a+b \tan (c+d x)}}{\tan ^{\frac {5}{2}}(c+d x)} \, dx=\frac {i \sqrt {-b+i a} \arctan \left (\frac {\sqrt {-b+i a} \sqrt {\tan (c+d x)}}{\sqrt {a+b \tan (c+d x)}}\right )}{d}+\frac {i \sqrt {b+i a} \text {arctanh}\left (\frac {\sqrt {b+i a} \sqrt {\tan (c+d x)}}{\sqrt {a+b \tan (c+d x)}}\right )}{d}-\frac {2 \sqrt {a+b \tan (c+d x)}}{3 d \tan ^{\frac {3}{2}}(c+d x)}-\frac {2 b \sqrt {a+b \tan (c+d x)}}{3 a d \sqrt {\tan (c+d x)}} \]

[In]

Int[Sqrt[a + b*Tan[c + d*x]]/Tan[c + d*x]^(5/2),x]

[Out]

(I*Sqrt[I*a - b]*ArcTan[(Sqrt[I*a - b]*Sqrt[Tan[c + d*x]])/Sqrt[a + b*Tan[c + d*x]]])/d + (I*Sqrt[I*a + b]*Arc
Tanh[(Sqrt[I*a + b]*Sqrt[Tan[c + d*x]])/Sqrt[a + b*Tan[c + d*x]]])/d - (2*Sqrt[a + b*Tan[c + d*x]])/(3*d*Tan[c
+ d*x]^(3/2)) - (2*b*Sqrt[a + b*Tan[c + d*x]])/(3*a*d*Sqrt[Tan[c + d*x]])

Rule 21

Int[(u_.)*((a_) + (b_.)*(v_))^(m_.)*((c_) + (d_.)*(v_))^(n_.), x_Symbol] :> Dist[(b/d)^m, Int[u*(c + d*v)^(m +
n), x], x] /; FreeQ[{a, b, c, d, n}, x] && EqQ[b*c - a*d, 0] && IntegerQ[m] && ( !IntegerQ[n] || SimplerQ[c +
d*x, a + b*x])

Rule 95

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 211

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

Rule 214

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

Rule 924

Int[((d_.) + (e_.)*(x_))^(m_)/(Sqrt[(f_.) + (g_.)*(x_)]*((a_.) + (c_.)*(x_)^2)), x_Symbol] :> Int[ExpandIntegr
and[1/(Sqrt[d + e*x]*Sqrt[f + g*x]), (d + e*x)^(m + 1/2)/(a + c*x^2), x], x] /; FreeQ[{a, c, d, e, f, g}, x] &
& NeQ[c*d^2 + a*e^2, 0] && IGtQ[m + 1/2, 0]

Rule 3649

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

Rule 3656

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

Rule 3730

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*Ta
n[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*} \text {integral}& = -\frac {2 \sqrt {a+b \tan (c+d x)}}{3 d \tan ^{\frac {3}{2}}(c+d x)}-\frac {2}{3} \int \frac {-\frac {b}{2}+\frac {3}{2} a \tan (c+d x)+b \tan ^2(c+d x)}{\tan ^{\frac {3}{2}}(c+d x) \sqrt {a+b \tan (c+d x)}} \, dx \\ & = -\frac {2 \sqrt {a+b \tan (c+d x)}}{3 d \tan ^{\frac {3}{2}}(c+d x)}-\frac {2 b \sqrt {a+b \tan (c+d x)}}{3 a d \sqrt {\tan (c+d x)}}+\frac {4 \int \frac {-\frac {3 a^2}{4}-\frac {3}{4} a b \tan (c+d x)}{\sqrt {\tan (c+d x)} \sqrt {a+b \tan (c+d x)}} \, dx}{3 a} \\ & = -\frac {2 \sqrt {a+b \tan (c+d x)}}{3 d \tan ^{\frac {3}{2}}(c+d x)}-\frac {2 b \sqrt {a+b \tan (c+d x)}}{3 a d \sqrt {\tan (c+d x)}}-\int \frac {\sqrt {a+b \tan (c+d x)}}{\sqrt {\tan (c+d x)}} \, dx \\ & = -\frac {2 \sqrt {a+b \tan (c+d x)}}{3 d \tan ^{\frac {3}{2}}(c+d x)}-\frac {2 b \sqrt {a+b \tan (c+d x)}}{3 a d \sqrt {\tan (c+d x)}}-\frac {\text {Subst}\left (\int \frac {\sqrt {a+b x}}{\sqrt {x} \left (1+x^2\right )} \, dx,x,\tan (c+d x)\right )}{d} \\ & = -\frac {2 \sqrt {a+b \tan (c+d x)}}{3 d \tan ^{\frac {3}{2}}(c+d x)}-\frac {2 b \sqrt {a+b \tan (c+d x)}}{3 a d \sqrt {\tan (c+d x)}}-\frac {\text {Subst}\left (\int \left (\frac {i a-b}{2 (i-x) \sqrt {x} \sqrt {a+b x}}+\frac {i a+b}{2 \sqrt {x} (i+x) \sqrt {a+b x}}\right ) \, dx,x,\tan (c+d x)\right )}{d} \\ & = -\frac {2 \sqrt {a+b \tan (c+d x)}}{3 d \tan ^{\frac {3}{2}}(c+d x)}-\frac {2 b \sqrt {a+b \tan (c+d x)}}{3 a d \sqrt {\tan (c+d x)}}-\frac {(i a-b) \text {Subst}\left (\int \frac {1}{(i-x) \sqrt {x} \sqrt {a+b x}} \, dx,x,\tan (c+d x)\right )}{2 d}-\frac {(i a+b) \text {Subst}\left (\int \frac {1}{\sqrt {x} (i+x) \sqrt {a+b x}} \, dx,x,\tan (c+d x)\right )}{2 d} \\ & = -\frac {2 \sqrt {a+b \tan (c+d x)}}{3 d \tan ^{\frac {3}{2}}(c+d x)}-\frac {2 b \sqrt {a+b \tan (c+d x)}}{3 a d \sqrt {\tan (c+d x)}}-\frac {(i a-b) \text {Subst}\left (\int \frac {1}{i-(a+i b) x^2} \, dx,x,\frac {\sqrt {\tan (c+d x)}}{\sqrt {a+b \tan (c+d x)}}\right )}{d}-\frac {(i a+b) \text {Subst}\left (\int \frac {1}{i-(-a+i b) x^2} \, dx,x,\frac {\sqrt {\tan (c+d x)}}{\sqrt {a+b \tan (c+d x)}}\right )}{d} \\ & = \frac {i \sqrt {i a-b} \arctan \left (\frac {\sqrt {i a-b} \sqrt {\tan (c+d x)}}{\sqrt {a+b \tan (c+d x)}}\right )}{d}+\frac {i \sqrt {i a+b} \text {arctanh}\left (\frac {\sqrt {i a+b} \sqrt {\tan (c+d x)}}{\sqrt {a+b \tan (c+d x)}}\right )}{d}-\frac {2 \sqrt {a+b \tan (c+d x)}}{3 d \tan ^{\frac {3}{2}}(c+d x)}-\frac {2 b \sqrt {a+b \tan (c+d x)}}{3 a d \sqrt {\tan (c+d x)}} \\ \end{align*}

Mathematica [A] (verified)

Time = 0.66 (sec) , antiderivative size = 161, normalized size of antiderivative = 0.89 \[ \int \frac {\sqrt {a+b \tan (c+d x)}}{\tan ^{\frac {5}{2}}(c+d x)} \, dx=\frac {-3 (-1)^{3/4} \sqrt {-a+i b} \arctan \left (\frac {\sqrt [4]{-1} \sqrt {-a+i b} \sqrt {\tan (c+d x)}}{\sqrt {a+b \tan (c+d x)}}\right )+3 (-1)^{3/4} \sqrt {a+i b} \arctan \left (\frac {\sqrt [4]{-1} \sqrt {a+i b} \sqrt {\tan (c+d x)}}{\sqrt {a+b \tan (c+d x)}}\right )-\frac {2 (a+b \tan (c+d x))^{3/2}}{a \tan ^{\frac {3}{2}}(c+d x)}}{3 d} \]

[In]

Integrate[Sqrt[a + b*Tan[c + d*x]]/Tan[c + d*x]^(5/2),x]

[Out]

(-3*(-1)^(3/4)*Sqrt[-a + I*b]*ArcTan[((-1)^(1/4)*Sqrt[-a + I*b]*Sqrt[Tan[c + d*x]])/Sqrt[a + b*Tan[c + d*x]]]
+ 3*(-1)^(3/4)*Sqrt[a + I*b]*ArcTan[((-1)^(1/4)*Sqrt[a + I*b]*Sqrt[Tan[c + d*x]])/Sqrt[a + b*Tan[c + d*x]]] -
(2*(a + b*Tan[c + d*x])^(3/2))/(a*Tan[c + d*x]^(3/2)))/(3*d)

Maple [B] (warning: unable to verify)

result has leaf size over 500,000. Avoiding possible recursion issues.

Time = 0.54 (sec) , antiderivative size = 1090997, normalized size of antiderivative = 6027.61

\[\text {output too large to display}\]

[In]

int((a+b*tan(d*x+c))^(1/2)/tan(d*x+c)^(5/2),x)

[Out]

result too large to display

Fricas [B] (verification not implemented)

Leaf count of result is larger than twice the leaf count of optimal. 2743 vs. \(2 (141) = 282\).

Time = 0.49 (sec) , antiderivative size = 2743, normalized size of antiderivative = 15.15 \[ \int \frac {\sqrt {a+b \tan (c+d x)}}{\tan ^{\frac {5}{2}}(c+d x)} \, dx=\text {Too large to display} \]

[In]

integrate((a+b*tan(d*x+c))^(1/2)/tan(d*x+c)^(5/2),x, algorithm="fricas")

[Out]

1/24*(3*a*d*sqrt(-(d^2*sqrt(-a^2/d^4) + b)/d^2)*log((2*(2*a^4*b + 4*a^2*b^3 + (a^5 + 3*a^3*b^2 + 4*a*b^4)*tan(
d*x + c) + (2*(a^3*b + 2*a*b^3)*d^2*tan(d*x + c) - (a^4 + 3*a^2*b^2 + 4*b^4)*d^2)*sqrt(-a^2/d^4))*sqrt(b*tan(d
*x + c) + a)*sqrt(tan(d*x + c)) + ((a^4*b + 4*a^2*b^3)*d*tan(d*x + c)^2 - 2*(a^5 + 3*a^3*b^2 + 4*a*b^4)*d*tan(
d*x + c) - (3*a^4*b + 4*a^2*b^3)*d + (a^4*d^3 - (a^4 + 6*a^2*b^2 + 8*b^4)*d^3*tan(d*x + c)^2 - 4*(a^3*b + 2*a*
b^3)*d^3*tan(d*x + c))*sqrt(-a^2/d^4))*sqrt(-(d^2*sqrt(-a^2/d^4) + b)/d^2))/(tan(d*x + c)^2 + 1))*tan(d*x + c)
^2 + 3*a*d*sqrt(-(d^2*sqrt(-a^2/d^4) + b)/d^2)*log(-(2*(2*a^4*b + 4*a^2*b^3 + (a^5 + 3*a^3*b^2 + 4*a*b^4)*tan(
d*x + c) + (2*(a^3*b + 2*a*b^3)*d^2*tan(d*x + c) - (a^4 + 3*a^2*b^2 + 4*b^4)*d^2)*sqrt(-a^2/d^4))*sqrt(b*tan(d
*x + c) + a)*sqrt(tan(d*x + c)) + ((a^4*b + 4*a^2*b^3)*d*tan(d*x + c)^2 - 2*(a^5 + 3*a^3*b^2 + 4*a*b^4)*d*tan(
d*x + c) - (3*a^4*b + 4*a^2*b^3)*d + (a^4*d^3 - (a^4 + 6*a^2*b^2 + 8*b^4)*d^3*tan(d*x + c)^2 - 4*(a^3*b + 2*a*
b^3)*d^3*tan(d*x + c))*sqrt(-a^2/d^4))*sqrt(-(d^2*sqrt(-a^2/d^4) + b)/d^2))/(tan(d*x + c)^2 + 1))*tan(d*x + c)
^2 - 3*a*d*sqrt(-(d^2*sqrt(-a^2/d^4) + b)/d^2)*log((2*(2*a^4*b + 4*a^2*b^3 + (a^5 + 3*a^3*b^2 + 4*a*b^4)*tan(d
*x + c) + (2*(a^3*b + 2*a*b^3)*d^2*tan(d*x + c) - (a^4 + 3*a^2*b^2 + 4*b^4)*d^2)*sqrt(-a^2/d^4))*sqrt(b*tan(d*
x + c) + a)*sqrt(tan(d*x + c)) - ((a^4*b + 4*a^2*b^3)*d*tan(d*x + c)^2 - 2*(a^5 + 3*a^3*b^2 + 4*a*b^4)*d*tan(d
*x + c) - (3*a^4*b + 4*a^2*b^3)*d + (a^4*d^3 - (a^4 + 6*a^2*b^2 + 8*b^4)*d^3*tan(d*x + c)^2 - 4*(a^3*b + 2*a*b
^3)*d^3*tan(d*x + c))*sqrt(-a^2/d^4))*sqrt(-(d^2*sqrt(-a^2/d^4) + b)/d^2))/(tan(d*x + c)^2 + 1))*tan(d*x + c)^
2 - 3*a*d*sqrt(-(d^2*sqrt(-a^2/d^4) + b)/d^2)*log(-(2*(2*a^4*b + 4*a^2*b^3 + (a^5 + 3*a^3*b^2 + 4*a*b^4)*tan(d
*x + c) + (2*(a^3*b + 2*a*b^3)*d^2*tan(d*x + c) - (a^4 + 3*a^2*b^2 + 4*b^4)*d^2)*sqrt(-a^2/d^4))*sqrt(b*tan(d*
x + c) + a)*sqrt(tan(d*x + c)) - ((a^4*b + 4*a^2*b^3)*d*tan(d*x + c)^2 - 2*(a^5 + 3*a^3*b^2 + 4*a*b^4)*d*tan(d
*x + c) - (3*a^4*b + 4*a^2*b^3)*d + (a^4*d^3 - (a^4 + 6*a^2*b^2 + 8*b^4)*d^3*tan(d*x + c)^2 - 4*(a^3*b + 2*a*b
^3)*d^3*tan(d*x + c))*sqrt(-a^2/d^4))*sqrt(-(d^2*sqrt(-a^2/d^4) + b)/d^2))/(tan(d*x + c)^2 + 1))*tan(d*x + c)^
2 + 3*a*d*sqrt((d^2*sqrt(-a^2/d^4) - b)/d^2)*log((2*(2*a^4*b + 4*a^2*b^3 + (a^5 + 3*a^3*b^2 + 4*a*b^4)*tan(d*x
+ c) - (2*(a^3*b + 2*a*b^3)*d^2*tan(d*x + c) - (a^4 + 3*a^2*b^2 + 4*b^4)*d^2)*sqrt(-a^2/d^4))*sqrt(b*tan(d*x
+ c) + a)*sqrt(tan(d*x + c)) + ((a^4*b + 4*a^2*b^3)*d*tan(d*x + c)^2 - 2*(a^5 + 3*a^3*b^2 + 4*a*b^4)*d*tan(d*x
+ c) - (3*a^4*b + 4*a^2*b^3)*d - (a^4*d^3 - (a^4 + 6*a^2*b^2 + 8*b^4)*d^3*tan(d*x + c)^2 - 4*(a^3*b + 2*a*b^3
)*d^3*tan(d*x + c))*sqrt(-a^2/d^4))*sqrt((d^2*sqrt(-a^2/d^4) - b)/d^2))/(tan(d*x + c)^2 + 1))*tan(d*x + c)^2 +
3*a*d*sqrt((d^2*sqrt(-a^2/d^4) - b)/d^2)*log(-(2*(2*a^4*b + 4*a^2*b^3 + (a^5 + 3*a^3*b^2 + 4*a*b^4)*tan(d*x +
c) - (2*(a^3*b + 2*a*b^3)*d^2*tan(d*x + c) - (a^4 + 3*a^2*b^2 + 4*b^4)*d^2)*sqrt(-a^2/d^4))*sqrt(b*tan(d*x +
c) + a)*sqrt(tan(d*x + c)) + ((a^4*b + 4*a^2*b^3)*d*tan(d*x + c)^2 - 2*(a^5 + 3*a^3*b^2 + 4*a*b^4)*d*tan(d*x +
c) - (3*a^4*b + 4*a^2*b^3)*d - (a^4*d^3 - (a^4 + 6*a^2*b^2 + 8*b^4)*d^3*tan(d*x + c)^2 - 4*(a^3*b + 2*a*b^3)*
d^3*tan(d*x + c))*sqrt(-a^2/d^4))*sqrt((d^2*sqrt(-a^2/d^4) - b)/d^2))/(tan(d*x + c)^2 + 1))*tan(d*x + c)^2 - 3
*a*d*sqrt((d^2*sqrt(-a^2/d^4) - b)/d^2)*log((2*(2*a^4*b + 4*a^2*b^3 + (a^5 + 3*a^3*b^2 + 4*a*b^4)*tan(d*x + c)
- (2*(a^3*b + 2*a*b^3)*d^2*tan(d*x + c) - (a^4 + 3*a^2*b^2 + 4*b^4)*d^2)*sqrt(-a^2/d^4))*sqrt(b*tan(d*x + c)
+ a)*sqrt(tan(d*x + c)) - ((a^4*b + 4*a^2*b^3)*d*tan(d*x + c)^2 - 2*(a^5 + 3*a^3*b^2 + 4*a*b^4)*d*tan(d*x + c)
- (3*a^4*b + 4*a^2*b^3)*d - (a^4*d^3 - (a^4 + 6*a^2*b^2 + 8*b^4)*d^3*tan(d*x + c)^2 - 4*(a^3*b + 2*a*b^3)*d^3
*tan(d*x + c))*sqrt(-a^2/d^4))*sqrt((d^2*sqrt(-a^2/d^4) - b)/d^2))/(tan(d*x + c)^2 + 1))*tan(d*x + c)^2 - 3*a*
d*sqrt((d^2*sqrt(-a^2/d^4) - b)/d^2)*log(-(2*(2*a^4*b + 4*a^2*b^3 + (a^5 + 3*a^3*b^2 + 4*a*b^4)*tan(d*x + c) -
(2*(a^3*b + 2*a*b^3)*d^2*tan(d*x + c) - (a^4 + 3*a^2*b^2 + 4*b^4)*d^2)*sqrt(-a^2/d^4))*sqrt(b*tan(d*x + c) +
a)*sqrt(tan(d*x + c)) - ((a^4*b + 4*a^2*b^3)*d*tan(d*x + c)^2 - 2*(a^5 + 3*a^3*b^2 + 4*a*b^4)*d*tan(d*x + c) -
(3*a^4*b + 4*a^2*b^3)*d - (a^4*d^3 - (a^4 + 6*a^2*b^2 + 8*b^4)*d^3*tan(d*x + c)^2 - 4*(a^3*b + 2*a*b^3)*d^3*t
an(d*x + c))*sqrt(-a^2/d^4))*sqrt((d^2*sqrt(-a^2/d^4) - b)/d^2))/(tan(d*x + c)^2 + 1))*tan(d*x + c)^2 - 16*(b*
tan(d*x + c) + a)^(3/2)*sqrt(tan(d*x + c)))/(a*d*tan(d*x + c)^2)

Sympy [F]

\[ \int \frac {\sqrt {a+b \tan (c+d x)}}{\tan ^{\frac {5}{2}}(c+d x)} \, dx=\int \frac {\sqrt {a + b \tan {\left (c + d x \right )}}}{\tan ^{\frac {5}{2}}{\left (c + d x \right )}}\, dx \]

[In]

integrate((a+b*tan(d*x+c))**(1/2)/tan(d*x+c)**(5/2),x)

[Out]

Integral(sqrt(a + b*tan(c + d*x))/tan(c + d*x)**(5/2), x)

Maxima [F]

\[ \int \frac {\sqrt {a+b \tan (c+d x)}}{\tan ^{\frac {5}{2}}(c+d x)} \, dx=\int { \frac {\sqrt {b \tan \left (d x + c\right ) + a}}{\tan \left (d x + c\right )^{\frac {5}{2}}} \,d x } \]

[In]

integrate((a+b*tan(d*x+c))^(1/2)/tan(d*x+c)^(5/2),x, algorithm="maxima")

[Out]

integrate(sqrt(b*tan(d*x + c) + a)/tan(d*x + c)^(5/2), x)

Giac [F(-1)]

Timed out. \[ \int \frac {\sqrt {a+b \tan (c+d x)}}{\tan ^{\frac {5}{2}}(c+d x)} \, dx=\text {Timed out} \]

[In]

integrate((a+b*tan(d*x+c))^(1/2)/tan(d*x+c)^(5/2),x, algorithm="giac")

[Out]

Timed out

Mupad [F(-1)]

Timed out. \[ \int \frac {\sqrt {a+b \tan (c+d x)}}{\tan ^{\frac {5}{2}}(c+d x)} \, dx=\int \frac {\sqrt {a+b\,\mathrm {tan}\left (c+d\,x\right )}}{{\mathrm {tan}\left (c+d\,x\right )}^{5/2}} \,d x \]

[In]

int((a + b*tan(c + d*x))^(1/2)/tan(c + d*x)^(5/2),x)

[Out]

int((a + b*tan(c + d*x))^(1/2)/tan(c + d*x)^(5/2), x)