\(\int \frac {\sqrt {a+b \tan (c+d x)} (A+B \tan (c+d x))}{\tan ^{\frac {7}{2}}(c+d x)} \, dx\) [432]

Optimal result

Integrand size = 35, antiderivative size = 250 \[ \int \frac {\sqrt {a+b \tan (c+d x)} (A+B \tan (c+d x))}{\tan ^{\frac {7}{2}}(c+d x)} \, dx=\frac {\sqrt {i a-b} (A+i B) \arctan \left (\frac {\sqrt {i a-b} \sqrt {\tan (c+d x)}}{\sqrt {a+b \tan (c+d x)}}\right )}{d}-\frac {\sqrt {i a+b} (A-i B) \text {arctanh}\left (\frac {\sqrt {i a+b} \sqrt {\tan (c+d x)}}{\sqrt {a+b \tan (c+d x)}}\right )}{d}-\frac {2 A \sqrt {a+b \tan (c+d x)}}{5 d \tan ^{\frac {5}{2}}(c+d x)}-\frac {2 (A b+5 a B) \sqrt {a+b \tan (c+d x)}}{15 a d \tan ^{\frac {3}{2}}(c+d x)}+\frac {2 \left (15 a^2 A+2 A b^2-5 a b B\right ) \sqrt {a+b \tan (c+d x)}}{15 a^2 d \sqrt {\tan (c+d x)}} \]

[Out]

(A+I*B)*arctan((I*a-b)^(1/2)*tan(d*x+c)^(1/2)/(a+b*tan(d*x+c))^(1/2))*(I*a-b)^(1/2)/d-(A-I*B)*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+2/15*(15*A*a^2+2*A*b^2-5*B*a*b)*(a+b*tan(d*x+c)
)^(1/2)/a^2/d/tan(d*x+c)^(1/2)-2/5*A*(a+b*tan(d*x+c))^(1/2)/d/tan(d*x+c)^(5/2)-2/15*(A*b+5*B*a)*(a+b*tan(d*x+c
))^(1/2)/a/d/tan(d*x+c)^(3/2)

Rubi [A] (verified)

Time = 1.27 (sec) , antiderivative size = 250, normalized size of antiderivative = 1.00, number of steps used = 10, number of rules used = 7, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.200, Rules used = {3689, 3730, 3697, 3696, 95, 209, 212} \[ \int \frac {\sqrt {a+b \tan (c+d x)} (A+B \tan (c+d x))}{\tan ^{\frac {7}{2}}(c+d x)} \, dx=\frac {2 \left (15 a^2 A-5 a b B+2 A b^2\right ) \sqrt {a+b \tan (c+d x)}}{15 a^2 d \sqrt {\tan (c+d x)}}+\frac {\sqrt {-b+i a} (A+i B) \arctan \left (\frac {\sqrt {-b+i a} \sqrt {\tan (c+d x)}}{\sqrt {a+b \tan (c+d x)}}\right )}{d}-\frac {\sqrt {b+i a} (A-i B) \text {arctanh}\left (\frac {\sqrt {b+i a} \sqrt {\tan (c+d x)}}{\sqrt {a+b \tan (c+d x)}}\right )}{d}-\frac {2 (5 a B+A b) \sqrt {a+b \tan (c+d x)}}{15 a d \tan ^{\frac {3}{2}}(c+d x)}-\frac {2 A \sqrt {a+b \tan (c+d x)}}{5 d \tan ^{\frac {5}{2}}(c+d x)} \]

[In]

Int[(Sqrt[a + b*Tan[c + d*x]]*(A + B*Tan[c + d*x]))/Tan[c + d*x]^(7/2),x]

[Out]

(Sqrt[I*a - b]*(A + I*B)*ArcTan[(Sqrt[I*a - b]*Sqrt[Tan[c + d*x]])/Sqrt[a + b*Tan[c + d*x]]])/d - (Sqrt[I*a +
b]*(A - I*B)*ArcTanh[(Sqrt[I*a + b]*Sqrt[Tan[c + d*x]])/Sqrt[a + b*Tan[c + d*x]]])/d - (2*A*Sqrt[a + b*Tan[c +
d*x]])/(5*d*Tan[c + d*x]^(5/2)) - (2*(A*b + 5*a*B)*Sqrt[a + b*Tan[c + d*x]])/(15*a*d*Tan[c + d*x]^(3/2)) + (2
*(15*a^2*A + 2*A*b^2 - 5*a*b*B)*Sqrt[a + b*Tan[c + d*x]])/(15*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 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 \sqrt {a+b \tan (c+d x)}}{5 d \tan ^{\frac {5}{2}}(c+d x)}-\frac {2}{5} \int \frac {\frac {1}{2} (-A b-5 a B)+\frac {5}{2} (a A-b B) \tan (c+d x)+2 A b \tan ^2(c+d x)}{\tan ^{\frac {5}{2}}(c+d x) \sqrt {a+b \tan (c+d x)}} \, dx \\ & = -\frac {2 A \sqrt {a+b \tan (c+d x)}}{5 d \tan ^{\frac {5}{2}}(c+d x)}-\frac {2 (A b+5 a B) \sqrt {a+b \tan (c+d x)}}{15 a d \tan ^{\frac {3}{2}}(c+d x)}+\frac {4 \int \frac {\frac {1}{4} \left (-15 a^2 A-2 A b^2+5 a b B\right )-\frac {15}{4} a (A b+a B) \tan (c+d x)-\frac {1}{2} b (A b+5 a B) \tan ^2(c+d x)}{\tan ^{\frac {3}{2}}(c+d x) \sqrt {a+b \tan (c+d x)}} \, dx}{15 a} \\ & = -\frac {2 A \sqrt {a+b \tan (c+d x)}}{5 d \tan ^{\frac {5}{2}}(c+d x)}-\frac {2 (A b+5 a B) \sqrt {a+b \tan (c+d x)}}{15 a d \tan ^{\frac {3}{2}}(c+d x)}+\frac {2 \left (15 a^2 A+2 A b^2-5 a b B\right ) \sqrt {a+b \tan (c+d x)}}{15 a^2 d \sqrt {\tan (c+d x)}}-\frac {8 \int \frac {\frac {15}{8} a^2 (A b+a B)-\frac {15}{8} a^2 (a A-b B) \tan (c+d x)}{\sqrt {\tan (c+d x)} \sqrt {a+b \tan (c+d x)}} \, dx}{15 a^2} \\ & = -\frac {2 A \sqrt {a+b \tan (c+d x)}}{5 d \tan ^{\frac {5}{2}}(c+d x)}-\frac {2 (A b+5 a B) \sqrt {a+b \tan (c+d x)}}{15 a d \tan ^{\frac {3}{2}}(c+d x)}+\frac {2 \left (15 a^2 A+2 A b^2-5 a b B\right ) \sqrt {a+b \tan (c+d x)}}{15 a^2 d \sqrt {\tan (c+d x)}}-\frac {\left (4 \left (\frac {15}{8} a^2 (A b+a B)-\frac {15}{8} i a^2 (a A-b B)\right )\right ) \int \frac {1-i \tan (c+d x)}{\sqrt {\tan (c+d x)} \sqrt {a+b \tan (c+d x)}} \, dx}{15 a^2}-\frac {\left (4 \left (\frac {15}{8} a^2 (A b+a B)+\frac {15}{8} i a^2 (a A-b B)\right )\right ) \int \frac {1+i \tan (c+d x)}{\sqrt {\tan (c+d x)} \sqrt {a+b \tan (c+d x)}} \, dx}{15 a^2} \\ & = -\frac {2 A \sqrt {a+b \tan (c+d x)}}{5 d \tan ^{\frac {5}{2}}(c+d x)}-\frac {2 (A b+5 a B) \sqrt {a+b \tan (c+d x)}}{15 a d \tan ^{\frac {3}{2}}(c+d x)}+\frac {2 \left (15 a^2 A+2 A b^2-5 a b B\right ) \sqrt {a+b \tan (c+d x)}}{15 a^2 d \sqrt {\tan (c+d x)}}-\frac {\left (4 \left (\frac {15}{8} a^2 (A b+a B)-\frac {15}{8} i a^2 (a A-b B)\right )\right ) \text {Subst}\left (\int \frac {1}{(1+i x) \sqrt {x} \sqrt {a+b x}} \, dx,x,\tan (c+d x)\right )}{15 a^2 d}-\frac {\left (4 \left (\frac {15}{8} a^2 (A b+a B)+\frac {15}{8} i a^2 (a A-b B)\right )\right ) \text {Subst}\left (\int \frac {1}{(1-i x) \sqrt {x} \sqrt {a+b x}} \, dx,x,\tan (c+d x)\right )}{15 a^2 d} \\ & = -\frac {2 A \sqrt {a+b \tan (c+d x)}}{5 d \tan ^{\frac {5}{2}}(c+d x)}-\frac {2 (A b+5 a B) \sqrt {a+b \tan (c+d x)}}{15 a d \tan ^{\frac {3}{2}}(c+d x)}+\frac {2 \left (15 a^2 A+2 A b^2-5 a b B\right ) \sqrt {a+b \tan (c+d x)}}{15 a^2 d \sqrt {\tan (c+d x)}}-\frac {\left (8 \left (\frac {15}{8} a^2 (A b+a B)-\frac {15}{8} i a^2 (a A-b B)\right )\right ) \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 )}{15 a^2 d}-\frac {\left (8 \left (\frac {15}{8} a^2 (A b+a B)+\frac {15}{8} i a^2 (a A-b B)\right )\right ) \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 )}{15 a^2 d} \\ & = \frac {\sqrt {i a-b} (A+i B) \arctan \left (\frac {\sqrt {i a-b} \sqrt {\tan (c+d x)}}{\sqrt {a+b \tan (c+d x)}}\right )}{d}-\frac {\sqrt {i a+b} (A-i 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 A \sqrt {a+b \tan (c+d x)}}{5 d \tan ^{\frac {5}{2}}(c+d x)}-\frac {2 (A b+5 a B) \sqrt {a+b \tan (c+d x)}}{15 a d \tan ^{\frac {3}{2}}(c+d x)}+\frac {2 \left (15 a^2 A+2 A b^2-5 a b B\right ) \sqrt {a+b \tan (c+d x)}}{15 a^2 d \sqrt {\tan (c+d x)}} \\ \end{align*}

Mathematica [A] (verified)

Time = 3.43 (sec) , antiderivative size = 226, normalized size of antiderivative = 0.90 \[ \int \frac {\sqrt {a+b \tan (c+d x)} (A+B \tan (c+d x))}{\tan ^{\frac {7}{2}}(c+d x)} \, dx=\frac {15 \sqrt [4]{-1} \sqrt {-a+i b} (A-i B) \arctan \left (\frac {\sqrt [4]{-1} \sqrt {-a+i b} \sqrt {\tan (c+d x)}}{\sqrt {a+b \tan (c+d x)}}\right )+15 \sqrt [4]{-1} \sqrt {a+i b} (A+i B) \arctan \left (\frac {\sqrt [4]{-1} \sqrt {a+i b} \sqrt {\tan (c+d x)}}{\sqrt {a+b \tan (c+d x)}}\right )+\frac {2 \sqrt {a+b \tan (c+d x)} \left (-3 a^2 A-a (A b+5 a B) \tan (c+d x)+\left (15 a^2 A+2 A b^2-5 a b B\right ) \tan ^2(c+d x)\right )}{a^2 \tan ^{\frac {5}{2}}(c+d x)}}{15 d} \]

[In]

Integrate[(Sqrt[a + b*Tan[c + d*x]]*(A + B*Tan[c + d*x]))/Tan[c + d*x]^(7/2),x]

[Out]

(15*(-1)^(1/4)*Sqrt[-a + I*b]*(A - I*B)*ArcTan[((-1)^(1/4)*Sqrt[-a + I*b]*Sqrt[Tan[c + d*x]])/Sqrt[a + b*Tan[c
+ d*x]]] + 15*(-1)^(1/4)*Sqrt[a + I*b]*(A + I*B)*ArcTan[((-1)^(1/4)*Sqrt[a + I*b]*Sqrt[Tan[c + d*x]])/Sqrt[a
+ b*Tan[c + d*x]]] + (2*Sqrt[a + b*Tan[c + d*x]]*(-3*a^2*A - a*(A*b + 5*a*B)*Tan[c + d*x] + (15*a^2*A + 2*A*b^
2 - 5*a*b*B)*Tan[c + d*x]^2))/(a^2*Tan[c + d*x]^(5/2)))/(15*d)

Maple [B] (warning: unable to verify)

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

Time = 1.40 (sec) , antiderivative size = 2183172, normalized size of antiderivative = 8732.69

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

[In]

int((a+b*tan(d*x+c))^(1/2)*(A+B*tan(d*x+c))/tan(d*x+c)^(7/2),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. 7964 vs. \(2 (204) = 408\).

Time = 1.42 (sec) , antiderivative size = 7964, normalized size of antiderivative = 31.86 \[ \int \frac {\sqrt {a+b \tan (c+d x)} (A+B \tan (c+d x))}{\tan ^{\frac {7}{2}}(c+d x)} \, dx=\text {Too large to display} \]

[In]

integrate((a+b*tan(d*x+c))^(1/2)*(A+B*tan(d*x+c))/tan(d*x+c)^(7/2),x, algorithm="fricas")

[Out]

Too large to include

Sympy [F]

\[ \int \frac {\sqrt {a+b \tan (c+d x)} (A+B \tan (c+d x))}{\tan ^{\frac {7}{2}}(c+d x)} \, dx=\int \frac {\left (A + B \tan {\left (c + d x \right )}\right ) \sqrt {a + b \tan {\left (c + d x \right )}}}{\tan ^{\frac {7}{2}}{\left (c + d x \right )}}\, dx \]

[In]

integrate((a+b*tan(d*x+c))**(1/2)*(A+B*tan(d*x+c))/tan(d*x+c)**(7/2),x)

[Out]

Integral((A + B*tan(c + d*x))*sqrt(a + b*tan(c + d*x))/tan(c + d*x)**(7/2), x)

Maxima [F(-1)]

Timed out. \[ \int \frac {\sqrt {a+b \tan (c+d x)} (A+B \tan (c+d x))}{\tan ^{\frac {7}{2}}(c+d x)} \, dx=\text {Timed out} \]

[In]

integrate((a+b*tan(d*x+c))^(1/2)*(A+B*tan(d*x+c))/tan(d*x+c)^(7/2),x, algorithm="maxima")

[Out]

Timed out

Giac [F(-1)]

Timed out. \[ \int \frac {\sqrt {a+b \tan (c+d x)} (A+B \tan (c+d x))}{\tan ^{\frac {7}{2}}(c+d x)} \, dx=\text {Timed out} \]

[In]

integrate((a+b*tan(d*x+c))^(1/2)*(A+B*tan(d*x+c))/tan(d*x+c)^(7/2),x, algorithm="giac")

[Out]

Timed out

Mupad [F(-1)]

Timed out. \[ \int \frac {\sqrt {a+b \tan (c+d x)} (A+B \tan (c+d x))}{\tan ^{\frac {7}{2}}(c+d x)} \, dx=\int \frac {\left (A+B\,\mathrm {tan}\left (c+d\,x\right )\right )\,\sqrt {a+b\,\mathrm {tan}\left (c+d\,x\right )}}{{\mathrm {tan}\left (c+d\,x\right )}^{7/2}} \,d x \]

[In]

int(((A + B*tan(c + d*x))*(a + b*tan(c + d*x))^(1/2))/tan(c + d*x)^(7/2),x)

[Out]

int(((A + B*tan(c + d*x))*(a + b*tan(c + d*x))^(1/2))/tan(c + d*x)^(7/2), x)