3.10.75 \(\int (a+b \sec (c+d x))^{3/2} (a b B-a^2 C+b^2 B \sec (c+d x)+b^2 C \sec ^2(c+d x)) \, dx\) [975]

Optimal. Leaf size=448 \[ -\frac {2 (a-b) \sqrt {a+b} \left (35 a b B-12 a^2 C+9 b^2 C\right ) \cot (c+d x) E\left (\text {ArcSin}\left (\frac {\sqrt {a+b \sec (c+d x)}}{\sqrt {a+b}}\right )|\frac {a+b}{a-b}\right ) \sqrt {\frac {b (1-\sec (c+d x))}{a+b}} \sqrt {-\frac {b (1+\sec (c+d x))}{a-b}}}{15 d}-\frac {2 \sqrt {a+b} \left (a b^2 (35 B-12 C)-b^3 (5 B-9 C)+30 a^3 C-3 a^2 b (15 B+4 C)\right ) \cot (c+d x) F\left (\text {ArcSin}\left (\frac {\sqrt {a+b \sec (c+d x)}}{\sqrt {a+b}}\right )|\frac {a+b}{a-b}\right ) \sqrt {\frac {b (1-\sec (c+d x))}{a+b}} \sqrt {-\frac {b (1+\sec (c+d x))}{a-b}}}{15 d}-\frac {2 a^2 \sqrt {a+b} (b B-a C) \cot (c+d x) \Pi \left (\frac {a+b}{a};\text {ArcSin}\left (\frac {\sqrt {a+b \sec (c+d x)}}{\sqrt {a+b}}\right )|\frac {a+b}{a-b}\right ) \sqrt {\frac {b (1-\sec (c+d x))}{a+b}} \sqrt {-\frac {b (1+\sec (c+d x))}{a-b}}}{d}+\frac {2 b^2 (5 b B+3 a C) \sqrt {a+b \sec (c+d x)} \tan (c+d x)}{15 d}+\frac {2 b^2 C (a+b \sec (c+d x))^{3/2} \tan (c+d x)}{5 d} \]

[Out]

-2/15*(a-b)*(35*B*a*b-12*C*a^2+9*C*b^2)*cot(d*x+c)*EllipticE((a+b*sec(d*x+c))^(1/2)/(a+b)^(1/2),((a+b)/(a-b))^
(1/2))*(a+b)^(1/2)*(b*(1-sec(d*x+c))/(a+b))^(1/2)*(-b*(1+sec(d*x+c))/(a-b))^(1/2)/d-2/15*(a*b^2*(35*B-12*C)-b^
3*(5*B-9*C)+30*a^3*C-3*a^2*b*(15*B+4*C))*cot(d*x+c)*EllipticF((a+b*sec(d*x+c))^(1/2)/(a+b)^(1/2),((a+b)/(a-b))
^(1/2))*(a+b)^(1/2)*(b*(1-sec(d*x+c))/(a+b))^(1/2)*(-b*(1+sec(d*x+c))/(a-b))^(1/2)/d-2*a^2*(B*b-C*a)*cot(d*x+c
)*EllipticPi((a+b*sec(d*x+c))^(1/2)/(a+b)^(1/2),(a+b)/a,((a+b)/(a-b))^(1/2))*(a+b)^(1/2)*(b*(1-sec(d*x+c))/(a+
b))^(1/2)*(-b*(1+sec(d*x+c))/(a-b))^(1/2)/d+2/5*b^2*C*(a+b*sec(d*x+c))^(3/2)*tan(d*x+c)/d+2/15*b^2*(5*B*b+3*C*
a)*(a+b*sec(d*x+c))^(1/2)*tan(d*x+c)/d

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Rubi [A]
time = 0.62, antiderivative size = 448, normalized size of antiderivative = 1.00, number of steps used = 8, number of rules used = 8, integrand size = 50, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.160, Rules used = {4126, 4003, 4141, 4143, 4006, 3869, 3917, 4089} \begin {gather*} -\frac {2 (a-b) \sqrt {a+b} \left (-12 a^2 C+35 a b B+9 b^2 C\right ) \cot (c+d x) \sqrt {\frac {b (1-\sec (c+d x))}{a+b}} \sqrt {-\frac {b (\sec (c+d x)+1)}{a-b}} E\left (\text {ArcSin}\left (\frac {\sqrt {a+b \sec (c+d x)}}{\sqrt {a+b}}\right )|\frac {a+b}{a-b}\right )}{15 d}-\frac {2 a^2 \sqrt {a+b} (b B-a C) \cot (c+d x) \sqrt {\frac {b (1-\sec (c+d x))}{a+b}} \sqrt {-\frac {b (\sec (c+d x)+1)}{a-b}} \Pi \left (\frac {a+b}{a};\text {ArcSin}\left (\frac {\sqrt {a+b \sec (c+d x)}}{\sqrt {a+b}}\right )|\frac {a+b}{a-b}\right )}{d}-\frac {2 \sqrt {a+b} \left (30 a^3 C-3 a^2 b (15 B+4 C)+a b^2 (35 B-12 C)-b^3 (5 B-9 C)\right ) \cot (c+d x) \sqrt {\frac {b (1-\sec (c+d x))}{a+b}} \sqrt {-\frac {b (\sec (c+d x)+1)}{a-b}} F\left (\text {ArcSin}\left (\frac {\sqrt {a+b \sec (c+d x)}}{\sqrt {a+b}}\right )|\frac {a+b}{a-b}\right )}{15 d}+\frac {2 b^2 (3 a C+5 b B) \tan (c+d x) \sqrt {a+b \sec (c+d x)}}{15 d}+\frac {2 b^2 C \tan (c+d x) (a+b \sec (c+d x))^{3/2}}{5 d} \end {gather*}

Antiderivative was successfully verified.

[In]

Int[(a + b*Sec[c + d*x])^(3/2)*(a*b*B - a^2*C + b^2*B*Sec[c + d*x] + b^2*C*Sec[c + d*x]^2),x]

[Out]

(-2*(a - b)*Sqrt[a + b]*(35*a*b*B - 12*a^2*C + 9*b^2*C)*Cot[c + d*x]*EllipticE[ArcSin[Sqrt[a + b*Sec[c + d*x]]
/Sqrt[a + b]], (a + b)/(a - b)]*Sqrt[(b*(1 - Sec[c + d*x]))/(a + b)]*Sqrt[-((b*(1 + Sec[c + d*x]))/(a - b))])/
(15*d) - (2*Sqrt[a + b]*(a*b^2*(35*B - 12*C) - b^3*(5*B - 9*C) + 30*a^3*C - 3*a^2*b*(15*B + 4*C))*Cot[c + d*x]
*EllipticF[ArcSin[Sqrt[a + b*Sec[c + d*x]]/Sqrt[a + b]], (a + b)/(a - b)]*Sqrt[(b*(1 - Sec[c + d*x]))/(a + b)]
*Sqrt[-((b*(1 + Sec[c + d*x]))/(a - b))])/(15*d) - (2*a^2*Sqrt[a + b]*(b*B - a*C)*Cot[c + d*x]*EllipticPi[(a +
b)/a, ArcSin[Sqrt[a + b*Sec[c + d*x]]/Sqrt[a + b]], (a + b)/(a - b)]*Sqrt[(b*(1 - Sec[c + d*x]))/(a + b)]*Sqr
t[-((b*(1 + Sec[c + d*x]))/(a - b))])/d + (2*b^2*(5*b*B + 3*a*C)*Sqrt[a + b*Sec[c + d*x]]*Tan[c + d*x])/(15*d)
+ (2*b^2*C*(a + b*Sec[c + d*x])^(3/2)*Tan[c + d*x])/(5*d)

Rule 3869

Int[1/Sqrt[csc[(c_.) + (d_.)*(x_)]*(b_.) + (a_)], x_Symbol] :> Simp[2*(Rt[a + b, 2]/(a*d*Cot[c + d*x]))*Sqrt[b
*((1 - Csc[c + d*x])/(a + b))]*Sqrt[(-b)*((1 + Csc[c + d*x])/(a - b))]*EllipticPi[(a + b)/a, ArcSin[Sqrt[a + b
*Csc[c + d*x]]/Rt[a + b, 2]], (a + b)/(a - b)], x] /; FreeQ[{a, b, c, d}, x] && NeQ[a^2 - b^2, 0]

Rule 3917

Int[csc[(e_.) + (f_.)*(x_)]/Sqrt[csc[(e_.) + (f_.)*(x_)]*(b_.) + (a_)], x_Symbol] :> Simp[-2*(Rt[a + b, 2]/(b*
f*Cot[e + f*x]))*Sqrt[(b*(1 - Csc[e + f*x]))/(a + b)]*Sqrt[(-b)*((1 + Csc[e + f*x])/(a - b))]*EllipticF[ArcSin
[Sqrt[a + b*Csc[e + f*x]]/Rt[a + b, 2]], (a + b)/(a - b)], x] /; FreeQ[{a, b, e, f}, x] && NeQ[a^2 - b^2, 0]

Rule 4003

Int[(csc[(e_.) + (f_.)*(x_)]*(b_.) + (a_))^(m_)*(csc[(e_.) + (f_.)*(x_)]*(d_.) + (c_)), x_Symbol] :> Simp[(-b)
*d*Cot[e + f*x]*((a + b*Csc[e + f*x])^(m - 1)/(f*m)), x] + Dist[1/m, Int[(a + b*Csc[e + f*x])^(m - 2)*Simp[a^2
*c*m + (b^2*d*(m - 1) + 2*a*b*c*m + a^2*d*m)*Csc[e + f*x] + b*(b*c*m + a*d*(2*m - 1))*Csc[e + f*x]^2, x], x],
x] /; FreeQ[{a, b, c, d, e, f}, x] && NeQ[b*c - a*d, 0] && GtQ[m, 1] && NeQ[a^2 - b^2, 0] && IntegerQ[2*m]

Rule 4006

Int[(csc[(e_.) + (f_.)*(x_)]*(d_.) + (c_))/Sqrt[csc[(e_.) + (f_.)*(x_)]*(b_.) + (a_)], x_Symbol] :> Dist[c, In
t[1/Sqrt[a + b*Csc[e + f*x]], x], x] + Dist[d, Int[Csc[e + f*x]/Sqrt[a + b*Csc[e + f*x]], x], x] /; FreeQ[{a,
b, c, d, e, f}, x] && NeQ[b*c - a*d, 0] && NeQ[a^2 - b^2, 0]

Rule 4089

Int[(csc[(e_.) + (f_.)*(x_)]*(csc[(e_.) + (f_.)*(x_)]*(B_.) + (A_)))/Sqrt[csc[(e_.) + (f_.)*(x_)]*(b_.) + (a_)
], x_Symbol] :> Simp[-2*(A*b - a*B)*Rt[a + b*(B/A), 2]*Sqrt[b*((1 - Csc[e + f*x])/(a + b))]*(Sqrt[(-b)*((1 + C
sc[e + f*x])/(a - b))]/(b^2*f*Cot[e + f*x]))*EllipticE[ArcSin[Sqrt[a + b*Csc[e + f*x]]/Rt[a + b*(B/A), 2]], (a
*A + b*B)/(a*A - b*B)], x] /; FreeQ[{a, b, e, f, A, B}, x] && NeQ[a^2 - b^2, 0] && EqQ[A^2 - B^2, 0]

Rule 4126

Int[((A_.) + csc[(e_.) + (f_.)*(x_)]*(B_.) + csc[(e_.) + (f_.)*(x_)]^2*(C_.))*(csc[(e_.) + (f_.)*(x_)]*(b_.) +
(a_))^(m_.), x_Symbol] :> Dist[1/b^2, Int[(a + b*Csc[e + f*x])^(m + 1)*Simp[b*B - a*C + b*C*Csc[e + f*x], x],
x], x] /; FreeQ[{a, b, e, f, A, B, C, m}, x] && EqQ[A*b^2 - a*b*B + a^2*C, 0]

Rule 4141

Int[((A_.) + csc[(e_.) + (f_.)*(x_)]*(B_.) + csc[(e_.) + (f_.)*(x_)]^2*(C_.))*(csc[(e_.) + (f_.)*(x_)]*(b_.) +
(a_))^(m_.), x_Symbol] :> Simp[(-C)*Cot[e + f*x]*((a + b*Csc[e + f*x])^m/(f*(m + 1))), x] + Dist[1/(m + 1), I
nt[(a + b*Csc[e + f*x])^(m - 1)*Simp[a*A*(m + 1) + ((A*b + a*B)*(m + 1) + b*C*m)*Csc[e + f*x] + (b*B*(m + 1) +
a*C*m)*Csc[e + f*x]^2, x], x], x] /; FreeQ[{a, b, e, f, A, B, C}, x] && NeQ[a^2 - b^2, 0] && IGtQ[2*m, 0]

Rule 4143

Int[((A_.) + csc[(e_.) + (f_.)*(x_)]*(B_.) + csc[(e_.) + (f_.)*(x_)]^2*(C_.))/Sqrt[csc[(e_.) + (f_.)*(x_)]*(b_
.) + (a_)], x_Symbol] :> Int[(A + (B - C)*Csc[e + f*x])/Sqrt[a + b*Csc[e + f*x]], x] + Dist[C, Int[Csc[e + f*x
]*((1 + Csc[e + f*x])/Sqrt[a + b*Csc[e + f*x]]), x], x] /; FreeQ[{a, b, e, f, A, B, C}, x] && NeQ[a^2 - b^2, 0
]

Rubi steps

\begin {align*} \int (a+b \sec (c+d x))^{3/2} \left (a b B-a^2 C+b^2 B \sec (c+d x)+b^2 C \sec ^2(c+d x)\right ) \, dx &=\frac {\int (a+b \sec (c+d x))^{5/2} \left (b^2 (b B-a C)+b^3 C \sec (c+d x)\right ) \, dx}{b^2}\\ &=\frac {2 b^2 C (a+b \sec (c+d x))^{3/2} \tan (c+d x)}{5 d}+\frac {2 \int \sqrt {a+b \sec (c+d x)} \left (\frac {5}{2} a^2 b^2 (b B-a C)+\frac {1}{2} b^3 \left (10 a b B-5 a^2 C+3 b^2 C\right ) \sec (c+d x)+\frac {1}{2} b^4 (5 b B+3 a C) \sec ^2(c+d x)\right ) \, dx}{5 b^2}\\ &=\frac {2 b^2 (5 b B+3 a C) \sqrt {a+b \sec (c+d x)} \tan (c+d x)}{15 d}+\frac {2 b^2 C (a+b \sec (c+d x))^{3/2} \tan (c+d x)}{5 d}+\frac {4 \int \frac {\frac {15}{4} a^3 b^2 (b B-a C)+\frac {1}{4} b^3 \left (45 a^2 b B+5 b^3 B-30 a^3 C+12 a b^2 C\right ) \sec (c+d x)+\frac {1}{4} b^4 \left (35 a b B-12 a^2 C+9 b^2 C\right ) \sec ^2(c+d x)}{\sqrt {a+b \sec (c+d x)}} \, dx}{15 b^2}\\ &=\frac {2 b^2 (5 b B+3 a C) \sqrt {a+b \sec (c+d x)} \tan (c+d x)}{15 d}+\frac {2 b^2 C (a+b \sec (c+d x))^{3/2} \tan (c+d x)}{5 d}+\frac {4 \int \frac {\frac {15}{4} a^3 b^2 (b B-a C)+\left (-\frac {1}{4} b^4 \left (35 a b B-12 a^2 C+9 b^2 C\right )+\frac {1}{4} b^3 \left (45 a^2 b B+5 b^3 B-30 a^3 C+12 a b^2 C\right )\right ) \sec (c+d x)}{\sqrt {a+b \sec (c+d x)}} \, dx}{15 b^2}+\frac {1}{15} \left (b^2 \left (35 a b B-12 a^2 C+9 b^2 C\right )\right ) \int \frac {\sec (c+d x) (1+\sec (c+d x))}{\sqrt {a+b \sec (c+d x)}} \, dx\\ &=-\frac {2 (a-b) \sqrt {a+b} \left (35 a b B-12 a^2 C+9 b^2 C\right ) \cot (c+d x) E\left (\sin ^{-1}\left (\frac {\sqrt {a+b \sec (c+d x)}}{\sqrt {a+b}}\right )|\frac {a+b}{a-b}\right ) \sqrt {\frac {b (1-\sec (c+d x))}{a+b}} \sqrt {-\frac {b (1+\sec (c+d x))}{a-b}}}{15 d}+\frac {2 b^2 (5 b B+3 a C) \sqrt {a+b \sec (c+d x)} \tan (c+d x)}{15 d}+\frac {2 b^2 C (a+b \sec (c+d x))^{3/2} \tan (c+d x)}{5 d}+\left (a^3 (b B-a C)\right ) \int \frac {1}{\sqrt {a+b \sec (c+d x)}} \, dx-\frac {1}{15} \left (b \left (a b^2 (35 B-12 C)-b^3 (5 B-9 C)+30 a^3 C-3 a^2 b (15 B+4 C)\right )\right ) \int \frac {\sec (c+d x)}{\sqrt {a+b \sec (c+d x)}} \, dx\\ &=-\frac {2 (a-b) \sqrt {a+b} \left (35 a b B-12 a^2 C+9 b^2 C\right ) \cot (c+d x) E\left (\sin ^{-1}\left (\frac {\sqrt {a+b \sec (c+d x)}}{\sqrt {a+b}}\right )|\frac {a+b}{a-b}\right ) \sqrt {\frac {b (1-\sec (c+d x))}{a+b}} \sqrt {-\frac {b (1+\sec (c+d x))}{a-b}}}{15 d}-\frac {2 \sqrt {a+b} \left (a b^2 (35 B-12 C)-b^3 (5 B-9 C)+30 a^3 C-3 a^2 b (15 B+4 C)\right ) \cot (c+d x) F\left (\sin ^{-1}\left (\frac {\sqrt {a+b \sec (c+d x)}}{\sqrt {a+b}}\right )|\frac {a+b}{a-b}\right ) \sqrt {\frac {b (1-\sec (c+d x))}{a+b}} \sqrt {-\frac {b (1+\sec (c+d x))}{a-b}}}{15 d}-\frac {2 a^2 \sqrt {a+b} (b B-a C) \cot (c+d x) \Pi \left (\frac {a+b}{a};\sin ^{-1}\left (\frac {\sqrt {a+b \sec (c+d x)}}{\sqrt {a+b}}\right )|\frac {a+b}{a-b}\right ) \sqrt {\frac {b (1-\sec (c+d x))}{a+b}} \sqrt {-\frac {b (1+\sec (c+d x))}{a-b}}}{d}+\frac {2 b^2 (5 b B+3 a C) \sqrt {a+b \sec (c+d x)} \tan (c+d x)}{15 d}+\frac {2 b^2 C (a+b \sec (c+d x))^{3/2} \tan (c+d x)}{5 d}\\ \end {align*}

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Mathematica [A]
time = 20.21, size = 689, normalized size = 1.54 \begin {gather*} \frac {2 \sqrt {\cos (c+d x) \sec ^2\left (\frac {1}{2} (c+d x)\right )} \sqrt {\cos ^2\left (\frac {1}{2} (c+d x)\right ) \sec (c+d x)} (a+b \sec (c+d x))^{5/2} (b B-a C+b C \sec (c+d x)) \left (-b (a+b) \left (35 a b B-12 a^2 C+9 b^2 C\right ) E\left (\text {ArcSin}\left (\tan \left (\frac {1}{2} (c+d x)\right )\right )|\frac {a-b}{a+b}\right ) \sec ^2\left (\frac {1}{2} (c+d x)\right ) \sqrt {\frac {(b+a \cos (c+d x)) \sec ^2\left (\frac {1}{2} (c+d x)\right )}{a+b}}+b (a+b) \left (-15 a^2 C+3 a b (10 B+C)+b^2 (5 B+9 C)\right ) F\left (\text {ArcSin}\left (\tan \left (\frac {1}{2} (c+d x)\right )\right )|\frac {a-b}{a+b}\right ) \sec ^2\left (\frac {1}{2} (c+d x)\right ) \sqrt {\frac {(b+a \cos (c+d x)) \sec ^2\left (\frac {1}{2} (c+d x)\right )}{a+b}}-15 a^2 (b B-a C) \left ((a-b) F\left (\text {ArcSin}\left (\tan \left (\frac {1}{2} (c+d x)\right )\right )|\frac {a-b}{a+b}\right )-2 a \Pi \left (-1;\text {ArcSin}\left (\tan \left (\frac {1}{2} (c+d x)\right )\right )|\frac {a-b}{a+b}\right )\right ) \sec ^2\left (\frac {1}{2} (c+d x)\right ) \sqrt {\frac {(b+a \cos (c+d x)) \sec ^2\left (\frac {1}{2} (c+d x)\right )}{a+b}}-b \left (35 a b B-12 a^2 C+9 b^2 C\right ) (b+a \cos (c+d x)) \left (\cos (c+d x) \sec ^2\left (\frac {1}{2} (c+d x)\right )\right )^{3/2} \sec (c+d x) \tan \left (\frac {1}{2} (c+d x)\right )\right )}{15 d (b+a \cos (c+d x))^3 (b C+b B \cos (c+d x)-a C \cos (c+d x)) \sec ^2\left (\frac {1}{2} (c+d x)\right )^{3/2} \sec ^{\frac {7}{2}}(c+d x)}+\frac {\cos ^3(c+d x) (a+b \sec (c+d x))^{5/2} (b B-a C+b C \sec (c+d x)) \left (\frac {2}{15} b \left (35 a b B-12 a^2 C+9 b^2 C\right ) \sin (c+d x)+\frac {2}{15} \sec (c+d x) \left (5 b^3 B \sin (c+d x)+6 a b^2 C \sin (c+d x)\right )+\frac {2}{5} b^3 C \sec (c+d x) \tan (c+d x)\right )}{d (b+a \cos (c+d x))^2 (b C+b B \cos (c+d x)-a C \cos (c+d x))} \end {gather*}

Warning: Unable to verify antiderivative.

[In]

Integrate[(a + b*Sec[c + d*x])^(3/2)*(a*b*B - a^2*C + b^2*B*Sec[c + d*x] + b^2*C*Sec[c + d*x]^2),x]

[Out]

(2*Sqrt[Cos[c + d*x]*Sec[(c + d*x)/2]^2]*Sqrt[Cos[(c + d*x)/2]^2*Sec[c + d*x]]*(a + b*Sec[c + d*x])^(5/2)*(b*B
- a*C + b*C*Sec[c + d*x])*(-(b*(a + b)*(35*a*b*B - 12*a^2*C + 9*b^2*C)*EllipticE[ArcSin[Tan[(c + d*x)/2]], (a
- b)/(a + b)]*Sec[(c + d*x)/2]^2*Sqrt[((b + a*Cos[c + d*x])*Sec[(c + d*x)/2]^2)/(a + b)]) + b*(a + b)*(-15*a^
2*C + 3*a*b*(10*B + C) + b^2*(5*B + 9*C))*EllipticF[ArcSin[Tan[(c + d*x)/2]], (a - b)/(a + b)]*Sec[(c + d*x)/2
]^2*Sqrt[((b + a*Cos[c + d*x])*Sec[(c + d*x)/2]^2)/(a + b)] - 15*a^2*(b*B - a*C)*((a - b)*EllipticF[ArcSin[Tan
[(c + d*x)/2]], (a - b)/(a + b)] - 2*a*EllipticPi[-1, ArcSin[Tan[(c + d*x)/2]], (a - b)/(a + b)])*Sec[(c + d*x
)/2]^2*Sqrt[((b + a*Cos[c + d*x])*Sec[(c + d*x)/2]^2)/(a + b)] - b*(35*a*b*B - 12*a^2*C + 9*b^2*C)*(b + a*Cos[
c + d*x])*(Cos[c + d*x]*Sec[(c + d*x)/2]^2)^(3/2)*Sec[c + d*x]*Tan[(c + d*x)/2]))/(15*d*(b + a*Cos[c + d*x])^3
*(b*C + b*B*Cos[c + d*x] - a*C*Cos[c + d*x])*(Sec[(c + d*x)/2]^2)^(3/2)*Sec[c + d*x]^(7/2)) + (Cos[c + d*x]^3*
(a + b*Sec[c + d*x])^(5/2)*(b*B - a*C + b*C*Sec[c + d*x])*((2*b*(35*a*b*B - 12*a^2*C + 9*b^2*C)*Sin[c + d*x])/
15 + (2*Sec[c + d*x]*(5*b^3*B*Sin[c + d*x] + 6*a*b^2*C*Sin[c + d*x]))/15 + (2*b^3*C*Sec[c + d*x]*Tan[c + d*x])
/5))/(d*(b + a*Cos[c + d*x])^2*(b*C + b*B*Cos[c + d*x] - a*C*Cos[c + d*x]))

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Maple [B] Leaf count of result is larger than twice the leaf count of optimal. \(3699\) vs. \(2(409)=818\).
time = 0.38, size = 3700, normalized size = 8.26

method	result	size
default	\(\text {Expression too large to display}\)	\(3700\)

Verification of antiderivative is not currently implemented for this CAS.

[In]

int((a+b*sec(d*x+c))^(3/2)*(a*b*B-a^2*C+b^2*B*sec(d*x+c)+b^2*C*sec(d*x+c)^2),x,method=_RETURNVERBOSE)

[Out]

2/15/d*(1+cos(d*x+c))^2*((b+a*cos(d*x+c))/cos(d*x+c))^(1/2)*(-1+cos(d*x+c))^2*(15*B*sin(d*x+c)*cos(d*x+c)^2*El
lipticF((-1+cos(d*x+c))/sin(d*x+c),((a-b)/(a+b))^(1/2))*(cos(d*x+c)/(1+cos(d*x+c)))^(1/2)*((b+a*cos(d*x+c))/(1
+cos(d*x+c))/(a+b))^(1/2)*a^3*b+30*C*cos(d*x+c)^3*(cos(d*x+c)/(1+cos(d*x+c)))^(1/2)*((b+a*cos(d*x+c))/(1+cos(d
*x+c))/(a+b))^(1/2)*EllipticF((-1+cos(d*x+c))/sin(d*x+c),((a-b)/(a+b))^(1/2))*sin(d*x+c)*a^3*b+12*C*cos(d*x+c)
^3*(cos(d*x+c)/(1+cos(d*x+c)))^(1/2)*((b+a*cos(d*x+c))/(1+cos(d*x+c))/(a+b))^(1/2)*EllipticF((-1+cos(d*x+c))/s
in(d*x+c),((a-b)/(a+b))^(1/2))*sin(d*x+c)*a^2*b^2-12*C*cos(d*x+c)^3*(cos(d*x+c)/(1+cos(d*x+c)))^(1/2)*((b+a*co
s(d*x+c))/(1+cos(d*x+c))/(a+b))^(1/2)*EllipticF((-1+cos(d*x+c))/sin(d*x+c),((a-b)/(a+b))^(1/2))*sin(d*x+c)*a*b
^3-12*C*cos(d*x+c)^3*(cos(d*x+c)/(1+cos(d*x+c)))^(1/2)*((b+a*cos(d*x+c))/(1+cos(d*x+c))/(a+b))^(1/2)*EllipticE
((-1+cos(d*x+c))/sin(d*x+c),((a-b)/(a+b))^(1/2))*sin(d*x+c)*a^3*b-12*C*cos(d*x+c)^3*(cos(d*x+c)/(1+cos(d*x+c))
)^(1/2)*((b+a*cos(d*x+c))/(1+cos(d*x+c))/(a+b))^(1/2)*EllipticE((-1+cos(d*x+c))/sin(d*x+c),((a-b)/(a+b))^(1/2)
)*sin(d*x+c)*a^2*b^2+9*C*cos(d*x+c)^3*(cos(d*x+c)/(1+cos(d*x+c)))^(1/2)*((b+a*cos(d*x+c))/(1+cos(d*x+c))/(a+b)
)^(1/2)*EllipticE((-1+cos(d*x+c))/sin(d*x+c),((a-b)/(a+b))^(1/2))*sin(d*x+c)*a*b^3-5*B*sin(d*x+c)*cos(d*x+c)^3
*(cos(d*x+c)/(1+cos(d*x+c)))^(1/2)*((b+a*cos(d*x+c))/(1+cos(d*x+c))/(a+b))^(1/2)*EllipticF((-1+cos(d*x+c))/sin
(d*x+c),((a-b)/(a+b))^(1/2))*b^4-15*C*cos(d*x+c)^3*(cos(d*x+c)/(1+cos(d*x+c)))^(1/2)*((b+a*cos(d*x+c))/(1+cos(
d*x+c))/(a+b))^(1/2)*EllipticF((-1+cos(d*x+c))/sin(d*x+c),((a-b)/(a+b))^(1/2))*sin(d*x+c)*a^4+9*C*cos(d*x+c)^3
*(cos(d*x+c)/(1+cos(d*x+c)))^(1/2)*((b+a*cos(d*x+c))/(1+cos(d*x+c))/(a+b))^(1/2)*EllipticE((-1+cos(d*x+c))/sin
(d*x+c),((a-b)/(a+b))^(1/2))*sin(d*x+c)*b^4+30*C*cos(d*x+c)^3*(cos(d*x+c)/(1+cos(d*x+c)))^(1/2)*((b+a*cos(d*x+
c))/(1+cos(d*x+c))/(a+b))^(1/2)*EllipticPi((-1+cos(d*x+c))/sin(d*x+c),-1,((a-b)/(a+b))^(1/2))*sin(d*x+c)*a^4-1
5*C*cos(d*x+c)^2*(cos(d*x+c)/(1+cos(d*x+c)))^(1/2)*((b+a*cos(d*x+c))/(1+cos(d*x+c))/(a+b))^(1/2)*EllipticF((-1
+cos(d*x+c))/sin(d*x+c),((a-b)/(a+b))^(1/2))*sin(d*x+c)*a^4+9*C*cos(d*x+c)^2*(cos(d*x+c)/(1+cos(d*x+c)))^(1/2)
*((b+a*cos(d*x+c))/(1+cos(d*x+c))/(a+b))^(1/2)*EllipticE((-1+cos(d*x+c))/sin(d*x+c),((a-b)/(a+b))^(1/2))*sin(d
*x+c)*b^4+30*C*cos(d*x+c)^2*(cos(d*x+c)/(1+cos(d*x+c)))^(1/2)*((b+a*cos(d*x+c))/(1+cos(d*x+c))/(a+b))^(1/2)*El
lipticPi((-1+cos(d*x+c))/sin(d*x+c),-1,((a-b)/(a+b))^(1/2))*sin(d*x+c)*a^4+3*C*b^4+35*B*(cos(d*x+c)/(1+cos(d*x
+c)))^(1/2)*((b+a*cos(d*x+c))/(1+cos(d*x+c))/(a+b))^(1/2)*sin(d*x+c)*cos(d*x+c)^2*EllipticE((-1+cos(d*x+c))/si
n(d*x+c),((a-b)/(a+b))^(1/2))*a^2*b^2-45*B*(cos(d*x+c)/(1+cos(d*x+c)))^(1/2)*((b+a*cos(d*x+c))/(1+cos(d*x+c))/
(a+b))^(1/2)*sin(d*x+c)*cos(d*x+c)^2*EllipticF((-1+cos(d*x+c))/sin(d*x+c),((a-b)/(a+b))^(1/2))*a^2*b^2-35*B*(c
os(d*x+c)/(1+cos(d*x+c)))^(1/2)*((b+a*cos(d*x+c))/(1+cos(d*x+c))/(a+b))^(1/2)*sin(d*x+c)*cos(d*x+c)^2*Elliptic
F((-1+cos(d*x+c))/sin(d*x+c),((a-b)/(a+b))^(1/2))*a*b^3-12*C*(cos(d*x+c)/(1+cos(d*x+c)))^(1/2)*((b+a*cos(d*x+c
))/(1+cos(d*x+c))/(a+b))^(1/2)*sin(d*x+c)*cos(d*x+c)^2*EllipticE((-1+cos(d*x+c))/sin(d*x+c),((a-b)/(a+b))^(1/2
))*a^3*b-12*C*(cos(d*x+c)/(1+cos(d*x+c)))^(1/2)*((b+a*cos(d*x+c))/(1+cos(d*x+c))/(a+b))^(1/2)*sin(d*x+c)*cos(d
*x+c)^2*EllipticE((-1+cos(d*x+c))/sin(d*x+c),((a-b)/(a+b))^(1/2))*a^2*b^2+9*C*(cos(d*x+c)/(1+cos(d*x+c)))^(1/2
)*((b+a*cos(d*x+c))/(1+cos(d*x+c))/(a+b))^(1/2)*sin(d*x+c)*cos(d*x+c)^2*EllipticE((-1+cos(d*x+c))/sin(d*x+c),(
(a-b)/(a+b))^(1/2))*a*b^3+30*C*(cos(d*x+c)/(1+cos(d*x+c)))^(1/2)*((b+a*cos(d*x+c))/(1+cos(d*x+c))/(a+b))^(1/2)
*sin(d*x+c)*cos(d*x+c)^2*EllipticF((-1+cos(d*x+c))/sin(d*x+c),((a-b)/(a+b))^(1/2))*a^3*b+12*C*(cos(d*x+c)/(1+c
os(d*x+c)))^(1/2)*((b+a*cos(d*x+c))/(1+cos(d*x+c))/(a+b))^(1/2)*sin(d*x+c)*cos(d*x+c)^2*EllipticF((-1+cos(d*x+
c))/sin(d*x+c),((a-b)/(a+b))^(1/2))*a^2*b^2-12*C*(cos(d*x+c)/(1+cos(d*x+c)))^(1/2)*((b+a*cos(d*x+c))/(1+cos(d*
x+c))/(a+b))^(1/2)*sin(d*x+c)*cos(d*x+c)^2*EllipticF((-1+cos(d*x+c))/sin(d*x+c),((a-b)/(a+b))^(1/2))*a*b^3-5*B
*(cos(d*x+c)/(1+cos(d*x+c)))^(1/2)*((b+a*cos(d*x+c))/(1+cos(d*x+c))/(a+b))^(1/2)*EllipticF((-1+cos(d*x+c))/sin
(d*x+c),((a-b)/(a+b))^(1/2))*cos(d*x+c)^2*sin(d*x+c)*b^4-9*C*(cos(d*x+c)/(1+cos(d*x+c)))^(1/2)*((b+a*cos(d*x+c
))/(1+cos(d*x+c))/(a+b))^(1/2)*EllipticF((-1+cos(d*x+c))/sin(d*x+c),((a-b)/(a+b))^(1/2))*cos(d*x+c)^2*sin(d*x+
c)*b^4+12*C*cos(d*x+c)^4*a^3*b-6*C*cos(d*x+c)^4*a^2*b^2-9*C*cos(d*x+c)^4*a*b^3-12*C*cos(d*x+c)^3*a^3*b-6*C*cos
(d*x+c)^2*a^2*b^2+9*C*cos(d*x+c)*a*b^3-35*B*cos(d*x+c)^3*a*b^3-30*B*cos(d*x+c)^2*(cos(d*x+c)/(1+cos(d*x+c)))^(
1/2)*((b+a*cos(d*x+c))/(1+cos(d*x+c))/(a+b))^(1/2)*EllipticPi((-1+cos(d*x+c))/sin(d*x+c),-1,((a-b)/(a+b))^(1/2
))*sin(d*x+c)*a^3*b-30*B*cos(d*x+c)^3*(cos(d*x+c)/(1+cos(d*x+c)))^(1/2)*((b+a*cos(d*x+c))/(1+cos(d*x+c))/(a+b)
)^(1/2)*EllipticPi((-1+cos(d*x+c))/sin(d*x+c),-1,((a-b)/(a+b))^(1/2))*sin(d*x+c)*a^3*b-5*B*cos(d*x+c)^3*b^4+5*
B*cos(d*x+c)*b^4+40*B*cos(d*x+c)^2*a*b^3-35*B*c...

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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*sec(d*x+c))^(3/2)*(a*b*B-a^2*C+b^2*B*sec(d*x+c)+b^2*C*sec(d*x+c)^2),x, algorithm="maxima")

[Out]

integrate((C*b^2*sec(d*x + c)^2 + B*b^2*sec(d*x + c) - C*a^2 + B*a*b)*(b*sec(d*x + c) + a)^(3/2), x)

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Fricas [F]
time = 0.00, size = 0, normalized size = 0.00 \begin {gather*} \text {could not integrate} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate((a+b*sec(d*x+c))^(3/2)*(a*b*B-a^2*C+b^2*B*sec(d*x+c)+b^2*C*sec(d*x+c)^2),x, algorithm="fricas")

[Out]

integral((C*b^3*sec(d*x + c)^3 - C*a^3 + B*a^2*b + (C*a*b^2 + B*b^3)*sec(d*x + c)^2 - (C*a^2*b - 2*B*a*b^2)*se
c(d*x + c))*sqrt(b*sec(d*x + c) + a), x)

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Sympy [F]
time = 0.00, size = 0, normalized size = 0.00 \begin {gather*} - \int C a^{3} \sqrt {a + b \sec {\left (c + d x \right )}}\, dx - \int \left (- B a^{2} b \sqrt {a + b \sec {\left (c + d x \right )}}\right )\, dx - \int \left (- B b^{3} \sqrt {a + b \sec {\left (c + d x \right )}} \sec ^{2}{\left (c + d x \right )}\right )\, dx - \int \left (- C b^{3} \sqrt {a + b \sec {\left (c + d x \right )}} \sec ^{3}{\left (c + d x \right )}\right )\, dx - \int \left (- 2 B a b^{2} \sqrt {a + b \sec {\left (c + d x \right )}} \sec {\left (c + d x \right )}\right )\, dx - \int \left (- C a b^{2} \sqrt {a + b \sec {\left (c + d x \right )}} \sec ^{2}{\left (c + d x \right )}\right )\, dx - \int C a^{2} b \sqrt {a + b \sec {\left (c + d x \right )}} \sec {\left (c + d x \right )}\, dx \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate((a+b*sec(d*x+c))**(3/2)*(a*b*B-a**2*C+b**2*B*sec(d*x+c)+b**2*C*sec(d*x+c)**2),x)

[Out]

-Integral(C*a**3*sqrt(a + b*sec(c + d*x)), x) - Integral(-B*a**2*b*sqrt(a + b*sec(c + d*x)), x) - Integral(-B*
b**3*sqrt(a + b*sec(c + d*x))*sec(c + d*x)**2, x) - Integral(-C*b**3*sqrt(a + b*sec(c + d*x))*sec(c + d*x)**3,
x) - Integral(-2*B*a*b**2*sqrt(a + b*sec(c + d*x))*sec(c + d*x), x) - Integral(-C*a*b**2*sqrt(a + b*sec(c + d
*x))*sec(c + d*x)**2, x) - Integral(C*a**2*b*sqrt(a + b*sec(c + d*x))*sec(c + d*x), x)

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Giac [F]
time = 0.00, size = 0, normalized size = 0.00 \begin {gather*} \text {could not integrate} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate((a+b*sec(d*x+c))^(3/2)*(a*b*B-a^2*C+b^2*B*sec(d*x+c)+b^2*C*sec(d*x+c)^2),x, algorithm="giac")

[Out]

integrate((C*b^2*sec(d*x + c)^2 + B*b^2*sec(d*x + c) - C*a^2 + B*a*b)*(b*sec(d*x + c) + a)^(3/2), x)

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Mupad [F]
time = 0.00, size = -1, normalized size = -0.00 \begin {gather*} \int {\left (a+\frac {b}{\cos \left (c+d\,x\right )}\right )}^{3/2}\,\left (\frac {B\,b^2}{\cos \left (c+d\,x\right )}-C\,a^2+\frac {C\,b^2}{{\cos \left (c+d\,x\right )}^2}+B\,a\,b\right ) \,d x \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

int((a + b/cos(c + d*x))^(3/2)*((B*b^2)/cos(c + d*x) - C*a^2 + (C*b^2)/cos(c + d*x)^2 + B*a*b),x)

[Out]

int((a + b/cos(c + d*x))^(3/2)*((B*b^2)/cos(c + d*x) - C*a^2 + (C*b^2)/cos(c + d*x)^2 + B*a*b), x)

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