3.7.14 \(\int \frac {\sqrt {a+b \tan (c+d x)}}{\tan ^{\frac {7}{2}}(c+d x)} \, dx\) [614]

Optimal. Leaf size=221 \[ \frac {\sqrt {i a-b} \text {ArcTan}\left (\frac {\sqrt {i a-b} \sqrt {\tan (c+d x)}}{\sqrt {a+b \tan (c+d x)}}\right )}{d}-\frac {\sqrt {i a+b} \tanh ^{-1}\left (\frac {\sqrt {i a+b} \sqrt {\tan (c+d x)}}{\sqrt {a+b \tan (c+d x)}}\right )}{d}-\frac {2 \sqrt {a+b \tan (c+d x)}}{5 d \tan ^{\frac {5}{2}}(c+d x)}-\frac {2 b \sqrt {a+b \tan (c+d x)}}{15 a d \tan ^{\frac {3}{2}}(c+d x)}+\frac {2 \left (15 a^2+2 b^2\right ) \sqrt {a+b \tan (c+d x)}}{15 a^2 d \sqrt {\tan (c+d x)}} \]

[Out]

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

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Rubi [A]
time = 0.54, antiderivative size = 221, normalized size of antiderivative = 1.00, number of steps used = 10, number of rules used = 7, integrand size = 25, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.280, Rules used = {3649, 3730, 3697, 3696, 95, 209, 212} \begin {gather*} \frac {2 \left (15 a^2+2 b^2\right ) \sqrt {a+b \tan (c+d x)}}{15 a^2 d \sqrt {\tan (c+d x)}}+\frac {\sqrt {-b+i a} \text {ArcTan}\left (\frac {\sqrt {-b+i a} \sqrt {\tan (c+d x)}}{\sqrt {a+b \tan (c+d x)}}\right )}{d}-\frac {2 b \sqrt {a+b \tan (c+d x)}}{15 a d \tan ^{\frac {3}{2}}(c+d x)}-\frac {2 \sqrt {a+b \tan (c+d x)}}{5 d \tan ^{\frac {5}{2}}(c+d x)}-\frac {\sqrt {b+i a} \tanh ^{-1}\left (\frac {\sqrt {b+i a} \sqrt {\tan (c+d x)}}{\sqrt {a+b \tan (c+d x)}}\right )}{d} \end {gather*}

Antiderivative was successfully verified.

[In]

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

[Out]

(Sqrt[I*a - b]*ArcTan[(Sqrt[I*a - b]*Sqrt[Tan[c + d*x]])/Sqrt[a + b*Tan[c + d*x]]])/d - (Sqrt[I*a + b]*ArcTanh
[(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]])/(5*d*Tan[c + d
*x]^(5/2)) - (2*b*Sqrt[a + b*Tan[c + d*x]])/(15*a*d*Tan[c + d*x]^(3/2)) + (2*(15*a^2 + 2*b^2)*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 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 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*} \int \frac {\sqrt {a+b \tan (c+d x)}}{\tan ^{\frac {7}{2}}(c+d x)} \, dx &=-\frac {2 \sqrt {a+b \tan (c+d x)}}{5 d \tan ^{\frac {5}{2}}(c+d x)}-\frac {2}{5} \int \frac {-\frac {b}{2}+\frac {5}{2} a \tan (c+d x)+2 b \tan ^2(c+d x)}{\tan ^{\frac {5}{2}}(c+d x) \sqrt {a+b \tan (c+d x)}} \, dx\\ &=-\frac {2 \sqrt {a+b \tan (c+d x)}}{5 d \tan ^{\frac {5}{2}}(c+d x)}-\frac {2 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^2-2 b^2\right )-\frac {15}{4} a b \tan (c+d x)-\frac {1}{2} b^2 \tan ^2(c+d x)}{\tan ^{\frac {3}{2}}(c+d x) \sqrt {a+b \tan (c+d x)}} \, dx}{15 a}\\ &=-\frac {2 \sqrt {a+b \tan (c+d x)}}{5 d \tan ^{\frac {5}{2}}(c+d x)}-\frac {2 b \sqrt {a+b \tan (c+d x)}}{15 a d \tan ^{\frac {3}{2}}(c+d x)}+\frac {2 \left (15 a^2+2 b^2\right ) \sqrt {a+b \tan (c+d x)}}{15 a^2 d \sqrt {\tan (c+d x)}}-\frac {8 \int \frac {\frac {15 a^2 b}{8}-\frac {15}{8} a^3 \tan (c+d x)}{\sqrt {\tan (c+d x)} \sqrt {a+b \tan (c+d x)}} \, dx}{15 a^2}\\ &=-\frac {2 \sqrt {a+b \tan (c+d x)}}{5 d \tan ^{\frac {5}{2}}(c+d x)}-\frac {2 b \sqrt {a+b \tan (c+d x)}}{15 a d \tan ^{\frac {3}{2}}(c+d x)}+\frac {2 \left (15 a^2+2 b^2\right ) \sqrt {a+b \tan (c+d x)}}{15 a^2 d \sqrt {\tan (c+d x)}}-\frac {1}{2} (-i a+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} (i a+b) \int \frac {1+i \tan (c+d x)}{\sqrt {\tan (c+d x)} \sqrt {a+b \tan (c+d x)}} \, dx\\ &=-\frac {2 \sqrt {a+b \tan (c+d x)}}{5 d \tan ^{\frac {5}{2}}(c+d x)}-\frac {2 b \sqrt {a+b \tan (c+d x)}}{15 a d \tan ^{\frac {3}{2}}(c+d x)}+\frac {2 \left (15 a^2+2 b^2\right ) \sqrt {a+b \tan (c+d x)}}{15 a^2 d \sqrt {\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 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 d}\\ &=-\frac {2 \sqrt {a+b \tan (c+d x)}}{5 d \tan ^{\frac {5}{2}}(c+d x)}-\frac {2 b \sqrt {a+b \tan (c+d x)}}{15 a d \tan ^{\frac {3}{2}}(c+d x)}+\frac {2 \left (15 a^2+2 b^2\right ) \sqrt {a+b \tan (c+d x)}}{15 a^2 d \sqrt {\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 )}{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 )}{d}\\ &=\frac {\sqrt {i a-b} \tan ^{-1}\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} \tanh ^{-1}\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)}}{5 d \tan ^{\frac {5}{2}}(c+d x)}-\frac {2 b \sqrt {a+b \tan (c+d x)}}{15 a d \tan ^{\frac {3}{2}}(c+d x)}+\frac {2 \left (15 a^2+2 b^2\right ) \sqrt {a+b \tan (c+d x)}}{15 a^2 d \sqrt {\tan (c+d x)}}\\ \end {align*}

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Mathematica [A]
time = 1.50, size = 197, normalized size = 0.89 \begin {gather*} \frac {15 \sqrt [4]{-1} \sqrt {-a+i b} \text {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} \text {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 b \tan (c+d x)+\left (15 a^2+2 b^2\right ) \tan ^2(c+d x)\right )}{a^2 \tan ^{\frac {5}{2}}(c+d x)}}{15 d} \end {gather*}

Antiderivative was successfully verified.

[In]

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

[Out]

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

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Maple [B] result has leaf size over 500,000. Avoiding possible recursion issues.
time = 0.40, size = 1092009, normalized size = 4941.22 \[\text {output too large to display}\]

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

result too large to display

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Maxima [F]
time = 0.00, size = 0, normalized size = 0.00 \begin {gather*} \text {Failed to integrate} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

integrate(sqrt(b*tan(d*x + c) + a)/tan(d*x + c)^(7/2), x)

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Fricas [F(-1)] Timed out
time = 0.00, size = 0, normalized size = 0.00 \begin {gather*} \text {Timed out} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

Timed out

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Sympy [F]
time = 0.00, size = 0, normalized size = 0.00 \begin {gather*} \int \frac {\sqrt {a + b \tan {\left (c + d x \right )}}}{\tan ^{\frac {7}{2}}{\left (c + d x \right )}}\, dx \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

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

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Giac [F(-1)] Timed out
time = 0.00, size = 0, normalized size = 0.00 \begin {gather*} \text {Timed out} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

Timed out

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Mupad [F]
time = 0.00, size = -1, normalized size = -0.00 \begin {gather*} \int \frac {\sqrt {a+b\,\mathrm {tan}\left (c+d\,x\right )}}{{\mathrm {tan}\left (c+d\,x\right )}^{7/2}} \,d x \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

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

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