Optimal. Leaf size=323 \[ \frac {(i a-b)^{3/2} (A+i B) \text {ArcTan}\left (\frac {\sqrt {i a-b} \sqrt {\tan (c+d x)}}{\sqrt {a+b \tan (c+d x)}}\right )}{d}+\frac {\left (6 a^2 A b-16 A b^3-a^3 B-24 a b^2 B\right ) \tanh ^{-1}\left (\frac {\sqrt {b} \sqrt {\tan (c+d x)}}{\sqrt {a+b \tan (c+d x)}}\right )}{8 b^{3/2} d}+\frac {(i a+b)^{3/2} (A-i B) \tanh ^{-1}\left (\frac {\sqrt {i a+b} \sqrt {\tan (c+d x)}}{\sqrt {a+b \tan (c+d x)}}\right )}{d}+\frac {\left (6 a A b-a^2 B-8 b^2 B\right ) \sqrt {\tan (c+d x)} \sqrt {a+b \tan (c+d x)}}{8 b d}+\frac {(6 A b-a B) \sqrt {\tan (c+d x)} (a+b \tan (c+d x))^{3/2}}{12 b d}+\frac {B \sqrt {\tan (c+d x)} (a+b \tan (c+d x))^{5/2}}{3 b d} \]
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Rubi [A]
time = 1.82, antiderivative size = 323, normalized size of antiderivative = 1.00, number of steps
used = 15, number of rules used = 10, integrand size = 35, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.286, Rules used = {3688, 3728,
3736, 6857, 65, 223, 212, 95, 211, 214} \begin {gather*} \frac {\left (a^2 (-B)+6 a A b-8 b^2 B\right ) \sqrt {\tan (c+d x)} \sqrt {a+b \tan (c+d x)}}{8 b d}+\frac {\left (a^3 (-B)+6 a^2 A b-24 a b^2 B-16 A b^3\right ) \tanh ^{-1}\left (\frac {\sqrt {b} \sqrt {\tan (c+d x)}}{\sqrt {a+b \tan (c+d x)}}\right )}{8 b^{3/2} d}+\frac {(-b+i a)^{3/2} (A+i B) \text {ArcTan}\left (\frac {\sqrt {-b+i a} \sqrt {\tan (c+d x)}}{\sqrt {a+b \tan (c+d x)}}\right )}{d}+\frac {(6 A b-a B) \sqrt {\tan (c+d x)} (a+b \tan (c+d x))^{3/2}}{12 b d}+\frac {(b+i a)^{3/2} (A-i B) \tanh ^{-1}\left (\frac {\sqrt {b+i a} \sqrt {\tan (c+d x)}}{\sqrt {a+b \tan (c+d x)}}\right )}{d}+\frac {B \sqrt {\tan (c+d x)} (a+b \tan (c+d x))^{5/2}}{3 b d} \end {gather*}
Antiderivative was successfully verified.
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Rule 65
Rule 95
Rule 211
Rule 212
Rule 214
Rule 223
Rule 3688
Rule 3728
Rule 3736
Rule 6857
Rubi steps
\begin {align*} \int \tan ^{\frac {3}{2}}(c+d x) (a+b \tan (c+d x))^{3/2} (A+B \tan (c+d x)) \, dx &=\frac {B \sqrt {\tan (c+d x)} (a+b \tan (c+d x))^{5/2}}{3 b d}+\frac {\int \frac {(a+b \tan (c+d x))^{3/2} \left (-\frac {a B}{2}-3 b B \tan (c+d x)+\frac {1}{2} (6 A b-a B) \tan ^2(c+d x)\right )}{\sqrt {\tan (c+d x)}} \, dx}{3 b}\\ &=\frac {(6 A b-a B) \sqrt {\tan (c+d x)} (a+b \tan (c+d x))^{3/2}}{12 b d}+\frac {B \sqrt {\tan (c+d x)} (a+b \tan (c+d x))^{5/2}}{3 b d}+\frac {\int \frac {\sqrt {a+b \tan (c+d x)} \left (-\frac {3}{4} a (2 A b+a B)-6 b (A b+a B) \tan (c+d x)+\frac {3}{4} \left (6 a A b-a^2 B-8 b^2 B\right ) \tan ^2(c+d x)\right )}{\sqrt {\tan (c+d x)}} \, dx}{6 b}\\ &=\frac {\left (6 a A b-a^2 B-8 b^2 B\right ) \sqrt {\tan (c+d x)} \sqrt {a+b \tan (c+d x)}}{8 b d}+\frac {(6 A b-a B) \sqrt {\tan (c+d x)} (a+b \tan (c+d x))^{3/2}}{12 b d}+\frac {B \sqrt {\tan (c+d x)} (a+b \tan (c+d x))^{5/2}}{3 b d}+\frac {\int \frac {-\frac {3}{8} a \left (10 a A b+a^2 B-8 b^2 B\right )-6 b \left (2 a A b+a^2 B-b^2 B\right ) \tan (c+d x)+\frac {3}{8} \left (6 a^2 A b-16 A b^3-a^3 B-24 a b^2 B\right ) \tan ^2(c+d x)}{\sqrt {\tan (c+d x)} \sqrt {a+b \tan (c+d x)}} \, dx}{6 b}\\ &=\frac {\left (6 a A b-a^2 B-8 b^2 B\right ) \sqrt {\tan (c+d x)} \sqrt {a+b \tan (c+d x)}}{8 b d}+\frac {(6 A b-a B) \sqrt {\tan (c+d x)} (a+b \tan (c+d x))^{3/2}}{12 b d}+\frac {B \sqrt {\tan (c+d x)} (a+b \tan (c+d x))^{5/2}}{3 b d}+\frac {\text {Subst}\left (\int \frac {-\frac {3}{8} a \left (10 a A b+a^2 B-8 b^2 B\right )-6 b \left (2 a A b+a^2 B-b^2 B\right ) x+\frac {3}{8} \left (6 a^2 A b-16 A b^3-a^3 B-24 a b^2 B\right ) x^2}{\sqrt {x} \sqrt {a+b x} \left (1+x^2\right )} \, dx,x,\tan (c+d x)\right )}{6 b d}\\ &=\frac {\left (6 a A b-a^2 B-8 b^2 B\right ) \sqrt {\tan (c+d x)} \sqrt {a+b \tan (c+d x)}}{8 b d}+\frac {(6 A b-a B) \sqrt {\tan (c+d x)} (a+b \tan (c+d x))^{3/2}}{12 b d}+\frac {B \sqrt {\tan (c+d x)} (a+b \tan (c+d x))^{5/2}}{3 b d}+\frac {\text {Subst}\left (\int \left (\frac {3 \left (6 a^2 A b-16 A b^3-a^3 B-24 a b^2 B\right )}{8 \sqrt {x} \sqrt {a+b x}}-\frac {6 \left (b \left (a^2 A-A b^2-2 a b B\right )+b \left (2 a A b+a^2 B-b^2 B\right ) x\right )}{\sqrt {x} \sqrt {a+b x} \left (1+x^2\right )}\right ) \, dx,x,\tan (c+d x)\right )}{6 b d}\\ &=\frac {\left (6 a A b-a^2 B-8 b^2 B\right ) \sqrt {\tan (c+d x)} \sqrt {a+b \tan (c+d x)}}{8 b d}+\frac {(6 A b-a B) \sqrt {\tan (c+d x)} (a+b \tan (c+d x))^{3/2}}{12 b d}+\frac {B \sqrt {\tan (c+d x)} (a+b \tan (c+d x))^{5/2}}{3 b d}-\frac {\text {Subst}\left (\int \frac {b \left (a^2 A-A b^2-2 a b B\right )+b \left (2 a A b+a^2 B-b^2 B\right ) x}{\sqrt {x} \sqrt {a+b x} \left (1+x^2\right )} \, dx,x,\tan (c+d x)\right )}{b d}+\frac {\left (6 a^2 A b-16 A b^3-a^3 B-24 a b^2 B\right ) \text {Subst}\left (\int \frac {1}{\sqrt {x} \sqrt {a+b x}} \, dx,x,\tan (c+d x)\right )}{16 b d}\\ &=\frac {\left (6 a A b-a^2 B-8 b^2 B\right ) \sqrt {\tan (c+d x)} \sqrt {a+b \tan (c+d x)}}{8 b d}+\frac {(6 A b-a B) \sqrt {\tan (c+d x)} (a+b \tan (c+d x))^{3/2}}{12 b d}+\frac {B \sqrt {\tan (c+d x)} (a+b \tan (c+d x))^{5/2}}{3 b d}-\frac {\text {Subst}\left (\int \left (\frac {i b \left (a^2 A-A b^2-2 a b B\right )-b \left (2 a A b+a^2 B-b^2 B\right )}{2 (i-x) \sqrt {x} \sqrt {a+b x}}+\frac {i b \left (a^2 A-A b^2-2 a b B\right )+b \left (2 a A b+a^2 B-b^2 B\right )}{2 \sqrt {x} (i+x) \sqrt {a+b x}}\right ) \, dx,x,\tan (c+d x)\right )}{b d}+\frac {\left (6 a^2 A b-16 A b^3-a^3 B-24 a b^2 B\right ) \text {Subst}\left (\int \frac {1}{\sqrt {a+b x^2}} \, dx,x,\sqrt {\tan (c+d x)}\right )}{8 b d}\\ &=\frac {\left (6 a A b-a^2 B-8 b^2 B\right ) \sqrt {\tan (c+d x)} \sqrt {a+b \tan (c+d x)}}{8 b d}+\frac {(6 A b-a B) \sqrt {\tan (c+d x)} (a+b \tan (c+d x))^{3/2}}{12 b d}+\frac {B \sqrt {\tan (c+d x)} (a+b \tan (c+d x))^{5/2}}{3 b d}-\frac {\left ((a+i b)^2 (i A-B)\right ) \text {Subst}\left (\int \frac {1}{(i-x) \sqrt {x} \sqrt {a+b x}} \, dx,x,\tan (c+d x)\right )}{2 d}-\frac {\left ((a-i b)^2 (i A+B)\right ) \text {Subst}\left (\int \frac {1}{\sqrt {x} (i+x) \sqrt {a+b x}} \, dx,x,\tan (c+d x)\right )}{2 d}+\frac {\left (6 a^2 A b-16 A b^3-a^3 B-24 a b^2 B\right ) \text {Subst}\left (\int \frac {1}{1-b x^2} \, dx,x,\frac {\sqrt {\tan (c+d x)}}{\sqrt {a+b \tan (c+d x)}}\right )}{8 b d}\\ &=\frac {\left (6 a^2 A b-16 A b^3-a^3 B-24 a b^2 B\right ) \tanh ^{-1}\left (\frac {\sqrt {b} \sqrt {\tan (c+d x)}}{\sqrt {a+b \tan (c+d x)}}\right )}{8 b^{3/2} d}+\frac {\left (6 a A b-a^2 B-8 b^2 B\right ) \sqrt {\tan (c+d x)} \sqrt {a+b \tan (c+d x)}}{8 b d}+\frac {(6 A b-a B) \sqrt {\tan (c+d x)} (a+b \tan (c+d x))^{3/2}}{12 b d}+\frac {B \sqrt {\tan (c+d x)} (a+b \tan (c+d x))^{5/2}}{3 b d}-\frac {\left ((a+i b)^2 (i A-B)\right ) \text {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 {\left ((a-i b)^2 (i A+B)\right ) \text {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 a-b)^{3/2} (A+i B) \tan ^{-1}\left (\frac {\sqrt {i a-b} \sqrt {\tan (c+d x)}}{\sqrt {a+b \tan (c+d x)}}\right )}{d}+\frac {\left (6 a^2 A b-16 A b^3-a^3 B-24 a b^2 B\right ) \tanh ^{-1}\left (\frac {\sqrt {b} \sqrt {\tan (c+d x)}}{\sqrt {a+b \tan (c+d x)}}\right )}{8 b^{3/2} d}+\frac {(i a+b)^{3/2} (A-i B) \tanh ^{-1}\left (\frac {\sqrt {i a+b} \sqrt {\tan (c+d x)}}{\sqrt {a+b \tan (c+d x)}}\right )}{d}+\frac {\left (6 a A b-a^2 B-8 b^2 B\right ) \sqrt {\tan (c+d x)} \sqrt {a+b \tan (c+d x)}}{8 b d}+\frac {(6 A b-a B) \sqrt {\tan (c+d x)} (a+b \tan (c+d x))^{3/2}}{12 b d}+\frac {B \sqrt {\tan (c+d x)} (a+b \tan (c+d x))^{5/2}}{3 b d}\\ \end {align*}
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Mathematica [A]
time = 2.80, size = 347, normalized size = 1.07 \begin {gather*} \frac {24 \sqrt [4]{-1} (-a+i b)^{3/2} b (i A+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 )+24 (-1)^{3/4} (a+i b)^{3/2} b (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 )-3 \left (-6 a A b+a^2 B+8 b^2 B\right ) \sqrt {\tan (c+d x)} \sqrt {a+b \tan (c+d x)}+2 (6 A b-a B) \sqrt {\tan (c+d x)} (a+b \tan (c+d x))^{3/2}+8 B \sqrt {\tan (c+d x)} (a+b \tan (c+d x))^{5/2}-\frac {3 \sqrt {a} \left (-6 a^2 A b+16 A b^3+a^3 B+24 a b^2 B\right ) \sinh ^{-1}\left (\frac {\sqrt {b} \sqrt {\tan (c+d x)}}{\sqrt {a}}\right ) \sqrt {1+\frac {b \tan (c+d x)}{a}}}{\sqrt {b} \sqrt {a+b \tan (c+d x)}}}{24 b d} \end {gather*}
Antiderivative was successfully verified.
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Maple [B] result has leaf size over 500,000. Avoiding possible recursion issues.
time = 0.53, size = 2401996, normalized size = 7436.52 \[\text {output too large to display}\]
Verification of antiderivative is not currently implemented for this CAS.
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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.
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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.
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Sympy [F(-2)]
time = 0.00, size = 0, normalized size = 0.00 \begin {gather*} \text {Exception raised: SystemError} \end {gather*}
Verification of antiderivative is not currently implemented for this CAS.
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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.
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Mupad [F]
time = 0.00, size = -1, normalized size = -0.00 \begin {gather*} \int {\mathrm {tan}\left (c+d\,x\right )}^{3/2}\,\left (A+B\,\mathrm {tan}\left (c+d\,x\right )\right )\,{\left (a+b\,\mathrm {tan}\left (c+d\,x\right )\right )}^{3/2} \,d x \end {gather*}
Verification of antiderivative is not currently implemented for this CAS.
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