3.532 \(\int \frac{\sec ^5(c+d x)}{(a+b \sin (c+d x))^{5/2}} \, dx\)

Optimal. Leaf size=339 \[ -\frac{a b \left (-16 a^2 b^2+3 a^4-127 b^4\right )}{8 d \left (a^2-b^2\right )^4 \sqrt{a+b \sin (c+d x)}}-\frac{b \left (-81 a^2 b^2+18 a^4-77 b^4\right )}{48 d \left (a^2-b^2\right )^3 (a+b \sin (c+d x))^{3/2}}-\frac{\left (12 a^2-54 a b+77 b^2\right ) \tanh ^{-1}\left (\frac{\sqrt{a+b \sin (c+d x)}}{\sqrt{a-b}}\right )}{32 d (a-b)^{9/2}}+\frac{\left (12 a^2+54 a b+77 b^2\right ) \tanh ^{-1}\left (\frac{\sqrt{a+b \sin (c+d x)}}{\sqrt{a+b}}\right )}{32 d (a+b)^{9/2}}-\frac{\sec ^4(c+d x) (b-a \sin (c+d x))}{4 d \left (a^2-b^2\right ) (a+b \sin (c+d x))^{3/2}}+\frac{\sec ^2(c+d x) \left (2 a \left (3 a^2-10 b^2\right ) \sin (c+d x)+b \left (3 a^2+11 b^2\right )\right )}{16 d \left (a^2-b^2\right )^2 (a+b \sin (c+d x))^{3/2}} \]

[Out]

-((12*a^2 - 54*a*b + 77*b^2)*ArcTanh[Sqrt[a + b*Sin[c + d*x]]/Sqrt[a - b]])/(32*(a - b)^(9/2)*d) + ((12*a^2 +
54*a*b + 77*b^2)*ArcTanh[Sqrt[a + b*Sin[c + d*x]]/Sqrt[a + b]])/(32*(a + b)^(9/2)*d) - (b*(18*a^4 - 81*a^2*b^2
- 77*b^4))/(48*(a^2 - b^2)^3*d*(a + b*Sin[c + d*x])^(3/2)) - (Sec[c + d*x]^4*(b - a*Sin[c + d*x]))/(4*(a^2 -
b^2)*d*(a + b*Sin[c + d*x])^(3/2)) - (a*b*(3*a^4 - 16*a^2*b^2 - 127*b^4))/(8*(a^2 - b^2)^4*d*Sqrt[a + b*Sin[c
+ d*x]]) + (Sec[c + d*x]^2*(b*(3*a^2 + 11*b^2) + 2*a*(3*a^2 - 10*b^2)*Sin[c + d*x]))/(16*(a^2 - b^2)^2*d*(a +
b*Sin[c + d*x])^(3/2))

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Rubi [A]  time = 0.644966, antiderivative size = 339, normalized size of antiderivative = 1., number of steps used = 9, number of rules used = 7, integrand size = 23, \(\frac{\text{number of rules}}{\text{integrand size}}\) = 0.304, Rules used = {2668, 741, 823, 829, 827, 1166, 206} \[ -\frac{a b \left (-16 a^2 b^2+3 a^4-127 b^4\right )}{8 d \left (a^2-b^2\right )^4 \sqrt{a+b \sin (c+d x)}}-\frac{b \left (-81 a^2 b^2+18 a^4-77 b^4\right )}{48 d \left (a^2-b^2\right )^3 (a+b \sin (c+d x))^{3/2}}-\frac{\left (12 a^2-54 a b+77 b^2\right ) \tanh ^{-1}\left (\frac{\sqrt{a+b \sin (c+d x)}}{\sqrt{a-b}}\right )}{32 d (a-b)^{9/2}}+\frac{\left (12 a^2+54 a b+77 b^2\right ) \tanh ^{-1}\left (\frac{\sqrt{a+b \sin (c+d x)}}{\sqrt{a+b}}\right )}{32 d (a+b)^{9/2}}-\frac{\sec ^4(c+d x) (b-a \sin (c+d x))}{4 d \left (a^2-b^2\right ) (a+b \sin (c+d x))^{3/2}}+\frac{\sec ^2(c+d x) \left (2 a \left (3 a^2-10 b^2\right ) \sin (c+d x)+b \left (3 a^2+11 b^2\right )\right )}{16 d \left (a^2-b^2\right )^2 (a+b \sin (c+d x))^{3/2}} \]

Antiderivative was successfully verified.

[In]

Int[Sec[c + d*x]^5/(a + b*Sin[c + d*x])^(5/2),x]

[Out]

-((12*a^2 - 54*a*b + 77*b^2)*ArcTanh[Sqrt[a + b*Sin[c + d*x]]/Sqrt[a - b]])/(32*(a - b)^(9/2)*d) + ((12*a^2 +
54*a*b + 77*b^2)*ArcTanh[Sqrt[a + b*Sin[c + d*x]]/Sqrt[a + b]])/(32*(a + b)^(9/2)*d) - (b*(18*a^4 - 81*a^2*b^2
- 77*b^4))/(48*(a^2 - b^2)^3*d*(a + b*Sin[c + d*x])^(3/2)) - (Sec[c + d*x]^4*(b - a*Sin[c + d*x]))/(4*(a^2 -
b^2)*d*(a + b*Sin[c + d*x])^(3/2)) - (a*b*(3*a^4 - 16*a^2*b^2 - 127*b^4))/(8*(a^2 - b^2)^4*d*Sqrt[a + b*Sin[c
+ d*x]]) + (Sec[c + d*x]^2*(b*(3*a^2 + 11*b^2) + 2*a*(3*a^2 - 10*b^2)*Sin[c + d*x]))/(16*(a^2 - b^2)^2*d*(a +
b*Sin[c + d*x])^(3/2))

Rule 2668

Int[cos[(e_.) + (f_.)*(x_)]^(p_.)*((a_) + (b_.)*sin[(e_.) + (f_.)*(x_)])^(m_.), x_Symbol] :> Dist[1/(b^p*f), S
ubst[Int[(a + x)^m*(b^2 - x^2)^((p - 1)/2), x], x, b*Sin[e + f*x]], x] /; FreeQ[{a, b, e, f, m}, x] && Integer
Q[(p - 1)/2] && NeQ[a^2 - b^2, 0]

Rule 741

Int[((d_) + (e_.)*(x_))^(m_)*((a_) + (c_.)*(x_)^2)^(p_), x_Symbol] :> -Simp[((d + e*x)^(m + 1)*(a*e + c*d*x)*(
a + c*x^2)^(p + 1))/(2*a*(p + 1)*(c*d^2 + a*e^2)), x] + Dist[1/(2*a*(p + 1)*(c*d^2 + a*e^2)), Int[(d + e*x)^m*
Simp[c*d^2*(2*p + 3) + a*e^2*(m + 2*p + 3) + c*e*d*(m + 2*p + 4)*x, x]*(a + c*x^2)^(p + 1), x], x] /; FreeQ[{a
, c, d, e, m}, x] && NeQ[c*d^2 + a*e^2, 0] && LtQ[p, -1] && IntQuadraticQ[a, 0, c, d, e, m, p, x]

Rule 823

Int[((d_.) + (e_.)*(x_))^(m_)*((f_.) + (g_.)*(x_))*((a_) + (c_.)*(x_)^2)^(p_), x_Symbol] :> -Simp[((d + e*x)^(
m + 1)*(f*a*c*e - a*g*c*d + c*(c*d*f + a*e*g)*x)*(a + c*x^2)^(p + 1))/(2*a*c*(p + 1)*(c*d^2 + a*e^2)), x] + Di
st[1/(2*a*c*(p + 1)*(c*d^2 + a*e^2)), Int[(d + e*x)^m*(a + c*x^2)^(p + 1)*Simp[f*(c^2*d^2*(2*p + 3) + a*c*e^2*
(m + 2*p + 3)) - a*c*d*e*g*m + c*e*(c*d*f + a*e*g)*(m + 2*p + 4)*x, x], x], x] /; FreeQ[{a, c, d, e, f, g}, x]
&& NeQ[c*d^2 + a*e^2, 0] && LtQ[p, -1] && (IntegerQ[m] || IntegerQ[p] || IntegersQ[2*m, 2*p])

Rule 829

Int[(((d_.) + (e_.)*(x_))^(m_)*((f_.) + (g_.)*(x_)))/((a_) + (c_.)*(x_)^2), x_Symbol] :> Simp[((e*f - d*g)*(d
+ e*x)^(m + 1))/((m + 1)*(c*d^2 + a*e^2)), x] + Dist[1/(c*d^2 + a*e^2), Int[((d + e*x)^(m + 1)*Simp[c*d*f + a*
e*g - c*(e*f - d*g)*x, x])/(a + c*x^2), x], x] /; FreeQ[{a, c, d, e, f, g, m}, x] && NeQ[c*d^2 + a*e^2, 0] &&
FractionQ[m] && LtQ[m, -1]

Rule 827

Int[((f_.) + (g_.)*(x_))/(Sqrt[(d_.) + (e_.)*(x_)]*((a_) + (c_.)*(x_)^2)), x_Symbol] :> Dist[2, Subst[Int[(e*f
- d*g + g*x^2)/(c*d^2 + a*e^2 - 2*c*d*x^2 + c*x^4), x], x, Sqrt[d + e*x]], x] /; FreeQ[{a, c, d, e, f, g}, x]
&& NeQ[c*d^2 + a*e^2, 0]

Rule 1166

Int[((d_) + (e_.)*(x_)^2)/((a_) + (b_.)*(x_)^2 + (c_.)*(x_)^4), x_Symbol] :> With[{q = Rt[b^2 - 4*a*c, 2]}, Di
st[e/2 + (2*c*d - b*e)/(2*q), Int[1/(b/2 - q/2 + c*x^2), x], x] + Dist[e/2 - (2*c*d - b*e)/(2*q), Int[1/(b/2 +
q/2 + c*x^2), x], x]] /; FreeQ[{a, b, c, d, e}, x] && NeQ[b^2 - 4*a*c, 0] && NeQ[c*d^2 - a*e^2, 0] && PosQ[b^
2 - 4*a*c]

Rule 206

Int[((a_) + (b_.)*(x_)^2)^(-1), x_Symbol] :> Simp[(1*ArcTanh[(Rt[-b, 2]*x)/Rt[a, 2]])/(Rt[a, 2]*Rt[-b, 2]), x]
/; FreeQ[{a, b}, x] && NegQ[a/b] && (GtQ[a, 0] || LtQ[b, 0])

Rubi steps

\begin{align*} \int \frac{\sec ^5(c+d x)}{(a+b \sin (c+d x))^{5/2}} \, dx &=\frac{b^5 \operatorname{Subst}\left (\int \frac{1}{(a+x)^{5/2} \left (b^2-x^2\right )^3} \, dx,x,b \sin (c+d x)\right )}{d}\\ &=-\frac{\sec ^4(c+d x) (b-a \sin (c+d x))}{4 \left (a^2-b^2\right ) d (a+b \sin (c+d x))^{3/2}}+\frac{b^3 \operatorname{Subst}\left (\int \frac{\frac{1}{2} \left (6 a^2-11 b^2\right )+\frac{9 a x}{2}}{(a+x)^{5/2} \left (b^2-x^2\right )^2} \, dx,x,b \sin (c+d x)\right )}{4 \left (a^2-b^2\right ) d}\\ &=-\frac{\sec ^4(c+d x) (b-a \sin (c+d x))}{4 \left (a^2-b^2\right ) d (a+b \sin (c+d x))^{3/2}}+\frac{\sec ^2(c+d x) \left (b \left (3 a^2+11 b^2\right )+2 a \left (3 a^2-10 b^2\right ) \sin (c+d x)\right )}{16 \left (a^2-b^2\right )^2 d (a+b \sin (c+d x))^{3/2}}-\frac{b \operatorname{Subst}\left (\int \frac{\frac{1}{4} \left (-12 a^4+19 a^2 b^2-77 b^4\right )-\frac{5}{2} a \left (3 a^2-10 b^2\right ) x}{(a+x)^{5/2} \left (b^2-x^2\right )} \, dx,x,b \sin (c+d x)\right )}{8 \left (a^2-b^2\right )^2 d}\\ &=-\frac{b \left (18 a^4-81 a^2 b^2-77 b^4\right )}{48 \left (a^2-b^2\right )^3 d (a+b \sin (c+d x))^{3/2}}-\frac{\sec ^4(c+d x) (b-a \sin (c+d x))}{4 \left (a^2-b^2\right ) d (a+b \sin (c+d x))^{3/2}}+\frac{\sec ^2(c+d x) \left (b \left (3 a^2+11 b^2\right )+2 a \left (3 a^2-10 b^2\right ) \sin (c+d x)\right )}{16 \left (a^2-b^2\right )^2 d (a+b \sin (c+d x))^{3/2}}+\frac{b \operatorname{Subst}\left (\int \frac{\frac{1}{4} a \left (12 a^4-49 a^2 b^2+177 b^4\right )+\frac{1}{4} \left (18 a^4-81 a^2 b^2-77 b^4\right ) x}{(a+x)^{3/2} \left (b^2-x^2\right )} \, dx,x,b \sin (c+d x)\right )}{8 \left (a^2-b^2\right )^3 d}\\ &=-\frac{b \left (18 a^4-81 a^2 b^2-77 b^4\right )}{48 \left (a^2-b^2\right )^3 d (a+b \sin (c+d x))^{3/2}}-\frac{\sec ^4(c+d x) (b-a \sin (c+d x))}{4 \left (a^2-b^2\right ) d (a+b \sin (c+d x))^{3/2}}-\frac{a b \left (3 a^4-16 a^2 b^2-127 b^4\right )}{8 \left (a^2-b^2\right )^4 d \sqrt{a+b \sin (c+d x)}}+\frac{\sec ^2(c+d x) \left (b \left (3 a^2+11 b^2\right )+2 a \left (3 a^2-10 b^2\right ) \sin (c+d x)\right )}{16 \left (a^2-b^2\right )^2 d (a+b \sin (c+d x))^{3/2}}-\frac{b \operatorname{Subst}\left (\int \frac{\frac{1}{4} \left (-12 a^6+67 a^4 b^2-258 a^2 b^4-77 b^6\right )-\frac{1}{2} a \left (3 a^4-16 a^2 b^2-127 b^4\right ) x}{\sqrt{a+x} \left (b^2-x^2\right )} \, dx,x,b \sin (c+d x)\right )}{8 \left (a^2-b^2\right )^4 d}\\ &=-\frac{b \left (18 a^4-81 a^2 b^2-77 b^4\right )}{48 \left (a^2-b^2\right )^3 d (a+b \sin (c+d x))^{3/2}}-\frac{\sec ^4(c+d x) (b-a \sin (c+d x))}{4 \left (a^2-b^2\right ) d (a+b \sin (c+d x))^{3/2}}-\frac{a b \left (3 a^4-16 a^2 b^2-127 b^4\right )}{8 \left (a^2-b^2\right )^4 d \sqrt{a+b \sin (c+d x)}}+\frac{\sec ^2(c+d x) \left (b \left (3 a^2+11 b^2\right )+2 a \left (3 a^2-10 b^2\right ) \sin (c+d x)\right )}{16 \left (a^2-b^2\right )^2 d (a+b \sin (c+d x))^{3/2}}-\frac{b \operatorname{Subst}\left (\int \frac{\frac{1}{2} a^2 \left (3 a^4-16 a^2 b^2-127 b^4\right )+\frac{1}{4} \left (-12 a^6+67 a^4 b^2-258 a^2 b^4-77 b^6\right )-\frac{1}{2} a \left (3 a^4-16 a^2 b^2-127 b^4\right ) x^2}{-a^2+b^2+2 a x^2-x^4} \, dx,x,\sqrt{a+b \sin (c+d x)}\right )}{4 \left (a^2-b^2\right )^4 d}\\ &=-\frac{b \left (18 a^4-81 a^2 b^2-77 b^4\right )}{48 \left (a^2-b^2\right )^3 d (a+b \sin (c+d x))^{3/2}}-\frac{\sec ^4(c+d x) (b-a \sin (c+d x))}{4 \left (a^2-b^2\right ) d (a+b \sin (c+d x))^{3/2}}-\frac{a b \left (3 a^4-16 a^2 b^2-127 b^4\right )}{8 \left (a^2-b^2\right )^4 d \sqrt{a+b \sin (c+d x)}}+\frac{\sec ^2(c+d x) \left (b \left (3 a^2+11 b^2\right )+2 a \left (3 a^2-10 b^2\right ) \sin (c+d x)\right )}{16 \left (a^2-b^2\right )^2 d (a+b \sin (c+d x))^{3/2}}-\frac{\left (12 a^2-54 a b+77 b^2\right ) \operatorname{Subst}\left (\int \frac{1}{a-b-x^2} \, dx,x,\sqrt{a+b \sin (c+d x)}\right )}{32 (a-b)^4 d}+\frac{\left (12 a^2+54 a b+77 b^2\right ) \operatorname{Subst}\left (\int \frac{1}{a+b-x^2} \, dx,x,\sqrt{a+b \sin (c+d x)}\right )}{32 (a+b)^4 d}\\ &=-\frac{\left (12 a^2-54 a b+77 b^2\right ) \tanh ^{-1}\left (\frac{\sqrt{a+b \sin (c+d x)}}{\sqrt{a-b}}\right )}{32 (a-b)^{9/2} d}+\frac{\left (12 a^2+54 a b+77 b^2\right ) \tanh ^{-1}\left (\frac{\sqrt{a+b \sin (c+d x)}}{\sqrt{a+b}}\right )}{32 (a+b)^{9/2} d}-\frac{b \left (18 a^4-81 a^2 b^2-77 b^4\right )}{48 \left (a^2-b^2\right )^3 d (a+b \sin (c+d x))^{3/2}}-\frac{\sec ^4(c+d x) (b-a \sin (c+d x))}{4 \left (a^2-b^2\right ) d (a+b \sin (c+d x))^{3/2}}-\frac{a b \left (3 a^4-16 a^2 b^2-127 b^4\right )}{8 \left (a^2-b^2\right )^4 d \sqrt{a+b \sin (c+d x)}}+\frac{\sec ^2(c+d x) \left (b \left (3 a^2+11 b^2\right )+2 a \left (3 a^2-10 b^2\right ) \sin (c+d x)\right )}{16 \left (a^2-b^2\right )^2 d (a+b \sin (c+d x))^{3/2}}\\ \end{align*}

Mathematica [C]  time = 3.36129, size = 296, normalized size = 0.87 \[ \frac{\frac{1}{2} \left (-81 a^2 b^2+18 a^4-77 b^4\right ) \left ((a+b) \, _2F_1\left (-\frac{3}{2},1;-\frac{1}{2};\frac{a+b \sin (c+d x)}{a-b}\right )+(b-a) \, _2F_1\left (-\frac{3}{2},1;-\frac{1}{2};\frac{a+b \sin (c+d x)}{a+b}\right )\right )-15 a \left (3 a^2-10 b^2\right ) (a+b \sin (c+d x)) \left ((a+b) \, _2F_1\left (-\frac{1}{2},1;\frac{1}{2};\frac{a+b \sin (c+d x)}{a-b}\right )+(b-a) \, _2F_1\left (-\frac{1}{2},1;\frac{1}{2};\frac{a+b \sin (c+d x)}{a+b}\right )\right )-3 (a-b) (a+b) \sec ^2(c+d x) \left (\left (6 a^3-20 a b^2\right ) \sin (c+d x)+3 a^2 b+11 b^3\right )-12 (a-b)^2 (a+b)^2 \sec ^4(c+d x) (a \sin (c+d x)-b)}{48 d \left (a^2-b^2\right )^2 \left (b^2-a^2\right ) (a+b \sin (c+d x))^{3/2}} \]

Antiderivative was successfully verified.

[In]

Integrate[Sec[c + d*x]^5/(a + b*Sin[c + d*x])^(5/2),x]

[Out]

(((18*a^4 - 81*a^2*b^2 - 77*b^4)*((a + b)*Hypergeometric2F1[-3/2, 1, -1/2, (a + b*Sin[c + d*x])/(a - b)] + (-a
+ b)*Hypergeometric2F1[-3/2, 1, -1/2, (a + b*Sin[c + d*x])/(a + b)]))/2 - 12*(a - b)^2*(a + b)^2*Sec[c + d*x]
^4*(-b + a*Sin[c + d*x]) - 15*a*(3*a^2 - 10*b^2)*((a + b)*Hypergeometric2F1[-1/2, 1, 1/2, (a + b*Sin[c + d*x])
/(a - b)] + (-a + b)*Hypergeometric2F1[-1/2, 1, 1/2, (a + b*Sin[c + d*x])/(a + b)])*(a + b*Sin[c + d*x]) - 3*(
a - b)*(a + b)*Sec[c + d*x]^2*(3*a^2*b + 11*b^3 + (6*a^3 - 20*a*b^2)*Sin[c + d*x]))/(48*(a^2 - b^2)^2*(-a^2 +
b^2)*d*(a + b*Sin[c + d*x])^(3/2))

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Maple [B]  time = 0.795, size = 682, normalized size = 2. \begin{align*} \text{result too large to display} \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

int(sec(d*x+c)^5/(a+b*sin(d*x+c))^(5/2),x)

[Out]

-3/16/d*b/(a-b)^4/(b*sin(d*x+c)+b)^2*(a+b*sin(d*x+c))^(3/2)*a+17/32/d*b^2/(a-b)^4/(b*sin(d*x+c)+b)^2*(a+b*sin(
d*x+c))^(3/2)+3/16/d*b/(a-b)^4/(b*sin(d*x+c)+b)^2*(a+b*sin(d*x+c))^(1/2)*a^2-25/32/d*b^2/(a-b)^4/(b*sin(d*x+c)
+b)^2*(a+b*sin(d*x+c))^(1/2)*a+19/32/d*b^3/(a-b)^4/(b*sin(d*x+c)+b)^2*(a+b*sin(d*x+c))^(1/2)+3/8/d/(a-b)^4/(-a
+b)^(1/2)*arctan((a+b*sin(d*x+c))^(1/2)/(-a+b)^(1/2))*a^2-27/16/d*b/(a-b)^4/(-a+b)^(1/2)*arctan((a+b*sin(d*x+c
))^(1/2)/(-a+b)^(1/2))*a+77/32/d*b^2/(a-b)^4/(-a+b)^(1/2)*arctan((a+b*sin(d*x+c))^(1/2)/(-a+b)^(1/2))+2/3/d*b^
5/(a-b)^3/(a+b)^3/(a+b*sin(d*x+c))^(3/2)+12/d*b^5*a/(a-b)^4/(a+b)^4/(a+b*sin(d*x+c))^(1/2)-3/16/d*b/(a+b)^4/(b
*sin(d*x+c)-b)^2*(a+b*sin(d*x+c))^(3/2)*a-17/32/d*b^2/(a+b)^4/(b*sin(d*x+c)-b)^2*(a+b*sin(d*x+c))^(3/2)+3/16/d
*b/(a+b)^4/(b*sin(d*x+c)-b)^2*(a+b*sin(d*x+c))^(1/2)*a^2+25/32/d*b^2/(a+b)^4/(b*sin(d*x+c)-b)^2*(a+b*sin(d*x+c
))^(1/2)*a+19/32/d*b^3/(a+b)^4/(b*sin(d*x+c)-b)^2*(a+b*sin(d*x+c))^(1/2)+3/8/d/(a+b)^(9/2)*arctanh((a+b*sin(d*
x+c))^(1/2)/(a+b)^(1/2))*a^2+27/16/d*b/(a+b)^(9/2)*arctanh((a+b*sin(d*x+c))^(1/2)/(a+b)^(1/2))*a+77/32/d*b^2/(
a+b)^(9/2)*arctanh((a+b*sin(d*x+c))^(1/2)/(a+b)^(1/2))

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

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate(sec(d*x+c)^5/(a+b*sin(d*x+c))^(5/2),x, algorithm="maxima")

[Out]

Exception raised: ValueError

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Fricas [F]  time = 0., size = 0, normalized size = 0. \begin{align*}{\rm integral}\left (-\frac{\sqrt{b \sin \left (d x + c\right ) + a} \sec \left (d x + c\right )^{5}}{3 \, a b^{2} \cos \left (d x + c\right )^{2} - a^{3} - 3 \, a b^{2} +{\left (b^{3} \cos \left (d x + c\right )^{2} - 3 \, a^{2} b - b^{3}\right )} \sin \left (d x + c\right )}, x\right ) \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate(sec(d*x+c)^5/(a+b*sin(d*x+c))^(5/2),x, algorithm="fricas")

[Out]

integral(-sqrt(b*sin(d*x + c) + a)*sec(d*x + c)^5/(3*a*b^2*cos(d*x + c)^2 - a^3 - 3*a*b^2 + (b^3*cos(d*x + c)^
2 - 3*a^2*b - b^3)*sin(d*x + c)), x)

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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(sec(d*x+c)**5/(a+b*sin(d*x+c))**(5/2),x)

[Out]

Timed out

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Giac [B]  time = 1.21919, size = 861, normalized size = 2.54 \begin{align*} \text{result too large to display} \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate(sec(d*x+c)^5/(a+b*sin(d*x+c))^(5/2),x, algorithm="giac")

[Out]

1/96*b^5*(3*(12*a^2 - 54*a*b + 77*b^2)*arctan(sqrt(b*sin(d*x + c) + a)/sqrt(-a + b))/((a^4*b^5*d - 4*a^3*b^6*d
+ 6*a^2*b^7*d - 4*a*b^8*d + b^9*d)*sqrt(-a + b)) - 3*(12*a^2 + 54*a*b + 77*b^2)*arctan(sqrt(b*sin(d*x + c) +
a)/sqrt(-a - b))/((a^4*b^5*d + 4*a^3*b^6*d + 6*a^2*b^7*d + 4*a*b^8*d + b^9*d)*sqrt(-a - b)) + 64*(18*(b*sin(d*
x + c) + a)*a + a^2 - b^2)/((a^8*d - 4*a^6*b^2*d + 6*a^4*b^4*d - 4*a^2*b^6*d + b^8*d)*(b*sin(d*x + c) + a)^(3/
2)) - 6*(6*(b*sin(d*x + c) + a)^(7/2)*a^5 - 18*(b*sin(d*x + c) + a)^(5/2)*a^6 + 18*(b*sin(d*x + c) + a)^(3/2)*
a^7 - 6*sqrt(b*sin(d*x + c) + a)*a^8 - 32*(b*sin(d*x + c) + a)^(7/2)*a^3*b^2 + 95*(b*sin(d*x + c) + a)^(5/2)*a
^4*b^2 - 104*(b*sin(d*x + c) + a)^(3/2)*a^5*b^2 + 41*sqrt(b*sin(d*x + c) + a)*a^6*b^2 - 62*(b*sin(d*x + c) + a
)^(7/2)*a*b^4 + 260*(b*sin(d*x + c) + a)^(5/2)*a^2*b^4 - 310*(b*sin(d*x + c) + a)^(3/2)*a^3*b^4 + 125*sqrt(b*s
in(d*x + c) + a)*a^4*b^4 + 15*(b*sin(d*x + c) + a)^(5/2)*b^6 + 44*(b*sin(d*x + c) + a)^(3/2)*a*b^6 - 141*sqrt(
b*sin(d*x + c) + a)*a^2*b^6 - 19*sqrt(b*sin(d*x + c) + a)*b^8)/((a^8*b^4*d - 4*a^6*b^6*d + 6*a^4*b^8*d - 4*a^2
*b^10*d + b^12*d)*((b*sin(d*x + c) + a)^2 - 2*(b*sin(d*x + c) + a)*a + a^2 - b^2)^2))