3.639 \(\int \frac{1}{\tan ^{\frac{7}{2}}(c+d x) \sqrt{a+b \tan (c+d x)}} \, dx\)

Optimal. Leaf size=229 \[ \frac{2 \left (15 a^2-8 b^2\right ) \sqrt{a+b \tan (c+d x)}}{15 a^3 d \sqrt{\tan (c+d x)}}+\frac{8 b \sqrt{a+b \tan (c+d x)}}{15 a^2 d \tan ^{\frac{3}{2}}(c+d x)}+\frac{i \tan ^{-1}\left (\frac{\sqrt{-b+i a} \sqrt{\tan (c+d x)}}{\sqrt{a+b \tan (c+d x)}}\right )}{d \sqrt{-b+i a}}-\frac{2 \sqrt{a+b \tan (c+d x)}}{5 a d \tan ^{\frac{5}{2}}(c+d x)}-\frac{i \tanh ^{-1}\left (\frac{\sqrt{b+i a} \sqrt{\tan (c+d x)}}{\sqrt{a+b \tan (c+d x)}}\right )}{d \sqrt{b+i a}} \]

[Out]

(I*ArcTan[(Sqrt[I*a - b]*Sqrt[Tan[c + d*x]])/Sqrt[a + b*Tan[c + d*x]]])/(Sqrt[I*a - b]*d) - (I*ArcTanh[(Sqrt[I
*a + b]*Sqrt[Tan[c + d*x]])/Sqrt[a + b*Tan[c + d*x]]])/(Sqrt[I*a + b]*d) - (2*Sqrt[a + b*Tan[c + d*x]])/(5*a*d
*Tan[c + d*x]^(5/2)) + (8*b*Sqrt[a + b*Tan[c + d*x]])/(15*a^2*d*Tan[c + d*x]^(3/2)) + (2*(15*a^2 - 8*b^2)*Sqrt
[a + b*Tan[c + d*x]])/(15*a^3*d*Sqrt[Tan[c + d*x]])

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Rubi [A]  time = 0.500721, antiderivative size = 229, normalized size of antiderivative = 1., number of steps used = 11, number of rules used = 9, integrand size = 25, \(\frac{\text{number of rules}}{\text{integrand size}}\) = 0.36, Rules used = {3569, 3649, 3650, 12, 3575, 910, 93, 205, 208} \[ \frac{2 \left (15 a^2-8 b^2\right ) \sqrt{a+b \tan (c+d x)}}{15 a^3 d \sqrt{\tan (c+d x)}}+\frac{8 b \sqrt{a+b \tan (c+d x)}}{15 a^2 d \tan ^{\frac{3}{2}}(c+d x)}+\frac{i \tan ^{-1}\left (\frac{\sqrt{-b+i a} \sqrt{\tan (c+d x)}}{\sqrt{a+b \tan (c+d x)}}\right )}{d \sqrt{-b+i a}}-\frac{2 \sqrt{a+b \tan (c+d x)}}{5 a d \tan ^{\frac{5}{2}}(c+d x)}-\frac{i \tanh ^{-1}\left (\frac{\sqrt{b+i a} \sqrt{\tan (c+d x)}}{\sqrt{a+b \tan (c+d x)}}\right )}{d \sqrt{b+i a}} \]

Antiderivative was successfully verified.

[In]

Int[1/(Tan[c + d*x]^(7/2)*Sqrt[a + b*Tan[c + d*x]]),x]

[Out]

(I*ArcTan[(Sqrt[I*a - b]*Sqrt[Tan[c + d*x]])/Sqrt[a + b*Tan[c + d*x]]])/(Sqrt[I*a - b]*d) - (I*ArcTanh[(Sqrt[I
*a + b]*Sqrt[Tan[c + d*x]])/Sqrt[a + b*Tan[c + d*x]]])/(Sqrt[I*a + b]*d) - (2*Sqrt[a + b*Tan[c + d*x]])/(5*a*d
*Tan[c + d*x]^(5/2)) + (8*b*Sqrt[a + b*Tan[c + d*x]])/(15*a^2*d*Tan[c + d*x]^(3/2)) + (2*(15*a^2 - 8*b^2)*Sqrt
[a + b*Tan[c + d*x]])/(15*a^3*d*Sqrt[Tan[c + d*x]])

Rule 3569

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

Rule 3649

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*T
an[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])))

Rule 3650

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

Rule 12

Int[(a_)*(u_), x_Symbol] :> Dist[a, Int[u, x], x] /; FreeQ[a, x] &&  !MatchQ[u, (b_)*(v_) /; FreeQ[b, x]]

Rule 3575

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 910

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 93

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 205

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

Rule 208

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

Rubi steps

\begin{align*} \int \frac{1}{\tan ^{\frac{7}{2}}(c+d x) \sqrt{a+b \tan (c+d x)}} \, dx &=-\frac{2 \sqrt{a+b \tan (c+d x)}}{5 a d \tan ^{\frac{5}{2}}(c+d x)}-\frac{2 \int \frac{2 b+\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}{5 a}\\ &=-\frac{2 \sqrt{a+b \tan (c+d x)}}{5 a d \tan ^{\frac{5}{2}}(c+d x)}+\frac{8 b \sqrt{a+b \tan (c+d x)}}{15 a^2 d \tan ^{\frac{3}{2}}(c+d x)}+\frac{4 \int \frac{\frac{1}{4} \left (-15 a^2+8 b^2\right )+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^2}\\ &=-\frac{2 \sqrt{a+b \tan (c+d x)}}{5 a d \tan ^{\frac{5}{2}}(c+d x)}+\frac{8 b \sqrt{a+b \tan (c+d x)}}{15 a^2 d \tan ^{\frac{3}{2}}(c+d x)}+\frac{2 \left (15 a^2-8 b^2\right ) \sqrt{a+b \tan (c+d x)}}{15 a^3 d \sqrt{\tan (c+d x)}}-\frac{8 \int -\frac{15 a^3 \sqrt{\tan (c+d x)}}{8 \sqrt{a+b \tan (c+d x)}} \, dx}{15 a^3}\\ &=-\frac{2 \sqrt{a+b \tan (c+d x)}}{5 a d \tan ^{\frac{5}{2}}(c+d x)}+\frac{8 b \sqrt{a+b \tan (c+d x)}}{15 a^2 d \tan ^{\frac{3}{2}}(c+d x)}+\frac{2 \left (15 a^2-8 b^2\right ) \sqrt{a+b \tan (c+d x)}}{15 a^3 d \sqrt{\tan (c+d x)}}+\int \frac{\sqrt{\tan (c+d x)}}{\sqrt{a+b \tan (c+d x)}} \, dx\\ &=-\frac{2 \sqrt{a+b \tan (c+d x)}}{5 a d \tan ^{\frac{5}{2}}(c+d x)}+\frac{8 b \sqrt{a+b \tan (c+d x)}}{15 a^2 d \tan ^{\frac{3}{2}}(c+d x)}+\frac{2 \left (15 a^2-8 b^2\right ) \sqrt{a+b \tan (c+d x)}}{15 a^3 d \sqrt{\tan (c+d x)}}+\frac{\operatorname{Subst}\left (\int \frac{\sqrt{x}}{\sqrt{a+b x} \left (1+x^2\right )} \, dx,x,\tan (c+d x)\right )}{d}\\ &=-\frac{2 \sqrt{a+b \tan (c+d x)}}{5 a d \tan ^{\frac{5}{2}}(c+d x)}+\frac{8 b \sqrt{a+b \tan (c+d x)}}{15 a^2 d \tan ^{\frac{3}{2}}(c+d x)}+\frac{2 \left (15 a^2-8 b^2\right ) \sqrt{a+b \tan (c+d x)}}{15 a^3 d \sqrt{\tan (c+d x)}}+\frac{\operatorname{Subst}\left (\int \left (-\frac{1}{2 (i-x) \sqrt{x} \sqrt{a+b x}}+\frac{1}{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)}}{5 a d \tan ^{\frac{5}{2}}(c+d x)}+\frac{8 b \sqrt{a+b \tan (c+d x)}}{15 a^2 d \tan ^{\frac{3}{2}}(c+d x)}+\frac{2 \left (15 a^2-8 b^2\right ) \sqrt{a+b \tan (c+d x)}}{15 a^3 d \sqrt{\tan (c+d x)}}-\frac{\operatorname{Subst}\left (\int \frac{1}{(i-x) \sqrt{x} \sqrt{a+b x}} \, dx,x,\tan (c+d x)\right )}{2 d}+\frac{\operatorname{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)}}{5 a d \tan ^{\frac{5}{2}}(c+d x)}+\frac{8 b \sqrt{a+b \tan (c+d x)}}{15 a^2 d \tan ^{\frac{3}{2}}(c+d x)}+\frac{2 \left (15 a^2-8 b^2\right ) \sqrt{a+b \tan (c+d x)}}{15 a^3 d \sqrt{\tan (c+d x)}}+\frac{\operatorname{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{\operatorname{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 \tan ^{-1}\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{i \tanh ^{-1}\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 \sqrt{a+b \tan (c+d x)}}{5 a d \tan ^{\frac{5}{2}}(c+d x)}+\frac{8 b \sqrt{a+b \tan (c+d x)}}{15 a^2 d \tan ^{\frac{3}{2}}(c+d x)}+\frac{2 \left (15 a^2-8 b^2\right ) \sqrt{a+b \tan (c+d x)}}{15 a^3 d \sqrt{\tan (c+d x)}}\\ \end{align*}

Mathematica [A]  time = 2.73527, size = 197, normalized size = 0.86 \[ \frac{\frac{2 \sqrt{a+b \tan (c+d x)} \left (\left (15 a^2-8 b^2\right ) \tan ^2(c+d x)-3 a^2+4 a b \tan (c+d x)\right )}{a^3 \tan ^{\frac{5}{2}}(c+d x)}+\frac{15 \sqrt [4]{-1} \tanh ^{-1}\left (\frac{\sqrt [4]{-1} \sqrt{-a-i b} \sqrt{\tan (c+d x)}}{\sqrt{a+b \tan (c+d x)}}\right )}{\sqrt{-a-i b}}-\frac{15 \sqrt [4]{-1} \tanh ^{-1}\left (\frac{\sqrt [4]{-1} \sqrt{a-i b} \sqrt{\tan (c+d x)}}{\sqrt{a+b \tan (c+d x)}}\right )}{\sqrt{a-i b}}}{15 d} \]

Antiderivative was successfully verified.

[In]

Integrate[1/(Tan[c + d*x]^(7/2)*Sqrt[a + b*Tan[c + d*x]]),x]

[Out]

((15*(-1)^(1/4)*ArcTanh[((-1)^(1/4)*Sqrt[-a - I*b]*Sqrt[Tan[c + d*x]])/Sqrt[a + b*Tan[c + d*x]]])/Sqrt[-a - I*
b] - (15*(-1)^(1/4)*ArcTanh[((-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]]*(-3*a^2 + 4*a*b*Tan[c + d*x] + (15*a^2 - 8*b^2)*Tan[c + d*x]^2))/(a^3*Tan[c
+ d*x]^(5/2)))/(15*d)

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Maple [B]  time = 0.351, size = 946113, normalized size = 4131.5 \begin{align*} \text{output too large to display} \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

result too large to display

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Maxima [F]  time = 0., size = 0, normalized size = 0. \begin{align*} \int \frac{1}{\sqrt{b \tan \left (d x + c\right ) + a} \tan \left (d x + c\right )^{\frac{7}{2}}}\,{d x} \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

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

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

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

Timed out

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Sympy [F(-1)]  time = 0., size = 0, normalized size = 0. \begin{align*} \text{Timed out} \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

Timed out

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Giac [F(-2)]  time = 0., size = 0, normalized size = 0. \begin{align*} \text{Exception raised: TypeError} \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

Exception raised: TypeError