\(\int \frac {A+B \tan (c+d x)}{\tan ^{\frac {5}{2}}(c+d x) \sqrt {a+b \tan (c+d x)}} \, dx\) [456]

Optimal result

Integrand size = 35, antiderivative size = 203 \[ \int \frac {A+B \tan (c+d x)}{\tan ^{\frac {5}{2}}(c+d x) \sqrt {a+b \tan (c+d x)}} \, dx=-\frac {(A+i B) \arctan \left (\frac {\sqrt {i a-b} \sqrt {\tan (c+d x)}}{\sqrt {a+b \tan (c+d x)}}\right )}{\sqrt {i a-b} d}-\frac {(A-i B) \text {arctanh}\left (\frac {\sqrt {i a+b} \sqrt {\tan (c+d x)}}{\sqrt {a+b \tan (c+d x)}}\right )}{\sqrt {i a+b} d}-\frac {2 A \sqrt {a+b \tan (c+d x)}}{3 a d \tan ^{\frac {3}{2}}(c+d x)}+\frac {2 (2 A b-3 a B) \sqrt {a+b \tan (c+d x)}}{3 a^2 d \sqrt {\tan (c+d x)}} \]

[Out]

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

Rubi [A] (verified)

Time = 1.00 (sec) , antiderivative size = 203, normalized size of antiderivative = 1.00, number of steps used = 9, number of rules used = 7, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.200, Rules used = {3690, 3730, 3697, 3696, 95, 209, 212} \[ \int \frac {A+B \tan (c+d x)}{\tan ^{\frac {5}{2}}(c+d x) \sqrt {a+b \tan (c+d x)}} \, dx=\frac {2 (2 A b-3 a B) \sqrt {a+b \tan (c+d x)}}{3 a^2 d \sqrt {\tan (c+d x)}}-\frac {(A+i B) \arctan \left (\frac {\sqrt {-b+i a} \sqrt {\tan (c+d x)}}{\sqrt {a+b \tan (c+d x)}}\right )}{d \sqrt {-b+i a}}-\frac {(A-i B) \text {arctanh}\left (\frac {\sqrt {b+i a} \sqrt {\tan (c+d x)}}{\sqrt {a+b \tan (c+d x)}}\right )}{d \sqrt {b+i a}}-\frac {2 A \sqrt {a+b \tan (c+d x)}}{3 a d \tan ^{\frac {3}{2}}(c+d x)} \]

[In]

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

[Out]

-(((A + I*B)*ArcTan[(Sqrt[I*a - b]*Sqrt[Tan[c + d*x]])/Sqrt[a + b*Tan[c + d*x]]])/(Sqrt[I*a - b]*d)) - ((A - I
*B)*ArcTanh[(Sqrt[I*a + b]*Sqrt[Tan[c + d*x]])/Sqrt[a + b*Tan[c + d*x]]])/(Sqrt[I*a + b]*d) - (2*A*Sqrt[a + b*
Tan[c + d*x]])/(3*a*d*Tan[c + d*x]^(3/2)) + (2*(2*A*b - 3*a*B)*Sqrt[a + b*Tan[c + d*x]])/(3*a^2*d*Sqrt[Tan[c +
d*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 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 3690

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

Rule 3696

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 3697

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

Mathematica [A] (verified)

Time = 2.32 (sec) , antiderivative size = 195, normalized size of antiderivative = 0.96 \[ \int \frac {A+B \tan (c+d x)}{\tan ^{\frac {5}{2}}(c+d x) \sqrt {a+b \tan (c+d x)}} \, dx=\frac {\frac {3 \sqrt [4]{-1} (i A+B) \arctan \left (\frac {\sqrt [4]{-1} \sqrt {-a+i b} \sqrt {\tan (c+d x)}}{\sqrt {a+b \tan (c+d x)}}\right )}{\sqrt {-a+i b}}+\frac {3 (-1)^{3/4} (A+i B) \arctan \left (\frac {\sqrt [4]{-1} \sqrt {a+i b} \sqrt {\tan (c+d x)}}{\sqrt {a+b \tan (c+d x)}}\right )}{\sqrt {a+i b}}-\frac {2 \sqrt {a+b \tan (c+d x)} (a A+(-2 A b+3 a B) \tan (c+d x))}{a^2 \tan ^{\frac {3}{2}}(c+d x)}}{3 d} \]

[In]

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

[Out]

((3*(-1)^(1/4)*(I*A + B)*ArcTan[((-1)^(1/4)*Sqrt[-a + I*b]*Sqrt[Tan[c + d*x]])/Sqrt[a + b*Tan[c + d*x]]])/Sqrt
[-a + I*b] + (3*(-1)^(3/4)*(A + I*B)*ArcTan[((-1)^(1/4)*Sqrt[a + I*b]*Sqrt[Tan[c + d*x]])/Sqrt[a + b*Tan[c + d
*x]]])/Sqrt[a + I*b] - (2*Sqrt[a + b*Tan[c + d*x]]*(a*A + (-2*A*b + 3*a*B)*Tan[c + d*x]))/(a^2*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 = 0.89 (sec) , antiderivative size = 1888895, normalized size of antiderivative = 9304.90

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

[In]

int((A+B*tan(d*x+c))/(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. 10067 vs. \(2 (163) = 326\).

Time = 3.63 (sec) , antiderivative size = 10067, normalized size of antiderivative = 49.59 \[ \int \frac {A+B \tan (c+d x)}{\tan ^{\frac {5}{2}}(c+d x) \sqrt {a+b \tan (c+d x)}} \, dx=\text {Too large to display} \]

[In]

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

[Out]

Too large to include

Sympy [F]

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

[In]

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

[Out]

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

Maxima [F]

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

[In]

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

[Out]

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

Giac [F(-1)]

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

[In]

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

[In]

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

[Out]

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