\(\int \frac {\sqrt {\tan (c+d x)} (A+B \tan (c+d x))}{(a+b \tan (c+d x))^{5/2}} \, dx\) [465]

Optimal result

Integrand size = 35, antiderivative size = 244 \[ \int \frac {\sqrt {\tan (c+d x)} (A+B \tan (c+d x))}{(a+b \tan (c+d x))^{5/2}} \, dx=-\frac {(i A-B) \arctan \left (\frac {\sqrt {i a-b} \sqrt {\tan (c+d x)}}{\sqrt {a+b \tan (c+d x)}}\right )}{(i a-b)^{5/2} d}+\frac {(i A+B) \text {arctanh}\left (\frac {\sqrt {i a+b} \sqrt {\tan (c+d x)}}{\sqrt {a+b \tan (c+d x)}}\right )}{(i a+b)^{5/2} d}-\frac {2 (A b-a B) \sqrt {\tan (c+d x)}}{3 \left (a^2+b^2\right ) d (a+b \tan (c+d x))^{3/2}}-\frac {2 \left (5 a^2 A b-A b^3-2 a^3 B+4 a b^2 B\right ) \sqrt {\tan (c+d x)}}{3 a \left (a^2+b^2\right )^2 d \sqrt {a+b \tan (c+d x)}} \]

[Out]

-(I*A-B)*arctan((I*a-b)^(1/2)*tan(d*x+c)^(1/2)/(a+b*tan(d*x+c))^(1/2))/(I*a-b)^(5/2)/d+(I*A+B)*arctanh((I*a+b)
^(1/2)*tan(d*x+c)^(1/2)/(a+b*tan(d*x+c))^(1/2))/(I*a+b)^(5/2)/d-2/3*(5*A*a^2*b-A*b^3-2*B*a^3+4*B*a*b^2)*tan(d*
x+c)^(1/2)/a/(a^2+b^2)^2/d/(a+b*tan(d*x+c))^(1/2)-2/3*(A*b-B*a)*tan(d*x+c)^(1/2)/(a^2+b^2)/d/(a+b*tan(d*x+c))^
(3/2)

Rubi [A] (verified)

Time = 1.15 (sec) , antiderivative size = 244, 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 = {3689, 3730, 3697, 3696, 95, 209, 212} \[ \int \frac {\sqrt {\tan (c+d x)} (A+B \tan (c+d x))}{(a+b \tan (c+d x))^{5/2}} \, dx=-\frac {2 (A b-a B) \sqrt {\tan (c+d x)}}{3 d \left (a^2+b^2\right ) (a+b \tan (c+d x))^{3/2}}-\frac {2 \left (-2 a^3 B+5 a^2 A b+4 a b^2 B-A b^3\right ) \sqrt {\tan (c+d x)}}{3 a d \left (a^2+b^2\right )^2 \sqrt {a+b \tan (c+d x)}}-\frac {(-B+i A) \arctan \left (\frac {\sqrt {-b+i a} \sqrt {\tan (c+d x)}}{\sqrt {a+b \tan (c+d x)}}\right )}{d (-b+i a)^{5/2}}+\frac {(B+i A) \text {arctanh}\left (\frac {\sqrt {b+i a} \sqrt {\tan (c+d x)}}{\sqrt {a+b \tan (c+d x)}}\right )}{d (b+i a)^{5/2}} \]

[In]

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

[Out]

-(((I*A - B)*ArcTan[(Sqrt[I*a - b]*Sqrt[Tan[c + d*x]])/Sqrt[a + b*Tan[c + d*x]]])/((I*a - b)^(5/2)*d)) + ((I*A
+ B)*ArcTanh[(Sqrt[I*a + b]*Sqrt[Tan[c + d*x]])/Sqrt[a + b*Tan[c + d*x]]])/((I*a + b)^(5/2)*d) - (2*(A*b - a*
B)*Sqrt[Tan[c + d*x]])/(3*(a^2 + b^2)*d*(a + b*Tan[c + d*x])^(3/2)) - (2*(5*a^2*A*b - A*b^3 - 2*a^3*B + 4*a*b^
2*B)*Sqrt[Tan[c + d*x]])/(3*a*(a^2 + b^2)^2*d*Sqrt[a + b*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 3689

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

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 b-a B) \sqrt {\tan (c+d x)}}{3 \left (a^2+b^2\right ) d (a+b \tan (c+d x))^{3/2}}-\frac {2 \int \frac {-\frac {1}{2} b (A b-a B)-\frac {3}{2} b (a A+b B) \tan (c+d x)+b (A b-a B) \tan ^2(c+d x)}{\sqrt {\tan (c+d x)} (a+b \tan (c+d x))^{3/2}} \, dx}{3 b \left (a^2+b^2\right )} \\ & = -\frac {2 (A b-a B) \sqrt {\tan (c+d x)}}{3 \left (a^2+b^2\right ) d (a+b \tan (c+d x))^{3/2}}-\frac {2 \left (5 a^2 A b-A b^3-2 a^3 B+4 a b^2 B\right ) \sqrt {\tan (c+d x)}}{3 a \left (a^2+b^2\right )^2 d \sqrt {a+b \tan (c+d x)}}-\frac {4 \int \frac {-\frac {3}{4} a b \left (2 a A b-a^2 B+b^2 B\right )-\frac {3}{4} a b \left (a^2 A-A b^2+2 a b B\right ) \tan (c+d x)}{\sqrt {\tan (c+d x)} \sqrt {a+b \tan (c+d x)}} \, dx}{3 a b \left (a^2+b^2\right )^2} \\ & = -\frac {2 (A b-a B) \sqrt {\tan (c+d x)}}{3 \left (a^2+b^2\right ) d (a+b \tan (c+d x))^{3/2}}-\frac {2 \left (5 a^2 A b-A b^3-2 a^3 B+4 a b^2 B\right ) \sqrt {\tan (c+d x)}}{3 a \left (a^2+b^2\right )^2 d \sqrt {a+b \tan (c+d x)}}+\frac {(i A-B) \int \frac {1-i \tan (c+d x)}{\sqrt {\tan (c+d x)} \sqrt {a+b \tan (c+d x)}} \, dx}{2 (a+i b)^2}-\frac {(i A+B) \int \frac {1+i \tan (c+d x)}{\sqrt {\tan (c+d x)} \sqrt {a+b \tan (c+d x)}} \, dx}{2 (a-i b)^2} \\ & = -\frac {2 (A b-a B) \sqrt {\tan (c+d x)}}{3 \left (a^2+b^2\right ) d (a+b \tan (c+d x))^{3/2}}-\frac {2 \left (5 a^2 A b-A b^3-2 a^3 B+4 a b^2 B\right ) \sqrt {\tan (c+d x)}}{3 a \left (a^2+b^2\right )^2 d \sqrt {a+b \tan (c+d x)}}+\frac {(i A-B) \text {Subst}\left (\int \frac {1}{(1+i x) \sqrt {x} \sqrt {a+b x}} \, dx,x,\tan (c+d x)\right )}{2 (a+i b)^2 d}-\frac {(i A+B) \text {Subst}\left (\int \frac {1}{(1-i x) \sqrt {x} \sqrt {a+b x}} \, dx,x,\tan (c+d x)\right )}{2 (a-i b)^2 d} \\ & = -\frac {2 (A b-a B) \sqrt {\tan (c+d x)}}{3 \left (a^2+b^2\right ) d (a+b \tan (c+d x))^{3/2}}-\frac {2 \left (5 a^2 A b-A b^3-2 a^3 B+4 a b^2 B\right ) \sqrt {\tan (c+d x)}}{3 a \left (a^2+b^2\right )^2 d \sqrt {a+b \tan (c+d x)}}+\frac {(i A-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 )}{(a+i b)^2 d}-\frac {(i A+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 )}{(a-i b)^2 d} \\ & = \frac {(i A-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} (a+i b)^2 d}+\frac {(i A+B) \text {arctanh}\left (\frac {\sqrt {i a+b} \sqrt {\tan (c+d x)}}{\sqrt {a+b \tan (c+d x)}}\right )}{(i a+b)^{5/2} d}-\frac {2 (A b-a B) \sqrt {\tan (c+d x)}}{3 \left (a^2+b^2\right ) d (a+b \tan (c+d x))^{3/2}}-\frac {2 \left (5 a^2 A b-A b^3-2 a^3 B+4 a b^2 B\right ) \sqrt {\tan (c+d x)}}{3 a \left (a^2+b^2\right )^2 d \sqrt {a+b \tan (c+d x)}} \\ \end{align*}

Mathematica [A] (verified)

Time = 3.74 (sec) , antiderivative size = 320, normalized size of antiderivative = 1.31 \[ \int \frac {\sqrt {\tan (c+d x)} (A+B \tan (c+d x))}{(a+b \tan (c+d x))^{5/2}} \, dx=\frac {\frac {2 b (A b-a B) \tan ^{\frac {3}{2}}(c+d x)}{(a+b \tan (c+d x))^{3/2}}+\frac {6 b \left (2 a A b-a^2 B+b^2 B\right ) \tan ^{\frac {3}{2}}(c+d x)}{\left (a^2+b^2\right ) \sqrt {a+b \tan (c+d x)}}+\frac {3 \left (-\frac {\sqrt [4]{-1} a (a+i b)^2 (A-i B) \arctan \left (\frac {\sqrt [4]{-1} \sqrt {-a+i b} \sqrt {\tan (c+d x)}}{\sqrt {a+b \tan (c+d x)}}\right )}{\sqrt {-a+i b}}+\frac {\sqrt [4]{-1} a (a-i b)^2 (A+i B) \arctan \left (\frac {\sqrt [4]{-1} \sqrt {a+i b} \sqrt {\tan (c+d x)}}{\sqrt {a+b \tan (c+d x)}}\right )}{\sqrt {a+i b}}+2 \left (-2 a A b+a^2 B-b^2 B\right ) \sqrt {\tan (c+d x)} \sqrt {a+b \tan (c+d x)}\right )}{a^2+b^2}}{3 a \left (a^2+b^2\right ) d} \]

[In]

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

[Out]

((2*b*(A*b - a*B)*Tan[c + d*x]^(3/2))/(a + b*Tan[c + d*x])^(3/2) + (6*b*(2*a*A*b - a^2*B + b^2*B)*Tan[c + d*x]
^(3/2))/((a^2 + b^2)*Sqrt[a + b*Tan[c + d*x]]) + (3*(-(((-1)^(1/4)*a*(a + I*b)^2*(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]) + ((-1)^(1/4)*a*(a - I*b)^2*(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*(-2*a*
A*b + a^2*B - b^2*B)*Sqrt[Tan[c + d*x]]*Sqrt[a + b*Tan[c + d*x]]))/(a^2 + b^2))/(3*a*(a^2 + b^2)*d)

Maple [B] (warning: unable to verify)

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

Time = 2.54 (sec) , antiderivative size = 2978176, normalized size of antiderivative = 12205.64

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

[In]

int(tan(d*x+c)^(1/2)*(A+B*tan(d*x+c))/(a+b*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. 27593 vs. \(2 (200) = 400\).

Time = 14.52 (sec) , antiderivative size = 27593, normalized size of antiderivative = 113.09 \[ \int \frac {\sqrt {\tan (c+d x)} (A+B \tan (c+d x))}{(a+b \tan (c+d x))^{5/2}} \, dx=\text {Too large to display} \]

[In]

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

[Out]

Too large to include

Sympy [F]

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

[In]

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

[Out]

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

Maxima [F]

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

[In]

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

[Out]

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

Giac [F(-1)]

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

[In]

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

[Out]

Timed out

Mupad [F(-1)]

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

[In]

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

[Out]

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